formulaplugins for stats 103013

Probability of dependent events see explanation An automobile manufacturing plant produced vehicles today: were sedans, were do it on papervans, and were trucks. Plant managers are going to choose two of these vehicles forthorough inspection. The first vehicle will be chosen at random, and then the second vehiclewill be chosen at random from the remaining vehicles. What is the probability that two sedanswill be selected?Report your answer as an exact fraction.

Identifying degrees of freedom

Selecting a distribution for inferences on the population mean

A business school is preparing an informational booklet for entering graduate students. Part of this material is a report on how various factors influence a student's GPA (grade point average). In particular, the school used a multiple regression model to study the relationship between graduate GPA and several independent variables . If data from current graduate students are used to develop the regression model, what are the degrees of freedom (df) for the regression sum of squares, the error sum of squares, and the total sum of squares?df for the regression sum of squares?df for the error sum of squares?df for the total sum of squares?

Suppose that we want to estimate the number of holding penalties assessed during a college football game. The sample of games we pick has a mean of 3.3 penalties per game and a standard deviation of 0.8 penalties per game. For each of the following sampling scenarios, determine which test statistic is appropriate to use when making inference statements about the population mean.(In the table, Z refers to a variable having a standard normal distribution, and T refers to avariable having a t distribution.)

Acceptance sampling

p actual proportionn number of sample Binary trials distributionc acceptance numberx number selected/found in the sample

40c 3p 0.16

Err:502

Personal computer manufacturers often buy many of the components in their computers from outside suppliers. One such company purchases memory chips from a supplier who ships the chips in large lots containing several hundred chips each. When a lot is received, a random sample of the chips is selected and tested on an automated machine to see if they can perform several tasks. If too many of the chips in the sample fail to pass the testing, the company rejects the entire lot and returns it to the supplier.

Suppose that the computer company wants to use a sampling plan that has a probability of of rejecting lots that consist of defective or worse. If the company uses an acceptance number of , find the minimum sample size that will give a probability of at least of rejecting lots that are defective or worse.

(If necessary, consult a list of formulas. Also note that the ALEKS calculator allows you to compute binomial probabilities.)

Personal computer manufacturers often buy many of the components in their computers from outside suppliers. One such company purchases memory chips from a supplier who ships the chips in large lots containing several hundred chips each. When a lot is received, a random sample of the chips is selected and tested on an automated machine to see if they can perform several tasks. If too many of the chips in the sample fail to pass the testing, the company

Suppose that the computer company wants to use a sampling plan that has a probability of of rejecting lots that consist of defective or worse. If the company uses an acceptance number of , find the minimum sample size that will

(If necessary, consult a list of formulas. Also note that the ALEKS calculator allows you to compute binomial

Bayes theorem

p(defective | A 0.2p(defective | B 0.19P(B) 0.75P(A) 0.25

ANSWER 0.26

0.74

p(blind | men 6 P(M) 100 p(blind | women 25 P(F) 10,000

0.96 male is colorblind CHOOSE ONE0.04 female is colorblind

In a factory, only two machines, A and B, manufacture washers. Neither machine is perfect: machine A produces defective washers 20% of the time, while machine B produces defectives 17% of the time. Machine B is more efficient than machine A and accounts for 80% of the total output of washers. For purposes of quality control, a sample of washers is taken and examined for defectives. Compute the probability that a randomly chosen washer found to be defective was manufactured by machine A. Round your answer to two decimal places.

probability that a randomly chosen washer found to be defective was manufactured by machine A

probability that a randomly chosen washer found to be defective was manufactured by machine B

Suppose that 6 men in 100 are colorblind, while 24 women in 10,000 are colorblind. Compute the probability that a colorblind individual is male. Assume that the population contains an equal number of males and females. Round your answer to two decimal places

A certain disease occurs in 47% of the population. A test for the disease is fairly accurate: it misclassifies people with the disease as healthy 4% of the time and reports that a healthy person is diseased just 6% of the time. Suppose that a person tests positive for the disease. Compute the probability that the person does indeed have the disease. Round your answer to two decimal places.

P(desease) 0.27p( negative | desease 0.09p(people | desease 0.09p(positive | desease 0.91P(no desease) 0.73

0.79 does have it 0.21 does not have it

p(correct| female 0.76p(incorrect| female 0.24p(correct| male 0.7p(incorrect| male 0.3P(F) 0.61P(M) 0.39ANSWER 0.44 to find a man 0.56 to find a woman

p(ON TIME | TOOK CAR ) 0.7p( NOT ON TIME | took bus ) 0.4P(on time | took bus 0.6P (took car) 0.2p(took car | on time 0.54 answer P took bus | on time 0.46

A psychologist interested in gender differences examined whether people can determine the sex of an individual from a sample of his/her handwriting. In a study of this matter, female handwriting samples were correctly identified as such by 70% of volunteer subjects, while male handwriting samples were correctly identified by 70% of subjects. In this study, 64% of the handwriting samples were from females. Suppose you know that a subject incorrectly identified a sample. What is the probability that this handwriting sample is from a man? Round your answer to two decimal places.

Archie has to go to school this morning for an important test, but he woke up late. He can either take the bus or take his unreliable car. If he takes the car, Archie knows from experience that he will make it to school without breaking down with probability 0.4 . However, the bus to school runs late 25 % of the time. Archie decides to choose between these options by tossing a coin. Suppose that Archie does, in fact, make it to the test on time. What is the probability that he took his car? Round your answer to two decimal places.(If necessary, consult a list of for

probability that a randomly chosen washer found to be defective was

probability that a randomly chosen washer found to be defective was

are colorblind. Compute the probability that a colorblind individual is male. Assume that the population contains an equal number of males and females. Round your answer to two decimal places

f the population. A test for the disease is fairly accurate: it misclassifies people with the 6% of the time. Suppose that a

person tests positive for the disease. Compute the probability that the person does indeed have the disease. Round your

to find a woman

A psychologist interested in gender differences examined whether people can determine the sex of an individual from a sample of his/her handwriting. In a study of this matter, female handwriting samples were correctly identified as such by 70% of volunteer subjects, while male

of the handwriting samples were from females. Suppose you know that a subject incorrectly identified a sample. What is the probability that this handwriting sample is from a man? Round your answer

Archie has to go to school this morning for an important test, but he woke up late. He can either take the bus or take his unreliable car. If he takes the car, Archie knows from experience that he will make it to school without breaking down with probability 0.4 . However, the bus to

of the time. Archie decides to choose between these options by tossing a coin. Suppose that Archie does, in fact, make it to the test on time. What is the probability that he took his car? Round your answer to two decimal places.

Binomial problems: Mean and standard deviation

n 130 number of trials

a. 0.98 percentage

p 0.02 probability of success on each particular trial

np 2.6 answer questio 1

b. np 2.6 questions 20.020.98

2.5481.596 answer

13 h, 13 s, 13d , 13 cl

a. n 15 number of trials

p 0.25 percentage probability of success on each particular trial


b. np 3.75 questions 2

At a recent meeting at the American College of Asthma, Allergy, and Immunology, researchers reported that allergy vaccinations are effective in treating sinusitis (inflammation of the membrane lining the facial sinuses) in individuals predisposed to allergies. Specifically, the researchers reported that 98% of individuals predisposed to allergies find relief from their sinusitis after allergy vaccinations. Suppose that this number is accurate and that 140 individuals predisposed to allergies are chosen at random and given allergy vaccinations.

a. Estimate the number of individuals in the random sample who do not find relief from their sinusitis by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.

b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.(If necessary, consult a list of formulas.)

Suppose that we've decided to test Clara, who works at the Psychic Center, to see if she really has psychic abilities. While talking to her on the phone, we'll thoroughly shuffle a standard deck of 52 cards (which is made up of 13 hearts, 13spades, 13 diamonds, and 13clubs) and draw one card at random. We'll ask Clara to name the suit (heart, spade, diamond, or club) of the card we drew. After getting her guess, we'll return the card to the deck, thoroughly shuffle the deck, draw another card, and get her guess for the suit of this second card. We'll repeat this process until we've drawn a total of 18cards and gotten her suit guesses for each.Assume that Clara is not clairvoyant, that is, assume that she randomly guesses on each card.

a. Estimate the number of cards in the sample for which Clara correctly guesses the suit by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

0.250.75

2.81251.677 answer

a. n 30 number of trials

p 0.5 percentage probability of success on each particular trial


b. np 15 questions 20.50.57.5

2.739 answer

a np 903%2.7 answer

b 2.70.030.97

2.6191.618 answer

Suppose that the New England Colonials baseball team is equally likely to win any particular game as not to win it. Suppose also that we choose a random sample of 30 Colonials games.Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

A machine that manufactures automobile pistons is estimated to produce a defective piston of the 1% time. Suppose that this estimate is correct and that a random sample of 90pistons produced by this machine is taken.Estimate the number of pistons in the sample that are defective by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

a np 500.05

2.5 answer

b 2.50.050.95

2.3751.541 answer

a np 50

10%

0.9

45.0 answer

b 450.90.14.5

2.121 answer

a np 500.07

3.5 answer

Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go") so it can make decisions regarding the possible construction of in-store play areas, the attendance of its mascot Sammy at the franchise locations, and so on. Anita's reports that 48% of its customers order their food to go. Suppose that this proportion is correct and that a random sample of 50 individual customers is taken.

Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about 5%of all visitors to the website are looking for other websites. Assume that this estimate is correct and that a random sample of 40visitors to the website is taken.Estimate the number of visitors in the sample who actually are looking for the company's website by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.

A rainstorm in Portland, Oregon, wiped out the electricity in 5% of the households in the city. Suppose that a random sample of 50 Portland households is taken after the rainstorm.a. Estimate the number of households in the sample that lost electricity by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three decimal places.

