# Test 2 and Solutions

```NAME
MATH 461 - X13, Test 2, Spring 2015
April 17, 2015
Calculators, books, notes and extra papers are not allowed on this test
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
```
The text passage contains a series of mathematical questions and their corresponding solutions, primarily focusing on probability and statistics. Each question is followed by a detailed solution.
multiple_questions_with_answers
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
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First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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system
# Test 2 and Solutions

```NAME
MATH 461 - X13, Test 2, Spring 2015
April 17, 2015
Calculators, books, notes and extra papers are not allowed on this test
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
```
user
The text passage contains a series of mathematical questions and their corresponding solutions, primarily focusing on probability and statistics. Each question is followed by a detailed solution.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS