Topic 2 Exercises
Question # 00445177
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Updated on: 12/18/2016 07:46 AM Due on: 12/20/2016
Complete the following exercises located at the end of each chapter and put them into a word document to be submitted as directed by the instructor
Show all relevant work; use the equation editor in Microsoft Word when necessary.
F i n d i n g Proportions
5.11 Scores on the Wechsler Adult Intelligence Scale (WAIS) approximate a
normal curve with a mean of 100 and a standard deviation of 15. What
proportion of IQ scores are
(a) above Kristen’s 125?
(b) below 82?
(c) within 9 points of the mean?
(d) more than 40 points from the mean?
F i n d i n g Scores
5.13 IQ scores on the WAIS test approximate a normal curve with a mean of
100 and a standard deviation of 15. What IQ score is identified with
F i n d i n g Scores
5.13 IQ scores on the WAIS test approximate a normal curve with a mean of
100 and a standard deviation of 15. What IQ score is identified with
(a) the upper 2 percent, that is, 2 percent to the right (and 98 percent to the left)?
(b) the lower 10 percent?
(c) the upper 60 percent?
(d) the middle 95 percent? [Remember, the middle 95 percent straddles the
line perpendicular to the mean (or the 50th percentile), with half of 95
percent, or 47.5 percent, above this line and the remaining 47.5 percent
below this line.]
line perpendicular to the mean (or the 50th percentile), with half of 95
percent, or 47.5 percent, above this line and the remaining 47.5 percent
below this line.]
(e) the middle 99 percent?
*5. 15 An investigator polls common cold sufferers, asking them to estimate the
number of hours of physical discomfort caused by their most recent colds.
Assume that their estimates approximate a normal curve with a mean of
83 hours and a standard deviation of 20 hours.
(a) What is the estimated number of hours for the shortest suffering 5 percent?
(b) What proportion of sufferers estimate that their colds lasted longer than
48 hours?
48 hours?
(c) What proportion suffered for fewer than 61 hours?
(d) What is the estimated number of hours suffered by the extreme 1 percent
either above or below the mean?
either above or below the mean?
(e) What proportion suffered for between 1 and 3 days, that is, between 24
and 72 hours?
and 72 hours?
(f) What is the estimated number of hours suffered by the middle 95 percent?
[See the comment about “middle 95 percent” in Question 5.13(d) .]
[See the comment about “middle 95 percent” in Question 5.13(d) .]
(g) What proportion suffered for between 2 and 4 days?
(h) A medical researcher wishes to concentrate on the 20 percent who suffered
the most. She will work only with those who estimate that they suffered
for more than hours.
the most. She will work only with those who estimate that they suffered
for more than hours.
i) Another researcher wishes to compare those who suffered least with
those who suffered most. If each group is to consist of only the extreme 3
percent, the mild group will consist of those who suffered for fewer than
hours, and the severe group will consist of those who suffered for
more than hours.
those who suffered most. If each group is to consist of only the extreme 3
percent, the mild group will consist of those who suffered for fewer than
hours, and the severe group will consist of those who suffered for
more than hours.
(j) Another survey found that people with colds who took daily doses of vitamin
C suffered, on the average, for 61 hours. What proportion of the original
survey (with a mean of 83 hours and a standard deviation of 20 hours)
suffered for more than 61 hours?
C suffered, on the average, for 61 hours. What proportion of the original
survey (with a mean of 83 hours and a standard deviation of 20 hours)
suffered for more than 61 hours?
(k) What proportion of the original survey suffered for exactly 61 hours?
(Be careful!)
Answers on page 505 .
(Be careful!)
Answers on page 505 .
*5.18 The body mass index (BMI) measures body size in people by dividing
weight (in pounds) by the square of height (in inches) and then multiplying
by a factor of 703. A BMI less than 18.5 is defined as underweight; between
18.5 to 24.9 is normal; between 25 and 29.9 is overweight; and 30 or more
is obese. It is well-established that Americans have become heavier during
the last half century. Assume that the positively skewed distribution of BMIs
for adult American males has a mean of 28 with a standard deviation of 4.
(a) Would the median BMI score exceed, equal, or be exceeded by the mean
BMI score of 28?
(b) What z score defines overweight?
(c) What z score defines obese?
