PROBABILITY DISTRIBUTIONSTUTORIAL 4

Exercise 35 (page 209)A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course.a. Compute the probability that two or fewer will withdraw.b. Compute the probability that exactly four will withdraw.

I COMPULSORY HOMEWORK

Exercise 35 (page 209)

A university found that 20% of its students withdraw without completing the introductorystatistics course Assume that 20 students registered for the course

a Compute the probability that two or fewer will withdraw

b Compute the probability that exactly four will withdraw

c Compute the probability that more than three will withdraw

d Compute the expected number of withdrawals

SOLUTION

Let x = number of students withdraw without completing the introductory statisticscourse

Assume that x follows a binomial probability distribution

a Compute the probability that two or fewer will withdraw.

Notes: Excel: True = 1 (cumulative)

False =0 (not cumulative

Exercise 42 (page 213)

More than 50 million guests stay at bed and breakfast (B&B) each year The Web site forthe Bed and Breakfast Inns of North American (http://www.bestinns.net), which averagesapproximately seven visitors per minute, enables many B&Bs to attract guests (time,September 2001)

a Compute the probability of no Web site visitors in a one-minute period

b Compute the probability of two or more Web site visitors in a one-minute period

c Compute the probability of one or more Web site visitors in a 30-second period

d Compute the probability of five or more Web site visitors in a one-minute period

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SOLUTION

Let x = the number of Web site visitors in a one-minute period

Assume that x is a Poisson random variable with a mean of 7, the probability function is:

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(0,7,FALSE)

b Compute the probability of two or more Web site visitors in a one-minute period.

Using Excel =1-POISSON.DIST(1,7/2,TRUE) =1-0.1358 = 0.9927

c Compute the probability of one or more Web site visitors in a 30-second period.

Let y = the number of Web site visitors in a 30-second period

Assume that y is a Poisson random variable with a mean of 3.5, the probabilityfunction is:

Using Excel =1-POISSON.DIST(0,3.5,true)

d Compute the probability of five or more Web site visitors in a one-minute period.

Draw a graph for the standard normal distribution Label the horizontal axis at values of-3, -2, -1, 0, 1, 2, and 3 Then use the table of probabilities for the standard normaldistribution inside the front cover of the text to compute the following probabilities:

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Given that z is a standard normal random variable, find z for each situation:

a The area to the left of z is 0.9750

b The area between 0 and z is 0.4750

c The area to the left of z is 0.7291

d The area to the right of z is 0.1314

e The area to the left of z is 0.6700

SOLUTION

a The area to the left of z is 0.9750

P (z ≤ z1) = 0.9750

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Using the Standard Normal Probability Table, we look for the location of the cumulativeprobability of 0.9750, so have z1 = 1.96

Using Excel = NORMSINV(0.9750)

b The area between 0 and z is 0.4750

P(0 ≤ z ≤ z2) = 0.4750

So, P (z ≤ z2) = P(0 ≤ z) + P(0 ≤ z ≤ z2) = 0.5 + 0.4750 = 0.9750

Similarly to (a), we have z2 = 1.96

c The area to the left of z is 0.7291

P (z ≤ z3) = 0.7291

Using the Standard Normal Probability Table, we look for the location of thecumulative probability of 0.7291, so have z3 = 0.61

Using Excel = NORMSINV(0.7291) = 0.6101

d The area to the right of z is 0.1314

P(Z≥Z4) = 0.1314, then the area to the left of z is therefore P(Z≤Z4)=1 – 0.1314 = 0.8686

P (z ≤ z4) = 0.8686

Using the Standard Normal Probability Table, we look for the location of thecumulative probability of 0.8686, so have z4 = 1.12

Using Excel = NORMSINV(0.8686) = 1.12

e The area to the left of z is 0.6700

P (z ≤ z5) = 0.6700

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Using the Standard Normal Probability Table, we look for the location of thecumulative probability of 0.6700, so have z5 = 0.44

Using Excel = NORMSINV(0.6700) = 0.44

e The area to the right of z is 0.3300

The average stock price for companies making up the S&P 500 is $30, and the standarddeviation is $8.20 (Business Week, Special Annual Issue, Spring 2003) Assume the stockprices are normally distributed.

a What is the probability a company will have a stock price of at least $40?

b What is the probability a company will have a stock price no higher than $20?

c How high does a stock price have to be to put a company in the top 10%?

SOLUTION

a What is the probability a company will have a stock price of at least $40?

At least 40 = Minimum = 40, or From 40

Let x is the stock price of a company and x is normally distributed with the mean of $30and standard deviation is $8.20

The probability a company will have a stock price of at least $40:

40 30 ( 40) ( ) ( 1.22) 1 ( 1.22) 1 0.8888 0.1112

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c How high does a stock price have to be to put a company in the top 10%?

To put the company to the top, the stock price of the company has to increase from current price ($30) to x 1 Then we need to looking for a specific cut off value of x 1

Notes: P(Z≥(x1-30)/82) is the right area of the curve but the probability of the wholearea under the curve =1 so P(Z≥(x1-70)/20) = 1-P(Z≤(x1-70)/20) So the probability now

= 0.9>0.5 so Z must be positive or in the right of Bell Shaped curve

II ADDITIONAL EXCERCISES:

STA228 PROBABILITY DISTRIBUTIONS TUTORIAL 4 Review the formula:

Exercise 27 – page 208:

Consider a binomial experiment with n = 20 and p = 0.70

a Compute f(12)

12 (20 12) 20!