3.5b 0.07

0.933.2551.804 answer

100%4

0.25probability of success on each particular trial

At a recent meeting at the American College of Asthma, Allergy, and Immunology, researchers reported that allergy vaccinations are effective in treating sinusitis (inflammation of the membrane lining the facial sinuses) in individuals predisposed to allergies.

of individuals predisposed to allergies find relief from their sinusitis after allergy individuals predisposed to allergies are chosen at random and

a. Estimate the number of individuals in the random sample who do not find relief from their sinusitis by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.

b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least

Suppose that we've decided to test Clara, who works at the Psychic Center, to see if she really has psychic abilities. While talking to her on hich is made up of 13 hearts, 13spades, 13 diamonds, and 13clubs) and

rd at random. We'll ask Clara to name the suit (heart, spade, diamond, or club) of the card we drew. After getting her guess, we'll return the card to the deck, thoroughly shuffle the deck, draw another card, and get her guess for the suit of this second card. We'll

s and gotten her suit guesses for each.Assume that Clara is not clairvoyant, that is, assume that she randomly guesses on each card.

a. Estimate the number of cards in the sample for which Clara correctly guesses the suit by giving the mean of the relevant distribution

b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three

100%win - no win 2

probability of success on each particular trial 0.5

win any particular game as not to win it. Suppose also that we

Estimate the number of games in the sample that the Colonials win by giving the mean of the relevant distribution (that is, the expectation

Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three

A machine that manufactures automobile pistons is estimated to produce a defective piston of the 1% time. Suppose that this estimate is

Estimate the number of pistons in the sample that are defective by giving the mean of the relevant distribution (that is, the expectation of

Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three

Anita's, a fast-food chain specializing in hot dogs and garlic fries, keeps track of the proportion of its customers who decide to eat in the restaurant (as opposed to ordering the food "to go") so it can make decisions regarding the possible construction of in-store play areas, the

48% of its customers order their food to go. dividual customers is taken.

Not all visitors to a certain company's website are customers or potential customers. In fact, the company's executives estimate that about of all visitors to the website are looking for other websites. Assume that this estimate is correct and that a random sample of 40visitors

Estimate the number of visitors in the sample who actually are looking for the company's website by giving the mean of the relevant distribution (that is, the expectation of the relevant random variable). Do not round your response.

of the households in the city. Suppose that a random sample of 50

a. Estimate the number of households in the sample that lost electricity by giving the mean of the relevant distribution (that is, the

b. Quantify the uncertainty of your estimate by giving the standard deviation of the distribution. Round your response to at least three

Binomial Problem Basic

question 1

# of people that would be interested --> 65%Random Sample in power -- > 6

Answer 0.075 already rounded

question 2

the probability of a loss of electricity is --> 10%the probability of no loss of electricity is 90%Random Sample in power -- > 6


question 3

Estimates 7%the number of Combination 10

A TV executive is interested in the popularity of a particular cable TV show. She has been told that a whopping 65% of American households would be interested in tuning in to a new network version of the show. If this is correct, what is the probability that all of the households in her city being monitored by the TV industry would tune in to the new show? Assume that the 6 households constitute a random sample of American households.Round your response to at least three decimal places.

(If necessary, consult a list of formulas.)

A rainstorm in Portland, Oregon, has wiped out the electricity in about 65% of the households in the city. A management team in Portland has a big meeting tomorrow, and all 6 members of the team are hard at work in their separate households, preparing their presentations. What is the probability that none of them has lost electricity in his/her household? Assume that their locations are spread out so that loss of electricity is independent among their households.Round your response to at least three decimal places.

Not all visitors to a certain company's website are customers. In fact, the website administrator estimates that about 12% of all visitors to the website are looking for other websites. Assuming that this estimate is correct, find the probability that, in a random sample of 5 visitors to the website, exactly 3 actually are looking for the website

the number chosen 3 7120

probability looking for particular website 0.930.000 answer

question 4

P (O) obese 0.6P (D) diabetes 0.4

P (OND) 0.65

find P( O U D) = --->>>>>>>>> P(O)+P(D)-P(O П D) 0.35 answer

question 5

group 5 0.5chosen 3 2combination 10ssssf 0.3125 answer

question 6

Estimates 96%

Suppose that 41% of all adults in a certain community are obese and that 34% suffer from diabetes. If 24% of the adults in this community are both obese and suffering from diabetes, what is the probability that a randomly chosen adult in this community is obese or suffers from diabetes (or both)?

Suppose that the New England Colonials baseball team is equally likely to win a game as not to win it. If 5 Colonials games are chosen at random, what is the probability that exactly 4 of those games are won by the Colonials?Round your response to at least three decimal places.

The workers' union at a particular university is quite strong. About 96% of all workers employed by the university belong to the workers' union. Recently, the workers went on strike, and now a local TV station plans to interview 5 workers (chosen at random) at the university to get their opinions on the strike. What is the probability that exactly 3 of the workers interviewed are union members?

the number of Combination 4the number chosen 3 1

4probability looking for particular website 0.04

0.142 answer

Binomial problems: AdvancedA new surgical operation procedure is said to be successful 75% of the time. If the procedure isperformed on 9patients independently, what is the probability that fewer than 2operations aresuccessful? Round your answer to four decimal places.

n = number, p = propability or percentage, x is the # choosenp sometimes will be " a, b, c, d" then the equation will be 1/4 . It's 1 out of the total

P(Bn, p <x)

1-P(Bn, p <x)1-P(Bn, p <x)

see explanation alex cal

at most

at least

6

more than

Central Limit Theorem: Sample Mean

0.1 Original mean0.07 original dev

0.093 New Mean40 N

6.3245550.0111-0.007

-0.6324560.2635

0.736 ANSWER more than

112 Original mean15 original dev

115.8 New Mean35 N

5.916082.5355

3.8

1.49874

According to records, the amount of precipitation in a certain city on a November day has a 0.10 mean of inches, with a standard deviation of 0.07 inches. What is the probability that the mean daily precipitation will be 0.093 inches or more for a random sample of 40 November days (taken over many years)? Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal place

A worldwide organization of academics claims that the mean IQ score of its members is 112 , with a standard deviation of 15. A randomly selected group of 35 members of this organization is tested, and the results reveal that the mean IQ score in this sample is 115.8 . If the organization's claim is correct, what is the probability of having a sample mean of 115.8 or less for a random sample of this size?

0.933 ANSWER Less than


31.5 New Mean34 N

5.8309521.2005

-0.5

-0.4164970.339 ANSWER Less than

The producer of a weight-loss pill advertises that people who use the pill lose, after one week, an average (mean) of 1.85 pounds with a standard deviation of 0.95 pounds. In a recent study, a group of 50 people who used this pill were interviewed. The study revealed that these people lost a mean of pounds after one week. If the producer's claim is correct, what is the probability that the mean weight loss after one week on this pill for a random sample of 50 individuals will be 1.68 pounds or less?


25000 New Mean80 N

8.944272411.2129

770

1.8725090.9694


520 Original mean180 original dev502 New Mean

35 N5.9160830.4256

-18

-0.591608

0.2771

The mean salary offered to students who are graduating from Coastal State University this year is $24230, with a standard deviation of 3678 . A random sample of 80 Coastal State students graduating this year has been selected. What is the probability that the mean salary offer for 80 these students is 25000 or more?

For a standardized psychology examination intended for psychology majors, the historical data show that scores have a mean of 520 and a standard deviation of 180. The grading process of this year's exam has just begun. The average score of 35 the exams graded so far is 502 . What is the probability that a sample of 35 exams will have a mean score of 502 or more if the exam scores follow the same distribution as in the past?

Central limit theorem: Sample proportion

1,100 0.01

0.02

A. n 1100p 0.01

11E(X) = NP 0.01

B Var(X) 0.0000090.003

c 3.3333333330.9996 ANSWER greater than or less than0.0004 ANSWER greater than or equal to, Less than or equal to 1-p(z..)

1650 0.12

0.1

A. n 1650p 0.12

The Credit Risk Department of a major bank estimates the default rate on loans under 10000 to be 1%. The bank will make 1100 loans that are under 10000 next month.A. Find the mean of P, where P is the proportion of defaults on the 1100 loans under 10000 to be made next month.B Find the standard deviation of P.C. Compute an approximation for , which is the probability that 2% or more of next month's loans under 10000 will be defaulted on. Round your answer to four decimal places.

P ( Z ≤ � )

Empirical research on stock market data for two consecutive trading days indicates that 60% of the stocks that went up on the first day also went up on the second day. Yesterday, 600 stocks went up.a. Find the mean of P , where gives the proportion of the 600 stocks that went up yesterday that will go up today.b. Find the standard deviation of p .c. Compute an approximation for , which is the probability that fewer than of the stocks that went up yesterday will go up again today. Round your answer to four decimal places.

P ( Z ≤ � )

198E(X) = NP 0.12

B Var(X) 0.0000640.008

c -2.50.0062 ANSWER greater than or less than0.9938 ANSWER greater than or equal to, Less than or equal to 1-p(z..)

1100 0.01

0.02

A. n 1100p 0.01

11E(X) = NP 0.01

B Var(X) 0.0000090.003


Withdrawal symptoms may occur when a person using a painkiller suddenly stops using it. For a special type of painkiller, withdrawal symptoms occur in 1% of the cases. Consider a random sample of 1100 people who have stopped using the painkillera. Find the mean of P , where P is the proportion of people in the sample who experience withdrawal symptoms.b Find the standard deviation of P.c Compute an approximation for , which is the probability that more than 2% of those sampled experience withdrawal symptoms. Round your answer to four decimal places

P ( Z ≤ � )

4400 0.01

0.015

A. n 4400p 0.01

44E(X) = NP 0.01

B Var(X) 0.000002250.0015


1650 0.12

0.1

A. n 1650p 0.12

198

A researcher wants to investigate the effects of environmental factors on IQ scores. For an initial study, she takes a sample of 400 people who grew up as the only child. She finds that 47.5% of them have an IQ score over 100 . It is known that of the general population has an IQ score exceeding .a. Find the mean of P , where is the proportion of people with IQ scores over 100 in a random sample of 400 people.b. Find the standard deviation of .Compute an approximation for , which is the probability that there will be 47.5% or more individuals with IQ scores over 100 in a random sample of 400. Round your answer to four decimal places.

P ( Z ≤ � )

In a memory test, the test subjects are shown 600 objects and are asked after two hours to name the objects. The historical records show that test subjects remember 60% of the objects after two hours.a. Find the mean of P , where is the proportion of objects that are remembered by a test subject.b. Find the standard deviation of P .c. Compute an approximation for , which is the probability that fewer than 63% of the objects are remembered by a test subject. Round your answer to four decimal places.

