Economics 506Sections M &M1
Solution to Homework 5
(The problem numbers refer to numbers in 7
th
edition)
7.1
:
a.
–
c.
Answers vary.
d.
The histogram exhibits a mound shape.
The sample mean should be close to 3.5 = μ
e.
The standard deviation should be close to σ/
3 = 1.708/
3 = .9860.
f.
Very similar pictures.
7.2
:
a.
P
(
Y
= 2) =
P
(
W
= 6) =
p
(4, 1, 1) +
p
(1, 4, 1) +
p
(1, 1, 4) +
p
(3, 2, 1) +
p
(3, 1, 2)
=
p
(2, 3, 1) +
p
(2, 1, 3) +
p
(1, 3, 2)+
p
(1, 2, 3) +
p
(2, 2, 2)
=
216
10
.
b.
Answers vary, but the relative frequency should be fairly close.
c.
The relative frequency should be even closer than what was observed in part b.
7.9
:
a.
P
(
Y
– μ ≤ .3) =
P
(–1.2 ≤
Z
≤ 1.2) = .7698.
b.
P
(
Y
– μ ≤ .3) =
P
(–.3
n
≤
Z
≤ .3
n
) = 1 – 2
P
(
Z
> .3
n
).
For
n
= 25, 36, 69, and
64, the probabilities are (respectively) .8664, .9284, .9642, and .9836.
c.
The probabilities increase with
n
, which is intuitive since the variance of
Y
decreases
with
n
.
d.
Yes, these results are consistent since the probability was less than .95 for values of
n
less than 43.
7.10
:
a.
P
(
Y
– μ ≤ .3) =
P
(–.15
n
≤
Z
≤ .15
n
) = 1 – 2
P
(
Z
> .15
n
).
For
n
= 9, the
probability is .3472 (a smaller value).
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 Fall '08
 Staff
 Economics, Normal Distribution, Standard Deviation, Binomial distribution, 1 degree, c. Y

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