MATH 461 Test 2, Spring 2007: Probability Distributions and Random Variables, Exams of Probability and Statistics

Solutions to test 2 of math 461 - probability distributions and random variables, held in spring 2007. The test covers various probability distributions, including exponential, normal, and poisson distributions, as well as concepts like memoryless property and independence. Students are expected to understand concepts related to probability distributions, random variables, and their properties.

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Pre 2010

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MATH 461- C13, Test 2, Spring 2007
April 20, 2007
Calculators, books, notes and extra papers are not allowed on this test
Show all of work to qualify for full credit
1. (10 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 λ= 1/5.
(a) If there are no customers when you enter the bank, find the probability that your
serving time will not exceed 4 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 7 minutes?
Solution: Let Xbe the amount of time that a customer spends being served by a teller.
Then Xis exponential with parameter 1/5.
(a) P(X4) = 1 e4/5
(b) Let tbe amount of time the customer has been served before you entered the
bank. Then by the memoryless property of the exponential distribution P(X >
7 + t|X > t) = P(X > 7) = e7/5.
2. (10 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 90 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 Xis hypergeometric. Since the population is large, hypergeometric
distribution can be approximated by a binomial distribution with parameters (100,0.8).
We approximate Xwith a normal variable with parameters (100(0.8),100(0.8)(0.2)) =
(80,16).
P(X90) = (continuity correction) = P(X89.5) = PX80
16 89.580
16
P(Z9.5
4) = P(Z2.375) = 1 Φ(2.375) = (normal table)
= 1 0.9912 = 0.0088 .
3. (10 points) Teams Aand Bplay a series of games. The series will end when one of
the teams wins 5 games. Suppose that team Awins each game with probability 1/4,
independently of the outcomes of other games. Find the probability that a total of 7
games are played.
1
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MATH 461- C13, Test 2, Spring 2007

April 20, 2007 Calculators, books, notes and extra papers are not allowed on this test Show all of work to qualify for full credit

  1. (10 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 λ = 1/5.

(a) If there are no customers when you enter the bank, find the probability that your serving time will not exceed 4 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 7 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 ≤ 4) = 1 − e−^4 /^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 > 7 + t | X > t) = P(X > 7) = e−^7 /^5.

  1. (10 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 90 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).

P(X ≥ 90) = (continuity correction) = P(X ≥ 89 .5) = P

X − 80

≈ P(Z ≥

) = P(Z ≥ 2 .375) = 1 − Φ(2.375) = (normal table) = 1 − 0 .9912 = 0. 0088.

  1. (10 points) Teams A and B play a series of games. The series will end when one of the teams wins 5 games. Suppose that team A wins each game with probability 1/4, independently of the outcomes of other games. Find the probability that a total of 7 games are played.

Solution; Let C be the event that a total of 7 games are played, let A = C ∩ {team A wins}, B = C ∩ {team B wins}. Then

P(A) =

P(B) =

Hence

P(C) = P(A) + P(B) = 15

  1. (10 points) Suppose that X and Y are independent random variables, X is Poisson with parameter λ, and Y is Poisson with parameter μ. Find the conditional distribution of the random variable Y given that X + Y = 50. Solution: See Example 4b, pp. 289-
  2. (20 points) Let X and Y be independent random variables each uniformly distributed over (0, 2). Find P(Y ≥ X | Y ≥ 23 ). Solution: P(Y ≥ X | Y ≥ 2 /3) = P(Y P^ ≥(YX,Y ≥ 2 ≥/3)^2 / 3). Clearly, P(Y ≥ 2 /3) = 1243 = 23.

P(Y ≥ X, Y ≥ 2 /3) =

y≥x,y≥ 2 / 3

f (x, y) dx dy

2 / 3

∫ (^) y

0

dx dy

2 / 3

y dy =

y^2 8

∣^22 / 3

Therefore, P(Y ≥ X | Y ≤ 2 /3) = (4/9)/(2/3) = 2/3.

  1. (10 points) The time it takes Ann to solve a problem is exponentially distributed with mean 15 minutes, and the time it takes Bridget to solve the same problem is also an exponential random variable with mean 20 minutes. Assuming that Ann and Bridget work on their problems independently of each other, find the probability that Bridget solves the problem before Ann does. Solution: X is exponential(λ), with λ = 1/15, Y is exponential(μ) with μ = 1/20. By independence, the joint density function of X and Y is f (x, y) = λe−λxμe−μy. Hence

P(Y ≤ X) =

y≤x

λe−λxμe−μy^ dx dy =

0

y

λe−λxμe−μy^ dx dy