Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 7: The Central Limit Theorem

49 problems · 1 figure

7-1 §7.1 source ↗

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let X ¯ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

What is the mean, standard deviation, and sample size?

Answer

mean = 4 hours, standard deviation = 1.2 hours, sample size = 16

7-3 §7.1 source ↗

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let X ¯ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.

a.

b. P(________ < x < ________) = _______

Figure 7.16 in the book ↗

Answer

a. Check student's solution.
b. 3.5, 4.25, 0.2441

7-5 §7.1 source ↗

Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let X ¯ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.

What causes the probabilities in Exercise 7.3 and Exercise 7.4 to be different?

Answer

The fact that the two distributions are different accounts for the different probabilities.

7-7 §7.2 source ↗

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Find the probability that the sum of the 95 values is greater than 7,650.

Answer

0.3345

7-9 §7.2 source ↗

Use the following information to answer the next four exercises: An unknown distribution has a mean of 80 and a standard deviation of 12. A sample size of 95 is drawn randomly from the population.

Find the sum that is two standard deviations above the mean of the sums.

Answer

7833.92

7-11 §7.2 source ↗


Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Find the probability that the sum of the 40 values is greater than 7,500.

Answer

0.0089

7-13 §7.2 source ↗


Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Find the sum that is one standard deviation above the mean of the sums.

Answer

7326.49

7-15 §7.2 source ↗


Use the following information to answer the next five exercises: The distribution of results from a cholesterol test has a mean of 180 and a standard deviation of 20. A sample size of 40 is drawn randomly.

Find the percentage of sums between 1.5 standard deviations below the mean of the sums and one standard deviation above the mean of the sums.

Answer

77.45%

7-17 §7.2 source ↗


Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Find the probability that the sum of the 100 values is less than 3,900.

Answer

0.4207

7-19 §7.2 source ↗


Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Find the sum with a z-score of –2.5.

Answer

3,888.5

7-21 §7.2 source ↗


Use the following information to answer the next six exercises: A researcher measures the amount of sugar in several cans of the same type of soda. The mean is 39.01 with a standard deviation of 0.5. The researcher randomly selects a sample of 100.

Find the probability that the sums will fall between the z-scores –2 and 1.

Answer

0.8186

7-23 §7.2 source ↗


Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

What is the standard deviation of ΣX?

Answer

5

7-25 §7.2 source ↗


Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

What is P(Σx > 290)?

Answer

0.9772

7-27 §7.2 source ↗


Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

In order for the sums of a distribution to approach a normal distribution, what must be true?

Answer

The sample size, n, gets larger.

7-29 §7.2 source ↗


Use the following information to answer the next four exercises: An unknown distribution has a mean 12 and a standard deviation of one. A sample size of 25 is taken. Let X = the object of interest.

An unknown distribution has a mean of 25 and a standard deviation of six. Let X = one object from this distribution. What is the sample size if the standard deviation of ΣX is 42?

Answer

49

7-31 §7.2 source ↗


Use the following information to answer the next three exercises:
A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

What is the z-score for Σx = 840?

Answer

26.00

7-33 §7.2 source ↗


Use the following information to answer the next three exercises:
A market researcher analyzes how many electronics devices customers buy in a single purchase. The distribution has a mean of three with a standard deviation of 0.7. She samples 400 customers.

What is P(Σx < 1186)?

Answer

0.1587

7-35 §7.2 source ↗


Use the following information to answer the next three exercises:
An unkwon distribution has a mean of 100, a standard deviation of 100, and a sample size of 100. Let X = one object of interest.

What is the standard deviation of ΣX?

Answer

1000

7-37 §7.3 source ↗

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

a. What is the distribution for the weights of one 25-pound lifting weight? What are the mean and standard deivation?

b. What is the distribution for the mean weight of 100 25-pound lifting weights?

c. Find the probability that the mean actual weight for the 100 weights is less than 24.9.

Answer

a. U(24, 26), 25, 0.5774

b. N(25, 0.0577)

c. 0.0416

7-39 §7.3 source ↗

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

Find the probability that the mean actual weight for the 100 weights is greater than 25.2.

Answer

0.0003

7-41 §7.3 source ↗

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

Find the 90th percentile for the mean weight for the 100 weights.

Answer

25.07

7-43 §7.3 source ↗

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

a. What is the distribution for the sum of the weights of 100 25-pound lifting weights?

b. Find P(Σx < 2450).

Answer

a. N(2500, 5.7735)

b. 0

7-45 §7.3 source ↗

Use the following information to answer the next 10 exercises: A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely, so the distribution of weights is uniform. A sample of 100 weights is taken.

Find the 90th percentile for the total weight of the 100 weights.