198 POPULATIONS, SAMPLES, AND PROBABILITY
Key Equations
ADDITION RULE
Pr(A or B) = Pr(A) + Pr(B)
MULTIPLICATION RULE
Pr(A and B) = [Pr(A)][Pr(B)]
REVIEW QUESTIONS
Key Equations
ADDITION RULE
Pr(A or B) = Pr(A) + Pr(B)
MULTIPLICATION RULE
Pr(A and B) = [Pr(A)][Pr(B)]
REVIEW QUESTIONS
8.10 Television stations sometimes solicit feedback volunteered by viewers
about a televised event. Following a televised debate between Barack
Obama and Mitt Romney in the 2012 U.S. presidential election campaign,
a TV station conducted a telephone poll to determine the “winner.” Callers
were given two phone numbers, one for Obama and the other for Romney,
to register their opinions automatically.
about a televised event. Following a televised debate between Barack
Obama and Mitt Romney in the 2012 U.S. presidential election campaign,
a TV station conducted a telephone poll to determine the “winner.” Callers
were given two phone numbers, one for Obama and the other for Romney,
to register their opinions automatically.
(a) Comment on whether or not this was a random sample.
(b) How might this poll have been improved?
REVIEW QUESTIONS 199
*8.14 The probability of a boy being born equals .50, or 1 / 2 , as does the probability
of a girl being born. For a randomly selected family with two
children, what’s the probability of
*8.14 The probability of a boy being born equals .50, or 1 / 2 , as does the probability
of a girl being born. For a randomly selected family with two
children, what’s the probability of
(a) two boys, that is, a boy and a boy? (Reminder: Before using either the
addition or multiplication rule, satisfy yourself that the various events
are either mutually exclusive or independent, respectively.)
addition or multiplication rule, satisfy yourself that the various events
are either mutually exclusive or independent, respectively.)
(b) two girls?
(c) either two boys or two girls?
Answers on page 509.
Answers on page 509.
200 POPULATIONS, SAMPLES, AND PROBABILITY
8.19 A sensor is used to monitor the performance of a nuclear reactor. The
sensor accurately reflects the state of the reactor with a probability of .97.
But with a probability of .02, it gives a false alarm (by reporting excessive
radiation even though the reactor is performing normally), and with a
probability of .01, it misses excessive radiation (by failing to report excessive
radiation even though the reactor is performing abnormally).
8.19 A sensor is used to monitor the performance of a nuclear reactor. The
sensor accurately reflects the state of the reactor with a probability of .97.
But with a probability of .02, it gives a false alarm (by reporting excessive
radiation even though the reactor is performing normally), and with a
probability of .01, it misses excessive radiation (by failing to report excessive
radiation even though the reactor is performing abnormally).
(a) What is the probability that a sensor will give an incorrect report, that is,
either a false alarm or a miss?
either a false alarm or a miss?
(b) To reduce costly shutdowns caused by false alarms, management introduces
a second completely independent sensor, and the reactor is shut
down only when both sensors report excessive radiation. (According to
this perspective, solitary reports of excessive radiation should be viewed
as false alarms and ignored, since both sensors provide accurate information
much of the time.) What is the new probability that the reactor
will be shut down because of simultaneous false alarms by both the first
and second sensors?
a second completely independent sensor, and the reactor is shut
down only when both sensors report excessive radiation. (According to
this perspective, solitary reports of excessive radiation should be viewed
as false alarms and ignored, since both sensors provide accurate information
much of the time.) What is the new probability that the reactor
will be shut down because of simultaneous false alarms by both the first
and second sensors?
(c) Being more concerned about failures to detect excessive radiation,
someone who lives near the nuclear reactor proposes an entirely
different strategy: Shut down the reactor whenever either sensor
reports excessive radiation. (According to this point of view, even a
solitary report of excessive radiation should trigger a shutdown,
since a failure to detect excessive radiation is potentially catastrophic.)
If this policy were adopted, what is the new probability
that excessive radiation will be missed simultaneously by both the
first and second sensors?
someone who lives near the nuclear reactor proposes an entirely
different strategy: Shut down the reactor whenever either sensor
reports excessive radiation. (According to this point of view, even a
solitary report of excessive radiation should trigger a shutdown,
since a failure to detect excessive radiation is potentially catastrophic.)
If this policy were adopted, what is the new probability
that excessive radiation will be missed simultaneously by both the
first and second sensors?
