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Assume the likelihood that any flight on Northwest Airlines arrives within 15 minutes of the scheduled time is .90. We select four flights from yesterday for study.

a. What is the likelihood all four of the selected flights arrived within 15 minutes of the scheduled time?

b. What is the likelihood that none of the selected flights arrived within 15 minutes of the scheduled time?

c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes of the scheduled time?

a. What is the likelihood all four of the selected flights arrived within 15 minutes of the

scheduled time?

 

p^n

0.9^4 = 0.6561

b. What is the likelihood that none of the selected flights arrived within 15 minutes of

the scheduled time?

 

(1-p)^n

(1-0.9)^4

0.1^4

= 0.0001

 

c. What is the likelihood at least one of the selected flights did not arrive within 15 minutes

of the scheduled time?

 

at least one = 1 - prob(all arrived)

= 1 - 0.6561

= 0.3439

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of two emails per hour. Assume the arrival of these emails is approximated by the Poisson distribution.

a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday?

b. What is the probability she received 5 or more email during the same period?

c. What is the probability she did not receive any email during the period?

The average is 2 per hour.

I'll use the Poisson calculator here:

 

a. What is the probability Linda Lahey, company president, received exactly 1 email between 4 P.M. and 5 P.M. yesterday?

0.270670566473225

 

Using the formula:

f(x) = e-λλx / x!

= e^(-2)*(2)^1 / 1

same result as the above with a calculator

b. What is the probability she received 5 or more email during the same period?

= 1 - 4 or less

= 1 - 0.947346982656289

= 0.052653017343711

 

Or with the formula:

1 - e^(-2)*(2)^0 / 1 - e^(-2)*(2)^1 / 1 - e^(-2)*(2)^2 / 2 - e^(-2)*(2)^3 / 6 - e^(-2)*(2)^4 / 24

which evaluates to the above

c. What is the probability she did not receive any email during the period?

0.135335283236613

Using the formula:

f(x) = e-λλx / x!

= e^(-2)*(2)^0 / 1

Again, the same result as the above...

Fast Service Truck Lines uses the Ford Super Duty F-750 exclusively. Management made a study of the maintenance costs and determined the number of miles traveled during the year followed the normal distribution. The mean of the distribution was 60,000 miles and the standard deviation 2,000 miles.

a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?

b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?

c. What percent of the Fords traveled 62,000 miles or less during the year?

d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles?

Explain.

a. What percent of the Ford Super Duty F-750s logged 65,200 miles or more?

z table:

z score:

(65200-60000)/2000 = 2.6

prob(z > 2.6) from the table:

0.0047

= 0.47%

b. What percent of the trucks logged more than 57,060 but less than 58,280 miles?

z(57060) = (57060-60000)/2000 = -1.47

z(58280) = (58280-60000)/2000 = -0.86

prob(-1.47 < z < -0.86) from a table:

12.41%

c. What percent of the Fords traveled 62,000 miles or less during the year?

z = (62000-60000)/2000 = 1

prob(z < 1) from the table:

84.13%

d. Is it reasonable to conclude that any of the trucks were driven more than 70,000 miles?

Explain.

z(70000) = (70000-60000)/2000 = 5

It is unlikely that one truck was above 5 stdevs above the mean. The prob is almost 0.

Families USA, a monthly magazine that discusses issues related to health and health costs, surveyed 20 of its subscribers. It found that the annual health insurance premiums for a family with coverage through an employer averaged $10,979. The standard deviation of the sample was $1,000.

a. Based on this sample information, develop a 90 percent confidence interval for the population mean yearly premium.

b. How large a sample is needed to find the population mean within $250 at 99 percent confidence?

a. Based on this sample information, develop a 90 percent confidence interval for the

population mean yearly premium.

The t value for df = n-1 = 19, with 90% confidence is:

1.7291

 

The interval goes from:

mean - t*sd/sqrt(N) to mean + t*sd/sqrt(N)

10979 - 1.7291*1000/sqrt(20) to 10979 - 1.7291*1000/sqrt(20)

 

With a calculator:

10592.36 to 11365.64

b. How large a sample is needed to find the population mean within $250 at 99 percent

confidence?

The z value for 99% confidence is 2.5758

The formula for sample size is:

N = (z*sd/E)^2

N = (2.5758*1000/250)^2

N = 106.16

Round up to:

N = 107

An insurance company, based on past experience, estimates the mean damage for a natural disaster in its area is $5,000. After introducing several plans to prevent loss, it randomly samples 200 policyholders and finds the mean amount per claim was $4,800 with a standard deviation of $1,300. Does it appear the prevention plans were effective in reducing the mean amount of a claim? Use the .05 significance level.

