) and then set its
derivative equal to zero. This is obtained below:
log
f
(
y
) = log
1
√
2
πσ
2

(
y

μ
)
2
2
σ
2
Taking the derivative of log
f
(
y
) yields the following:
d
dy
log
f
(
y
) =

(
y

μ
)
σ
2
Setting
d
dy
log
f
(
y
) = 0 we see that the maximum can only occur when
y
=
μ
. If
we now substitute
y
=
μ
into
f
(
y
) we can immediately see that at the maximum,
f
(
μ
) =
1
√
2
πσ
2
.
14
10
.
[5 points]
One method of arriving at economic forecasts is to use a consensus
approach.
A forecast is obtained from each of a large number of analysts; the
average of these individual forecasts is the consensus forecast.
Suppose that the
individual 1996 January prime interest rate forecasts of all economic analysts are
approximately normally distributed with mean 7% and standard deviation 2.6%.
If a single analyst is randomly selected from among this group, what is the proba
bility that the analysts forecast of the prime interest rate will exceed 11%? What
interval would the analyst have to provide so as to be 90% confident his interest
rate forecast will lie in this interval. (Assume a symmetric interval.)
Suggested Answers:
We know right away that
Y
∼
N
(0
.
07
,
0
.
026
2
). We are interested in
P
{
Y >
0
.
11
}
.
Since
Y
is not a standard normal variable, we must first standardize it. However, if
we do this to
Y
, we must also do this to 0.11. Accordingly, we have the following:
P
{
Y >
0
.
11
}
=
P
Y

μ
σ
>
0
.
11

μ
σ
=
P
Y

0
.
07
0
.
026
>
0
.
11

0
.
07
0
.
026
=
P
Y

0
.
07
0
.
026
>
1
.
54
Note now that
Y

0
.
07
0
.
026
is a standard normal variable and so we can go into the table
and find what is the probability that this quantity is greater that 1
.
54. We see that
this probability is 0.0618 or about 6.18%.
To obtain the 90% symmetric confidence interval, we note that this interval says
P
n
z
1
≤
Y

μ
σ
≤
z
2
o
= 0
.
9. Since the interval is symmetric, this means that
z
1
=

z
2
and
z
2
is computed as the value above which 5% of the data lies. Looking at
the table, we see that this number is 1.645. Thus, to obtain the values for
Y
rather
than the standardized version of
Y
, we proceed as follows.
P

1
.
645
≤
Y

μ
σ
≤
1
.
645
= 0
.
9
P

1
.
645
≤
Y

0
.
07
0
.
026
≤
1
.
645
= 0
.
9
P
{
0
.
07

1
.
645
×
0
.
026
≤
Y
≤
0
.
07 + 1
.
645
×
0
.
026
}
= 0
.
9
P
{
0
.
02723
≤
Y
≤
0
.
11277
}
= 0
.
9
Accordingly, the interval we seek is
Y
∈
[0
.
02723
,
0
.
11277]
15
11
.
[5 points]
Used photocopy machines are returned to the supplier, cleaned, and
then sent back out on lease agreements.
Major repairs are not made, however,
and as a result, some customers receive malfunctioning machines.
Among eight
used photocopiers available today, three are malfunctioning.
A customer wants
to lease four machines immediately. To meet the customers deadline, four of the
eight machines are randomly selected and, without further checking, shipped to the
customer. What is the probability that the customer receives
a)
No malfunctioning machines.
b)
At least one malfunctioning machine.
Suggested Answers:
Defined
Y
= the number of malfunctioning copiers selected. In this case,
Y
has a
hypergeometric distribution of the form
p
(
y
) =
(
3
y
)(
5
4

y
)
(
8
4
)
Accordingly, we have the following probabilities:
P
{
Y
= 0
}
=
p
(0) =
1
14
P
{
Y
≥
1
}
= 1

P
{
Y
= 0
}
= 1

p
(0) =
13
14
16
12
.
[10 points]
The magnitude of earthquakes recorded in a region of North America
can be modeled as having an exponential distribution with mean 2.4, as measured
on the Richter scale. Find the probability that an earthquake striking this region
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 Summer '14
 MirzaTrokic
 Normal Distribution, Variance, Probability distribution, Probability theory, probability density function