# Times of Populations Ranges

Please see attached file.

Exercises 80, 82, and 87 (Ch. 3)

80).

Creek Ratz is a very popular restaurant located along the coast of northern Florida. They serve a variety of steak and seafood dinners. During the summer beach season, they do not take reservations or accept "call ahead" seating. Management of the restaurant is concerned with the time a patron must wait before being seated for dinner. Listed below is the wait time, in minutes, for the 25 tables seated last Saturday night.

28 39 23 67 37 28 56 40 28 50

51 45 44 65 61 27 24 61 34 44

64 25 24 27 29 Ã‚ Ã‚ Ã‚ Ã‚ Ã‚

1. Explain why the times are a population.

2. Find the mean and median of the times.

3. Find the range and the standard deviation of the times.

82). The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Albuquerque, New Mexico during the month of May last year.

Electricity Cost Frequency

$ 80 up to $100 3

100 up to 120 8

120 up to 140 12

140 up to 160 16

160 up to 180 7

180 up to 200 4

Total 50

1. Estimate the mean cost.

2. Estimate the standard deviation.

3. Use the Empirical Rule to estimate the proportion of costs within two standard deviations of the mean. What are these limits?

87). Refer to the Real Estate data, which reports information on homes sold in the Denver, Colorado, area last year.

a. Select the variable selling price.

1. Find the mean, median, and the standard deviation.

2. Write a brief summary of the distribution of selling prices.

b. Select the variable referring to the area of the home in square feet.

3. Find the mean, median, and the standard deviation.

4. Write a brief summary of the distribution of the area of homes.

Exercises 34, 36, and 38 (Ch. 5)

34). P(A1) = .20, P(A2) = .40, and P(A2) = .40. P(B1|A2) = .25. P(B1|A2) = .05, and P(B1|A3) = .10. Use Bayes' theorem to determine P(A3 | B1).

36). Dr. Stallter has been teaching basic statistics for many years. She knows that 80 percent of the students will complete the assigned problems. She has also determined that among those who do their assignments, 90 percent will pass the course. Among those students who do not do their homework, 60 percent will pass. Mike Fishbaugh took statistics last semester from Dr. Stallter and received a passing grade. What is the probability that he completed the assignments?

38). One-fourth of the residents of the Burning Ridge Estates leave their garage doors open when they are away from home. The local chief of police estimates that 5 percent of the garages with open doors will have something stolen, but only 1 percent of those closed will have something stolen. If a garage is robbed, what is the probability the doors were left open?

Exercises 45 and 62 (Ch. 6)

45). A Tamiami shearing machine is producing 10 percent defective pieces, which is abnormally high. The quality control engineer has been checking the output by almost continuous sampling since the abnormal condition began. What is the probability that in a sample of 10 pieces:

1. Exactly 5 will be defective?

2. 5 or more will be defective?

62). Suppose 1.5 percent of the antennas on new Nokia cell phones are defective. For a random sample of 200 antennas, find the probability that:

1. None of the antennas is defective.

2. Three or more of the antennas are defective.

Exercises 42 and 45 (Ch. 7)

42). The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution.

1). Determine the z values for 29 and 34 hours. What percent of the garages take between 32 hours and 34 hours to erect?

2) What percent of the garages take between 29 hours and 34 hours to erect?

3) What percent of the garages take 28.7 hours or less to erect?

4) Of the garages, 5 percent take how many hours or more to erect?

45) Shaver Manufacturing, Inc., offers dental insurance to its employees. A recent study by the human resource director

shows the annual cost per employee per year followed the normal probability distribution, with a mean of $1,280 and a

standard deviation of $420 per year.

1). What fraction of the employees cost more than $1,500 per year for dental expenses?

2). What fraction of the employees cost between $1,500 and $2,000 per year?

3). Estimate the percent that did not have any dental expense.

4). What was the cost for the 10 percent of employees who incurred the highest dental expense?

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#### Solution Preview

The solutions are given in the attached file.

Exercise (80):

Question (1) : Explain why the times are a population.

Answer : In the description of the question it is not said that the given numbers in the table are a part of the group, but it is said that the numbers represent the waiting time of the entire group (of 25 customers). Therefore, it is not a sample but represent the waiting time of entire population of customers on last Saturday.

Question (2): Find the mean and median of the time.

Answer : For computing the mean and standard deviation we do the required computations as below.

Waiting times

x x2

1 28 784

2 39 1521

3 23 529

4 67 4489

5 37 1369

6 28 784

7 56 3136

8 40 1600

9 28 784

10 50 2500

11 51 2601

12 45 2025

13 44 1936

14 65 4225

15 61 3721

16 27 729

17 24 576

18 61 3721

19 34 1156

20 44 1936

21 64 4096

22 25 625

23 24 576

24 27 729

25 29 841

SUM = 1021 46989

From the above table, we have ,

Mean = = 40.84

For finding the median, we arranged the given waiting times in increasing order as below.

23 24 24 25 27 27 28 28 28 29 34 37 39 40 44 44 45 50 51 56 61 61 64 65 67

Median = The number in the Center place = 39

Question (3): Find the range and the standard deviation of the times.

Answer : Range is the difference between the highest and the lowest number.

Therefore, Range = 67 - 23 = 44

Standard Deviation = = 14.55

Exercise (82): To find the mean and the standard deviation, we compute the required values as shown in the following table.

Answer (1): ( Finding Mean Cost )

Cost Frequency Mid value

f X f Ã— X

80 to 100 3 90 270

100 to 120 8 110 880

120 to 140 12 130 1560

140 to 160 16 150 2400

160 to 180 7 170 1190

180 to 200 4 190 760

SUM = 50 7060

From the above table, we have N = 50 and

Mean ...

#### Solution Summary

The expert explains why the times are a population. The mean and median of the times are found.