Unit 4 Starters

Starter 7.1.1

• Suppose a fair coin is tossed 4 times. Find the probability that heads comes up exactly two times.

Answer• A tree diagram will show that there are 16

possible outcomes.

• The tree diagram also shows that 6 branches contain exactly two heads.

• Therefore P(two H) = 6/16

• Another approach:– There are 2x2x2x2=16 outcomes from 4 flips

– Choose any two flips to be heads: 4C2=6

– So P(two H) = 6/16

Starter 7.1.2

• A 9-sided die has three faces that show 1, two faces that show 2, and one face each showing 3, 4, 5, 6.

• Let X be the number that shows face-up.

• Draw the PDF histogram of X.

Answer

• What is the total area under the histogram?• (3/9) + (2/9) + 4 x (1/9) = 9/9 = 1

Starter 7.1.3• What is the difference between a discrete random variable and a continuous random variable?

– There are two important ideas here

• Assume a certain continuous random variable X is known to be N(55, 1.2). If you randomly choose one observation of X, what is the probability that:

a) X < 55b) X < 53c) X > 53

Hint: normalcdf(LB, UB, mean, s.d.)

Answer• A discrete random variable has a finite number of

outcomes that can only take on specific (usually integer) values. It can be represented by a table or histogram.

• A continuous random variable has an infinite number of outcomes. We can never list them all, and they typically include rational and irrational decimals. It can be represented by a density curve (only).

• P(X<55) = .5 by definition: 55 is the median• P(X<53) = normalcdf(0, 53, 55, 1.2) = .0478• P(X>53) = 1 – .0478 = .9521

Starter 7.2.1

• An unfair six-sided die comes up 1 on half of its rolls. The other half of its rolls are evenly spread through the other 5 outcomes.

• If X is the number that comes up, write PDF for X.

Answer

X 1 2 3 4 5 6

P(X) .5 .1 .1 .1 .1 .1

Starter 7.2.2

• Calculate from formula the variance of these data:

1, 2, 3, 4, 5

Answer

2 2 2 2 22

1 2 3 4 53

5

(1 3) (2 3) (3 3) (4 3) (5 3)2

5X

x

Starter 7.2.3

• Suppose I measure the heights of a class of fourth-graders and find the distribution of heights to be N(100, 2 cm).

• Then I have them all stand on top of a 10 cm high step and I measure again.

• What change would you expect in the mean?

• What change would you expect in the standard deviation?

Answer

• Since we added 10 to all values, the mean should increase by 10 cm.

• Adding 10 to all values did not change the spread; it just changed the location if we plotted on an axis. So we expect the standard deviation to remain unchanged at 2.

Starter 7.2.4• A bag contains 4 red marbles and one

white marble. Two marbles are chosen without replacement. What is the probability that they are both red?

• A worker is paid each week in the following manner: He draws two bills from a bag which contains four $20 bills and one $100 bill. What is the expected outcome of his average weekly pay over the long run?

Answer• Let X = the amount of weekly pay

– Notice that X can only be $40 or $120

• There are 5C2 (= 10) ways to draw two bills from the five in the bag

• There are 4C2 (= 6) ways to draw two $20 bills from the four in the bag– So P(40) = 6/10

• There are 4 ways to draw one $100 and one $20– So P(120) = 4/10

• E(X) = $40 (6/10) + $120 (4/10) = $72

Starter 8.1.1

• Here’s a game you will like: Let’s bet a dollar on this proposition: I will roll a fair die once. If it comes up 1 or 2, I win. If it comes up 3, 4, 5, or 6, you win!– What is the probability that I win?– What is the probability that I lose?– How much should we each bet to make this a

fair game?

Answer

• P(win) = 2/6 or 1/3

• P(lose) = 4/6 or 2/3

• Odds that I win are 2:4 (or 1:2) against, so you should bet $2 and I bet $1– Note: “fair” in this context means E(x) =0,

where x represents my net winnings.– E(x) = (+2)(1/3) + (-1)(2/3) = 0, so it’s fair

Starter 8.1.2

• Five white marbles are on a table. Two of them are to be painted with a “W” and the rest will be painted with a “L”.

• How many ways are there to choose the two marbles to be painted “W”?

Starter 8.1.3• A baseball player bats .325 over a full

season. If he bats 5 times today, find the probability he gets at least 3 hits.

Answer

• This is a binomial setting, so use the binomial CDF of at most 2 hits, then subtract from 1

• 1 – binomcdf(5, .325, 2) = .197

Starter 8.1.4• A manufacturer produces a large number of

toasters. From past experience, he knows that about 2% are defective. In a quality control procedure, we randomly select 20 toasters for testing. We want to determine the probability that no more than one of these toasters is defective.– Is this a binomial setting? Justify your answer.– Find the probability that exactly one toaster is defective.– Find the probability that at most one toaster is defective.– Find the mean and standard deviation for the problem.

Answer

• Two outcomes (good / defective), a fixed number of trials (20), independent trials, probability is fixed (2%), so it is binomial.

• P(X=1) = binompdf(20, .02, 1) = 27%

• P(X1) = binomcdf(20, .02, 1) = 94%

• μ = np = (20)(.02) = .4– So we expect on average .4 defectives in

each group of 20

• σ = √npq = √(20)(.02)(.98) = .626

Starter 8.2.1

• Fred Funk hits his tee shots straight most of the time. In fact, last year he put 78% of his tee shots in the fairway.

• In yesterday’s round, he hit only 7 of 14 shots in the fairway.

• What is the likelihood that he would have that bad a day (or worse) from the tee?

Answer• This is a binomial setting

– There are 14 trials where he succeeds or fails at hitting the fairway

– All trials are independent with fixed probability

• To do this poorly (or worse) means he hits at most 7 fairways.– He could hit 7 or 6 or 5 or …

• Binomcdf(14, .78, 7) = .02– So there is only a 2% chance he will do that

poorly.

Starter 8.2.2

• The SAT Math and Verbal sections are both designed to be approximately normal with a mean of 500 and standard deviation of 100.

• If we defined a new measure (TOTAL) by adding the scores on the two sections, what would you expect the mean and standard deviation of TOTAL to be?

(Assume math and verbal are independent)

Answer

500 500 1000TOTAL Math Verbal

2 2 2TOTAL Math Verbal

2 10000 10000 20000TOTAL

20000 141.4TOTAL

Starter

• Write the characteristics of the binomial setting.

• What is the difference between the binomial setting and the geometric setting?

Answer• The binomial setting has 4 characteristics:

– There are only two possible outcomes– Each trial has fixed probability of success– Each trial is independent of all other trials– There are a fixed number of trials; the variable of

interest is the number of successes

• The geometric setting has the same first three characteristics. The difference is in the fourth: there are an unknown number of trials; trials stop after the first success; the variable of interest is the number of trials until the first success