Student: _____________________ Instructor: Amanda Talley

Assignment: Practice Test 2

Date: _____________________ Course: Math 134 - Elementary Statistics

1. Express the indicated degree of likelihood as a probability value between 0 and 1.

When using a computer to randomly generate the last digit of a phone number to be called for a survey, there is 1 chance in
10 that the last digit is zero.

The probability is .
(Type an integer or a decimal. Do not round.)

Answer: 0.1

ID: 4.1.17

2. Refer to the sample data for polygraph tests shown below. If one of the test subjects is randomly selected, what is the
probability that the subject is not lying? Is the result close to the probability of 0.430 for a negative test result?
Did the Subject Actually Lie?
No (Did Not Lie) Yes (Lied)
Positive test results 12 45
Negative test results 33 10

The probability that a randomly selected polygraph test subject was not lying is .
(Type an integer or decimal rounded to three decimal places as needed.)

Is the result close to the probability, rounded to three decimal places, of 0.430 for a negative test result?

(1) because there is (2) a 0.050 absolute difference between the probability of a true
response and the probability of a negative test result.

(1) No, (2) exactly

Yes, more than
less than

Answers 0.450

(1) Yes,
(2) less than

ID: 4.1.24
3. A modified roulette wheel has 40 slots. One slot is 0, another is 00, and the others are numbered 1 through 38, respectively.
You are placing a bet that the outcome is an odd number. (In roulette, 0 and 00 are neither odd nor even.)

a. What is your probability of winning?

The probability of winning is .

(Type an integer or a simplified fraction.)

b. What are the actual odds against winning?

The actual odds against winning are : .

c. When you bet that the outcome is an odd number, the payoff odds are 1:1. How much profit do you make if you bet $12
and win?

If you win, the payoff is $ .

d. How much profit should you make on the $12 bet if you could somehow convince the casino to change its payoff odds so
that they are the same as the actual odds against winning?

$ (Round to the nearest cent as needed.)

Answers 19
40
21

19
12
13.26

ID: 4.1.42

4. A research center poll showed that 75% of people believe that it is morally wrong to not report all income on tax returns.
What is the probability that someone does not have this belief?

The probability that someone does not believe that it is morally wrong to not report all income on tax returns is
.
(Type an integer or a decimal.)

Answer: 0.25

ID: 4.2.5
5. Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are
randomly selected from those included in the table.

Drive-thru Restaurant
A B C D
Order Accurate 324 272 248 146
Order Not Accurate 40 50 38 15

If one order is selected, find the probability of getting an order that is not accurate or is from Restaurant C. Are the events of
selecting an order that is not accurate and selecting an order from Restaurant C disjoint events?

The probability of getting an order from Restaurant C or an order that is not accurate is .
(Round to three decimal places as needed.)

Are the events of selecting an order from Restaurant C and selecting an inaccurate order disjoint events?

The events (1) disjoint because it (2) possible to (3)

(1) are not (2) is (3) receive an inaccurate order from Restaurant C.
are is not pick an order from Restaurant A, B, or D.
pick an accurate order.

Answers 0.345
(1) are not
(2) is
(3) receive an inaccurate order from Restaurant C.

ID: 4.2.12
6. Use the data in the following table, which lists drive-thru order accuracy at popular fast food chains. Assume that orders are
randomly selected from those included in the table.

Drive-thru Restaurant
A B C D
Order Accurate 311 280 233 140
Order Not Accurate 32 59 31 15

If two orders are selected, find the probability that they are both accurate. Complete parts (a) and (b) below.

a. Assume that the selections are made with replacement. Are the events independent?

The probability is . The events (1) independent.

(Do not round until the final answer. Round to four decimal places as needed.)

b. Assume that the selections are made without replacement. Are the events independent?

The probability is . The events (2) independent.

(Do not round until the final answer. Round to four decimal places as needed.)

(1) are (2) are not

are not are

Answers 0.7666

(1) are
0.7665

(2) are not

ID: 4.2.15

7. In a study of helicopter usage and patient survival, among the 50,536 patients transported by helicopter, 224 of them left the
treatment center against medical advice, and the other 50,312 did not leave against medical advice. If 50 of the subjects
transported by helicopter are randomly selected without replacement, what is the probability that none of them left the
treatment center against medical advice?