P ( Z ≤ � )

E(X) = NP 0.12

B Var(X) 0.0000640.008

c -2.50.0062 ANSWER greater than or less than0.9938 ANSWER greater than or equal to, Less than or equal to 1-p(z..)

more than

10000 to be 1%. The bank will

10000 to be made next

2% or more of next

of the stocks that went up

stocks that went up yesterday that will go up

fewer than of the stocks that

fewer thanmore than

more than

Withdrawal symptoms may occur when a person using a painkiller suddenly stops using it. For a special type of people who have stopped

a. Find the mean of P , where P is the proportion of people in the sample who experience withdrawal symptoms.

more than 2% of those

more than

A researcher wants to investigate the effects of environmental factors on IQ scores. For an initial study, she takes a of them have an IQ score over 100 . It is

in a random sample of 400

47.5% or more ound your answer to four decimal places.

ects and are asked after two hours to name the objects. The

a. Find the mean of P , where is the proportion of objects that are remembered by a test subject.

fewer than 63% of the

fewer thanmore than

Chebyshev's theorem and the empirical rule

theorem [100(1-(1/k^2)]

empirical rule 68% stand dev is 195% stand dev is 299.7% stand dev is 3

PROBLEM TYPE 1mean 1137std devia 203

question

dis 1 28.4 dis 2 66.8lies below mean --> -5.46 & lies above the mean --> -5.27

Answer 96%

question

% from problem 84% 0.84 0.16[100(1-(1/k^2)] --> +/- 2.5 <---lies with that of value of standard deviation of the mean

ANSWER 629.51644.5

question

dis 1 731 dis 2 1543lies below mean --> -2 & lies above the mean --> 2

Answer 75%

question

According to Chebyshev's theorem, at least _____% of the measurements lie between 545.5 cc and 1710.5 cc.

thus, <--- of the measurements lie between dis 1 and dis 2

According to Chebyshev's theorem, at least 84% of the fares lie between _____dollars and ____dollars.

According to Chebyshev's theorem, at least _____% of the easurements lie between 662 cc and 1594 cc.

dis 1 731 dis 2 1543lies below mean --> -2 & lies above the mean --> 2empirical rule

68% stand dev is 195% stand dev is 299.7% stand dev is 3

question

mean 1137deviation level 1 using the Empirical rule, choose the appropriate deviationstd deviation 203answer 1 934answer 2 1340

PROBLEM TYPE 2MEAN 50.3

question

std deviation 3Answer 89%

question

value 1 31.1 value 2 69.5lies below mean --> -19.2 lies above the mean --> 19.2

Suppose that the distribution is bell-shaped. According to the empirical rule, approximately _____% of the measurements lie between 662 cc and 1594 cc.

Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 68% of the measurements lie between _____cc and _____cc.

According to Chebyshev's theorem, at least _____% of the scores lie within 3 standard deviations of the mean, 50.3

Suppose that the distribution is bell-shaped. If approximately 95% of the scores lie between 31.1 and 69.5 , then the approximate value of the standard deviation for the distribution, according to the empirical rule, is _____.

empirical rule68% stand dev is 195% stand dev is 299.7% stand dev is 3

dis of the two values 19.2Std deviation 2 using the Empirical rule, choose the appropriate deviationanswer 9.6

<---lies with that of value of standard deviation of the mean

using the Empirical rule, choose the appropriate deviation

Choosing an appropriate sample sizePercentage

Keyconfidence

0.950.99

0.9

deviation 750margin error 135confidence 1.645 USE THE CONFIDENCE VALUE IN ORANGE FOR THE PROPER %

1233.759.138889

answer 83.52

Suppose that a researcher is interested in estimating the mean systolic blood pressure, M , ofexecutives of major corporations. He plans to use the blood pressures of a random sample ofexecutives of major corporations to estimate M . Assuming that the standard deviation of thepopulation of systolic blood pressures of executives of major corporations is 26 mm Hg, what isthe minimum sample size needed for the researcher to be 95% confident that his estimate iswithin 5 mm Hg of M ?Carry your intermediate computations to at least three decimal places. Write your answer as awhole number (and make sure that it is the minimum whole number that satisfies the requirements).

round up to the MINIMUM WHOLE NUMBER value, not WHOLE NUMBER

A personality test has a subsection designed to assess the "honesty" of the test-taker. Suppose that you're interested in the mean score, , on this subsection among the general population. You decide that you'll use the mean of a random sample of scores on this subsection to estimate . What is the minimum sample size needed in order for you to be confident that your estimate is within of ? Use the value for the population standard deviation of scores on this subsection.

Managers at an automobile manufacturing plant would like to estimate the mean completion time of an assembly line operation, . The managers plan to choose a random sample of completion times and estimate via the sample. Assuming that the standard deviation of the population of completion times is minutes, what is the minimum sample size needed for the managers to be confident that their estimate is within minutes of ?

Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time, , of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate . Assuming that the standard deviation of the population of shopping times at the supermarkets is minutes, what is the minimum sample size she must collect in order for her to be confident that her estimate is within minutes of ?

A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more that that on average, the corporation may lose money, and if it dispenses less, the customers may complain.BIG Corporation would like to estimate the mean amount of coffee, , dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate . Assuming that the standard deviation of cup amounts dispensed by this machine is ounces, what is the minimum sample size needed in order for BIG to be confident that its estimate is within ounces of ?

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be confident that its estimate is within of ? Suppose that the standard deviation of wages of foodservice workers in the U.S. is about .

A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like to estimate the mean monthly mileage, , of cars rented in the U.S. over the past year. The consumer group plans to choose a random sample of monthly U.S. rental car mileages and then estimate using the mean of the sample.Using the value miles per month as the standard deviation of monthly U.S. rental car mileages from the past year, what is the minimum sample size needed in order for the consumer group to be confident that its estimate is within miles per month of ?

Percentage Use this value 0.05 0.025 -1.959964 1.9600.01 0.005 -2.575829 2.576

0.1 0.05 -1.644854 1.645

g, what is

Carry your intermediate computations to at least three decimal places. Write your answer as awhole number (and make sure that it is the minimum whole number that satisfies the requirements).

MINIMUM WHOLE NUMBER value, not WHOLE NUMBER

A personality test has a subsection designed to assess the "honesty" of the test-taker. Suppose that you're interested in the mean score, , on this subsection among the general population. You decide that you'll use the mean of a random sample of scores on this subsection to estimate . What is the minimum sample size needed in order for you to be confident that your estimate is within of ? Use the value for

Managers at an automobile manufacturing plant would like to estimate the mean completion time of an assembly line operation, . The managers plan to choose a random sample of completion times and estimate via the sample. Assuming that the standard deviation of the population of completion times is minutes, what is the minimum sample size needed for the managers to be confident that their

Executives of a supermarket chain are interested in the amount of time that customers spend in the stores during shopping trips. The executives hire a statistical consultant and ask her to determine the mean shopping time, , of customers at the supermarkets. The consultant will collect a random sample of shopping times at the supermarkets and use the mean of these shopping times to estimate . Assuming that the standard deviation of the population of shopping times at the supermarkets is minutes, what is the minimum sample

A coin-operated coffee machine made by BIG Corporation was designed to discharge a mean of eight ounces of coffee per cup. If it dispenses more that that on average, the corporation may lose money, and if it dispenses less, the customers may complain.BIG Corporation would like to estimate the mean amount of coffee, , dispensed per cup by this machine. BIG will choose a random sample of cup amounts dispensed by this machine and use this sample to estimate . Assuming that the standard deviation of cup amounts dispensed by this machine is ounces, what is the minimum sample size needed in order for BIG to be confident that its

A union of restaurant and foodservice workers would like to estimate the mean hourly wage, , of foodservice workers in the U.S. The union will choose a random sample of wages and then estimate using the mean of the sample. What is the minimum sample size needed in order for the union to be confident that its estimate is within of ? Suppose that the standard deviation of wages of foodservice

A consumer advocacy group is doing a large study on car rental practices. Among other things, the consumer group would like to estimate the mean monthly mileage, , of cars rented in the U.S. over the past year. The consumer group plans to choose a random

Using the value miles per month as the standard deviation of monthly U.S. rental car mileages from the past year, what is the minimum sample size needed in order for the consumer group to be confident that its estimate is within miles per month of ?

combinations

If you have n objects and chose r of them, the number of combinations is:

n! / ( r! (n-r)! )

here we have:

21 C 19 = 21! / (19! (21-19)!) = 210

n 24r 22combinati 276

You can answer any 19 questions from the 21 questions on an exam. In how many different ways can you choose the 19 questions, assuming that the order in which you choose the questions is irrelevant?

this can be written as nCr

p(college grad) 88% fP(internet user) 7% ep(college grad П internet user) 6% both

p(college grad | internet user) p(college grad П internet user) 0.06P(internet user 0.07 0.857143 answer

p(college grad | internet user) p(college grad П internet user) 0.06p(college grad) 0.07 0.857143 answer

Conditional probability: Basic

It is estimated that 27% of all California adults are college graduates and that 31% of California adults are regular internet users. It is also estimated that 20% of

California adults are both college graduates and regular internet users.

(a) Among California adults, what is the probability that a randomly chosen internet user is a college graduate? Round your answer to 2 decimal places

(b) What is the probability that a California adult is an internet user, given that he or she is a college graduate? Round your answer to 2 decimal places.

Confidence interval for the population mean: Use of the standard normal

DEV 22N 70

M 1251.645

8.3666ANSWER 120.7 Lower LimitANSWER 129.3 Upper Limit

The lifetime of a certain brand of electric light bulb is known to have a standard deviation of 51 hours. Suppose that a random sample of 50 bulbs of this brand has a mean lifetime of 476 hours. Find a 90% confidence interval for the true mean lifetime of all light bulbs of this brand. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.

What is the lower limit of the 90% confidence interval?What is the upper limit of the 90% confidence interval?

