Math 5010-1, Spring 2005
Assignment 6
Problems
#2, p. 171. The possible values of X are 1, 2, 3, . . . , 36. The probabilities are routine computations.
For instance, P {X = 1} = P (1, 1) = 1/36, P {X = 2} = p(1, 2) + p(2, 1) = 2/36,
P {X = 3} = p(1, 3) + p(3, 1) = 2/36, P {X = 4} = p(2, 2) + p(1, 4) + p(4, 1) = 3/36,
P {X = 5} = p(1, 5) + p(5, 1) = 2/36, P {X = 6} = p(2, 3) + p(3, 2) + p(1, 6) + p(6, 1) =
4/36, etc.
#3, p. 171. The possible values of X are 3, 4, 5, . . . , 18. The probabilities are routine calculations.
For instance, P {X = 3} = p(1, 1, 1) = 3/63 , P {X = 4} = p(1, 1, 2) + p(1, 2, 1) +
p(2, 1, 1) = 3/63 , etc.
#4, p. 171. The problem can be interpreted in the following way: Lay down 10 seats in a row.
Number them from left to right (1 to 10), each number denoting a rank. Then seat,
completely at random, the five men and the five women. In this way you will find
that
5
,
P {X = 1} = P (W1 ) =
10
5
5
P {X = 2} = P (M1 ∩ W2 ) = P (M1 )P (W2 |M1 ) =
× ,
10 9
5
4 5
P {X = 3} = P (M1 ∩ M2 ∩ W3 ) =
× × ,
10 9 8
5
4 3 5
P {X = 4} = P (M1 ∩ M2 ∩ M3 ∩ W4 ) =
× × × ,
10 9 8 7
5
4 3 2 5
P {X = 5} = · · · =
× × × × ,
10 9 8 7 6
5
4 3 2 1
P {X = 6} = · · · =
× × × × .
10 9 8 7 6
where Wj = {woman in seat j} and Mj = {man in seat j}. The probability that
X = j, for any other value of j, is zero.
R1
#7, p. 229. We know two things: First 0 (a + bx2 ) dx = 1. Therefore,
a+
Also, we know that
R1
0
b
= 1.
3
x(a + bx) dx = (3/5). But
a b
3
+ = .
2 3
5
(1)
R1
0
x(a + bx) dx =
a
2
+ 3b . Therefore,
(2)
Subtract (2) from (1) to find that (a/2) = (2/5), or a = (4/5). This shows that
b = (3/5) too.
1
#9, p. 229. (Example 4b of Chapter 4—Please note the typo)
As in Chapter 4 (p. 134),
bX − (s − X)`, if X ≤ s,
P (s) =
sb,
otherwise.
Therefore,
Z s
Z ∞
E [P (s)] =
{bx − (s − x)`} f (x) dx +
sbf (x) dx
0
s
Z s
Z s
Z ∞
= (b + `)
xf (x) dx − s`
f (x) dx + s`
f (x) dx
0
0
s
Z s
= (b + `)
xf (x) dx − s`F (s) + sb [1 − F (s)] .
0
Therefore, by the fundamental theorem of calculus,
d
E [P (s)] = (b + `)sf (s) − ` {sf (s) + F (s)} + b {1 − F (s) − sf (s)}
ds
= −`F (s) + b − bF (s) = b − (b + `)F (s).
Differentiate again to see that
d2
E [P (s)] = −(` + b)f (s) ≤ 0.
ds2
So if we set (d/ds)E[P (s)] to zero we have a max. The rest follows immediately from these
computations.
#12, p. 229. Let X denote the position of the breakdown-point. Let g(x) denote the distance to the
nearest service station, if the breakdown occurs at x. [X is random; x is a number.]
If the service stations are in A, B, and at the 50-mile mark then
Z ∞
Z 100
1
E[g(X)] =
g(x)f (x) dx =
g(x) dx
100 0
−∞
Z 25
Z 50
Z 75
Z 100
1
=
x dx +
(50 − x) dx +
(x − 50) dx +
(100 − x) dx
100
0
25
50
75
Z 25
1
=
4×
x dx
100
0
= 12.5.
If we put the service stations at 25, 50, and 75-mile positions (as suggested), then
Z 25
Z 37.5
1
(25 − x) dx +
(x − 25) dx
E[g(X)] =
100
0
25
Z 50
Z 67.5
Z 75
+
(50 − x) dx +
(x − 50) dx +
(75 − x) dx
37.5
50
67.5
Z 100
+
(x − 75) dx
75
Z 25
Z 37.5
1
=
(25 − x) dx + 4
(x − 25) dx = 9.375.
2
100
0
25
2
The second method is better.
#14, p. 229. Evidently,
n
1
Z
xn dx =
E [X ] =
0
1
.
n+1
To do this directly, we need first the pdf of X n . But g(x) = xn is increasing and
differentiable, so the pdf of Y = g(X) = X n is
d −1
g (y)
(y = g(x) some x)
fY (y) = fX g −1 (y)
dy
1
= y (1/n)−1 , if 0 ≤ y ≤ 1.
n
I have used the fact that g −1 (y) = y 1/n , so that (d/dy)g −1 (y) = (1/n)y (1/n)−1 .
Therefore,
Z
E[Y ] =
0
1
1
yfY (y) dy =
n
1
Z
y 1/n dy =
0
1
1
=
,
n(1 + (1/n))
n+1
as before.
#15, p. 229. (b) Written in terms of N (0, 1), we have
4−µ
16 − µ
P {4 < X < 16} = P
< N (0, 1) <
σ
σ
16 − 10
4 − 10
=Φ
−Φ
6
6
= Φ(1) − Φ(−1) = 2Φ(1) − 1
≈ (2 × 0.8413) − 1 ≈ 0.6826.
3