Math 511 – Review for Exam 3
1. Suppose that Y is a random variable having mean 10, variance 8, and moment
generating function H(t). Suppose that the moment generating function for X
is M (t) = e4t H (3t) . Find the mean and variance for X.
Solution: µ X = 34, σ X2 = 72
2. Suppose that X is a continuous random variable with pdf f(x) given by f (x) =
k
x >1.
x4
(a). What is the value of k?
(b). What is the value of µ = E(X) ?
2
(c). What is the value of Var(X) = σ ? (d). What is the value of P(X > 6 | X > 2) ?
Solution:
(a). k = 3
(b). µ =
3
2
(c). σ 2 =
3
4
(d).
1
27
3. Suppose that X is a Normal Random Variable with µ = 12, σ 2 = 4 .
(a). What is P(X ≥ 6 | X ≤ 12) ? (b). What is P(10 ≤ X ≤ 14) ?
Solution: (a). 0.9974
(b). 0.6826
4. In each problem below, the moment generating function for X is given. Find the mean
and variance of X, and give the name of the distribution.
et
(a). The moment generating function for X is M (t) =
.
3 − 2et
18
⎛ 6 ⎞
(b). The moment generating function for X is M (t) = ⎜
.
⎝ 6 − 2t ⎟⎠
Solution:
(a). Geometric with mean 3 and variance 6.
(b). Gamma with mean 6 and variance 2.
5. The time needed to repair the transmission of a Honda Civic is exponentially
distributed with the average time of repair being 2 hours.
(a). What is the probability that a particular repair job takes more than 3 hours?
(b.) Given that the mechanics have been working on a car for more than 2 hours
already, what is the probability that the entire repair takes more than 5 hours?
Solution: (a). e−
3
2
(b). e−
3
2
6. Suppose that a train arrives sometime between noon and 1:00 PM (exclusive) and at
random. Let X denote the time of arrival, in minutes after noon, of the train. So if the
train arrives at 12:32 then X = 32. You may assume that X is a continuous random
variable that is uniform on (0, 60).
(a). What is the expected time of arrival of the train?
(b). What is the Variance of X?
Solution: (a). 12:30
(c). What is P( 10 ≤ X ≤ 25)?
(b). Variance is 300
(c). P(10 ≤ X ≤ 25) = 0.25.
7. Let X equal the number customers that enter a store every hour. Assume X is Poisson
with mean 8. Let W be the time in hours before the 5th customer enters.
(a). What is the mean and variance of W?
W is a gamma random variable with α = 5, β = 1 8 .
So, µW = αβ = 5 8 , and σ W2 = αβ 2 = 5 64 .
(b). What is P (W > 1) ?
Solution:
(a). W is a gamma random variable with
α = 5, β = 1 8 , and so µ = αβ = 5 8 , and Var(W ) = αβ 2 = 5 64 .
8k
(b). P (W > 1) = P ( X ≤ 4 ) = e ∑ .
k=0 k!
−8
4
8. Let X be a random variable with mean 60 and variance 16.
(a). Find a number A such that P(50 ≤ X ≤ 70) ≥ A
(b). Find a number B such that P( X − 60 ≥ 12) ≤ B .
Hint: Use Chebyshev’s Inequality.
Solution:
Using Chebyshev’s Inequality:
(a). P(50 ≤ X ≤ 70) = P(−10 ≤ X − 60 ≤ 10) = P ( X − 60 ≤ 10 ) =
= P ( X − µ ≤ 2.5σ ) ≥ 1 −
(b). P ( X − 60 ≥ 12 ) = P ( X − µ ≥ 3σ ) ≤
1
= 0.84
2.5 2
1
.
9
9. (a). Suppose that Z is a standard normal random variable.
What is the value of P(Z 2 < 1.44) ?
Solution: P(Z 2 < 1.44) = P (| Z |< 1.2 ) = P ( −1.2 < Z < 1.2 ) = 1 − 2P ( Z > 1.2 ) = 0.7698
(b). If the probability P ( −a < Z < b ) = 0.406 and a, b > 0, what is the
value of P(Z > a) + P(Z > b) ?
Solution: Since,
P ( −a < Z < b ) = P ( Z > −a ) − P ( Z > b ) = 1− P ( Z > a ) − P(Z > b) = 1− ⎡⎣ P(Z > a) + P ( Z > b ) ⎤⎦
We get,
P(Z > a) + P(Z > b) = 1 − P ( −a < Z < b ) = 1 − 0.406 = .594
10. Suppose that X has the pdf f X (x), x > 0. If Y = aX + b a > 0 , how is the pdf of Y related
to the pdf of X? (You may assume that both pdf’s are defined for all real numbers).
Solution: fY (x) =
1
a
f X ( x −b
a ) a > 0.
11. Suppose that a basketball player makes 80% of his free throws. Let X denote the number
of free throws he must attempt to make a total of 10 shots.
(a). What is an expression for P(X = k) , k =10, 11, 12, 13, …
(b). What is the variance of X?
Solution: X has a negative binomial distribution with r = 10 and p = 0.8.
⎛ k − 1⎞
(a). So we have P ( X = k ) = ⎜
(0.8)10 )(0.2)k −10 .
⎟
⎝ 9 ⎠
⎛ 0.2 ⎞ 25
(b). Var(X) = 10 × ⎜ 2 ⎟ =
.
⎝ .8 ⎠ 8
12. (X, Y) is distributed uniformly on the triangle with vertices (4,4), (4, 0), and (0, 4).