The union for a particular industry has determined that the standard deviation of the daily wages of its workers is $ 18. A random sample of 150 workers in this industry has a mean daily wage of $110 . Find a 90%confidence interval for the true mean daily wage of all union workers in the industry. Then complete the table below

The mean diastolic blood pressure for a random sample of 90 people was 80 millimeters of mercury. If the standard deviation of individual blood pressure readings is known to be 10 millimeters of mercury, find a 95% confidence interval for the true mean diastolic blood pressure of all people. Then complete the table below

A laboratory in New York is interested in finding the mean chloride level for a healthy resident in the state. A random sample of 80 healthy residents has a mean chloride level of 103 mEq/L. If it is known that the chloride levels in healthy individuals residing in New York have a standard deviation of 35 mEq/L, find a 90%confidence interval for the true mean chloride level of all healthy New York residents. Then complete the table below

An existing inventory for a test measuring self-esteem indicates that the scores have a standard deviation of 9 . A psychologist gave the self-esteem test to a random sample of 150 individuals, and their mean score was 65 . Construct a 95%confidence interval for the true mean of all test scores. Then complete the table below.

The standard deviation of test scores on a certain achievement test is 11.2 . A random sample of 80scores on this test had a mean of71.9. Based on this sample, find a 95% confidence interval for the true mean of all scores. Then complete the table below

The breaking strengths of cables produced by a certain manufacturer have a standard deviation of 92 pounds. A random sample of 150 newly manufactured cables has a mean breaking strength of 1700 pounds. Based on this sample, find a 90% confidence interval for the true mean breaking strength of all cables produced by this manufacturer. Then complete the table below.

Confidence interval for the population mean: Use of the standard normal

90% 1.64595% 1.96

Confidence interval for the population standard deviation

l u 95 n 125 125 1.00 5 mean 103.5 103.5 0.950 upperdev 18.4 18.4 0.050 lowerdfreedm 124 124 lower uppern-1 1 1 confidence 99% 0.005 0.995

0.05 0.95 95% 0.025 0.975chisqd 150.989 99.283 90% 0.050 0.950

124 124338.56 338.56

41981.44 41981.44204.89373 204.893712.287776 9.964067

ANSWER 16.67 20.56Lower Upper

416424457282512

Lifetimes of AAA batteries are approximately normally distributed. A manufacturer wants to estimatethe standard deviation of the lifetime of the AAA batteries it produces. A random sample of 21 AAAbatteries produced by this manufacturer lasted a mean of 9.6 hours with a standard deviation of 2.3hours. Find a 99% confidence interval for the population standard deviation of the lifetimes of AAAbatteries produced by the manufacturer. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to at leasttwo decimal places.


The scores on an examination in economics are approximately normally distributed with mean 500 and an unknown standard deviation. The following is a random sample of scores from this examination.Find a 90% confidence interval for the population standard deviation. Then complete the table below.

select all of these and choose S in Aleks cal for the Sample Mean

Pilots who cannot maintain regular sleep hours due to their work schedule often suffer from insomnia. A recent study on sleeping patterns of pilots focused on quantifying deviations from regular sleep hours. A random sample of 24 commercial airline pilots was interviewed, and the pilots in the sample reported the time at which they went to sleep on their most recent working day. The study gave the sample mean and standard deviation of the times reported by pilots, with these times measured in hours after midnight. (Thus, if the pilot reported going to sleep at 11 p.m., the measurement was -1.) The sample mean was 0.7 hours, and the standard deviation was 1.8 hours. Assume that the sample is drawn from a normally distributed population. Find a 90% confidence interval for the population standard deviation, that is, the standard deviation of the time (hours after midnight) at which pilots go to sleep on their work days. Then complete the table below.

A psychologist wants to estimate the standard deviation of IQ scores. It is widely believed that IQ scores follow a normal distribution. Her random sample of 26 IQ scores has a mean of 102.5 and a standard deviation of 17.4 . Find the 90% confidence interval for the population standard deviation based on this sample. Then complete the table below.

The standard deviation of the daily demand for a product is an important factor for inventory control for the product. Suppose that a pharmacy wants to estimate the standard deviation of the daily demand for a certain antibiotic. It is known that the daily demand for this antibiotic follows an approximately normal distribution. A random sample of 18 days has a sample mean of 120 orders for this antibiotic with a standard deviation of 10 orders. Find a 90% confidence interval for the population standard deviation of the daily demand for this antibiotic. Then complete the table below

l un 5 5mean 22 22dev 4.06 4.06dfreedm 4 4n-1 1 1

0.025 0.975chisqd 11.143 0.484

4 416.4836 16.483665.9344 65.9344

8.12 8.123.338156 0.696002


The following data were randomly drawn from an approximately normal population.18, 18, 22, 25, 27Based on these data, find a 95% confidence interval for the population standard deviation. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

Confidence interval for the population mean: Use of the t distribution

l u 95 n 16 16 1.00 5 mean 92 92 0.950 upperdev 13 13 0.050 lowerdg fredom 15 15 lower uppert 0.025 0.025 confidence 99% 0.005 0.995

1.761 1.761 95% 0.025 0.9754 4 90% 0.050 0.950


calculate T in aleksenter T results here

A corporation that maintains a large fleet of company cars for the use of its sales staff is interested in the mean distance driven monthly per sales person. The following table gives the monthly distances in milesdriven by a random sample of 12 sales persons: 2667, 1943, 2152, 1937, 2184, 2399, 2641, 2317, 2541, 2076, 2285, 2637. Based on this sample, find a 95% confidence interval for the mean number of miles driven monthly by members of the sales staff, assuming that monthly driving distances are normally distributed. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.


l un 5 5mean 22 22dev 4.06 4.06dfreedm 4 4n-1 1 1

0.025 0.975chisqd 11.143 0.484

4 416.4836 16.4836

ANSWER #REF! #REF!Lower Upper

The following data were randomly drawn from an approximately normal population.18, 18, 22, 25, 27Based on these data, find a 95% confidence interval for the population standard deviation. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. (If necessary, consult a list of formulas.)

Confidence interval for a population proportion

^ 170 170 66p 200 200 275

0.85 0.85 confidence 99% 0.0051-p 0.1500 0.1500 95% 0.025

0.1275 0.1275 90% 0.0500.0006 0.0006

sqrt 0.0252 0.0252

z 2.5758 2.57580.0650 0.0650

answer0.785 0.915

LOWER UPPER

A researcher wishes to estimate the proportion of X-ray machines that malfunction. A random 275 sample of machines is taken, and 66 of the machines in the sample malfunction.Based upon this, compute a 90% confidence interval for the proportion of all X-ray machines that malfunction. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places.What is the lower limit of the 90% confidence interval?What is the upper limit of the 90% confidence interval?

Enter the value in aleks and calculate the Z

To estimate the percentage of a species of rodent that carries a specific viral infection, 125 rodents are randomly selected and examined, and 61 of them are found to be infected. Based upon this, find a 95% confidence interval for the proportion of all rodents of this species that are infected with the virus. Then complete the table below.

Confidence interval for the difference of population means: Use of the standard normal

l zm1 169.00 99% 2.58 0.005sdev 1 5 95% 1.96 0.025 1.96n1 12 90% 1.64 0.05 1.645m2 164.00 sdev 2 9n2 9z 1.9600

5.00 2581

2.0833333333339

11.083333333333.32916405924

-1.53 lower11.53 upper

In a current study, a random sample of teachers from Arizona and an independent random sample of 10 teachers from Nevada have been asked to report their annual income. The data obtained are as follows:Annual income in dollarsArizona 34022, 41515, 38444, 35184, 34074, 35435, 42621, 43101, 46582, 32199.Nevada 41120, 24940, 36266, 34307, 42230, 49416, 27985, 39965, 54428, 44120.The population standard deviations for the annual incomes of teachers in Arizona and in Nevada are estimated as and , respectively. It is also known that both populations are approximately normally distributed. Construct a confidence interval for the difference between the mean annual income of teachers from Arizona () and the mean annual income of teachers from Nevada (). Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

Confidence interval for the difference of population means: Use of the standard normal

In a current study, a random sample of teachers from Arizona and an independent random sample of 10 teachers from

The population standard deviations for the annual incomes of teachers in Arizona and in Nevada are estimated as and , Construct a confidence interval

between the mean annual income of teachers from Arizona () and the mean annual income of teachers

Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places.

l tn1 16 90% 0.05 1.7139m1 92.00 95% 0.025 2.1098sdev 1 13 99% 0.005 2.8453n2 10m2 87.00 sdev 2 16

0.005freedom 24t 2.7969 enter T HEREx1 - x2 5.00

253523044839

24201.625

1/n1 0.06251/n2 0.1

32.7640635.7239901

-11.01 lower21.01 upper

Confidence interval for the difference of population means: Use of the t distribution

The university data center has two main computers. The center wants to examine whether computer is receiving tasks that require processing times comparable to those of computer . A random sample of processing times from computer showed a mean of seconds with a standard deviation of seconds, while a random sample of processing times from computer (chosen independently of those for computer ) showed a mean of seconds with a standard deviation of seconds. Assume that the populations of processing times are normally distributed for each of the two computers and that the variances are equal. Construct a confidence interval for the difference between the mean processing time of computer , , and the mean processing time of computer , . Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. (If necessary, consult a list of formulas.)

Confidence interval for the difference of population proportions

n1 350n1#chose 290.00 n2 325.00 n2#chosen 289z 2.580 p1 0.829 p2 0.889 confidence 99% 0.005 2.580

60.000 95% 0.025 1.960 36.000 90% 0.050 1.645

n1p1 290.000 0.000 0.000

v+ 0.001 sqt 0.027

0.829 0.889

p1-p2 (0.061)-0.129 lower0.008 upper

In random, independent samples of adults and teenagers who watched a certain television show, adults and teens indicated that they liked the show. Let be the proportion of all adults watching the show who liked it, and let be the proportion of all teens watching the show who liked it. Find a confidence interval for . Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal places. (If necessary, consult a list of formulas.)

To determine the possible effect of a chemical treatment on the rate of seed germination, chemically treated seeds and untreated seeds were sown. It was found that of the chemically treated seeds and of the untreated seeds germinated. Let be the proportion of the population of all chemically treated seeds that germinate, and let be the proportion of the population of all untreated seeds that germinate. Find a confidence interval for . Then complete the table below

In a comparative study of two new drugs, A and B, patients were treated with drug A, and patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that patients were cured using drug A and patients were cured using drug B. Let be the proportion of the population of all patients who are cured using drug A, and let be the proportion of the population of all patients who are cured using drug B. Find a confidence interval for . Then complete the table below.