Answer

2507.40

7-47 §7.3 source ↗


Use the following information to answer the next five exercises:
The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

a. What is the standard deviation?

b. What is the parameter m?

Answer

a. 10

b. 1 10

7-49 §7.3 source ↗


Use the following information to answer the next five exercises:
The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

What is the distribution for the mean length of time 64 batteries last?

Answer

N ( 10,  10 8 )

7-51 §7.3 source ↗


Use the following information to answer the next five exercises:
The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

Find the probability that the sample mean is between 7 and 11.

Answer

0.7799

7-53 §7.3 source ↗


Use the following information to answer the next five exercises:
The length of time a particular smartphone's battery lasts follows an exponential distribution with a mean of ten months. A sample of 64 of these smartphones is taken.

Find the interquartile range (IQR) for the mean amount of time 64 batteries last.

Answer

1.69

7-55 §7.3 source ↗


Use the following information to answer the next six exercises:
A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.

Find P(Σx > 420).

Answer

0.0072

7-57 §7.3 source ↗


Use the following information to answer the next six exercises:
A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.

Find the 15th percentile for the sums.

Answer

391.54

7-59 §7.3 source ↗


Use the following information to answer the next six exercises:
A uniform distribution has a minimum of six and a maximum of ten. A sample of 50 is taken.

Find the third quartile for the sums.

Answer

405.51

7-61 §7.1 source ↗

Previously, De Anza's statistics students estimated that the amount of change daytime statistics students carry is exponentially distributed with a mean of $0.88. Suppose that we randomly pick 25 daytime statistics students.

a. In words, Χ = ____________.

b. Χ ~ _____(_____, _____)

c. In words, X ¯ = ____________.

d. X ¯ ~ ______ (______, ______)

e. Find the probability that an individual had between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.

f. Find the probability that the average amount of change of the 25 students was between $0.80 and $1.00. Graph the situation, and shade in the area to be determined.

g. Explain why there is a difference in part (e) and part (f).

Answer

a. Χ = amount of change students carry

b. Χ ~ Exp(1/0.88) or approximately Χ ~ Exp(1.1364)

c. X ¯ = average amount of change carried by a sample of 25 students.

d. X ¯ ~ N(0.88, 0.176)

e. 0.0819

f. 0.4276

g. The probability in part (e) represents the probability of an individual value. In part (f), the probability describes the mean of a sample of 25. Part (f) relies on the central limit theorem, so the distributions are different. Part (e) is exponential and part (f) is normal.

7-63 §7.1 source ↗

According to the Internal Revenue Service, the average length of time for an individual to complete (keep records for, learn, prepare, copy, assemble, and send) IRS Form 1040 is 10.53 hours (without any attached schedules). The distribution is unknown. Let us assume that the standard deviation is two hours. Suppose we randomly sample 36 taxpayers.

a. In words, Χ = _____________.

b. In words, X ¯ = _____________.

c. X ¯ ~ _____(_____, _____)

d. Would you be surprised if the 36 taxpayers finished their Form 1040s in an average of more than 12 hours? Explain why or why not in complete sentences.

e. Would you be surprised if one taxpayer finished his or her Form 1040 in more than 12 hours? In a complete sentence, explain why.

Answer

a. length of time for an individual to complete IRS form 1040, in hours

b. mean length of time for a sample of 36 taxpayers to complete IRS form 1040, in hours

c. N ( 10.53,  1 3 )

d. Yes, I would be surprised, because the probability is almost 0.

e. No, I would not be totally surprised because the probability is 0.2312.

7-65 §7.1 source ↗

The length of songs in a collector’s online album collection is uniformly distributed from 2 to 3.5 minutes. Suppose we randomly pick five albums from the collection. There are a total of 43 songs on the five albums.

a. In words, Χ = _________.

b. Χ ~ _____________

c. In words, X ¯ = _____________.

d. X ¯ ~ _____(_____, _____)

e. Find the first quartile for the average song length.

f. The IQR for the average song length is _______–_______.

Answer

a. the length of a song, in minutes, in the collection

b. U(2, 3.5)

c. the average length, in minutes, of the songs from a sample of five albums from the collection

d. N(2.75, 0.0220)

e. 2.74 minutes

f. 0.03 minutes

7-67 §7.1 source ↗

Determine which of the following are true and which are false. Then, in complete sentences, justify your answers.

a. When the sample size is large, the mean of X ¯ is approximately equal to the mean of Χ.

b. When the sample size is large, X ¯ is approximately normally distributed.

c. When the sample size is large, the standard deviation of X ¯ is approximately the same as the standard deviation of Χ.

