# Test 2 and Solutions

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MATH 461 - X13, Test 2, Spring 2015
April 17, 2015
Calculators, books, notes and extra papers are not allowed on this test
Please show all your work and explain all answers to qualify for full credit
1. (14 points) At a certain bank, the amount of time (in minutes) that a customer spends
being served by a teller is an exponential random variable with parameter 1/5.
(a) If there are no customers when you enter the bank, find the probability that your
serving time will not exceed 7 minutes.
(b) If there is a customer in service when you enter the bank, what is the probability
that he or she will still be with the teller after an additional 4 minutes?
Solution: Let X be the amount of time that a customer spends being served by a teller.
Then X is exponential with parameter 1/5.
(a) P(X ≤ 7) = 1 − e−7/5
(b) Let t be amount of time the customer has been served before you entered the
bank. Then by the memoryless property of the exponential distribution P(X &gt;
4 + t | X &gt; t) = P(X &gt; 4) = e−4/5 .
2. (12 points) Suppose that X is uniformly distributed over the interval (−2, 2). Find the
probability density function of Y = X 4 .
Solution: Since −2 &lt; X &lt; 2, we have 0 ≤ Y = X 4 &lt; 24 . For any y ∈ (0, 24 ),
FY (y) = P(Y ≤ y) = P(X 4 ≤ y) = P(−y 1/4 ≤ X ≤ y 1/4 )
1 1/4
=
y .
2
By differentiating FY we obtain the density
1
, 0 &lt; y &lt; 24
8y 3/4
fY (a) =
0,
otherwise
3. (15 points) If 80 percent of the population of a large community is in favor of a proposed
rise in school taxes, use normal approximation to find the probability that a random
sample of 100 people will contain at least 85 who are in favor of the proposition.
Solution: Let X = the number of people in the random sample who are in favor of the
proposition. Then X is hypergeometric. Since the population is large, hypergeometric
distribution can be approximated by a binomial distribution with parameters (100, 0.8).
We approximate X with a normal variable with parameters (100(0.8), 100(0.8)(0.2)) =
(80, 16).
84.5 − 80
X − 80
√
≥ √
P(X ≥ 85) = (continuity correction) = P(X ≥ 84.5) = P
16
16
4.5
≈ P(Z ≥
) = P(Z ≥ 1.125) = 1 − Φ(1.125) = (normal table)
4
= 1 − 0.8697 = 0.1303 .
4. (15 points) Suppose that X and Y are independent random variables, X is Poisson with
parameter λ, and Y is Poisson with parameter &micro;. Find the conditional distribution of
the random variable X given that X + Y = 50.
Solution: See Example 4b, p. 264
5. (15 points) Suppose that X and Y are independent geometric random variables with
parameter p = 31 . Find P(X &lt; Y ).
Solution: Using the independence of X and Y , we have
P(X &lt; Y ) =
=
=
∞
X
k=1
∞
X
P(X = k, X &lt; Y ) =
P(X = k, k &lt; Y )
k=1
∞ X
P(X = k)P(k &lt; Y ) =
k=1
∞ X
2
9
∞
X
k=1
k=1
4
9
k−1
=
2 1
91−
4
9
2
3
k−1
1
3
k
2
3
2
= .
5
6. (12 points) A man and a woman agree to meet at a certain location about 12:30 pm.
The man will arrive at a time uniformly distributed between 12:15 and 12:45, and the
woman will arrive at a time uniformly distributed between 12:00 and 1:00. Find the
probability that the first to arrive has to wait no longer than 10 minutes.
Solution: Let X be the time man arrives and Y the time the woman arrives. We may
assume that X is uniform on (15, 45) and Y uniform on (0, 60). We know that X
and Y are independent. By drawing a picture and calculating the areas of appropriate
regions we can conclude that the desired probability is 23 .
1
1
Alternatively, by using that the joint density of X and Y is f rac13 &times; 60
= 1800
, we
can compute
Z 45 Z x+10
1
P(|X − Y | ≤ 10) = P(−10 ≤ Y − X ≤ 10) =
dy dx
15
x−10 1800
Z 45
1
1
1
=
20 =
20 &times; 30 = .
1800 15
1800
3
7. (17 points) The joint density of X and Y is given by
1
(x + 1), 0 &lt; x &lt; 1, 0 &lt; y &lt; 2
3
f (x, y) =
0,
otherwise.
(a) Find the conditional density of Y given that X = 12 .
(b) Are X and Y independent?
(c) Find E[X] and Var(X).
Solution:
(a) We first compute the marginal density of X:
Z ∞
Z 2
1
2
fX (x) =
f (x, y) dy =
(x + 1) dy = (x + 1) .
3
−∞
0 3
Now,
f ( 1 , y)
1
fY |X (y| ) = 2 1 =
2
fX ( 2 )
1
3
1
3
1
2
1
2
+1
1
= .
2
+1
This means that the conditional density of Y given X =
1
2
is uniform.
(b) The joint density f (x, y) can be written as the product 23 (x + 1) &times; 2 of a function
of x and a function of y. Therefore, X and Y are independent.
(c)
∞
1
2
2 3 1 2 1 5
E[X] =
xfX (x) dx =
x (x + 1) dx =
x + x =
3
9
3
9
0
−∞
0
Z ∞
Z 1
1
2
2 3 2 2 14
7
E[X 2 ] =
x2 fX (x) dx =
x2 (x + 1) dx =
x + x =
=
3
12
9
36
18
0
−∞
0
7
25
63 − 50
13
Var(X) = E[X 2 ] − (EX)2 =
−
=
=
18 81
162
162
Z
Z
```