Chapter 4 Solutions
4.1.
Eleven of the Frst 20 digits on line 109 correspond
to “heads” so the proportion of heads is
11
20
=
0
.
55.
This is close to 0.5, but not exactly the same because of random variation.
36009
19365
15412
39638
HTHHT
HTHTT
HTHHH
HTTHT
4.3.
If you hear music (or talking) one time, you will almost certainly hear the same thing for
several more checks after that. (±or example, if you tune in at the beginning of a 5-minute
song and check back every 5 seconds, you’ll hear that same song over 30 times.)
4.4.
To estimate the probability, count the number of times the dice show 7 or 11, then divide
by 25. ±or “perfectly made” (fair) dice, the number of winning rolls will nearly always
(99.4% of the time) be between 1 and 11 out of 25.
4.5.
The table on the right shows information from
www.mms.com
as of this writing. The exercise speci-
Fed M&M’s Milk Chocolate Candies, but based on these
numbers, results will be similar for other popular varieties.
Of course, answers will vary, but students who take reason-
ably large samples should get results close to the numbers
in this table. (±or example, samples of size 50 will almost
always be within
±
12%, while size 75 should give results within
±
10%.)
M&M’s variety
Green %
Milk Chocolate
16%
Peanut
15%
Dark Chocolate
16%
Dark Choc. Peanut
15%
Almond
20%
Peanut Butter
20%
4.6.
Out of a very large number of patients taking this medication, the fraction who experience
this bad side effect is about 0.00001.
Note:
Student explanations will vary, but should make clear that 0.00001 is a long-run
average rate of occurrence. Because a probability of 0.00001 is often stated as “1 in in
100,000,” it is tempting to interpret this probability as meaning “exactly 1 out of every
100,000.” While we
expect
about 1 occurrence of side effects out of 100,000 patients, the
actual number of side effects patients is random; it might be 0, or 1, or 2,
...
.
4.7. (a)
Most answers will be between 35% and 65%.
(b)
Based on 10,000 simulated
trials—more than students are expected to do—there is about an 80% chance of having a
longest run of four or more (i.e., either making or missing four shots in a row), a 54%
chance of getting Fve or more, a 31% chance of getting six or more, and a 16% chance of
getting seven or more. The average (“expected”) longest run length is about six.
4.8. (a)
–
(c)
Results will vary, but after
n
tosses, the
distribution of the proportion
ˆ
p
is approximately Nor-
mal with mean 0.5 and standard deviation 1
/(
2
√
n
)
,
while the distribution of the count of heads is ap-
proximately Normal with mean 0
.
5
n
and standard
deviation
√
n
/
2, so using the 68–95–99.7 rule, we
have the results shown in the table on the right. Note that the range for
ˆ
p
gets narrower,
while the range for the count gets wider.
99.7% Range
99.7% Range
n
for
ˆ
p
for count
40
0
.
5
±
0
.
237
20
±
9
.
5
120
0
.
5
±
0
.
137
60
±
16
.
4
240
0
.
5
±
0
.
097
120
±
23
.
2
480
0
.
5
±
0
.
068
240
±
32
.
9
149