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STAT 420
Spring 2012
Homework #1
(due Friday, January 27, by 4:00 p.m.)
1.
a)
At Anytown College, the heights of female students are normally distributed with
mean 66 inches and standard deviation 1.5 inches. The heights of male students are
also normally distributed with mean 69 inches and standard deviation 2 inches.
For Homecoming, a male student and a female student are selected independently
at random to be the King and the Queen.
What is the probability that the female
student selected to be the Queen is taller than the male student selected to be the King?
Need
P
(
Queen – King > 0
) = ?
E
(
Queen – King
) = 66 – 69 = –
3.
Var
(
Queen – King
) = 1.5
2
+ 2
2
=
6.25.
SD
(
Queen – King
) = 2.5.
(
Queen – King
)
is normally distributed.
P
(
Queen – King > 0
) =
( )


>
5
.
2
3
0
Z
P
= P
(
Z > 1.20
) = 1 –
Φ
(
1.20
) =
0.1151
.
b)
Suppose that a population of married couples in Anytown have heights in inches,
X for the wife, and Y for the husband.
Suppose that
(
X
, Y
) has a bivariate normal
distribution with parameters
μ
X
= 66,
σ
X
= 1.5,
μ
Y
= 69,
σ
Y
= 2,
ρ
= 0.44.
What is the probability that the wife is taller than her husband?
Need
P
(
Wife – Husband > 0
) = ?
E
(
Wife – Husband
) = 66 – 69 = –
3.
Var
(
Wife – Husband
) =
σ
X
2
– 2
ρ
σ
X
σ
Y
+
σ
Y
2
= 1.5
2
– 2
⋅
0.44
⋅
1.5
⋅
2 + 2
2
=
3.61.
SD
(
Wife – Husband
) = 1.9.
(
Wife – Husband
)
is normally distributed.
P
(
Wife – Husband > 0
) =
( )


>
9
.
1
3
0
Z
P
= P
(
Z > 1.58
) = 1 –
Φ
(
1.58
) =
0.0571
.
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View Full Document 2.
Suppose that company A and company B are in the same industry sector, and the
prices of their stocks, $X per share for company A and $Y per share for company
B, vary from day to day randomly according to a bivariate normal distribution with
parameters
μ
X
= 45,
σ
X
= 5.6,
μ
Y
= 25,
σ
Y
= 5,
ρ
= 0.8.
a)
What is the probability that on a given day the price of stock for company B
(
Y
)
exceeds $33?
Y
has Normal distribution with mean
μ
Y
= 25
and
standard deviation
σ
Y
= 5.
P
(
Y > 33
)
=

>
5
25
33
Z
P
=
P
(
Z > 1.60
)
=
1 –
Φ
(
1.60
)
=
1 – 0.9452
=
0.0548
.
b)
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This note was uploaded on 02/07/2012 for the course STAT 420 taught by Professor Stepanov during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Stepanov
 Standard Deviation

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