text: # 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
  ```