202 POPULATIONS, SAMPLES, AND PROBABILITY
*8.21 Assume that the probability of breast cancer equals .01 for women in
the 50–59 age group. Furthermore, if a women does have breast cancer,
the probability of a true positive mammogram (correct detection of
breast cancer) equals .80 and the probability of a false negative mammogram
(a miss) equals .20. On the other hand, if a women does not
have breast cancer, the probability of a true negative mammogram (correct
nondetection) equals .90 and the probability of a false positive
mammogram (a false alarm) equals .10. (Hint: Use a frequency analysis
to answer questions. To facilitate checking your answers with those in
the book, begin with a total of 1,000 women, then branch into the number
of women who do or do not have breast cancer, and finally, under
each of these numbers, branch into the number of women with positive
and negative mammograms.)
*8.21 Assume that the probability of breast cancer equals .01 for women in
the 50–59 age group. Furthermore, if a women does have breast cancer,
the probability of a true positive mammogram (correct detection of
breast cancer) equals .80 and the probability of a false negative mammogram
(a miss) equals .20. On the other hand, if a women does not
have breast cancer, the probability of a true negative mammogram (correct
nondetection) equals .90 and the probability of a false positive
mammogram (a false alarm) equals .10. (Hint: Use a frequency analysis
to answer questions. To facilitate checking your answers with those in
the book, begin with a total of 1,000 women, then branch into the number
of women who do or do not have breast cancer, and finally, under
each of these numbers, branch into the number of women with positive
and negative mammograms.)
(a) What is the probability that a randomly selected woman will have a
positive mammogram?
(b) What is the probability of having breast cancer, given a positive mammogram?
(c) What is the probability of not having breast cancer, given a negative
mammogram?
Answers on page 509.
mammogram?
Answers on page 509.
Answers:
ANSWERS TO SELECTED QUESTIONS 505
5.15 (a) 83 1 (–1.64)(20) 5 50.2
or 83 1 (–1.65)(20) 5 50
(b) .9599
(c) .1357
(d) 83 1 162.332 1202 5 e
129.6
36.4
(e) .2896
(f) 83 1 161.962 1202 5 e
122.2
43.8
(g) .7021
(h) 83 1 (0.84)(20) 5 99.8
(i) 83 1 161.882 1202 5 e
120.6
45.4
(j) .8643
(k) 0 since exactly 61 equals 61.000 etc. to infinity, a point along the base of the
normal curve that is associated with no area under the normal curve.
5.15 (a) 83 1 (–1.64)(20) 5 50.2
or 83 1 (–1.65)(20) 5 50
(b) .9599
(c) .1357
(d) 83 1 162.332 1202 5 e
129.6
36.4
(e) .2896
(f) 83 1 161.962 1202 5 e
122.2
43.8
(g) .7021
(h) 83 1 (0.84)(20) 5 99.8
(i) 83 1 161.882 1202 5 e
120.6
45.4
(j) .8643
(k) 0 since exactly 61 equals 61.000 etc. to infinity, a point along the base of the
normal curve that is associated with no area under the normal curve.
5.18 (a) mean exceeds median
(b) 20.75
(c) 0.50
(b) 20.75
(c) 0.50
ANSWERS TO SELECTED QUESTIONS 509
8.14 (a) A1–2 B A1–2 B 5 1–4
(b) A1–2) A1–2 B 5 1–4
(c) A1–4)1A1–4 B 5 2–4
8.21
1,000
Women
990
No Breast Cancer
.99
.90 .10 .20 .80
.01
891
True
Negative
99
False
Positive
10
Breast Cancer
2
False
Negative
8
True
Positive
(a)
99 1 8
1,000 5
107
1,000 5 .107
(b)
8
99 1 8 5
8
107 5 .075 .
(c)
891
891 1 2 5 .998
Copyright © 2015 John Wiley & Sons, Inc.
8.14 (a) A1–2 B A1–2 B 5 1–4
(b) A1–2) A1–2 B 5 1–4
(c) A1–4)1A1–4 B 5 2–4
8.21
1,000
Women
990
No Breast Cancer
.99
.90 .10 .20 .80
.01
891
True
Negative
99
False
Positive
10
Breast Cancer
2
False
Negative
8
True
Positive
(a)
99 1 8
1,000 5
107
1,000 5 .107
(b)
8
99 1 8 5
8
107 5 .075 .
(c)
891
891 1 2 5 .998
Copyright © 2015 John Wiley & Sons, Inc.
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Solution: Topic 2 Exercises