**1.** With seven different
categories, a pie chart would be very cluttered; in addition, it would not give
a visual picture of the width of each interval, only the number of banks in
each interval. A bar chart would be very misleading because the the intervals
are of very different widths; the first interval is only 25 million wide. The
next-to-last interval is nearly 100 times wider and its area will be correspondingly
exaggerated.

**2.** Van der Eb is appealing
to the fallacious law of averages, the mistaken belief that above-average returns
must be followed by below-average returns.

**3.** We can use the binomial
model to reason that the probability that the average success frequency will
be close to the probability of success increases as the number of trials increases.
Therefore, the general principle is that if your success probability is larger
than 0.5, then the more games you play, the more likely you are to win more
games than your opponent. On the other hand, it is often easier to win more
games than your opponent if you play an odd number of games than if you win
an even number; for example, it is easier to win 2 of 3 than to win 3 of 4.
For the question of 5 versus 10 games, 10 games has the disadvantage that a
5-5 tie is a possible outcome (indeed a likely outcome if the success probability
is close to 0.5). A tie is not possible if 5 games are played. It turns out
that 5 games is better is the success probability is less than about 2/3 and
10 games is better other wise. (I gave full credit to answers that used either
principle.)

**4.** Let Psat be the probability
of rain on Saturday and Psun be the probability of rain on Sunday. The probability
of no rain on Saturday and no rain on Sunday is given by the multiplication
rule: P[A and B] = P[A]P[B | A]. Here,

The candidate's answer is evidently

In addition to assuming independence, the job candidate is answering the wrong question.

**5.** We can use Bayes' Rule:

**6.** For B to win the $1000,
B will have to win three games in a row. The probability of this happening is
(1/2)^3 = 1/8. Thus, A has a 7/8 chance of winning. The expected values are:

B: $1000(1/8) + $0(7/8) = $125

**7.** Half of the country's
communities may not contain half of the country's people. If the communities
with nonfluoridated water are generally small, then 90 percent (or more) of
the population might live in the communities with fluoridated water supplies.
In addition, because of confounding effects, it may be that fluoridated water
and AIDS cases are more prevalent in large cities, without there being any causal
relationship between the two.

**8.** They should have asked
one person from each family; instead they asked every child and will consequently
have more responses from persons in a large family than from persons in a small
family. Thus they overestimated the average number of school-age children per
family having school-age children. Suppose that there are two families, one
with one child and one with five children. The average family has (1 + 5)/2
= 3 children. If they ask every child, they will get 6 responses and calculate
the average to be (1 + 5 + 5 + 5 + 5 + 5)/6 = 4.33.

**9.** For more up days than
down days, there must be more than 125 up days:

**10.** The artist used chartjunk
to enliven the graph, but did not identify which answer goes with which pie
slice. It is puzzling that the caption indicates 5 possible responses, yet the
pie has only four slices; either two categories were combined or there were
no responses in one of the categories. The 11% slice looks too big in relation
to the 9% slice. We told that "1 = not safe at all" and that "5 = completely
safe," but we are not told the words associated with the middle three numbers.
We are not told the number of people surveyed (it was a minuscule 27). We are
told that this poll "reveals little change in perceived safety," but we are
not shown the results of the earlier poll.