Find the joint pdf of (X, Y) and also the marginal probabilities for each of X and Y.
x
y
Solution: f ( x, y ) = 1 8 . f X ( x ) = , fY ( y ) = .
8
8
13. Suppose that a trail has an average of 2.5 defects per foot and that these defects satisfy a
Poisson distribution. Let W denote the number of feet before the first defect is found.
(a). What kind of random variable is W?
(b). What is the mean of W?
(c). What is the value of P(W < 0.2)?
Solution: (a). W is an exponential random variable.
(b). The mean of W is θ = 0.4 feet between defects .
(c). P(W < 0.2) = 1 − P(W > 0.2) = 1 − e
− 0.2 0.4
= 1− e
− 12
= 0.39
14. Cars arrive at a tollbooth at an average rate of two per minute according to a Poisson
process. What is the probability that the operator must wait more than 5 minutes to
collect 12 tolls?
(a). Express the answer as a finite sum.
(b). Express the answer as a definite integral.
Hint: The pdf for a gamma random variable is f (x) =
− xβ
xα −1
.
e
Γ (α ) β α
Solution:
(a). Let X be the Poisson random variable that denotes the number of tolls in 5 minutes.
11
10 k e−10
≈ 0.69677
Then X has mean λ = 10 and P (W > 5 ) = P ( X ≤ 11) = ∑
k!
k=0
∞
⌠ x11e−2 x
(b). W is gamma with α = 12, β = 2 and so, P (W > 5 ) = ⎮
12 dx .
⌡5 11!( 0.5 )
This integral also evaluates to ≈ 0.69677 .
1
15. Suppose that X is an exponential random variable with mean 6.
Let W = 3X – 2.
(a). Determine the mean and variance of W.
(b). Determine the pdf for W.
Solution:
µ = E(W ) = 3E(X) − 2 = 3 × 6 − 2 = 16 .
Var(W ) = 9Var(X) = 9 × 36 = 144 .
(b). Determine the pdf for W.
Solution:
The cdf for X is FX (t) = P(X ≤ t) = 1 − e
− t6
.
⎛ t +2⎞
−⎜
⎟
t + 2⎞
⎛
= 1 − e ⎝ 18 ⎠ .
So, FW (t) = P (W ≤ t ) = P ( 3X − 2 ≤ t ) = P ⎜ X ≤
⎟
⎝
3 ⎠
Taking the derivative of FW (t) = 1 − e
⎛ t +2⎞
−⎜
⎝ 18 ⎟⎠
, gives fW (t) =
1 − (t + 2 18 )
.
e
18
16. Suppose that we choose 7 numbers at random from the 50 digits in
{1, 2, …, 50} without replacement.
Let X denote the minimum and Y the maximum of the numbers drawn.
(a). Find the joint pmf of X and Y.
Solution:
⎧ ⎛ j − k − 1⎞
⎪⎜
⎟
⎪⎪ ⎝ 5 ⎠ , for k − j ≥ 6
P ( X = k, Y = j ) = ⎨ ⎛ 50 ⎞
⎪ ⎜⎝ 7 ⎟⎠
⎪
0
otherwise
⎪⎩
(b) Express the marginal pmf for X as a sum.
Solution:
50
44
⎛ j − k − 1⎞
⎛ j − k − 1⎞
∑ ⎜⎝ 5 ⎟⎠
∑
⎜⎝ 5 ⎟⎠
P ( X = k ) = j=k+6
, P (Y = j ) = k=1
.
⎛ 50 ⎞
⎛ 50 ⎞
⎜⎝ 7 ⎟⎠
⎜⎝ 7 ⎟⎠
17. Suppose that the continuous random variables X, Y are uniformly distributed over the
region: x 2 ≤ y ≤ 1, and 0 ≤ x ≤ 1 .
(a). Find the marginal distributions for each of X and Y.
Solution:
f ( x, y ) =
1
=
Area(region)
Area of region = Area of unit square {(0, 0), (0, 1), (1, 0), (1, 1) – Area under the curve y = x 2
1
= 1− ∫ x 2 dx = 1− 1 3 = 2 3
0
And so f (x, y) =
1
f X (x) = ∫ 2
x
3
1
3
= .
2
2
3
2 dy =
3(1− x 2 )
2
, 0 ≤ x ≤ 1 fY (y) = ∫
y
0
(b). Evaluate P ( 12 ≤ X ≤ 1) .Solution: P ( 12 ≤ X ≤ 1) =
3
2
3
2 dy =
∫
1
1
2
3 y
, 0 ≤ y ≤ 1.
2
1− x 2 dx = 5 16 .
18. Suppose that the life of a particular electronic tablet is exponentially distributed with a
mean of 4 years. How long a warrantee should the manufacturer give in order feel secure
that at most 10% of the tablets will fail will during the warrantee period?
Solution: We want e 4 = P ( X > t ) = 0.9. And so, t = −4 ln ( 0.9 ) ≈ 0.421 .
So, the warranty should be about 5 months.
−t
From the Text
3.93, 4.27, 4.33, 4.77, 4.91, 4.97, 5.29(a)
Previously Assigned from the text:
3.97,4.49, 4.51, 4.59. 4.62, 4.63(a), 4.69, 4.71, 4.73, 4.75
4.89, 4.93, 4.103, 4.109 - just find E(L)
5.3, 5.7, 5.9(a), 5.11, 5.13, 5.15, 5.17, 5.21, 5.23(a), 5.25(a)-(b), 5.27(a), 5.33(a)