to Solutions Homework 5
4.45
Let X denote the number of underfilled cartons in the selected 50 cartons, If 50
cartons are selected independently, then X can be modeled to have a binomial
distribution with p .
a. p = 0.05, We then have
2
P ( X 2) = p ( x)
x =0
2
50
= (0.05) x (1 0.05)50 x
x =0 x
= 0.5405
b. p = 0.1, We then have
2
P ( X 2) = p ( x )
x =0
2
50
= (0.1) x (1 0.1)50 x
x=0 x
= 0.1117
4.49
a. The assumption is that every inspection is an independent event.
b. P (a wing crack is detected) = p1 p2 p3 = 0.36
Let X denote the number of planes detected a wing crack, the binomial distribution is
a reasonable model for this experiment, p = 0.36 , then
P ( X 1) = 1 P ( X = 0)
3
0
3
= 1 ( 0.36 ) ( 1 0.36 )
0
= 0.7379
4.54
Let X denote the number of components last longer than 1000 hours, then X can be
modeled to have a binomial distribution with p = 1 0.15 = 0.85
a. We have
4
P ( X = 2) = (0.85) 2 (1 0.85) 2 = 0.0975
2
b. If the subsystem operates for longer than 1000 hours, any two or more of the four
components last for more than 1000 hours, so we get
4
4
4
P ( X 2) = (0.85) 2 (1 0.85) 2 + (0.85)3 (1 0.85)1 + (0.85) 4 (1 0.85) 0
2
3
4
= 0.988
4.58
a. X has a binomial distribution with p = 0.1
n
0
P ( X 1) 1 = P ( X = 0) = 1 ( 0.1) (1 0.1) n = 1 0.9n = 0.95
0
Thus n =
log 0.05
= 28
log 0.9
b. X has a binomial distribution with p = 0.1
n
0
n
P ( X 1) = 1 P ( X = 0) = 1 ( 0.05 ) ( 1 0.05 ) = 1 0.95n = 0.95
0
Thus n =
log 0.05
= 58
log 0.95
4.60
For each case,
3
3
x
3 x
P(innocent people receiving death penalty) = 0.01 ( 0.05 ) ( 1 0.05 )
x =2 x
= 0.0000725
Let X denote the number of innocent people receiving death penalty, X can be
modeled as a binomial distribution with n = 100, p = 0.0000725 , then
E ( X ) = np = 100 0.0000725 = 0.00725
V ( X ) = np ( 1 p ) = 100 0.0000725 0.9999275 = 0.007249
= V ( X ) = 0.0851
4.67
Let X denote the number of defective engines tested before a good engine is found,
X can be modeled as a geometric distribution with p = 1 0.1 = 0.9 , we need to
calculate P ( X = 2)
P ( X = 2) = q 2 p = ( 0.1) 0.9 = 0.009
2
4.68
Define
event A: the first two engines are defective
event B: at least four defective engines are tested before the first nondefective engine
is found.
We need to find P ( B | A) =
P ( B A)
P ( A)
The first and second test are independent, then we have
P ( A) = 0.1 0.1 = 0.01
Event B A , then
P ( B A) = P ( B ) = (0.1) k 0.9 = 0.0001
k =4
Therefore,
P ( B | A) =
0.0001
= 0.01
0.01