Download Quiz 4 Solution

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STT 441
Quiz 4
Apr. 1, 2015
1. The time X (in hours) required to repair a machine is an exponentially distributed random
variable with parameter λ = 1/3. The pdf of an exponential distribution is
f (x) =
1
exp(−x/λ)
λ
for x > 0.
What is
(a) the probability that a repair time exceeds 2 hours?
[5pts]
Let X denote the time (in hours) required to repair a machine. Then X ∼ Exponential( 13 ).
P(X > 2) =
Z
∞
3e
−3x
2
dx = −e
∞
−6
= e = 0.0025.
−3x 2
(b) the conditional probability that a repair takes at least 10 hours, given that its duration
exceeds 9 hours?
[5pts]
By the memoryless property of exponential random variables, we have
P (X ≥ 10|X > 9) = P (X ≥ 9 + 1|X > 9) = P (X ≥ 1)
Z ∞
∞
3e−3x dx = −e−3x = 0.0497.
=
1
1
2. Suppose that the height, in inches, of a 25-year-old man is a normal random variable with
parameters µ = 71 and σ 2 = 6.25. Using the attached standard normal table to compute the
following.
(a) What percentage of 25-year-old men are over 6 feet and 2 inches (=74 inches) tall ? [5pts]
Let X be the height of a a 25-year-old man. Then
X − 71
74 − 71 √
> √
6.25
6.25
74 − 71
= 1 − Φ( √
) = 1 − Φ(1.2) = 0.1151.
6.25
P (X > 74) = P
where Φ(·) is the CDF function of the standard normal distribution.
(b) What percentage of men in the 6-footer club are over 6 feet and 5 inches (=77 inches)?
(Here you need to compute the conditional probability)
[5pts]
Let X be the height of a a 25-year-old man. Then
P (X > 77|X > 72) = P ({X > 77} ∩ {X > 72})/P (X > 72) = P (X > 77)/P (X > 72).
Because,
X − 71
77 − 71 √
> √
6.25
6.25
77 − 71
= 1 − Φ( √
) = 1 − Φ(2.4) = 0.0082
6.25
P (X > 77) = P
and
X − 71
72 − 71 √
> √
6.25
6.25
72 − 71
) = 1 − Φ(0.4) = 0.3446
= 1 − Φ( √
6.25
P (X > 72) = P
where Φ(·) is the CDF function of the standard normal distribution, we have
P (X > 77|X > 72) = P (X > 77)/P (X > 72) = 0.0082/0.3446 = 0.0238.
2
3. If X is an exponential random variable with parameter λ = 1 (The expectation of X is 1).
(a) Find the CDF of Y = 3 log X.
[5pts]
The distribution function of Y is
FY (y) = P (Y ≤ y) = P (3 log X ≤ y) = P (X ≤ ey/3 )
Z ey/3
ey/3
e−x dx = −e−x = 1 − exp(−ey/3 ),
=
0
0
for all −∞ < y < ∞.
(b) Find the probability density function for Y defined in part (a).
The pdf of Y is
fY (y) =
d
1
FY (y) = exp(−ey/3 )ey/3
dy
3
3
− ∞ < y < ∞.
[5pts]