In random, independent samples of adults and teenagers who watched a certain television show, adults and teens indicated that they liked the show. Let be the proportion of all adults watching the show who liked it, and let be the

. Then complete the table below.Carry your intermediate computations to at least three decimal places. Round your responses to at least three decimal

To determine the possible effect of a chemical treatment on the rate of seed germination, chemically treated seeds and untreated seeds were sown. It was found that of the chemically treated seeds and of the untreated seeds germinated. Let be the proportion of the population of all chemically treated seeds that germinate, and let be the proportion of the population of all untreated seeds that germinate. Find a confidence interval for . Then complete the

In a comparative study of two new drugs, A and B, patients were treated with drug A, and patients were treated with drug B. (The two treatment groups were randomly and independently chosen.) It was found that patients were cured using drug A and patients were cured using drug B. Let be the proportion of the population of all patients who are cured using drug A, and let be the proportion of the population of all patients who are cured using drug B. Find a

Confidence interval for the ratio of population variances

n#1 10 10mean1 0.12 0.12 lower upperdev1 0.0036 0.0036 confidence 99% 0.005 0.995N#2 10 10 95% 0.025 0.975MEAN2 0.12 0.12 90% 0.050 0.950DEV2 0.0049 0.0049F DIS 4.026 0.248fredom Num 9 9 use ALEKS to calculate F DISTRIBUTTIONfreedom Den 9 9

0.73 0.730.25 4.03 ENTER F DISTRIBUTION RESULTS HERE0.18 2.96

LOWER LOWER

Random samples are drawn independently from two normally distributed populations, and the following statistics are obtained:Group 1 Group 2 (The first row gives the sample sizes, the second row gives the sample means, and the third row gives the sample variances.) Construct a confidence interval for , the ratio of the variances of these two populations. Then complete the table below.Carry your intermediate computations to at least three decimal places. Write your final responses to at least two decimal places.

What is the lower limit of the confidence interval? What is the upper limit of the confidence interval?

Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 25 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are summarized below:Current Year Last Year(The first row gives the sample means, and the second row gives the sample variances.) Assuming that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year, construct a confidence interval for , the ratio of the variance of the life span for the current year to the variance of the life span for last year. Then complete the table below.Carry your intermediate computations to at least three decimal places. Write your final responses to at least two decimal places. (If necessary, consult a list of formulas.)

An examination in psychology has been taken by psychology majors and also by some students from other majors. It is An examination in psychology has been taken by psychology majors and also by some students from other majors. It is widely believed that the scores for both groups are normally distributed. A random sample of examinations completed by psychology majors and an independent random sample of examinations completed by students from other majors are selected. Among sampled students, the psychology majors scored a mean of points with a variance of , and the students from other majors scored a mean of points with a variance of . Construct a confidence interval for , the ratio of the variance of all scores of psychology majors to the variance of all scores of other majors. Then complete the table below.

use ALEKS to calculate F DISTRIBUTTION

Random samples are drawn independently from two normally distributed populations, and the following statistics are

(The first row gives the sample sizes, the second row gives the sample means, and the third row gives the sample

two populations. Then complete the table

Carry your intermediate computations to at least three decimal places. Write your final responses to at least two

Actuaries use various parameters when evaluating the cost of a life insurance policy. The variance of the life spans of a population is one of the parameters used for the evaluation. Each year, the actuaries at a particular insurance company randomly sample 25 people who died during the year (with the samples chosen independently from year to year) to see whether the variance of life spans has changed. The life span data from this year and from last year are

Assuming that life spans are approximately normally distributed for each of the populations of people who died this year and people who died last year, construct a confidence interval for , the ratio of the variance of the life span for

Carry your intermediate computations to at least three decimal places. Write your final responses to at least two

An examination in psychology has been taken by psychology majors and also by some students from other majors. It is An examination in psychology has been taken by psychology majors and also by some students from other majors. It is widely believed that the scores for both groups are normally distributed. A random sample of examinations completed by psychology majors and an independent random sample of examinations completed by students from other majors are selected. Among sampled students, the psychology majors scored a mean of points with a variance of , and the students from other majors scored a mean of points with a variance of . Construct a confidence interval for , the ratio of the variance of all scores of psychology majors to the variance of all scores of other majors. Then

X Y XY do not touch12.9 224.3 2893.47 Sum ----> 147.819.1 217.9 4161.8911.7 275.6 3224.5214.5 271.9 3942.5512.6 198.8 2504.88

7.8 232.4 1812.724.2 304.8 1280.166.8 240.6 1636.08

14.4 188.9 2720.164.1 317 1299.7

13.9 196.1 2725.795.6 274 1534.46.5 227.3 1477.455.5 272.7 1499.858.2 280.9 2303.38

0

Total 147.8 3723.20 35017Average 9.8533 248.2133 2188.5625Count 15 15 16Total dev 9.8533 248.2133

-1668.93067294.00

sdevof x 4.583sdevof y 40.080

question

ANSWER -5.67669

question

ANSWER -0.649 already rounded

r -0.865 <---The value of the sample correlation coefficient

248.2133 the r will be given

9.8533

Computing the sample correlation coefficient and the coefficients for the least-squares regression line

What is the value of the slope of the least-squares regression line for these data? Round your answer to at least two decimal places

What is the value of the sample correlation coefficient for these data? Round your answer to at least three decimal places

Sy

Sx

b1 0.3 <--The least-squares regression line for these data has a slope of approximately

question

ANSWER -7.5654

yes question

ANSWER 245.2573

b0=b0=y-b1x y b1 x

yes question

ANSWER 0.034r=b1

yes question

ANSWER 322.7580

Find B1, with R given

What is the value of the slope of the least-squares regression line for these data? Round your answer to at least two decimal places

b1=r Sy

Sx

use this if B1 is given

What is the value of the y-intercept of the least-squares regression line for these data? Round youranswer to at least two decimal places.

Find R, with B1 given

What is the value of the sample correlation coefficient for these data? Round your answer to atleast three decimal places.

Sx

Sy

use this if B1 is not given

What is the value of the y-intercept of the least-squares regression line for these data? Round youranswer to at least two decimal places.

stdr devia of X stdr devia of Ysquareroot ---> 9.2822 3723.2 571.8475

85.5008 918.89823.4102 750.0295

21.5915 561.05827.5442 2441.67754.2162 250.0615

31.9602 3202.05089.3228 57.9628

20.6722 3518.071533.1008 4731.605516.3755 2715.799518.0908 664.952211.2448 437.367518.9515 599.5968

2.7335 1068.4182

Sum ---> 293.9973 22489.39733333320.9998 1606.3855

sqrt 4.5826 40.0797

question

question

worksheet # 3

<---The value of the sample correlation coefficient

the r will be given

of the least-squares regression for these data? Round your answer to at least two

What is the value of the sample correlation coefficient for these data? Round your answer to at least three decimal places

<--The least-squares regression line for these data has a slope of approximately

question

question

question

question

of the least-squares regression for these data? Round your answer to at least two

t of the least-squares regression line

What is the value of the sample correlation coefficient for these data?

t of the least-squares regression line

Cumulative distributions and ogives

sample # 25do not touch

X asis factor frequency Relative Freq Cum Relative Frequency500 700 6 0.24 0.24700 900 8 0.32 0.56 <<<------ Find the cumulative relative frequency polygon900 1100 6 0.24 0.80

1100 1300 2 0.08 0.881300 1500 3 0.12 1.00

Always add the frequecny FIRST, so I can clear the non-needed cells

<<<------ Find the cumulative relative frequency polygon

Discrete probability distribution: Basic

enter your value here

P (X=x)-5 0.11-1 0.123 0.1456

0.37

answer 0.63 split this # in two, like 20, 43 or 40, 23 or 10, 53, etc.

Value of x of X

example do not touch this book.

P (X=x)-5 0.11-1 0.123 0.1456

0.37

0.63 split this # in two, like 20, 43 or 40, 23 or 10, 53, etc. It does not matter, as long as it's split

Value of x of X

It does not matter, as long as it's split

Probabilities of draws with replacement

46

Power 5 rolling timesAnswer 0.132

p( at least one 2 among the 4 rolls ) = 1-p(no 2 amoongs the 4 rolls )

5 the probablility of rolling somehting different from 2 will 5/6power 4

answer 0.518

ball 1 8 red 15ball 2 7 pinkpower 4

8 probablity of pink drawned 8/13 to the 4 powertanswer 0.081

A fair die is rolled 4 times. What is the probability of having no 1 and no 3 among the rolls? Round your answer to three decimal places.(If necessary, consult a list of formulas.)

P( no 1 and no 3 among the 4 rolls)

A fair die is rolled 4 times. What is the probability that a 2 is obtained on at least one of the rolls? Round your answer to three decimal places.

An urn contains 5 red and 8 pink balls. Four balls are randomly drawn from the urn in succession, with replacement. That is, after each draw, the selected ball is returned to the urn. What is the probability that all 4 balls drawn from the urn are pink? Round your answer to three decimal places.(If necessary, consult a list of formulas.)

the probablility of rolling somehting different from 2 will 5/6

among the rolls? Round your answer to three decimal places.

of the rolls? Round your answer to three decimal places.

