Exam
Name___________________________________
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
In a large class, the professor has each person toss a coin several times and calculate the proportion of his or her tosses
that were heads. The students then report their results, and the professor plots a histogram of these several proportions.
Use the 68-95-99.7 Rule to provide the appropriate response.
1) If the students toss the coin 100 times each, about 99.7% should have proportions between what
two numbers?
1)
A) 0.35 and 0.65
B) 0.00075 and 0.99925
C) 0.15 and 0.2
D) 0.485 and 0.515
E) 0.0015 and 0.9985
Find the specified probability, from a table of Normal probabilities.
2) Researchers believe that 6% of children have a gene that may be linked to a certain childhood
disease. In an effort to track 50 of these children, researchers test 950 newborns for the presence of
this gene. What is the probability that they find enough subjects for their study?
A) 0.1685
B) 0.8315
C) 0.7912
D) 0.337
2)
E) 0.8507
Answer the question.
3) A national study reported that 75% of high school graduates pursue a college education
immediately after graduation. A private high school advertises that 156 of their 196 graduates last
year went on to college. Does this school have an unusually high proportion of students going to
college?
3)
A) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 0.89 standard deviations above the mean.
B) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 1.48 standard deviations above the mean.
C) This school can boast an unusually high proportion of students going to college. Their
proportion is 1.78 standard deviations above the mean.
D) This school cannot boast an unusually high proportion of students going to college. Their
proportion is only 1.19 standard deviations above the mean.
E) This school can boast an unusually high proportion of students going to college. Their
proportion is 1.19 standard deviations above the mean.
Use the given degree of confidence and sample data to construct a confidence interval for the population proportion.
4) Of 369 randomly selected medical students, 23 said that they planned to work in a rural
community. Construct a 95% confidence interval for the percentage of all medical students who
plan to work in a rural community.
A) (3.30%, 9.17%)
B) (4.16%, 8.30%)
C) (2.99%, 9.47%)
D) (5.32%, 7.14%)
E) (3.77%, 8.70%)
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4)
Write the null and alternative hypotheses you would use to test the following situation.
5) A weight loss center provided a loss for 72% of its participants. The center's leader decides to test a
new weight loss strategy to see if it's better. What are the null and alternative hypotheses?
5)
A) H0: p = 0.72
HA: p > 0.72
B) H0: p > 0.72
HA: p < 0.72
C) H0: p = 0.72
HA: p ≠ 0.72
D) H0: p > 0.72
HA: p = 0.72
E) H0: p = 0.72
HA: p < 0.72
Find the P-value for the indicated hypothesis test.
6) A manufacturer claims that fewer than 6% of its fax machines are defective. In a random sample of
97 such fax machines, 5% are defective. Find the P-value for a test of the manufacturer's claim.
A) 0.3264
B) 0.1591
C) 0.1736
6)
D) 0.3409
Provide an appropriate response.
7) The city management company claims that 75% of all low income housing is 1500 sq. ft. The
tenants believe the proportion of housing this size is smaller than the claim, and hire an
independent engineering firm to test a random sample size of 120 and found 1500 sq. ft. housing at
66%. Test an appropriate hypothesis and state your conclusion. Be sure the appropriate
assumptions and conditions are satisfied before you proceed.
7)
A) H0: p = 0.75; HA: p < 0.75; z = -2.28; P-value = 0.0113. This data shows that the proportion of
1500 sq. ft. housing is less than 75%
B) H0: p = 0.75; HA: p > 0.75; z = -2.32; P-value = 0.9898. This data does not show that the
proportion of 1500 sq. ft. housing is less than 75%
C) H0: p = 0.75; HA: p > 0.75; z = 2.32; P-value = 0.9898. This data shows that the proportion of
1500 sq. ft. housing is less than 75%
D) H0 : p = 0.75; HA: p < 0.075 z = 2.28; P-value = 0.0113. This data does not show that the
proportion of 1500 sq. ft. housing is less than 75%
E) H0 : p = 0.75; HA: p ≠ 0.75; z = -2.32; P-value = 0.0204. This data shows that the proportion of
1500 sq. ft. housing is less than 75%
Find the mean of the data.
8) The employees at Frank's Furniture earned the following amounts, in dollars, last week.
$400.45
$257.47
$384.12
$216.04
$405.49
$487.32
Round your answer to the nearest cent.
A) $418.18
B) $430.18
C) $537.72
2
D) $358.48
E) $487.32
8)
Find the median of the data.
9) The number of cars passing through a bank drive-up line during each 15-minute period was
recorded. The results are shown below.
9)
27 29 27 30
30 27 32 29
37 33 33 31
26 33 27 22
17 29 29 29
A) 30 cars
B) 29 cars
C) 28.85 cars
D) 33 cars
E) 27 cars
Solve the problem.
10) Here are the prices for 8 different CD players. Find the standard deviation.
$195
$358
$201
A) $565,441.0
$276
$161
B) $94.1
$301
$387
C) $144.5
10)
$128
D) $161
E) $503,506.1
Use summary statistics to answer the question.
11) Here are some summary statistics for annual snowfall in a certain town compiled over the last 15
years: lowest amount = 10 inches, mean = 41 inches, median = 34 inches, range = 90 inches,
IQR = 51, Q1 = 16, standard deviation = 10 inches. Between what two values are the middle 50% of
snowfall found?