Answer

a. True. The mean of a sampling distribution of the means is approximately the mean of the data distribution.

b. True. According to the central limit theorem, the larger the sample, the closer the sampling distribution of the means becomes normal.

c. The standard deviation of the sampling distribution of the means will decrease, making it approximately the same as the standard deviation of X as the sample size increases.

7-69 §7.1 source ↗

The distribution of income in some economically developing countries is considered wedge shaped (many very poor people, very few middle income people, and even fewer wealthy people). Suppose we pick a country with a wedge-shaped distribution. Let the average salary be $2,000 per year with a standard deviation of $8,000. We randomly survey 1,000 residents of that country.

a. In words, Χ = _____________.

b. In words, X ¯ = _____________.

c. X ¯ ~ _____(_____, _____)

d. How is it possible for the standard deviation to be greater than the average?

e. Why is it more likely that the average salary of the 1,000 residents will be from $2,000 to $2,100 than from $2,100 to $2,200?

Answer

a. X = the yearly income of someone in a Third World country

b. the average salary from samples of 1,000 residents of a Third World country

c. X ¯ ∼ N ( 2,000,  8,000 1,000 )

d. Very wide differences in data values can have averages smaller than standard deviations.

e. The distribution of the sample mean will have higher probabilities closer to the population mean.
P(2,000 < X ¯ < 2,100) = 0.1537
P(2,100 < X ¯ < 2,200) = 0.1317

7-71 §7.1 source ↗

The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations. The distribution to use for the average cost of gasoline for the 16 gas stations is:

a. X ¯ ~ N(4.59, 0.10)

b. X ¯ ~ N ( 4.59,  0.10 16 )

c. X ¯ ~ N ( 4.59,  16 0.10 )

d. X ¯ ~ N ( 4.59,  16 0.10 )

Answer

b

7-73 §7.2 source ↗

Suppose that the duration of a particular type of criminal trial is known to have a mean of 21 days and a standard deviation of seven days. We randomly sample nine trials.

a. In words, ΣX = ______________.

b. ΣX ~ _____(_____, _____)

c. Find the probability that the total length of the nine trials is at least 225 days.

d. Ninety percent of the total of nine of these types of trials will last at least how long?

Answer

a. the total length of time for nine criminal trials

b. N(189, 21)

c. 0.0432

d. 162.09; 90 percent of the total nine trials of this type will last 162 days or more.

7-75 §7.2 source ↗

Salaries for entry-level managers at a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from these restaurants.

a. In words, X = ______________.

b. X ~ _____(_____, _____)

c. In words, ΣX = _____________.

d. ΣX ~ _____(_____, _____)

e. Find the probability that the managers earn a total of over $400,000.

f. Find the 90th percentile for an individual manager's salary.

g. Find the 90th percentile for the sum of ten managers' salary.

h. If we surveyed 70 managers instead of ten, graphically, how would that change the distribution in part (d)?

i. If each of the 70 managers received a $3,000 raise, graphically, how would that change the distribution in part (b)?

Answer

a. X = the salary of one elementary school teacher in the district

b. X ~ N(44000, 6500)

c. ΣX ~ sum of the salaries of 10 elementary school teachers in the sample

d. ΣX ~ N(44,000, 20,554.80)

e. 0.9742

f. $52,330.09

g. 466,342.04

h. Sampling 70 teachers instead of 10 would cause the distribution to be more spread out. It would be a more symmetrical normal curve.

i. If every teacher received a $3,000 raise, the distribution of X would shift to the right by $3,000. In other words, it would have a mean of $47,000.

7-77 §7.3 source ↗

Table 7.7 in the book ↗
Company Closing Stock Prices
1 8.625
2 30.25
3 27.625
4 46.75
5 32.875
6 18.25
7 5
8 0.125
9 2.9375
10 6.875
11 28.25
12 24.25
13 21
14 1.5
15 30.25
16 71
17 43.5
18 49.25
19 2.5625
20 31
21 16.5
22 9.5
23 18.5
24 18
25 9
26 10.5
27 16.625
28 1.25
29 18
30 12.87
31 7
32 12.875
33 2.875
34 60.25
35 29.25

The closing stock prices of 35 U.S. semiconductor manufacturers are given as follows:

a. In words, Χ = ______________.

b.

i. x ¯ = _____

ii. sx = _____

iii. n = _____

c. Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of 10.

d. In words, describe the distribution of the stock prices.

e. Randomly average five stock prices together. (Use a random number generator.) Continue averaging five prices together until you have 10 averages. List those 10 averages.

f. Use the 10 averages from part (e) to calculate the following:

i. x ¯ = _____

ii. sx = _____

g. Construct a histogram of the distribution of the averages. Start at x = –0.0005. Use bar widths of 10.

h. Does this histogram look like the graph in Part (c)?

i. In one or two complete sentences, explain why the graphs either look the same or look different.

j. Based on the theory of the central limit theorem, X ¯ ~ _____(_____, ____).