The probability is .
(Round to three decimal places as needed.)

Answer: 0.801

ID: 4.2.30
8. Is the random variable given in the accompanying table discrete Number of
or continuous? Explain. Girls, x P(x)
0 0.063
1 0.250
2 0.375
3 0.250
4 0.063

The random variable given in the accompanying table is (1) because (2)

(1) discrete
continuous

(2) the probabilities are decimal values.

the probabilities are all between 0 and 1, inclusive.
there are infinitely many values, and they are countable.
there are infinitely many values, and they are not countable.
the x-values are numbers. there are a finite number of values.
at least one probability is less than 0 or greater than 1.
the probabilities do not add up to 1.
the probabilities add up to 1.

Answers (1) discrete

(2) there are a finite number of values.

ID: 5.1.2
9. Five males with an X-linked genetic disorder have one child each. The x P(x)
random variable x is the number of children among the five who inherit 0 0.032
the X-linked genetic disorder. Determine whether a probability distribution 1 0.152
is given. If a probability distribution is given, find its mean and standard 2 0.316
deviation. If a probability distribution is not given, identify the requirements 3 0.316
that are not satisfied. 4 0.152
5 0.032

Does the table show a probability distribution? Select all that apply.

A. Yes, the table shows a probability distribution.

B. No, the random variable x's number values are not associated with probabilities.
C. No, the random variable x is categorical instead of numerical.
D. No, the sum of all the probabilities is not equal to 1.
E. No, not every probability is between 0 and 1 inclusive.

Find the mean of the random variable x. Select the correct choice below and, if necessary, fill in the answer box to complete
your choice.

A. μ = child(ren) (Round to one decimal place as needed.)

B. The table does not show a probability distribution.

Find the standard deviation of the random variable x. Select the correct choice below and, if necessary, fill in the answer box
to complete your choice.

A. σ = child(ren) (Round to one decimal place as needed.)

B. The table does not show a probability distribution.

Answers A. Yes, the table shows a probability distribution.

A. μ = 2.5 child(ren) (Round to one decimal place as needed.)

A. σ = 1.1 child(ren) (Round to one decimal place as needed.)

ID: 5.1.7
10. The accompanying table describes results from groups of 10 births from 10 different sets of parents. The random variable
x represents the number of girls among 10 children. Use the range rule of thumb to determine whether 1 girl in 10 births is
a significantly low number of girls.
1
Click the icon to view the table.

Use the range rule of thumb to identify a range of values that are not significant.

The maximum value in this range is girls.

(Round to one decimal place as needed.)

The minimum value in this range is girls.

(Round to one decimal place as needed.)

Based on the result, is 1 girl in 10 births a significantly low number of girls? Explain.

A. Yes, 1 girl is a significantly low number of girls, because 1 girl is above the range of values that
are not significant.
B. No, 1 girl is not a significantly low number of girls, because 1 girl is within the range of values
that are not significant.
C. Yes, 1 girl is a significantly low number of girls, because 1 girl is below the range of values that
are not significant.
D. Not enough information is given.

1: Probability Distribution for x

Number of
Girls x P(x)
0 0.002
1 0.012
2 0.037
3 0.116
4 0.207
5 0.243
6 0.196
7 0.116
8 0.036
9 0.019
10 0.016

Answers 8.5
1.7
C.
Yes, 1 girl is a significantly low number of girls, because 1 girl is below the range of values that are not
significant.

ID: 5.1.16
11. The accompanying table describes the random variable x, the numbers of adults in groups of five who reported
sleepwalking. Complete parts (a) through (d) below.
2
Click the icon to view the table.

a. Find the probability of getting exactly 4 sleepwalkers among 5 adults.

(Type an integer or a decimal. Do not round.)

b. Find the probability of getting 4 or more sleepwalkers among 5 adults.

(Type an integer or a decimal. Do not round.)

c. Which probability is relevant for determining whether 4 is a significantly high number of sleepwalkers among 5 adults:
the result from part (a) or part (b)?

A. Since the probability of getting 5 sleepwalkers includes getting 4 sleepwalkers, the result from
part (b) is the relevant probability.
B. Since the probability of getting 5 sleepwalkers is less likely than getting 4 sleepwalkers, the
result from part (a) is the relevant probability.
C. Since the probability of getting 4 sleepwalkers is the result from part (a), this is the relevant
probability.
D. Since the probability of getting fewer than 4 sleepwalkers is the complement of the result from
part (b), this is the relevant probability.

d. Is 4 a significantly high number of 4 sleepwalkers among 5 adults? Why or why not? Use 0.05 as the threshold for a
significant event.

