h4 expectedvalue normaldistribution solution

Bus310 – Homework expectedvalue-normaldistribution

HOMEWORK

EXPECTED VALUE AND SAMPLING DISTRIBUTIONS

Due: Sunday, March 2nd

Your Name: Simina Fluture

EXPECTED VALUES: E(X) = XiP(Xi)

PROBLEM 1 (9 points):

What is the expected value of a lottery ticket where there are only 1000 chances in a 10 million of winning the

grand prize of $100 Million?

$10 millions * 1000/10millions = 1000 /10 * $100 = $10,000

An expensive ticket.


You have been offered a chance to purchase a lottery ticket with a 3% chance of making $1,000; 25% chance of

making $100 (and 72% chance of making $0). The price of the ticket is $5.

Should you buy it or you shouldn’t? Why?

$1000 * 3/100 + $100 * 25/100 = $55


Party Time Co. will run a promotional raffle that offers a chance to win either a lifetime discount on

Spring Festival Merchandises (which results in a $1,000 savings) or a 5-year limited discount on any

party-goods (which results in a $500 savings).

Winning a lifetime discount has a 1-in-200 chance.

Winning a 5-year limited discount has a 1-in-20 chance.

Is this promotion worth it if the tickets cost $15?

EV = $1000 * 1/200 + $500 * 1/20 = $30

Yes, the expected value is higher than the ticket’s cost.


(SAMPLING) NORMAL DISTRIBUTIONS

General thoughts for this problem set:

Any normal distribution can be converted into a standard normal distribution by transforming

the normal random variable into the standard normal random variable:

Z =

X

μ = the mean and σ = is the standard deviation.

For each question you need to ask yourself if you're looking for X or you’re looking for Z:

You're looking for X if you're given a percentile and asked to come up with a concrete

number (number of years or score on a test)

You're looking for Z if you're given a concrete number (like a score or number of years)

and asked to find out a probability or a percentile. Based on the Z value from the table

you can find the area under the curve.

Before solving the homework, again, carefully review the lecture notes. It is important to

also understand why you are doing the specific computations. The next step would be for

you to check and understand the solved problems from the lecture notes and the solution

of the following problem. It is always useful to draw the picture first.

I strongly suggest you to use the z-table of the Lecture Notes 4C.

PROBLEM 1:

Suppose that New York State high school average scores, for students who graduate, are normally distributed

with a population mean of 70 and a population standard deviation of 13.

a) The “middle” 95% of all NYS high school students have average scores between ______ and ________?

b) What proportion of NYS high school students have average scores between 60 and 75?

c) Calculate the 14th percentile.

d) Calculate the 92nd percentile.

PROBLEM 1 - SOLUTION

a) The “middle” 95% of all NYS high school students have average scores between ___ and ___ ?

.4750 .4750

? 70 ? HS average

________________________________

-1.96 0 1.96 Z

If you want to consider the middle 95%, it means that you will have to take 5% off, 2.5% from the left tail and

2.5% from the right tail.


On one direction, from the middle (value 0) you will have a distance of 0.5 – 2.5% = 0.4750. Looking on the

chart this will correspond to a z=1.96

X = μ +(± Z)x σ = 70 + (± 1.96)x13 = 70 ± 1.96 (13) 70 ± 25.48

ANS: 95% of all NYS high school students have average scores between 44.52 and 95.48.

b) What proportion of NYS high school students have average scores between 60 and 75?

.2794 .1480

60 70 75 HS average

________________________________

-.77 0 .38 Z

Z1 = (60 – 70) / 13 = -0.77 (it corresponds to the area under the curve from the mean to value 60; area on

the left of the average)

Z2 = (75 – 70) / 13 = .38 (it corresponds to the area under the curve from the mean to value 75; area on

the right of the average)

We are interested on the total area ANS: .2794 + .1480 = .4274

42.74% of NYS high school students have average scores between 60 and 75.

Percentiles:

Here is a definition of percentile from Wikipedia:

"In statistics, a percentile (or centile) is the value of a variable below which a certain percent of observations

fall. For example, the 20th percentile is the value (or score) below which 20 percent of the observations may be

found. "

So, a percentile represents a value. If for example somebody's GPA score puts him/her in the 93% (93

percentile) it means that 93% of the students who took the test received a lower score.

It is important to acknowledge the fact that the mean represents the 50th percentile. This will give you an idea

in terms of where the other values for other specific percentiles fall. For example any percentile > 50%, will be

represented by a value greater than the mean and it will be (on the x axis) to the right of the mean value. If we

consider the normalized value (z-score) it will be a positive value.

On the other hand any percentile < 50%, will be represented by a value smaller than the mean and it will be to

the left of the mean value. If we consider the normalized value (z-score) it will be a negative value.

80th percentile

You need to find the X value corresponding to the 80%.

80 percentile is to the right of 50%, you will have a corresponding positive Z and an X value greater than the

mean.

How we are going to get this value? There is a formula that gives us the correspondence between X (value) and

http://en.wikipedia.org/wiki/Statistics

http://en.wikipedia.org/wiki/Percentage


its normalized value Z.

Z = (X-miu)/stdev

In the above formula we know miu (mean), we know stdev, and we can find out Z. Once we find out z then we

can compute X which is what we are looking for.

How do we find Z? Z represents the distance from mean to our value X.

The distance from mean to X (representing 80%) will correspond to the area

under the curve between 50% and 80% = .8 - .5= .3000

once we have the area we can retrieve Z and next we can compute X.