Answer

a. X = the closing stock prices for U.S. semiconductor manufacturers

b. i. $20.71, ii. $17.31, iii. 35

c.

d. exponential distribution, Χ ~ Exp ( 1 20.71 )

e. Answers will vary.

f. i. $20.71, ii. $11.14

g. Answers will vary.

h. Answers will vary.

i. Answers will vary.

j. N ( 20.71,  17.31 5 )

7-79 §7.3 source ↗


Use the following information to answer the next three exercises:
Richard’s Furniture Company delivers furniture from 10 a.m. to 2 p.m. continuously and uniformly. We are interested in how long (in hours) past the 10 a.m. start time that individuals wait for their delivery.

The average wait time is:

a. one hour

b. two hours

c. two and a half hours

d. four hours

Answer

b

7-81 §7.3 source ↗

Use the following information to answer the next two exercises: The time to wait for a particular rural bus is distributed uniformly from zero to 75 minutes. One hundred riders are randomly sampled to learn how long they waited.

The 90th percentile sample average wait time (in minutes) for a sample of 100 riders is:

a. 315.0

b. 40.3

c. 38.5

d. 65.2

Answer

b

7-83 §7.3 source ↗


Use the following to answer the next two exercises:
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.

What's the approximate probability that the average price for 16 gas stations is more than $4.69?

a. almost zero

b. 0.1587

c. 0.0943

d. unknown

Answer

a

7-85 §7.3 source ↗


Use the following to answer the next two exercises:
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.

Suppose in a local kindergarten through 12th grade (K–12) school district, 53 percent of the population favor a charter school for grades K through five. A simple random sample of 300 is surveyed. Calculate the following using the normal approximation to the binomial distribtion.

a. Find the probability that less than 100 favor a charter school for grades K through 5.

b. Find the probability that 170 or more favor a charter school for grades K through 5.

c. Find the probability that no more than 140 favor a charter school for grades K through 5.

d. Find the probability that there are fewer than 130 that favor a charter school for grades K through 5.

e. Find the probability that exactly 150 favor a charter school for grades K through 5.

If you have access to an appropriate calculator or computer software, try calculating these probabilities using the technology.

Answer

a. 0

b. 0.1123

c. 0.0162

d. 0.0003

e. 0.0268

7-87 §7.3 source ↗


Use the following to answer the next two exercises:
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.

X ~ N(60, 9). Suppose that you form random samples of 25 from this distribution. Let X ¯ be the random variable of averages. Let ΣX be the random variable of sums. For parts (c) through (f), sketch the graph, shade the region, label and scale the horizontal axis for X ¯ , and find the probability.

a. Sketch the distributions of X and X ¯ on the same graph.

b. X ¯ ~ _____(_____, _____)

c. P( x ¯ < 60) = _____

d. Find the 30th percentile for the mean.

e. P(56 < x ¯ < 62) = _____

f. P(18 < x ¯ < 58) = _____

g. Σx ~ _____(_____, _____)

h. Find the minimum value for the upper quartile for the sum.

i. P(1400 < Σx < 1550) = _____

Answer

a. Check student’s solution.

b. X ¯ ~ N ( 60,  9 25 )

c. 0.5000

d. 59.06

e. 0.8536

f. 0.1333

g. N(1500, 45)

h. 1530.35

i. 0.6877

7-89 §7.3 source ↗


Use the following to answer the next two exercises:
The cost of unleaded gasoline in the Bay Area once followed an unknown distribution with a mean of $4.59 and a standard deviation of $0.10. Sixteen gas stations from the Bay Area are randomly chosen. We are interested in the average cost of gasoline for the 16 gas stations.

Salaries for managers in a restaurant chain are normally distributed with a mean of $44,000 and a standard deviation of $6,500. We randomly survey 10 managers from that district.

a. Find the 90th percentile for an individual manager's salary.

b. Find the 90th percentile for the average manager's salary.

Answer

a. $52,330

b. $46,634

7-91 §7.3 source ↗


For each problem, wherever possible, provide graphs and use a calculator.

NeverReady batteries has engineered a newer, longer-lasting AAA battery. The company claims this battery has an average life span of 17 hours with a standard deviation of 0.8 hours. Your statistics class questions this claim. As a class, you randomly select 30 batteries and find that the sample mean life span is 16.7 hours. If the process is working properly, what is the probability of getting a random sample of 30 batteries in which the sample mean life span is 16.7 hours or less? Is the company’s claim reasonable?