H0: mean = 5000

Ha: mean < 5000

The critical value for one tail, from a table, at 0.05 is:

Z = -1.6449

Get the test statistic:

Z = (x-mu)/(sigma/sqrt(N))

Z = (4800-5000)/(1300/sqrt(200))

Z = -2.1757

This value is below the test statistic, so we reject the null hypothesis, and conclude that the mean has decreased.

The amount of income spent on housing is an important component of the cost of living. The total costs of housing for homeowners might include mortgage payments, property taxes, and utility costs (water, heat, electricity). An economist selected a sample of 20 homeowners in New England and then calculated these total housing costs as a percent of monthly income, five years ago and now. The information is reported below. Is it reasonable to conclude the percent is less now than five years ago?

Homeowner Five Years Ago Now Homeowner Five Years Ago Now

1 17% 10% 11 35% 32%

2 20 39 12 16 32

3 29 37 13 23 21

4 43 27 14 33 12

5 36 12 15 44 40

6 43 41 16 44 42

7 45 24 17 28 22

8 19 26 18 29 19

9 49 28 19 39 35

10 49 26 20 22 12

SEE EXCEL FILE

Martin Motors has in stock three cars of the same make and model. The president would like to compare the gas consumption of the three cars (labeled car A, car B, and car C) using four different types of gasoline. For each trial, a gallon of gasoline was added to an empty tank, and the car was driven until it ran out of gas. The following table shows the number of miles driven in each trial.

Distance (miles)

Types of Gasoline Car A Car B Car C

Regular 22.4 20.8 21.5

Super regular 17.0 19.4 20.7

Unleaded 19.2 20.2 21.2

Premium unleaded 20.3 18.6 20.4

SEE EXCEL FILE

A suburban hotel derives its gross income from its hotel and restaurant operations. The owners are interested in the relationship between the number of rooms occupied on a nightly basis and the revenue per day in the restaurant. Below is a sample of 25 days

(Monday through Thursday) from last year showing the restaurant income and number of rooms occupied.

Day Income Occupied Day Income Occupied

1 $1,452 23 14 $1,425 27

2 1,361 47 15 1,445 34

3 1,426 21 16 1,439 15

4 1,470 39 17 1,348 19

5 1,456 37 18 1,450 38

6 1,430 29 19 1,431 44

7 1,354 23 20 1,446 47

8 1,442 44 21 1,485 43

9 1,394 45 22 1,405 38

10 1,459 16 23 1,461 51

11 1,399 30 24 1,490 61

12 1,458 42 25 1,426 39

13 1,537 54

SEE EXCEL FILE

Fran’s Convenience Marts are located throughout metropolitan Erie, Pennsylvania. Fran, the owner, would like to expand into other communities in northwestern Pennsylvania and southwestern New York, such as Jamestown, Corry, Meadville, and Warren. To prepare her presentation to the local bank, she would like to better understand the factors that make

a particular outlet profitable. She must do all the work herself, so she will not be able to study all her outlets. She selects a random sample of 15 marts and records the average daily sales (Y), the floor space (area), the number of parking spaces, and the median income of families in that ZIP code region for each. The sample information is reported below.

Sampled Daily Store Parking Income

Mart Sales Area Spaces ($ thousands)

1 $1,840 532 6 44

2 1,746 478 4 51

3 1,812 530 7 45

4 1,806 508 7 46

5 1,792 514 5 44

6 1,825 556 6 46

7 1,811 541 4 49

8 1,803 513 6 52

9 1,830 532 5 46

10 1,827 537 5 46

11 1,764 499 3 48

12 1,825 510 8 47

13 1,763 490 4 48

14 1,846 516 8 45

15 1,815 482 7 43

SEE EXCEL FILE

The use of cellular phones in automobiles has increased dramatically in the last few years. Of concern to traffic experts, as well as manufacturers of cellular phones, is the effect on accident rates. Is someone who is using a cellular phone more likely to be involved in a traffic accident? What is your conclusion from the following sample information? Use the .05 significance level.

Had Accident Did Not Have an Accident

in the Last Year in the Last Year

Uses a cell phone 25 300

Does not use a cell phone 50 400

SEE EXCEL FILE

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