Topic 2 Exercises
Grand Canyon University: PSY 520 (520-0101)
Mary Young
February 27, 2017
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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
The proportion of population with mean m and deviation d above k is
=
where Erfc(x) = and Erf(x) =
(a) above Kristen’s 125?
= 0.0478
(b) below 82?
(c) within 9 points of the mean?
(d) more than 40 points from the mean?
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 identifi ed with
(a) the upper 2 percent, that is, 2 percent to the right (and 98 percent to the left)?
Solve equation with respect to k:
where above = 2/100:
k = 130.8
(b) the lower 10 percent?
Solve equation with respect to k:
where below = 1/10:
k = 80.777
(c) the upper 60 percent?
Solve equation with respect to k:
where above = 6/10:
k = 96.2
(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.]
Solve equation with respect to
where between = 95/100
= 70.6,
(e) the middle 99 percent?
Solve equation with respect to
/2
where between = 99/100
= 61.36, = 138.64
5. 15 and 5.16, remember to decide fi rst whether a proportion or a score is to be found. *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?
where m = 83, d = 20, below = 5/100
k = 50.1, that is, these people suffer 50.1 hours or less.
(b) What proportion of sufferers estimate that their colds lasted longer than 48 hours?
above = = 0.96 = 96%
(c) What proportion suffered for fewer than 61 hours?
below = = 0.1357 = 13.57%
(d) What is the estimated number of hours suffered by the extreme 1 percent either above or below the
mean?
Solve equation with respect to
= outside/2
where outside = 1/100
= 31.48, = 134.52, below 31.48 and above 134.52 hours.
I assume that 1% is the sum of extreme low and extreme high.
(e) What proportion suffered for between 1 and 3 days, that is, between 24 and 72 hours?
=
0.2896 = 28.96%
(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) .]
= between/2
where between = 95/100
= 43.8, = 122.2, below 43.8 and 122.2 hours.
(g) What proportion suffered for between 2 and 4 days?
=
0.7406 = 74.06%
(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.
Solve equation with respect to k:
where above = 2/10:
k = 99.83 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.
Solve equation with respect to
= outside
where outside = 3/100
= 45.38, = 120.62, below 45.38 and above 120.62 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?
below = = 0.1357 = 13.57%
(k) What proportion of the original survey suffered for exactly 61 hours? (Be careful!)
Zero. *Exactly* 61 hours is zero-width stripe, and it integrates to zero.
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 defi ned 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?
At positively skewed distribution median < mean, or
median < 28.
(b) What z score defines overweight?
(c) What z score defines obese?
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.
(a) Comment on whether or not this was a random sample.
A similar poll of 1936. The Literary Digest“ conducted a poll, mailing out millions of postcards and
simply counting the returns. "Then, in 1936, its 2.3 million "voters" constituted a huge sample, but they
were generally more affluent Americans who tended to have Republican sympathies. The Literary Digest
was ignorant of this new bias; the week before election day, it reported that Alf Landon was far more
popular than Roosevelt. At the same time, George Gallup conducted a far smaller (but more scientifically
based) survey, in which he polled a demographically representative sample. Gallup correctly predicted
Roosevelt's landslide victory. The Literary Digest soon went out of business, while polling started to take
off.” - Wikipedia.
No, the poll was not random - it consisted of self-selected people who wanted to watch the debate, and
who wanted to call. On the other hand, similar self-selectiveness results in people going to vote, thus
being “likely voters". I wouldn’t be surprised, that the results of this pole were close to the results of the
election.
(b) How might this poll have been improved?
To do what Gallup, and others do - don’t get calls, but call to a balanced sample of voters, and likely
voters.
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.)
The events of equal probability: BB, BG, GB, GG. Pr(BB) = 1/4.
(b) two girls?
The events of equal probability: BB, BG, GB, GG. Pr(GG) = 1/4.
(c) either two boys or two girls?
The events of equal probability: BB, BG, GB, GG. Pr(BB or GG) = 2/4 = 1/2.
8.16 A traditional test for extra-sensory perception (ESP) involves a set of playing cards, each of which
shows a different symbol (circle, square, cross, star, or wavy lines). If C represents a correct guess and I
an incorrect guess, what is the probability of
(a) C?
1/5
(b) CI (in that order) for two guesses?
1/5 x 4/5 = 4/25 = 16%
4/5 x 4/5 x 4/5 = 64/125 = 51.2%
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?
There is not enough data for the answer.
Suppose the probability of reactor performing normally is p. Then the probability of sensor giving
incorrect report is
0.02p + 0.01(1-p) = 0.01(1+p), which depends on p and may be any number between 0.01 and 0.02.
NOTE. By comparison, the similar Problem 8.21 has equivalent information - the probability of breast
cancer for women of age 50-59.
(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 fi rst and
second sensors?
Here the probability does not depend on the probability of reactor performing normally - because “false
alarm” means that reactor performs normally. The probability of two false alarms is 0.02 x 0.02 = 0.0004.
(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 fi rst and second sensors? *
0.01 x 0.01 = 0.0001
*8.20 Continue to assume that people are equally likely to be born during any of the months. However,
just for the sake of this exercise, assume that there is a tendency for married couples to have been born
during the same month. Furthermore, we wish to calculate the probability of a husband and wife both
being born during December.
(a) It would be appropriate to use the multiplication rule for independent outcomes? True or false?
Copyright © 2015 John Wiley & Sons, Inc. 202 POPULATIONS, SAMPLES, AND PROBABILITY
Evidently false, since “there is a tendency for married couples to have been born during the same month”.
(b) The probability of a married couple both being born during December is smaller than, equal to, or
larger than (1/12)(1/12) = 1/144.
1/144 would be probability if the events of husband and wife being born in some month were
independent. But the tendency for married couples to have been born during the same month means that
such probability is more than that, more than 1/144. It can be anything between 1/144 and 1/12.
(c) With only the above information, it would be possible to calculate the actual probability of a married
couple both being born during December? True or false?
False. With no more information we can only say that such probability is greater than 1/144 and less or
equal to 1/12.
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 fi nally, 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?
Using formula from 8.19 with p = 0.99 (probability of “performing normally”):
0.99 x 0.1 + 0.01 x 0.8 = 0.107
(b) What is the probability of having breast cancer, given a positive mammogram?
Conditional probability
Pr(Cancer | Positive mammogram) = Pr(Cancer AND Positive mammogram) / Pr(Positive mammogram)
= 0.01 x 0.2 / 0.107 = 0.0187
(c) What is the probability of not having breast cancer, given a negative mammogram?
Probability of negative mammogram
0.99 x 0.9 + 0.01 x 0.2 = 0.893
Conditional probability
Pr(No cancer | Negative mammogram) = Pr(No cancer AND Negative mammogram) / Pr(Negative
mammogram) = 0.99 x 0.9 / 0.893 = 0.998