11)
A) 10.25 and 30.75
B) 16 and 67
C) 41 and 34
D) 8.5 and 25.5
E) 10 and 100
Solve the problem.
12) The mean weight of babies born in Central hospital last year was 6.3 pounds. Suppose the
standard deviation of the weights is 2.1 pounds. Which would be more unusual, a baby weighing
4 pounds or a baby weighing 8.5 pounds? Explain.
12)
A) A 4 pound baby is more unusual (z = -1.05) compared with an 8.5 pound baby (z = -1.10).
B) An 8.5 pound baby is more unusual (z = -1.05) compared with a 4 pound baby (z = -1.10).
C) A 4 pound baby is more unusual (z = -1.10) compared with an 8.5 pound baby (z = 1.05).
D) An 8.5 pound baby is more unusual (z = 1.90) compared with a 4 pound baby (z = 4.05).
E) An 8.5 pound baby is more unusual (z = -1.10) compared with a 4 pound baby (z = -1.05).
13) A town's average snowfall is 46 inches per year with a standard deviation of 10 inches. Using a
Normal model, what values should border the middle 68% of the model?
A) 51 inches and 41 inches
B) 48 inches and 44 inches
C) 66 inches and 26 inches
D) 56 inches and 36 inches
E) 46 inches and 42.6 inches
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13)
Solve the problem. Round to the nearest tenth.
14) Based on the Normal model for snowfall in a certain town N(57, 8), how many inches of snow
would represent the 75th percentile?
A) 65 inches
B) 62.4 inches
C) 42.8 inches
D) 51.6 inches
14)
E) 49 inches
Solve the problem.
15) An Imaginary Poll in April 2005 asked 931 U.S. adults what their main source of news was:
newspapers, television, internet, or radio? Here are the results:
Response
Number
Newspapers 242
Television 398
Internet
126
Radio
165
Total
931
If we select a person at random from this sample of 931 adults, what is the probability that the
person responded "Newspapers"?
A) 0.242
B) 0.177
C) 0.135
D) 0.427
15)
E) 0.260
Solve the problem. Round your answer, as needed.
16) A study conducted at a certain college shows that 64% of the school's graduates find a job in their
chosen field within a year after graduation. Find the probability that among 5 randomly selected
graduates, at least one finds a job in his or her chosen field within a year of graduating.
A) 0.8926
B) 0.2000
C) 0.9940
16)
D) 0.3200
Find the indicated probability.
17)
17) The table below describes the smoking habits of a group of asthma sufferers.
Light Heavy
Nonsmoker smoker smoker Total
Men
317
78
77
472
Women
342
64
69
475
Total
659
142
146
947
If one of the 947 subjects is randomly selected, find the probability that the person chosen is a
female nonsmoker.
A) 0.502
B) 0.361
C) 0.720
D) 0.519
E) 0.836
18) The table shows the political affiliation of voters in one city and their positions on stronger gun
control laws.
Stronger Gun Control
Favor Oppose
Republican
0.090
0.26
Democrat
0.22
0.2
Other
0.11
0.12
What is the probability that a voter who favors stronger gun control laws is a Republican?
A) 0.350
B) 0.257
C) 0.090
D) 0.420
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E) 0.214
18)
Find the expected value of the random variable.
19) You roll a pair of dice. If you get a sum greater than 10 you win $50. If you get a double you win
$25. If you get a double and a sum greater than 10 you win a $75. Otherwise you win nothing.
You pay $5 to play.
Find the expected amount you win at this game.
A) $8.33
B) $4.72
C) $4.03
D) $5.42
19)
E) $3.33
Find the indicated probability.
20) An archer is able to hit the bull's eye 76% of the time. If she shoots 10 arrows, what is the
probability that her first bull's-eye comes on the 4th arrow? Assume each shot is independent of
the others.
A) 0.01382
B) 0.00252
C) 0.10535
D) 0.01051
20)
E) 0.76
Solve.
21) Suppose that 12% of students at one college have high blood pressure. If you keep picking students
at random from this college, how many students do you expect to test before finding one with high
blood pressure?
A) 0.12
B) 0.88
C) 8.33
D) 1.14
21)
E) 12
Find the indicated probability.
22) A multiple choice test has 10 questions each of which has 4 possible answers, only one of which is
correct. If Judy, who forgot to study for the test, guesses on all questions, what is the probability
that she will answer exactly 3 questions correctly?
A) 0.2503
B) 0.2816
C) 0.0156
D) 0.7497
22)
E) 0.0021
Find the probability of the outcome described.
23) A tennis player makes a successful first serve 60% of the time. If she serves 10 times, what is the
probability that she gets at least 3 first serves in? Assume that each serve is independent of the
others.
A) 0.0123
B) 0.0425
C) 0.0548
D) 0.9452
23)
E) 0.9877
Find the specified probability, from a table of Normal probabilities.
24) A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a
mean of $54 and a standard deviation of $18. What is the probability that the next 100 customers
will spend an average of less than $56 on dinner?
A) 0.1335
B) 0.5442
C) 0.4558
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D) 0.8665
E) 0.1552
24)
Answer Key
Testname: SAMPLE FINAL
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