In San Francisco, 30% of workers take public transportation daily

a In a sample of 10 workers, what is the probability that exactly three workers take

public transportation daily?

Let x = number of workers take public transportation daily

Assume that x follows a binomial probability distribution

The probability that exactly three workers take public transportation daily is:

f(3) = =BINOMDIST(3,10,0.3,FALSE) = 0.2668

b In a sample of 10 workers, what is the probability that at least three workers take

public transportation daily?

The probability that at least three workers take public transportation daily is:

P(x≥3) = 1- P(x≤2) =1 - BINOMDIST(2,10,0.3,true) = 0.6172

Consider a Poisson distribution with µ=3.

a Write the appropriate Poisson probability function.

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(2,3,FALSE)

or = POISSON((2,3,FALSE) (depend on different Excel version)

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(2,3,FALSE)

Using Excel =1-POISSON.DIST(1,3,TRUE)

Note: True=1: Cumulative

False = 0: not cumulative

Exercise 39 – page 212:

Consider a Poisson distribution with a mean of two occurrences per time period

a Write the appropriate Poisson probability function.

With x is the number of occurrences in one time period, the probability function is:

With e = 2.71828 (the natural constant)

b What is the expected number of occurrences in three time periods?

µ = 2x3 = 6

c Write the appropriate Poisson probability function to determine the probability of x

occurrences in three time periods

With x is the number of occurrences in three time periods, the probability function is

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d Compute the probability of two occurrences in one time period.

Apply the function in (a), the probability of two occurrences in one time period is:

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(2,2,FALSE)

e Compute the probability of six occurrences in three time periods.

Apply the function in (c), the probability of six occurrences in three time periods is:

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(6,6,FALSE)

Ask students to compare the results of (d) and (e) Explain why they are different

f Compute the probability of five occurrences in two time periods.

Let x3 = the number of occurrences in two time periods

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(5,4,FALSE)

Exercise 44 – page 213:

An average of 15 aircraft accidents occur each year

a Compute the mean number of aircraft accidents per month.

Let x = the number of aircraft accidents in one month

Assume that the events of aircraft accidents are independent and their probabilities arethe same, so x is a Poisson random variable with the mean of 15/12 = 1.25

b Compute the probability of no accidents during a month.

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(0,1.25,FALSE)

c Compute the probability of exactly one accident during a month.

Using Poisson probabilities table page 602 (Appendix B)

Using Excel =POISSON.DIST(1,1.25,FALSE)

d Compute the probability of more than one accident during a month.

a Sketch a normal curve for the probability density function Label the horizontal axis

with values of 35, 40, 45, 50, 55, 60 and 65

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b What is the probability the random variable will assume a value between 45 and

55?

The range from 45 to 55 is one standard deviation of the mean (from 50 – 1.5 to 50 +1.5),therefore, according to the empirical rule applied for a normal distribution, there are 68%

of probability that the random variable will assume a value between 45 and 55

c What is the probability the random variable will assume a value between 40 and

60?

The range from 40 to 60 is two standard deviations of the mean (from 50 – 2.5 to 50+2.5), therefore, according to the empirical rule applied for a normal distribution, thereare 95.44% of probability that the random variable will assume a value between 40 and60

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= 0.2389

Exercise 20 – page 241:

In January 2003, the American worker spent an average of 77 hours logged on to theInternet while at work Assume the population mean is 77 hours, the times are normallydistributed, and that the standard deviation is 20 hours

a What is the probability that in January 2003 a randomly selected worker spent fewer than 50 hours logged on to the Internet?

Let x = number of hours that a worker spent logged on to the Internet in January 2003

X is normally distributed with the mean of 77 hours and the standard deviation of 20hours

The probability that in January 2003 a randomly selected worker spent fewer than 50hours logged on to the Internet is:

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100 70 ( 100) ( ) ( 1.5) 1 ( 1.5) 1 0.9332 0.0668

20

c A person is classified as a heavy user if he or she is in the upper 20% of usage In

January 2003, how many hours did a worker have to be logged on to the Internet to beconsidered a heavy user?

We need to looking for a specific cut off value of x1 where P(x ≥ x1 ) = 20% = 0.2

1 1 1 1

P(Z≥(x1-70)/20) is the right area of the curve but the probability of the whole area underthe curve =1 so P(Z≥(x1-70)/20) = 1-P(Z≤(x1-70)/20) So the probability now = 0.8>0.5

so Z must be positive or in the right of Bell Shaped curve

Exercise 22 – page 242:

The mean hourly pay rate for financial managers in the East North Central region is

$32.62, and the standard deviation is $2.32 Assume that pay rates are normallydistributed

a What is the probability a financial manager earns between $30 and $35 per hour?

Let x = hourly pay rate for a financial manager in the East North Central region

X is normally distributed with the mean of $32.62 and the standard deviation of $2.32.The probability a financial manager earns between $30 and $35 per hour is:

1.282.32

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Notes: P(Z≥(x1-32.62)/2.32) is the right area of the curve but the probability of the wholearea under the curve =1 so P(Z≥(x1-70)/20) = 1-P(Z≤(x1-70)/20) So the probability now

= 0.9>0.5 so Z must be positive or in the right of Bell Shaped curve

c For a randomly selected financial manager, what is the probability the manager earned less than $28 per hour?

The probability the manager earned less than $28 per hour is:

28 32.62

2.32