Chapter 9Chapter 9

Review ProblemsReview Problems

Problem #1Problem #1 Jack is taking Math 110 this semester Jack is taking Math 110 this semester

and wants to evaluate his progress and wants to evaluate his progress using data from quizzes he has taken using data from quizzes he has taken so far. His quiz scores are as follows: so far. His quiz scores are as follows: 75, 80, 60, 90, 100, 40, 75, 95, 60, 75.75, 80, 60, 90, 100, 40, 75, 95, 60, 75.

Find the following:Find the following:a.) meana.) meanb.) medianb.) medianc.) modec.) moded.) range d.) range e.) standard deviatione.) standard deviation

Problem #2Problem #2The normal monthly temperatures (in The normal monthly temperatures (in

degrees Fahrenheit) for the month of degrees Fahrenheit) for the month of July is listed for 20 different U.S. cities.July is listed for 20 different U.S. cities.95.195.195.795.7 77.177.1 83.483.4 90.990.981.081.079.479.4 88.388.3 78.978.9 87.287.285.685.696.296.2 89.589.5 79.879.8 77.377.393.393.390.890.8 86.186.1 88.788.7 79.879.8

Find the mean, median, and mode of Find the mean, median, and mode of this data.this data.

Problem #3Problem #3Using the Using the

employment employment information in the information in the table at right for table at right for Omega Omega Corporation, find Corporation, find the mean, mode, the mean, mode, and estimated and estimated median of the median of the grouped data.grouped data.

Years of Years of ServiceService

Number Number of of

EmployeeEmployeess

1 – 51 – 5 1515

6 – 106 – 10 3232

11 – 1511 – 15 4747

16 – 2016 – 20 2929

21 – 2521 – 25 1818

26 – 3026 – 30 66

Problem #4Problem #4

The following The following distribution of distribution of commuting distances commuting distances was obtained for a was obtained for a sample of Mutual of sample of Mutual of Nebraska employees.Nebraska employees.

Find the mean and Find the mean and standard deviation standard deviation for the commuting for the commuting distances.distances.

Distance Distance (miles)(miles)

FrequencFrequencyy

1.0 – 3.01.0 – 3.0 22

3.0 – 5.03.0 – 5.0 66

5.0 – 7.05.0 – 7.0 1212

7.0 – 9.07.0 – 9.0 5050

9.0 – 11.09.0 – 11.0 3535

11.0 – 11.0 – 13.013.0 1515

13.0 – 13.0 – 15.015.0 55

Problem #5Problem #5

Find the percent of the total area Find the percent of the total area under a normal curve that is under a normal curve that is contained in the interval between z = contained in the interval between z = -1.95 and z = -.25.-1.95 and z = -.25.

Problem #6Problem #6

Find the z-score such that 3.5% of Find the z-score such that 3.5% of the total area is to the left of z.the total area is to the left of z.

Problem #7Problem #7

Find a z-score such that 4% of the Find a z-score such that 4% of the total area is to the right of z.total area is to the right of z.

Problem #8Problem #8

Computers are shut down for certain Computers are shut down for certain periods of time for routine periods of time for routine maintenance, installation of new maintenance, installation of new hardware, and so on. The down hardware, and so on. The down times for a particular computer are times for a particular computer are normally distributed with a mean of normally distributed with a mean of 1.5 hours and a standard deviation of 1.5 hours and a standard deviation of 0.4 hour. What percentage of the 0.4 hour. What percentage of the down times exceed 2 hours?down times exceed 2 hours?

Problem #9Problem #9

According to the November 1993 issue According to the November 1993 issue of of Harper’sHarper’s magazine, kids spend from magazine, kids spend from 1200 to 1800 hours a year in front of 1200 to 1800 hours a year in front of the television set. Suppose the time the television set. Suppose the time spent by kids in front of the television spent by kids in front of the television set is normally distributed with a mean set is normally distributed with a mean equal to 1500 hours and a standard equal to 1500 hours and a standard deviation equal to 100 hours. What deviation equal to 100 hours. What percentage spend between 1400 and percentage spend between 1400 and 1600 hours?1600 hours?

Problem #10Problem #10

The length of useful life of a The length of useful life of a fluorescent tube used for indoor fluorescent tube used for indoor gardening is normally distributed. The gardening is normally distributed. The useful life has a mean of 600 hours useful life has a mean of 600 hours and a standard deviation of 40 hours. and a standard deviation of 40 hours. A company installs 10,000 of these A company installs 10,000 of these fluorescent tubes. Find the fluorescent tubes. Find the approximate number of tubes that will:approximate number of tubes that will:a.) last more than 700 hours.a.) last more than 700 hours.b.) last less than 600 hours.b.) last less than 600 hours.

AnswersAnswers(Solutions available in SSCC Library at the Reserve (Solutions available in SSCC Library at the Reserve

Desk.)Desk.)1.) 1.) mean = 75mean = 75

median = 75median = 75mode = 75mode = 75range = 60range = 60standard deviation = 18.105standard deviation = 18.105

2.) 2.) mean = 86.205mean = 86.205median = 86.65median = 86.65mode = 79.8mode = 79.8

3.) 3.) mean = 13.714mean = 13.714mode = occurs in interval 11-15 yearsmode = occurs in interval 11-15 years

estimated mode = 13 estimated mode = 13 median = occurs in interval 11-15 median = occurs in interval 11-15

yearsyears estimated median = 13 estimated median = 13

Answers continuedAnswers continued(Solutions available in SSCC Library at the Reserve (Solutions available in SSCC Library at the Reserve

Desk.)Desk.)

4.) 4.) mean = 8.8mean = 8.8standard deviation = 2.369standard deviation = 2.369

5.)5.) .3757.37576.) z = -1.816.) z = -1.817.) z = 1.757.) z = 1.758.) 10.56%8.) 10.56%9.) 68.26%9.) 68.26%10.) a.)10.) a.) 62 tubes62 tubes

b.) b.) 5000 tubes5000 tubes