A. No, since the appropriate probability is greater than 0.05, it is not a significantly high number.
B. Yes, since the appropriate probability is less than 0.05, it is a significantly high number.
C. Yes, since the appropriate probability is greater than 0.05, it is a significantly high number.
D. No, since the appropriate probability is less than 0.05, it is not a significantly high number.

2: Probability Distribution for x

x P(x)
0 0.188
1 0.407
2 0.246
3 0.131
4 0.021
5 0.007

Answers 0.021

0.028
A.
Since the probability of getting 5 sleepwalkers includes getting 4 sleepwalkers, the result from part (b) is the
relevant probability.

B. Yes, since the appropriate probability is less than 0.05, it is a significantly high number.

ID: 5.1.24
12. Determine whether or not the procedure described below results in a binomial distribution. If it is not binomial, identify at
least one requirement that is not satisfied.

Five hundred different voters in a region with two major political parties, A and B, are randomly selected from the
population of 5500 registered voters. Each is asked if he or she is a member of political party A, recording Yes or No.

Choose the correct answer below.

A. No, the trials are not independent and the sample is more than 5% of the population.
B. No, the number of trials is not fixed.
C. No, the probability of success is not the same in all trials.
D. No, there are more than two possible outcomes.
E. Yes, the result is a binomial probability distribution.

Answer: A. No, the trials are not independent and the sample is more than 5% of the population.

ID: 5.2.11

13. Multiple-choice questions each have four possible answers (a, b, c, d), one of which is correct. Assume that you guess the
answers to three such questions.

a. Use the multiplication rule to find P(CWW), where C denotes a correct answer and W denotes a wrong answer.

P(CWW) = (Type an exact answer.)

b. Beginning with CWW, make a complete list of the different possible arrangements of one correct answer and two wrong
answers, then find the probability for each entry in the list.

P(CWW) − see above

P(WWC) =
P(WCW) =
(Type exact answers.)

c. Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are
made?

(Type an exact answer.)

Answers 0.140625
0.140625
0.140625
0.421875

ID: 5.2.13
14. A survey showed that 79% of adults need correction (eyeglasses, contacts, surgery, etc.) for their eyesight. If 10 adults are
randomly selected, find the probability that at least 9 of them need correction for their eyesight. Is 9 a significantly high
number of adults requiring eyesight correction?

The probability that at least 9 of the 10 adults require eyesight correction is .

(Round to three decimal places as needed.)

Is 9 a significantly high number of adults requiring eyesight correction? Note that a small probability is one that is less than
0.05.

A. No, because the probability of this occurring is not small.

B. Yes, because the probability of this occurring is not small.
C. No, because the probability of this occurring is small.
D. Yes, because the probability of this occurring is small.

Answers 0.346

A. No, because the probability of this occurring is not small.

ID: 5.2.26
15. Assume that different groups of couples use a particular method of gender selection and each couple gives birth to one
baby. This method is designed to increase the likelihood that each baby will be a girl, but assume that the method has no
effect, so the probability of a girl is 0.5. Assume that the groups consist of 46 couples. Complete parts (a) through (c)
below.

a. Find the mean and the standard deviation for the numbers of girls in groups of 46 births.

The value of the mean is μ = .

(Type an integer or a decimal. Do not round.)

The value of the standard deviation is σ = .

(Round to one decimal place as needed.)

b. Use the range rule of thumb to find the values separating results that are significantly low or significantly high.

Values of girls or fewer are significantly low.

(Round to one decimal place as needed.)

Values of girls or greater are significantly high.

(Round to one decimal place as needed.)

c. Is the result of 37 girls a result that is significantly high? What does it suggest about the effectiveness of the method?

The result (1) significantly high, because 37 girls is (2) girls. A result of

37 girls would suggest that the method (3)

(Round to one decimal place as needed.)

(1) is not (2) greater than (3) is not effective.

is equal to is effective.
less than

Answers 23
3.4

16.2
29.8
(1) is
(2) greater than

29.8
(3) is effective.