The reasoning is similar for let’s say the 27%; Now your X will be to the left of the mean and the corresponding

Z will be considered negative. Your X value will be lower than the mean.

The distance from mean to the X (representing 27%) will have as corresponding area

under the curve the area between 50% and 27% = .5 - .27= .2300

Keep in mind that z will be negative in this situation.

c) Calculate the 14th percentile.

.1400 .3600

? 70 HS average

________________________________

-1.08 0 Z

We need to calculate X. For that, first we need to find the Z value.

In a normal distribution mean=median=mode. The value 0 corresponds to the median = 50%.

This means that the 14% is on the left of 0 (the 50%).

What is the value of Z, the distance from 0?

The area under the curve will be 50% - 14% = 36% = 0.3600

0.3600 corresponds (from the table) to a Z of 1.08. This Z is to the left of 0, therefore we will consider its

negative value (-1.08)

Z = -1.08 = (X- 70) / 13

X = 55.96 years

d) Calculate the 92nd percentile. .4200

70 ? HS average


________________________________

0 1.41 Z

The value 0 corresponds to the median = 50%.

This means that the 92% is on the right of 0 (the 50%).

The area under the curve, from 0 to Z is: 92%-50% = 42% = 0.4200 (from the table) Z = 1.41

Z = 1.41 = (X- 70) / 13

X = 88.33 years

------------------------------------------------------------------------------------------------------------------------------


Suppose SPS professors have an average life normally distributed of 82 years, with a population standard

deviation of 5 years.

a) What percent of SPS professors will not live more than 75 years?

Z = (75-82)/5 = -1.4 (to the left of the mean) , the corresponding area = .4192

.4192 represents the probability of living between 75 and 82.

The probability of living less than 75 years is : .5000 - .4192 = 8.08%

b) Calculate the 90th percentile.

90% from the mean value we have an area under the curve of .4000

To an area of .4000 corresponds a z = +1.28

+1.28 = (X – 82)/5 X = 82 + 1.28*5 = 88.4

90% of people live below 88.4 years.

c) What percent of SPS professors will make it past the age of 75?

From a, it should be 100% - 8.08% = 91.92%

d) What proportion of SPS professors will live between 85 and 90 years?

Z1 = (85-82)/5 = .6 corresponding area = .2257 (to the right of the mean)

.2257 represents the probability of living between mean and 85 years.

Z2 = (90-82)/5 = 1.6 corresponding area 2 = .4452 (to the right of the mean)

.4452 represents the probability of living between 82 (mean) and 90

Probability of living between 85 and 90 = .4452 - .2257 = 21.95%

e) Calculate the 45th percentile.

45% is 5% from the mean value; 45% is to the left of the mean 50%.

Area = .05 from the mean z = -0.13 = (X-82)/5 X = 82 - .13*5 = 81.35 years

PROBLEM 3 (20 points)

Overall, the amount of work-hours involved in the festival preparation is normally distributed around

50 hours with a standard deviation of 7 hours.

a) What’s the probability that the mean number of work-hours will be between 30 and 55?

Z1 = (30-50)/7 = -2.86 corresponding area1 = .4979 (to the left of the mean)

It represents the probability of working between 30 and 50 hours

Z2 = (55-50)/7 = + .71 area2= .2612 (to the right of the mean) It represents the probability of having

worked hours between 50 and 55.

Answer: area1 + area2 = .2612 + .4979 = .7591 = 75.91%


b) The members at or below the 15%ile of number of worked-hours must attend a one-on-one meeting

with their supervisor. At least how many work-hours should you have in order to avoid attending such

session?

15%ile is to the left of the mean. Between 15%tile and the mean values, under the curve there is an area

of .3500.

Corresponding z = - 1.04 = (X-50)/7 X = 50 – 1.04*7 = 42.72 work-hours

c) How likely (what is the probability) is it to have the number of involved work-hours below 20?

Z = (50-20)/7 = 4.29 corresponding area = very, very close to .5

The probability of having below 20 work hours is almost 0.

d) How likely is it (what is the probability) that some employee will have his/her involved work- hours

between 45 and 60?

Z1 = (45-50)/7 = .71 corresponding area1 = .2612

Z2 = (60-50)/7 = 1.42 corresponding area2 = .4222

Answer: .2612 + .4222 = 68.34%


Science scores for high school seniors in the United States are normally distributed with a mean of 75 and a

standard deviation of 18. Students scoring in the top 5% are eligible for a special prize consisting of a laptop

and $2,000. What is the approximate cutoff score a student must get in order to receive the prize?

top 5% means the 95%tile an area of .4500 from the mean value corresponding z = 1.645 = (X-75)/18

You need a score of : 75 + 18*1.645 = 104.61


The average individual monthly spending in US for paging and messaging services is $12.75. If the standard

deviation is $2.5 and the amounts are normally distributed:

What is the probability that a random user pays more than $15.00?

Z = (15 – 12.75)/2.5 = .9 corresponding area = .3159; it represents the probability of paying between 12.75

and 15;

Answer: .5000 = .3159 = 18.41%

What is the probability that a user pays below $10.00?

Z = (10 – 12.75)/2.5 = -1.1 corresponding area = .3643 ; it represents the probability of paying between 10

and 12.75 dollars.

Answers: .5 - .3643 = 13.57%

Compute the 75% percentile.

The area under the curve between 50%ile and 75%ile is .2500 corresponding z = .67 = (X-12.75)/2.5

X = 12.75 + 2.5*.67 = $14.43

h4 expectedvalue normaldistribution solution

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