Estimating the mean of Grouped Data

sample # 19do not touch

X asis factor frequency Midpoint (col a + col b divided 20 10 4 5 20

10 20 6 15 9020 30 6 25 15030 40 2 35 7040 50 1 45 45

0 0

sum 375

Answr 19.73684


Estimating the standard deviation of Grouped Data

50 <--- sample #do not touch

frequency X asis factor do not touch5 860 880 870 4350.000 1296 6480

15 880 900 890 13350.000 256 384020 900 920 910 18200.000 16 320

5 920 940 930 4650.000 576 28805 940 960 950 4750.000 1936 9680

0 8208360 820836

sum 45300 23200906 473.46939

Answr 21.8 already rounded


enter only the X asis Factor and frequencyalways enter the frequency 1st

Midpoint (col a + col b divided 2

Hypothesis test for the population mean: Z test

the Type of Test Statisticmean 35dev 7

sample# 80 TWO TAILEDnew mean 37 Z DIS

KNOWN DEVIATION2.0008.9440.7832.556 the Value of the Test Statistics:

-2.5560.9950.0050.0050.011 the P-Value0.011 compare this number to the Level of Significance

the Type of Test Statistic ONE TAILEDmean 35 Z DISdev 7 KNOWN DEVIATIONsample# 80new mean 37

2.000

8.9440.783

A manufacturer claims that the mean lifetime, , of its light bulbs is months. The standard deviation of these lifetimes is months. Thirty-three bulbs are selected at random, and their mean lifetime is found to be months. Assume that the population is normally distributed. Can we conclude, at the level of significance, that the mean lifetime of light bulbs made by this manufacturer differs from months?Perform a two-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.)

2.556 the Value of the Test Statistics:

0.05 Plug in the Level of Signicance in ALEKS to calculate ZAleks result Critical value at the level of significance

compare this number to the Level of Significance

A manufacturer claims that the mean lifetime, , of its light bulbs is months. The standard deviation of these lifetimes is months. Thirty-three bulbs are selected at random, and their mean lifetime is found to be months. Assume that the population is normally distributed. Can we conclude, at the level of significance, that the mean lifetime of light bulbs

Carry your intermediate computations to at least three decimal places. (If necessary, consult a list of formulas.)

Independent events: Basic

0.3 Use this, when the A is accentP(B) 0.4

Find P(A)1-P(A) 0.7

Since A and B are independent, we have

P (A) * P(B) 0.28 ANSWER

P ( A U B) P(A)+P(B)-P(AUB) 0.82 ANSWER

P(B) 0.2 use this, when the B is accentP(A) 0.5

Find P(B)1-P(B) 0.8

Since A and B are independent, we have

P (A) * P(B) 0.4 ANSWER

P ( A U B) P(A)+P(B)-P(AUB) 0.9 ANSWER

P(Ā)

P ( A П B)

P ( A П B)

Law of total probabilities

uneven prob even proba88% of them failed 0.9 0.88

0.1 0.12every 12 2out of 13 13number of search 13 13sucessful search 0.09 0.05Hence the probabilty of seach 1/2 11

answer 0.22 0.132727

P(grade of B or below | no TA ) AKA failed 0.88P(grade better than b | no TA ) 0.13every 1out of 6number of session 6Hence the probabilty of seach 1/2 5

0.12

While working for Life magazine, Archie is assigned to the Galapagos Islands to photograph a rare, giant white turtle. The turtle can be found either on Isabella, the largest of the 16 Galapagos islands, or on Santa Cruz, the next largest island. Many have tried to locate the giant turtle, but 88% of them have failed. In addition, 11 out of every 12 searches in the past were concentrated on Santa Cruz, though only 11% of them were actually successful. Archie intends to search on Isabella. What is the probability that Archie will successfully find the turtle? Round your answer to two decimal places.

Archie is taking a course in probability theory. Discussion sessions with the class TA (teaching assistant) conflict with Archie's trips to the beach. Archie has to decide between attending TA sessions, which might help him do better in the probability course, and going to the beach. Archie learns that94 % of past students who did not attend TA sessions received a grade of B or below in the course. Archie also learns that 14 % of past students received a grade higher than B. After some thought, Archie decides to attend the TA sessions. Not all of his classmates share his decision; in fact, only 8 out of every 11 of his classmates attend discussion sessions with the TA. Compute the probability that Archie will receive a grade higher than a B in the course. Round your answer to two decimal places.

P(grade better than b) answer 0.18

dies from flu 0.31every 5out of 17

conditional probability that a non-vaccinated person will die from the flu 0.36not vac 12answer 0.19

94% of them failed 0.650.35

8 out of 11 attends 25 LCD30 LCD

number of search 30sucessful search 0.6Hence the probabilty of seach 1/2 5

answer 0.3

Archie must travel to a foreign country where a deadly strain of flu has ravaged the population. Already,31 % of the population has succumbed to the flu and died. There is a vaccine, but because the flu is extremely contagious, it spreads quite rapidly, and to date only 4 out of every 8 people have been vaccinated. Thirty-five percent of those who are not vaccinated die from the flu. Archie has just enough time to be vaccinated before he leaves. What is the probability that Archie will succumb to the flu and die while he is away? Round your answer to two decimal places.

Archie is stranded in a war zone in the land of the Xhosas. His only chance of survival is to reach Sudland, which lies 100 miles away. There are just two routes to Sudland; one is through a desert, and the other is a long and perilous trek through the mountains. Both routes are terribly dangerous. In the past, 46% of all who tried to reach Sudland failed. Of all attempts to reach Sudland, 1 out of every 6 were by the desert route. Although it is the longer route, the mountains seem to provide more success: 55% of those who attempted to reach Sudland through the mountains were successful. To save time, Archie intends to go via the desert. What are his chances of making it through to Sudland? Round your answer to two decimal places.

even proba

While working for Life magazine, Archie is assigned to the Galapagos Islands to photograph a rare, giant white turtle. The turtle can Galapagos islands, or on Santa Cruz, the next largest island. Many have tried to

arches in the past were concentrated on of them were actually successful. Archie intends to search on Isabella. What is the probability that

Archie is taking a course in probability theory. Discussion sessions with the class TA (teaching assistant) conflict with Archie's trips to the beach. Archie has to decide between attending TA sessions, which might help him do better in the

of past students who did not attend TA sessions of past students received a grade higher than B.

After some thought, Archie decides to attend the TA sessions. Not all of his classmates share his decision; in fact, only 8 of his classmates attend discussion sessions with the TA. Compute the probability that Archie will receive a

0.310.69

377

0.784

0.57

Archie must travel to a foreign country where a deadly strain of flu has ravaged the population. Already,31 % of the population has succumbed to the flu and died. There is a vaccine, but because the flu is extremely contagious, it spreads

of those who are not vaccinated die from the flu. Archie has just enough time to be vaccinated before he leaves. What is the probability that

Archie is stranded in a war zone in the land of the Xhosas. His only chance of survival is to reach Sudland, which lies 100 miles away. There are just two routes to Sudland; one is through a desert, and the other is a long and perilous trek

of all who tried to reach Sudland failed. Of all were by the desert route. Although it is the longer route, the mountains seem

of those who attempted to reach Sudland through the mountains were successful. To save time, Archie intends to go via the desert. What are his chances of making it through to Sudland? Round your answer to

Linear relationship and the sample correlation coefficient

X Y u v w t m n1 1 1 7.4 1 3.5 1 7.62 2 2 9.1 2 4.3 2 4.93 3 3 7.2 3 3.5 3 10.64 4 4 5.4 4 4.6 4 5.95 5 5 8.3 5 4.2 5 2.16 6 6 5.4 6 7.4 6 57 7 7 4.6 7 5.8 7 8.78 8 8 6.5 8 8 8 3.99 9 9 6 9 7.4 9 9.3

10 10 10 4 10 8 10 6.6

ANSWER 0.0007 ANSWER -0.0003 ANSWER 0.0004 ANSWER 0.0000

0.0003 -0.0004

Moving averages

5 weeks 3 weeks1 80.00 1 80.002 83.80 2 83.803 71.40 81.78 3 71.40 78.404 86.20 81.36 4 86.20 80.475 87.50 82.14 5 87.50 81.706 77.90 85.04 6 77.90 83.877 87.70 84.04 7 87.70 84.378 85.90 85.06 8 85.90 83.839 81.20 86.96 9 81.20 84.93

10 92.60 85.82 10 92.60 86.5711 87.4 87.36 11 87.4 87.0712 82 90.16 12 82 87.3313 93.6 89.62 13 93.6 87.6714 95.2 90.175 14 95.2 90.2715 89.9 390.25 15 89.9 92.90

1282.3 1282.3

University officials at a major university would like to use -term moving averages to analyze enrollment. Enrollment data (in thousands of students) for the most recent terms are shown in the middle column of the table below. Also shown is a time series plot of the data.In the far-right column of the table are the -term moving averages for the data. Fill in both blanks in this column. Round your responses to at least two decimal places

Normal distribution raw scores

mean 500dev 95calcutate the Z in Aleks 5%

z 1.6449ANSWR 656.3

mean 9.3dev 0.3

10%0.9

calcutate the Z in Aleks

z -1.2816ANSWR 8.9

mean 64dev 2.4

75%0.25


z 0.6745ANSWR 65.6

The distribution of scores on a standardized aptitude test is approximately normal with a mean of 500 and a standard deviation of 105 . What is the minimum score needed to be in the top 5% on this test? Carry your intermediate computations to at least four decimal places, and round your answer to the nearest integer.

Z0.05

Calcium levels in people are normally distributed with a mean of 9.5mg/dL and a standard deviation of 0.5mg/dL. Individuals with calcium levels in the bottom 10% of the population are considered to have low calcium levels. Find the calcium level that is the borderline between low calcium levels and those not considered low. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place

Z0.90

Suppose that the heights of adult women in the United States are normally distributed with a mean of 63.5 inches and a standard deviation of 2.2 inches. Jennifer is taller than 75% of the population of U.S. women. How tall (in inches) is Jennifer? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Z0.25

dev 1.4mean 8lowest 10%top 0.9


z -1.28166.20576

mean 230dev 30

10%0.9


z -1.2816ANSWR 191.6

mean 130dev 18

Suppose that the quarterly sales levels among health care information systems companies are approximately normally distributed with a mean of 9 million dollars and a standard deviation of 1.3million dollars. One health care information systems company considers a quarter a "failure" if its sales level that quarter is in the bottom 10% of all quarterly sales levels. Determine the sales level (in millions of dollars) that is the cutoff between quarters that are considered "failures" by that company and quarters that are not. Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Z0.90

According to her doctor, Mrs. Brown's cholesterol level is higher than only 15%of the females aged 50 and over. The cholesterol levels among females aged 50 and over are approximately normally distributed with a mean of 230mg/dL and a standard deviation of 25mg/dL. What is Mrs. Brown's cholesterol level? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Z0.85

Suppose that scores on a particular test are normally distributed with a mean of 130 and a standard deviation of 18 . What is the minimum score needed to be in the top 15% of the scores on the test? Carry your intermediate computations to at least four decimal places, and round your answer to at least one decimal place.