ID: 5.2.29
16. In analyzing hits by certain bombs in a war, an area was partitioned into 562 regions, each with an area of 0.85 km2. A
total of 525 bombs hit the combined area of 562 regions. Assume that we want to find the probability that a randomly
x −μ
μ • e
selected region had exactly two hits. In applying the Poisson probability distribution formula, P(x) = , identify the
x!
values of μ, x, and e. Also, briefly describe what each of those symbols represents.

Identify the values of μ, x, and e.

μ = , x= , and e =
(Type integers or decimals rounded to five decimal places as needed.)

Briefly describe what the symbol μ represents. Choose the correct answer below.

A. The symbol μ is a variable that represents the number of occurrences of the event.
B. The symbol μ represents a static value.
C. The symbol μ is a variable that represents the number of occurrences of the event in an
interval.
D. The symbol μ is a variable that represents the area of each region.
E. The symbol μ is a variable that represents the mean number of occurrences of the event in the
intervals.

Briefly describe what the symbol x represents. Choose the correct answer below.

A. The symbol x is a variable that represents the area of each region.

B. The symbol x represents a static value.
C. The symbol x is a variable that represents the number of occurrences of the event.
D. The symbol x is a variable that represents the mean number of occurrences of the event in the
intervals.
E. The symbol x is a variable that represents the number of occurrences of the event in an
interval.

Briefly describe what the symbol e represents. Choose the correct answer below.

A. The symbol e is a variable that represents the number of occurrences of the event.
B. The symbol e is a variable that represents the number of occurrences of the event in an
interval.
C. The symbol e is a variable that represents the area of each region.
D. The symbol e represents a static value.
E. The symbol e is a variable that represents the mean number of occurrences of the event in the
intervals.

Answers 0.93416

2.71828
E. The symbol μ is a variable that represents the mean number of occurrences of the event in the intervals.

E. The symbol x is a variable that represents the number of occurrences of the event in an interval.

D. The symbol e represents a static value.

 ID: 5.3.1

17. When studying radioactive material, a nuclear engineer found that over 365 days, 1,000,000 radioactive atoms decayed to
973,222 radioactive atoms, so 26,778 atoms decayed during 365 days.
a. Find the mean number of radioactive atoms that decayed in a day.
b. Find the probability that on a given day, 50 radioactive atoms decayed.

a. The mean number of radioactive atoms that decay per day is .

(Round to three decimal places as needed.)

b. The probability that on a given day, 50 radioactive atoms decayed, is .

(Round to six decimal places as needed.)

Answers 73.364

0.000850

ID: 5.3.11-T

18. What does the notation zα indicate?

The expression zα denotes the z score with an area of α (1)

(1) to its right.

to its left.
between − zα and zα .

Answer: (1) to its right.

ID: 6.1.4
19. Find the area of the shaded region. The graph depicts the standard normal
distribution with mean 0 and standard deviation 1.
Click to view page 1 of the table.3 Click to view page 2 of the table.4

z = 0.28

The area of the shaded region is .

(Round to four decimal places as needed.)

3: Standard Normal Table (Page 1)

4: Standard Normal Table (Page 2)
Answer: 0.6103

ID: 6.1.9
20. Find the indicated z score. The graph depicts the standard normal distribution with
mean 0 and standard deviation 1.
Click to view page 1 of the table.5 Click to view page 2 of the table.6

0.9573

The indicated z score is .

(Round to two decimal places as needed.)

5: Standard Normal Table (Page 1)

6: Standard Normal Table (Page 2)
Answer: 1.72

ID: 6.1.13
21. Assume the readings on thermometers are normally distributed with a mean of 0°C and a standard deviation of 1.00°C.
Find the probability that a randomly selected thermometer reads between − 2.03 and − 1.46 and draw a sketch of the
region.
Click to view page 1 of the table.7 Click to view page 2 of the table.8

Sketch the region. Choose the correct graph below.

A. B. C.

-2.03 -1.46 -2.03 -1.46 -2.03 -1.46

The probability is .
(Round to four decimal places as needed.)

7: Standard Normal Table (Page 1)

8: Standard Normal Table (Page 2)
Answers

C. -2.03 -1.46

0.0510
 ID: 6.1.27

22. Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a
mean of 0 and a standard deviation of 1. Draw a graph and find the probability of a bone density test score greater than 0.

