Homework Set #7 solutions Stat 420 Fall 2010 1. #4.62 (a) P(Z 2 < 1 ...

Homework Set #7 solutions
Stat 420 Fall 2010
1. #4.62 (a) P (Z 2 < 1) = P (−1 < Z < +1) = .6826 by the tables. (b)
P (−1.96 < Z < 1.96) = .95.
2. #4.64(a) Note that $450 is 2.5 standard deviations above the mean, so
by the table the probability is .0062.
3. #4.81 (a) Γ(1) = ∫ ∞
0 e −y dy = 1.
(b) Γ(α) = ∫ ∞
0 y α−1 e −y dy. Integrating by parts, use dv = e −y dy and
u = y α−1 . Then du = (α − 1)y α−2 dy and v = −e −y , and
Γ(α) = −y α−1 e −y | ∞ 0 + (α − 1)
∫ ∞
0
y α−2 e −y dy = (α − 1)Γ(α − 1).
4. #4.88: Let Y have an exponential distribution with β = 2.4. (a)
P (Y > 3) = 1 ∫ ∞
e −y/2.4 dy = .2865.
2.4 3
(b)
P (2 < Y < 3) = 1 ∫ 3
e −y/2.4 dy = .1481.
2.4 2
5. #4.90: Let Y = magnitude of the earthquake which is exponential with
β = 2.4. Let X = # of earthquakes that exceed 5.0 on the Richter scale.
Therefore, X is binomial with n = 10 and
p = P (Y > 5) = 1
2.4
∫ ∞
5
e −y/2.4 dy = .1245.
Then if X is the number that exceed 5.0, this is binomial with n=10
and p=.1245, so
P (X ≥ 1) = 1 − P (X = 0) = 1 − (.8755) 10 = .7354
6. #4.104: Y has an exponential distribution with β = 100. Then, P (Y >
200) = e 200/100 = e 2 . Let the random variable X = # of componential
that operate in the equipment for more than 200 hours. Then, X has
a binomial distribution and
P (equipment operates) = P (X ≥ 2) = P (X = 2) + P (X = 3) = .05.

7. #4.110: Y has a gamma distribution with α = 3 and β = .5. Thus,
E(Y) = 1.5 and V(Y) = .75.
8. #4.126: (a) F (y) = 3y 2 − 2y 3 , =≤ y ≤ 1. F (y) = 0 for y < 0 and
F (y) = 1 for y > 1.
6 * x * (1 - x)
(b)
0.0 0.5 1.0 1.5
f(y)
fy
0.0 0.2 0.4 0.6 0.8 1.0
F(y)
0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.2 0.4 0.6 0.8 1.0
(c) P (.5 < Y < .8) = F (.8) − F (.5) = .396.
9. #4.128: On (0, 1), F (y) = 1 − (1 − y) 3 , so we need to solve the cubic
equation 1 − (1 − φ .9 ) 3 = .9 or φ .9 = .5358.... this means the budgeted
cost should be $53.58.
10. #4.130:
E(Y 2 ) =
so
∫
Γ(α + β) 1
y α+1 (1−y) β−1 dy =
Γ(α)Γ(β) 0
V (Y ) = E(Y 2 )−E(Y ) 2 =
Γ(α + β) Γ(α + 2)Γ(β)
Γ(α)Γ(β) Γ(α + 2 + β) = (α + 1)α
(α + β)(α + β + 1) .
( ) 2
(α + 1)α α
(α + β)(α + β + 1) − =
α + β
αβ
(α + β) 2 (α + β + 1) .