Chapter 13 Solutions
13.1.
Yes: (1) We have a fixed number of observations (
n
=
15). (2) It is reasonable to believe
that each call is independent of the others. (3) “Success” means reaching a live person,
“failure” is any other outcome. (4) Each randomly dialed number has chance
p
=
0
.
2 of
reaching a live person.
13.2.
Not binomial: we do not have a fixed number of observations.
13.3.
Not binomial: because the student receives instruction after incorrect answers, her
probability of success is likely to increase.
13.4.
The number in the sample who say they feel pressure to go to a good college has
approximately the binomial distribution with
n
=
1000 and
p
=
0
.
63.
13.5. (a)
C
, the number caught, is binomial with
n
=
10 and
p
=
0
.
7.
M
, the number missed,
is binomial with
n
=
10 and
p
=
0
.
3.
(b)
We find
P
(
M
=
3
)
=
(
10
3
)
(
0
.
3
)
3
(
0
.
7
)
7
=
120
(
0
.
027
)(
0
.
08235
)
.
=
0
.
2668. With software, we find
P
(
M
≥
3
)
.
=
0
.
6172.
13.6.
Let
N
be the number of live persons contacted among the
15 calls observed. Then
N
has the binomial distribution with
n
=
15 and
p
=
0
.
2 (see also the solution to Exercise 13.1).
(a)
P
(
N
=
3
)
=
(
15
3
)
(
0
.
2
)
3
(
0
.
8
)
12
.
=
0
.
2501.
(b)
P
(
N
≤
3
)
=
P
(
N
=
0
)
+ ··· +
P
(
N
=
3
)
.
=
0
.
6482.
(c)
P
(
N
≥
3
)
=
P
(
N
=
3
)
+ ··· +
P
(
N
=
15
)
.
=
0
.
6020.
(d)
P
(
N
<
3
)
=
P
(
N
=
0
)
+ ··· +
P
(
N
=
2
)
.
=
0
.
3980.
(e)
P
(
N
>
3
)
=
P
(
N
=
4
)
+ ··· +
P
(
N
=
15
)
.
=
0
.
3518.
The TI83 screen on the right illustrates the use of that calculator’s
binompdf
and
binomcdf
functions (found in the DISTR menu) to compute the first two probabilities. The
first of these finds individual binomial probabilities, and the second finds cumulative prob
abilities (that is, it sums the probability from 0 up to a given number). Excel offers similar
features with its BINOMDIST function. Calculators that do not have binomial probabilities
may have a builtin function to compute, e.g.,
(
15
3
)
.
163
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Chapter 13
Binomial Distributions*
13.7.
The screenshots below show Google’s answers at the time these solutions were prepared.
(a)
5 choose 2
returns 10.
(b)
500 choose 2
returns 124,750, and
500 choose 100
returns 2
.
04169424
×
10
107
.
(c)
(10 choose 1)*0.1*0.9^9
returns 0.387420489.
13.8. (a)
With
n
=
15 and
p
=
0
.
2, we have
µ
=
np
=
3 calls.
(b)
σ
=
√
np
(
1
−
p
)
=
√
2
.
4
.
=
1
.
5492 calls.
(c)
With
p
=
0
.
08,
σ
=
√
1
.
104
.
=
1
.
0507 calls; with
p
=
0
.
01,
σ
=
√
0
.
1485
.
=
0
.
3854
calls. As
p
approaches 0, the standard deviation decreases (that is, it also approaches 0).
13.9. (a)
X
is binomial with
n
=
10 and
p
=
0
.
3;
Y
is binomial with
n
=
10 and
p
=
0
.
7
(b)
The mean of
Y
is
(
10
)(
0
.
7
)
=
7 errors caught, and for
X
the mean is
(
10
)(
0
.
3
)
=
3
errors missed.
(c)
The standard deviation of
Y
(or
X
) is
σ
=
√
(
10
)(
0
.
7
)(
0
.
3
)
.
=
1
.
4491
errors.
13.10.
Let
X
be the number of 1’s and 2’s; then
X
has a binomial
distribution with
n
=
90 and
p
=
0
.
477 (in the absence of
fraud). This should have a mean of 42.93 and standard deviation
σ
.
=
√
22
.
4524
=
4
.
7384. Therefore,
P
(
X
≤
29
)
.
=
P
Z
≤
29
−
42
.
93
4
.
7384
=
P
(
Z
≤ −
2
.
94
)
=
0
.
0016
.
(Using software, we find that the exact value is 0.0021.) This probability is quite small, so
we have reason to be suspicious.
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 Spring '10
 T
 Normal Distribution, Poisson Distribution, Standard Deviation, Probability theory, Binomial distribution, Normal Normal Normal

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