Sketch the region. Choose the correct graph below.

A. B. C. D.

0 0 0 0

The probability is .
(Round to four decimal places as needed.)

Answers

A. 0

0.5000

ID: 6.1.35
23. Find the area of the shaded region. The graph to the right depicts IQ scores of adults,
and those scores are normally distributed with a mean of 100 and a standard deviation of
15.

Click to view page 1 of the table.9 Click to view page 2 of the table.10

85 110

The area of the shaded region is . (Round to four decimal places as needed.)

9: Standard Normal Table (Page 1)

10: Standard Normal Table (Page 2)
Answer: 0.5889

ID: 6.2.7
24. Find the indicated IQ score. The graph to the right depicts IQ scores of adults, and those
scores are normally distributed with a mean of 100 and a standard deviation of 15.

Click to view page 1 of the table.11 Click to view page 2 of the table.12

0.9

The indicated IQ score, x, is . (Round to one decimal place as needed.)

11: Standard Normal Table (Page 1)

12: Standard Normal Table (Page 2)
Answer: 119.2

ID: 6.2.9
25. A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.5 in. A branch of
the military requires women's heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join
this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and
the tallest 2%, what are the new height requirements?
Click to view page 1 of the table.13 Click to view page 2 of the table.14

a. The percentage of women who meet the height requirement is %.

(Round to two decimal places as needed.)

Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?

A. Yes, because the percentage of women who meet the height requirement is fairly large.
B. No, because the percentage of women who meet the height requirement is fairly small.
C. Yes, because a large percentage of women are not allowed to join this branch of the military
because of their height.
D. No, because only a small percentage of women are not allowed to join this branch of the
military because of their height.

b. For the new height requirements, this branch of the military requires women's heights to be at least in
and at most in.
(Round to one decimal place as needed.)

13: Standard Normal Table (Page 1)

14: Standard Normal Table (Page 2)
Answers 98.46
D.
No, because only a small percentage of women are not allowed to join this branch of the military because of
their height.

57.6
68.5

ID: 6.2.21
26. For the purposes of constructing modified boxplots, outliers are defined as data values that are above Q3 by an amount
greater than 1.5 × IQR or below Q1 by an amount greater than 1.5 × IQR, where IQR is the interquartile range. Using this
definition of outliers, find the probability that when a value is randomly selected from a normal distribution, it is an outlier.

The probability that a randomly selected value taken from a normal distribution is considered an outlier is .
(Round to four decimal places as needed.)

Answer: 0.0070

ID: 6.2.36
27. Three randomly selected households are surveyed. The numbers of people in the households are 3, 4, and 11. Assume
that samples of size n = 2 are randomly selected with replacement from the population of 3, 4, and 11. Listed below are the
nine different samples. Complete parts (a) through (c).
3,3 3,4 3,11 4,3 4,4 4,11 11,3 11,4 11,11

a. Find the variance of each of the nine samples, then summarize the sampling distribution of the variances in the format
of a table representing the probability distribution of the distinct variance values.
s2 Probability
(1)

(2)

(3)

(4)

(Type an integer or a fraction. Use ascending order of the sample variances.)

b. Compare the population variance to the mean of the sample variances. Choose the correct answer below.

A. The population variance is equal to the mean of the sample variances.

B. The population variance is equal to the square root of the mean of the sample variances.
C. The population variance is equal to the square of the mean of the sample variances.

c. Do the sample variances target the value of the population variance? In general, do sample variances make good
estimators of population variances? Why or why not?

A. The sample variances target the population variance, therefore, sample variances do not make
good estimators of population variances.
B. The sample variances target the population variances, therefore, sample variances make good
estimators of population variances.
C. The sample variances do not target the population variance, therefore, sample variances do
not make good estimators of population variances.
D. The sample variances do not target the population variance, therefore, sample variances make
good estimators of population variances.

(1) 0 (2) 1 (3) 24.5 (4) 16

1 12.3 12.3 64
0.3 0.5 49 32
Answers (1) 0
3
9
(2) 0.5

2
9
(3) 24.5
2
9
(4) 32
2
9
A. The population variance is equal to the mean of the sample variances.

B.
The sample variances target the population variances, therefore, sample variances make good estimators of
population variances.