Answer

• We have μ = 17, σ = 0.8, x ¯ = 16.7, and n = 30. To calculate the probability, we use normalcdf(lower, upper, μ, σ n ) = normalcdf ( E–99,16.7,17, 0.8 30 ) = 0.0200.

• If the process is working properly, then the probability that a sample of 30 batteries would have at most 16.7 life span hours is only 2%. Therefore, the class was justified to question the claim.

7-93 §7.3 source ↗


For each problem, wherever possible, provide graphs and use a calculator.

Large bags of a brand of multicolored candies have a claimed net weight of 396.9 g. The standard deviation for the weight of the individual candies is 0.017 g. The following table is from a stats experiment conducted by a statistics class.

The bag contained 465 candies and the listed weights in the table came from randomly selected candies. Count the weights.

a. Find the mean sample weight and the standard deviation of the sample weights of candies in the table.

b. Find the sum of the sample weights in the table and the standard deviation of the sum of the weights.

c. If 465 candies are randomly selected, find the probability that their weights sum to at least 396.9 g.

d. Is the candy company's labeling accurate?

Table 7.8 in the book ↗
Red (g) Orange (g) Yellow (g) Brown (g) Blue (g) Green (g)
0.751 0.735 0.883 0.696 0.881 0.925
0.841 0.895 0.769 0.876 0.863 0.914
0.856 0.865 0.859 0.855 0.775 0.881
0.799 0.864 0.784 0.806 0.854 0.865
0.966 0.852 0.824 0.840 0.810 0.865
0.859 0.866 0.858 0.868 0.858 1.015
0.857 0.859 0.848 0.859 0.818 0.876
0.942 0.838 0.851 0.982 0.868 0.809
0.873 0.863 0.803 0.865
0.809 0.888 0.932 0.848
0.890 0.925 0.842 0.940
0.878 0.793 0.832 0.833
0.905 0.977 0.807 0.845
0.850 0.841 0.852
0.830 0.932 0.778
0.856 0.833 0.814
0.842 0.881 0.791
0.778 0.818 0.810
0.786 0.864 0.881
0.853 0.825
0.864 0.855
0.873 0.942
0.880 0.825
0.882 0.869
0.931 0.912
0.887

Answer

a. For the sample, we have n = 100, x ¯ = 0.862, and s = 0.05.

b. Σ x ¯ = 85.65, Σs = 5.18

c. normalcdf(396.9,E99,(465)(0.8565),(0.05)( 465 )) ≈ 1

d. Because the probability of a sample of size of 465 having at least a mean sum of 396.9 is appproximately 1, we can conclude that the company is correctly labeling their candy packages.

7-95 §7.3 source ↗


For each problem, wherever possible, provide graphs and use a calculator.

Your company has a contract to perform preventive maintenance on thousands of air conditioners in a large city. Based on service records from previous years, the time that a technician spends servicing a unit averages one hour with a standard deviation of one hour. In the coming week, your company will service a simple random sample of 70 units in the city. You plan to budget an average of 1.1 hours per technician to complete the work. Will this be enough time?

Answer

Use normalcdf ( E–99,1.1,1, 1 70 ) = 0.7986. This means that there is an 80 percent chance that the service time will be less than 1.1 hours. It may be wise to schedule more time because there is an associated 20 percent chance that the maintenance time will be greater than 1.1 hours.

7-97 §7.3 source ↗


For each problem, wherever possible, provide graphs and use a calculator.

Certain coins have an average weight of 5.201 g with a standard deviation of 0.065 g. If a vending machine is designed to accept coins whose weights range from 5.111 g to 5.291 g, what is the expected number of rejected coins when 280 randomly selected coins are inserted into the machine?

Answer

Because we have normalcdf ( 5.111,5.291,5.201, 0.065 280 ) ≈ 1, we can conclude that practically all the coins are within the limits; therefore, there should be no rejected coins out of a well-selected sample size of 280.

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Problems and answers are adapted from Statistics by Barbara Illowsky and Susan Dean, attributed to the Texas Education Agency (TEA) and published by OpenStax, and are used under the CC BY 4.0 licence. Access for free at https://openstax.org/books/statistics/pages/1-introduction. The original material is available at https://www.texasgateway.org/book/tea-statistics. Changes were made to the original material, including updates to art, structure, and other content updates. Art is credited in its own caption wherever the book credits it. This adaptation is neither affiliated with nor endorsed by OpenStax, Rice University or the Texas Education Agency (TEA).