Z0.15

z 1.0364 calcutate the Z in Aleks

ANSWR 148.7

dev 2.1mean 6.5

10%

z 1.2816 calcutate the Z in Aleks and enter here

ANSWR 9.2

mean 51dev 15

15%0.85

z -1.0364 calcutate the Z in Aleks and enter here

ANSWR 35.5

A newspaper article reported that people spend a mean of 6.5 hours per day watching TV, with a standard deviation 2.1 of hours. A psychologist would like to conduct interviews with the10% of the population who spend the most time watching TV. She assumes that the daily time people spend watching TV is normally distributed. At least how many hours of daily TV watching are necessary for a person to be eligible for the interview? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place

A newspaper article reported that people spend a mean of hours per day watching TV, with a standard

Z0.10

Risk taking is an important part of investing. In order to make suitable investment decisions on behalf of their customers, portfolio managers give a questionnaire to new customers to measure their desire to take financial risks. The scores on the questionnaire are approximately normally distributed with a mean of 51 and a standard deviation of 15. The customers with scores in the bottom 15% are described as "risk averse." What is the questionnaire score that separates customers who are considered risk averse from those who are not? Carry your intermediate computations to at least four decimal places. Round your answer to at least one decimal place.

Z0.85

Normal distribution: Word problems

x 53000mean 40756dev 8000

12244p (Z > 1.5305 )1-P 0.93705349

0.0629 ANSWER

Since aleks only have for a given z, use the complement rule 1-

x 83mean 81dev 8

2p (Z > 0.25

0.5987 ANSWER

According to a recent survey, the salaries of assistant professors have a mean of $ 40,756 and a standard deviation of $8000. Assuming that the salaries of assistant professors follow a normal distribution, find the proportion of assistant professors who earn more than $ 53000. Round your answer to at least four decimal places.

Suppose that the annual rate of return for a common biotechnology stock is normally distributed with a mean of 4% and a standard deviation of 4% . Find the probability that the one-year return of this stock will be negative. Round your answer to at least four decimal places.

x 35mean 50dev 10

-15p (Z > -1.5

0.0668

x 128mean 100dev 20

28p (Z > 1.4 )

0.9192 ANSWER

x 83mean 81dev 8

2

A certain test is designed to measure the satisfaction of an individual with his/her relationship. Suppose that the scores on this test are approximately normally distributed with a mean 50 of and a standard deviation of 10 . An individual with a score of 35 or less is considered dissatisfied with his/her relationship. According to this criterion, what proportion of people in relationships are dissatisfied? Round your answer to at least four decimal places.

Suppose that the time required to complete a 1040R tax form is normally distributed with a mean of 100 minutes and a standard deviation of 20 minutes. What proportion of 1040R tax forms will be completed in at most 128 minutes? Round your answer to at least four decimal places.

Suppose that pulse rates among healthy adults are normally distributed with a mean 81of beats/second and a standard deviation of 8 beats/second. What proportion of healthy adults have pulse rates that are at least 83 beats/sec? Round your answer to at least four decimal places.

p (Z > 0.25 )1-P 0.59870633

0.4013 ANSWER

x 72mean 71dev 3

1p (Z > 0.33333333 )

0.6306 ANSWER

x 56mean 60dev 9

-4p (Z > -0.4444444 )

0.3284 ANSWER

Suppose that the heights of adult men in the United States are normally distributed with a mean of 71 inches and a standard deviation of 3 inches. What proportion of the adult men in United States are at most 6 feet tall? (Hint: 6 feet = 72 inches.) Round your answer to at least four decimal places.

Suppose that the scores on a reading ability test are normally distributed with a mean of 60 and a standard deviation of 9 . What proportion of individuals score at most 56 points on this test? Round your answer to at least four decimal places.

Normal approximation to binomial

at least Probabilty Typen 40 110p 55% 63%

25 69-0.5

np 221-P 45%

9.924.5

2.53.1464265445105

0.7946

0.7866 AT LEAST0.213 answer

n 85p 32.0%

24 fewer Than-0.5

np 27.21-P 68%

18.49623.5-3.7

4.300697617829-0.8603

0.195

Suppose that a new treatment is successful in curing a common ailment 63% of the time. If the treatment is tried on a random sample of 110 patients, approximate the probability that at least 69 will be cured. Use the normal approximation to the binomial with a correction for continuity.Round your answer to at least three decimal places. Do not round any intermediate steps.

Vnp(1-p)

A brokerage survey reports that 36% of all individual investors have used a discount broker (one that does not charge the full commission). If a random sample of 125 individual investors is taken, approximate the probability that fewer than 44 have used a discount broker. Use the normal approximation to the binomial with a correction for continuity.

Vnp(1-p)

n 250

p 6% AT MOST13

0.5np 15.251-P 94%

14.3197513.5

-1.753.7841445532643

-0.46250.322 answer

n 85

p 29% MORE THAN24

0.5np 24.651-P 71%

17.501524.5

-0.154.1834794131201

-0.03590.486

0.514 ANSWER

Suppose that 14% of the people in a large city have used a hospital emergency room in the past year. If a random sample of 155 people from the city is taken, approximate the probability that at most 23 used an emergency room in the past year. Use the normal approximation to the binomial with a correction for continuity.

Vnp(1-p)

Suppose that 10% of the population of the U.S. is left-handed. If a random sample of 220 peoplefrom the U.S. is chosen, approximate the probability that more than 21 are left-handed. Use thenormal approximation to the binomial with a correction for continuity.

Vnp(1-p)

Adjusted Probability

Greater than or equal to -0.5 1-P(z…)

Greater than 0.5

Less than or equal to 0.5 1-P(z…)

Less than -0.5


Greater than 0.5


Less than -0.5

of the time. If the treatment is tried on a l be cured. Use the normal approximation to

Round your answer to at least three decimal places. Do not round any intermediate steps.

of all individual investors have used a discount broker (one that does not charge the individual investors is taken, approximate the probability that fewer than 44

have used a discount broker. Use the normal approximation to the binomial with a correction for continuity.


Greater than 0.5


Less than -0.5


Greater than 0.5


Less than -0.5

he people in a large city have used a hospital emergency room in the past year. If a random sample of used an emergency room in the past year.

ple of 220 peopleare left-handed. Use the

at least

Population standard deviationdo not touch do not touch

27 Sum ----> 145 squareroot ---> 428 count ---> 5 127 Total 29 432 931 4

Sum ---> 224.4


Probability of union: Basic

P(B) 0.8P(C ) 0.18P(B)*P(C ) 0.144

0.836

0.98

et and be two events such that and .

(a) Determine , given that B and C are independent .

(b) Determine , given that B and C are mutually exclusive

Do not round your responses.

ANSWER

ANSWER

Predictions from the least-squares regression line

y= 41.07 -0.48 slope run the sum of the table in ALEKS

Observed Predicated

x y23.7 27.8 29.7

+ or - for slope Less than or Greater than?

Observed 1363.4 4990.80 Predictated difference -3627.40

An advertising firm wishes to demonstrate to its clients the effectiveness of the advertising campaigns it has conducted. The following bivariate data on fifteen recent campaigns, including the cost of each campaign (in millions of dollars) and the resulting percentage increase in sales following the campaign, were presented by the firm. Based on these data, we would compute the least-squares regression line to be , with representing campaign cost and representing the resulting percentage increase in sales. (This line is shown in Figure 1, along with a scatter plot of the data.)

The managers of an electric utility wish to examine the relationship between temperature and electricity use in the utility's service region during the summer months. In particular, the managers wish to be able to predict total electricity use for a day from the maximum temperature that day. The bivariate data below give the maximum temperature (in degrees Fahrenheit) and the electricity use (in thousands of kilowatt hours) of electricity generated and sold for a random sample of fifteen summer days. A best-fitting line for the data, obtained from least-squares regression, is given by , in which denotes the maximum temperature and denotes the electricity use. This line is shown in the Figure 1 scatter plot.

Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by , in dollars) and the maximum temperature (denoted by , in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below

run the sum of the table in ALEKS

Less than or Greater than?

An advertising firm wishes to demonstrate to its clients the effectiveness of the advertising campaigns it has conducted. The following bivariate data on fifteen recent campaigns, including the cost of each campaign (in millions of dollars) and the resulting percentage increase in sales following the campaign, were presented by the firm. Based on these data, we would compute the least-squares regression line to be , with representing campaign cost and representing the resulting percentage increase in sales. (This

The managers of an electric utility wish to examine the relationship between temperature and electricity use in the utility's service region during the summer months. In particular, the managers wish to be able to predict total electricity use for a day from the maximum temperature that day. The bivariate data below give the maximum temperature (in degrees Fahrenheit) and the electricity use (in thousands of kilowatt hours) of electricity generated and sold for a random sample of fifteen summer days. A best-fitting line for the data, obtained from least-squares regression, is given by , in which denotes the maximum temperature and denotes the electricity use. This line is

Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by , in dollars) and the maximum temperature (denoted by , in

Rejecting unreasonable claims based on average statistics9% FALSE

Last year at least one of their clients made a profit of exactly 9% FALSE9% FALSE9% FALSE

Last year at least one of their clients made a profit of more than 14% FALSElast years ago some of their clients made a profit of less than 12% TRUE

$ 38,500 FALSE $ 38,500 FALSE $ 38,500 TRUE

$ 38,500 FALSE

$ 41,000 TRUE $ 35,000 FALSE

$ 37,000 TRUE

Last month all of their patients lost at least 6lbs FALSE

FALSE6lbs FALSE

Last month some of their patients lost more than 3lbs TRUE2lbs TRUE

This month at least one of their patients will lose at least 6lbs FALSELast month at least one of their patients lost at least 8lbs TRUELast month all of their patients lost less than 14lbs FALSELast month some of their patients lost 7lbs or more 7lbs TRUE

1280 FALSE

1140 FALSE

1140 FALSE

1140 TRUE

1140 FALSE

Last year all of their clients made a profit of at least .