ID: 6.3.7
28. Three randomly selected households are surveyed. The numbers of people in the households are 3, 4, and 11. Assume
that samples of size n = 2 are randomly selected with replacement from the population of 3, 4, and 11. Construct a
probability distribution table that describes the sampling distribution of the proportion of even numbers when samples of
sizes n = 2 are randomly selected. Does the mean of the sample proportions equal the proportion of even numbers in the
population? Do the sample proportions target the value of the population proportion? Does the sample proportion make a
good estimator of the population proportion? Listed below are the nine possible samples.
3,3 3,4 3,11 4,3 4,4 4,11 11,3 11,4 11,11

Construct the probability distribution table.

Sample
Probability
Proportion
(1)

(2)

(3)

(Type an integer or fraction.)

Choose the correct answer below.

A. The proportion of odd numbers in the population is equal to the mean of the sample
proportions of even numbers.
B. The proportion of even numbers in the population is equal to the mean of the sample
proportions of odd numbers.
C. The proportion of even numbers in the population is not equal to the mean of the sample
proportions.
D. The proportion of even numbers in the population is equal to the mean of the sample
proportions.

Choose the correct answer below.

A. The sample proportions do not target the proportion of even numbers in the population, so
sample proportions make good estimators of the population proportion.
B. The sample proportions do not target the proportion of even numbers in the population, so
sample proportions do not make good estimators of the population proportion.
C. The sample proportions target the proportion of even numbers in the population, so sample
proportions make good estimators of the population proportion.
D. The sample proportions target the proportion of even numbers in the population, so sample
proportions do not make good estimators of the population proportion.

(1) 0.25 (2) 0.75 (3) 0.75

0 0.25 0.25
0.1 0.5 1
Answers (1) 0
4
9
(2) 0.5

4
9
(3) 1
1
9
D. The proportion of even numbers in the population is equal to the mean of the sample proportions.
C.
The sample proportions target the proportion of even numbers in the population, so sample proportions make
good estimators of the population proportion.

ID: 6.3.10
29. A genetics experiment involves a population of fruit flies consisting of 1 male named Carl and 3 females named
Diane, Elaine, and Fiona. Assume that two fruit flies are randomly selected with replacement.

a. After listing the possible samples and finding the proportion of females in each sample, use a table to describe the
sampling distribution of the proportion of females.

Proportion of females Probability

0
0.5
1
(Type integers or fractions.)

b. Find the mean of the sampling distribution.

μ = (Round to two decimal places as needed.)

c. Is the mean of the sampling distribution [from part (b)] equal to the population proportion of females? If so, does the
mean of the sampling distribution of proportions always equal the population proportion?

A. Yes, the sample mean is equal to the population proportion of females. These values are
always equal, because proportion is a biased estimator.
B. No, the sample mean is equal to the population proportion of females. These values are not
always equal, because proportion is an unbiased estimator.
C. No, the sample mean is equal to the population proportion of females. These values are not
always equal, because proportion is a biased estimator.
D. Yes, the sample mean is equal to the population proportion of females. These values are
always equal, because proportion is an unbiased estimator.

Answers 1
16
6
16
9
16
0.75

D.
Yes, the sample mean is equal to the population proportion of females. These values are always equal,
because proportion is an unbiased estimator.

ID: 6.3.18
30. An elevator has a placard stating that the maximum capacity is 1740 lb—10 passengers. So, 10 adult male passengers
can have a mean weight of up to 1740 / 10 = 174 pounds. If the elevator is loaded with 10 adult male passengers, find the
probability that it is overloaded because they have a mean weight greater than 174 lb. (Assume that weights of males are
normally distributed with a mean of 180 lb and a standard deviation of 25 lb.) Does this elevator appear to be safe?

The probability the elevator is overloaded is .

(Round to four decimal places as needed.)

Does this elevator appear to be safe?

A. Yes, there is a good chance that 10 randomly selected people will not exceed the elevator
capacity.
B. No, there is a good chance that 10 randomly selected adult male passengers will exceed the
elevator capacity.
C. No, 10 randomly selected people will never be under the weight limit.
D. Yes, 10 randomly selected adult male passengers will always be under the weight limit.

Answers 0.7761
B.
No, there is a good chance that 10 randomly selected adult male passengers will exceed the elevator capacity.

ID: 6.4.9