Answers to Final Exam Review Questions
Question 1
: This problem is a Bayes Law problem.
The forward tree branches first on “will”
and “won’t” date George; the second branch is “is free” and “isn’t free”.
When you reverse the
tree you find that the
()
P
will date George says"free"
=
+
=
.
..
.
075
075 225
25
.
Question 2
: Here we have
X
=
13, Median = 14,
s
=
592
.
; the 98% C.I. is
729
1871
., .
A normal quantile plot, sometimes called a normal probability plot, is the technique I am looking
for.
This is the one that creates a straight line if the data is normally distributed.
Question 3
: Part (a) is hypergeometric, with
N
r
n
x
or
==
=
=
10
6
3
2
3
,,,
.
The answer is two thirds.
Part (b) is poisson with
µ
=
33
5
/
and
x
=
1.
The answer is .0787.
Question 4
: This is a straightforward normal probability calculation.
The answer for part (a) is
z
=
−
=−
0
5
31
16
.
.
.
and
Pz
>−
=
16
5636
.
The answer to part (b) is
−=
−
z
X
.
.
.
.
35
39
305
31
, or
X
=
293
Question 5
: In part (a) the mean is
∑
xf x
25
and the variance is
(
)
σµ
2
2
105
=
∑
xf
x
.
In part (b) you use the formula for the mean and variance of a
linear transformation of a random variable; that is
Y
ab
X
X
=+ ≡
.10
.
Therefore the mean of the phone bill is $2.50, and the variance is 1.05.
In
part (c) we have a problem involving the distribution of the mean.
The question asks the
probability that
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Px
P
x
n
x
i
i
>=
=
⎛
⎝
⎜
⎜
⎞
⎠
⎟
⎟
∑
∑
72
72
30
240
.
.
We know that
Ex
==
µ
250
.
and that
σ
x
n
=
105
30
187
.
.
.
Therefore the z value is
z
=
−
=−
240 250
187
53
..
.
.
and the answer is
Pz
>−
=
53
7019
.
In part (d) the answer is yes; because the sample size is large (n=30) the
sample means have a normal distribution even though the original x’s obviously did not.
We
needed the normal distribution to do our probability calculation.
Question 6
Part (a) is a binomial with n = 5 and p = 18/38. Being a net winner means winning 3, 4, or 5 of
your 5 bets.
The probability of this is
5
3
474
526
5
4
474
526
5
5
474
526
4514
32
41
50
⎛
⎝
⎜
⎞
⎠
⎟
+
⎛
⎝
⎜
⎞
⎠
⎟
+
⎛
⎝
⎜
⎞
⎠
⎟
=
.
For part (b) you are asked to do a normal approximation to a binomial with n=25 and p = 18/38 =
.474.
The distribution of the number of successes in 25 trials is approximately normal with
() ( )
25 .474
11.84
EX
n
p
=
and
1
25 .474 .526
2.497
X
np
p
=
×
×
=
Being a net winner means winning 13 or more of your bets.
The z value is
12.5 11.84
.26
2.497
z
−
.
The answer is
( )
.26
.5000 .1026
.3974
>
=−=
.
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 Spring '08
 Staff
 Normal Distribution, Null hypothesis, µ

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