Two years ago some of their clients made a profit of at least Last year fewer than half of their clients made a profit of or less

Last year more than half of their clients had a starting salary of at least Last year some of their clients had a starting salary of exactly Last year some of their clients had a starting salary of at least

Two years ago at least one of their clients had a starting salary of at least

Last year at least one of their clients had a starting salary of less than Last year some of their clients had a starting salary of less than

Last year at least one of their clients had a starting salary of more than

Last month, the number of their clients who lost 6lbs or more was equal to the number of their clients who lost 6lbs or less.Last month some of their patients lost exactly

Last month at least one of their patients lost more than

Last year all of their freshman students had a score of less than 1280 on the exam

Last year, the number of their freshman students who had a score of more than on the exam was equal to the number of their freshman students who had a score of less than 1140 on the exam

Last year some of their freshman students had a score of exactly 1140 on the exam

Last year some of their freshman students had a score of 1140 or above on the exam

Next year at least one of their freshman students will have a score of at least on the exam

900 TRUE

Two months ago some of their patients lost at least 8 lbs 8lbs FALSE

Last year at least one of their freshman students had a score of more than 900 on the exam

Last year some of their freshman students had a score of at least 1180 on the exam

gain 9% profit averagegain 9% profit averagegain 9% profit averagegain 9% profit averagegain 9% profit averagegain7% profit average

starting salary $38,500 avgstarting salary $38,500 avgstarting salary $38,500 avg

starting salary $38,500 avg

starting salary $38,500 avgstarting salary $39,500 avg

starting salary $39,000 avg

Lost 6 lbs average

Lost 6 lbs averageLost 6 lbs averageLost 6 lbs averageLost 6 lbs averageLost 6 lbs averageLost 8 lbs averageLost 8 lbs averageLost 7 lbs average

1280 aveg

1280 aveg

1280 aveg

1280 aveg

1280 aveg

1180

1180

lost 8 lbs

sample standard deviationdo not touch do not touch

10 Sum ----> 50 squareroot ---> 07 count ---> 5 9 Devia Sum ---> 18

12 Total dev 10 4 4.59 1

12 4 Answer 2.12

already rounded

Sketching the least-squares regression line

X Y Least-Squares Regression7.7 4.56.8 4.59.4 2.72.6 8.56.2 4.58.3 3.71.6 10.22.1 9.7

9 2.65.2 7.13.4 7.7

4 7.52.8 9.910 1.7

5 6.48.2 3.26.1 4.84.6 7.9

Total 103 107.10x1-x --> -95.3 -102.6 <--y1-ypower 9082.09 10526.76squareroot 9082.09 102.6multi powe 9777.78multi sqrrt 10.129166answer 965.310

1 2 3 4 5 6 7 8 9 10 110

2

4

6

8

10

12

f(x) = − 1.00107756618672 x + 11.6783882954018

Least-Squares Regression

Least-Squares RegressionLinear (Least-Squares Regression)

1 2 3 4 5 6 7 8 9 10 110

2

4

6

8

10

12

f(x) = − 1.00107756618672 x + 11.6783882954018

Least-Squares Regression

Least-Squares RegressionLinear (Least-Squares Regression)

Let be X: the annual rate of return for a common biotechnology stock

P(X < 0) = P((X - 0.065)/0.05 < (0 - 0.065)/0.05) = P(Z < - 1.3) = 0.9032

x 0mean 4sd 5

-0.8

P(Z < - 1.3) = 0.9032 use ALEKS calculator TO FIND ANSWER

Standardized Normal Distribution. Yahoo name

Suppose that the annual rate of return for a common biotechnology stock is normally distributed with a mean of 6.5% and a standard deviation of 5%. Find the probability that the one-year return of this stock will be negative. Round your answer to at least four decimal places

since aleks do not have this, use 1-P(Z ≤ -1.3)

T Distribution

Use aleks calculator

1-P(t >-1.51 with df 26)

Consider a t distribution with 16 degrees of freedom. Find the value of such that . Round your answer to at least three decimal places.

Compute:

enter as 1-2P(t>1.83)

Find value of C

enter as is in ALEKS

Compute:

enter as 1-2P(t>1.67)

Find value of C

enter as is in ALEKS

code to use on P(z0

Greater than or equal to

Greater than

Consider a t distribution with 26 degrees of freedom. Compute .Round your answer to at least three decimal places

enter this in ALEKS Calcuator c = t

enter this in ALEKS Calcuator c = t

Less than or equal to

Less than

0.07155

0.95 170.05

0.025 df 17

0.1 200.9 df 20

-0.5 1-P(z…) more than

0.5 at mostat least

TX; DFY

0.5 1-P(z…)

-0.5

Tree diagrams for conditional probabilitiesMeans Ā is accents

P ( B | D ) = 0.6P(D П B) = 0.24 0.45

P(D) = 0.4 0.55

0.160.4

0.350.21

0.60.39

0.65

1.0000

D

D denote the event that a randomly chosen student did not regularly attend the discussion sessions

B denote the event that a randomly chosen student received a grade of "B" or higher in the cours

B denote the event that a randomly chosen student did not receive a grade of "B" or higher in the course

P(D П B) =P (B | D ) =

P ( B | D ) = P(D П B) =

P(D) =P(D П B) =

P ( B | D ) =

denote the event that a randomly chosen student (enrolled in the course) regularly attended the discussion sessions,

Means Ā is accents

denote the event that a randomly chosen student did not receive a grade of "B" or higher in the course

what is the probability that a randomly chosen student received a grade of "B" or higher in the coursewhat is the probability that a randomly chosen student did not receive a grade of "B" or higher in the course

denote the event that a randomly chosen student (enrolled in the course) regularly attended the discussion sessions,

a grade of "B" or higher in the course

Sample # --> 32

Column 1 Colunm 2 Col 1 * Col b --->3 6 186 7.75 46.5

14 10 1409 11.75 105.75

0000000

Sum ----> 310.25

ANSWER 9.695313

Trend lines for yearly data

t= 34.443.13

beginning year 1993ending year 2012

19 each year = 1 quarter20

total 97.04

t= 21.110.44

beginning year 1995ending year 2012

q1 q22 q3 q4

4 quarters per year17 1 2 3 4

69 70 71 72t 51.47 51.91 52.35 52.79

total 208.52

The following time series data represent the yearly amounts spent on advertising (in millions of dollars) by a large toy company:45.4, 41.5, 45.0, 50.6, 60.2, 63.7, 73.9, 73.8This series of data begins in year (i.e., time period corresponds to ). Using regression analysis, a linear trend line of the form was fit to the data. Using this information, generate a forecast for the total yearly amount of money that will be spent on advertising in

The following time series data represent the quarterly amounts spent on advertising (in millions of dollars) by a large toy company (read across):Quarter 1 Quarter 2 Quarter 3 Quarter 410.9 12.0 10.4 12.111.0 11.4 11.0 11.111.1 13.2 14.1 13.812.0 14.0 14.8 13.111.9 13.8 13.7 15.212.9 16.2 13.3 17.614.1 This series of data begins in Quarter of (i.e., time period corresponds to the first quarter of ). Using regression analysis, a linear trend line of the form was fit to the data. Using this information, generate a forecast for the total yearly amount of money that will be spent on advertising in 2012

Explained and unexplained variation and the least-squares regression line

the value of the residual isy = 46.94 0.82 x 257.926 SSR 3148.968 0.88

( 257.30 252.70 ) -5.23 SST 3590.248

or

or

Bivariate data obtained for the paired variables and are shown below, in the table labelled "Sample data." These data are plotted in the scatter plot in Figure 1, which also displays the least-squares regression line for the data. The equation for this line is .In the "Calculations" table are calculations involving the observed values, the mean of these values, and the values predicted from the regression equation.

The least-squares regression line given above is said to be a line which "best fits" the sample data. The term "best fits" is used because the line has an equation that minimizes the error sum of square

the value of R2 is the proportion

The variation in the sample values that is explained by the estimated linear relationship between X and Y is given by the regression sum of squares (SSR

The total variation in the sample y values is given by the total sum of squares

The variation in the sample values that is not explained by the estimated linear relationship between X and Y is given by the error sum of squares (SSE)

Bivariate data obtained for the paired variables and are shown below, in the table labelled "Sample data." These data are plotted in the scatter plot in Figure 1, which also displays the least-squares regression line for the data. The

In the "Calculations" table are calculations involving the observed values, the mean of these values, and the values

The least-squares regression line given above is said to be a line which "best fits" the sample data. The term "best fits"

by the estimated linear relationship between X and Y is given by

by the estimated linear relationship between X and Y is given

Seasonal indexes: Multiplicative model

Tt= 739 15 t Quarter 4 per yearstarting quarter # 1 corresponds to quarter #forecast quarter # 1.23ending year 2005starting year 2000

52019

1024Answer 1259.52

Tt= 516 3 t Monthlyforecast quarter # 0.49ending year 2006starting year 2003remaining months 3

33640

636Answer 311.64

The quarterly sales volume (in thousands of dollars) for a manufacturer of snow skis is seasonal (there tend to be more skis sold during fall and winter months than during warmer months). The following table shows the quarterly seasonal indexes for the company's sales:Quarter 1 Quarter 2 Quarter 3 Quarter 41.22 0.71 0.57 1.51Using several years of sales data, beginning in Quarter 2 of 1999 (i.e., corresponds to Quarter of ), the manufacturer summarized that the trend line that describes the quarterly sales (in thousands of dollars) has the form . Assuming that a multiplicative model can be used to describe the sales data, generate a forecast for the total dollar sales that the company can expect to receive in Quarter 4 of 2004 .Express your answer in thousands of dollars.

The quarterly sales volume (in thousands of dollars) for a manufacturer of snow skis is seasonal (there tend to be more skis sold during fall and winter months than during warmer months). The following table shows the quarterly seasonal indexes for the

Using several years of sales data, beginning in Quarter 2 of 1999 (i.e., corresponds to Quarter of ), the manufacturer summarized that the trend line that describes the quarterly sales (in thousands of dollars) has the form . Assuming that a multiplicative model can be used to describe the sales data, generate a forecast for the total dollar sales that the company can

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