Final Mathematics 5010, Section 1, Fall 2004 Instructor: D.A. Levin

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1 Final Mathematics 51, Section 1, Fall 24 Instructor: D.A. Levin Name YOU MUST SHOW YOUR WORK TO RECEIVE CREDIT. A CORRECT ANSWER WITHOUT SHOWING YOUR REASONING WILL NOT RECEIVE CREDIT. Problem Points Possible Points Earned

A CORRECT ANSWER WITHOUT SHOWING YOUR REASONING WILL NOT RECEIVE

2 Problem 1. An instructor gives her class a set of 2 problems with the information the final exam will consist of a random selection of 1 of them. If a student has figured out how to do 12 of the problems, what is the probability that he or she will correctly answer (a) all 1 problems on the exam; (b) at least 8 of the problems on the exam. Solution. Imagine that an urn is filled with 2 balls, each representing one of the possible exam problems. Those balls corresponding to problems studied by the student are colored red; there are 12 such balls. The remaining balls are colored black. The instructor draws 1 of these balls at random without replacement. [Obviously, an instructor would not put the same problem twice on the exam! Thus, he samples without replacement.] Let X be the number of red balls drawn; this corresponds to the number of questions which the student can answer correctly. (a) (b) PX 1} ( 12 8 ) 1)( ). ( 2 1 PX 8} PX 8} + PX 9} + PX 1} ( 12 8 )( 8 2) ( 2 ) + 1 ( 12 9 )( 8 1) ( 2 ) + 1 ( 12 8 ) 1)( ). ( 2 1 2

Solution. Imagine that an urn is filled with 2 balls, each representing one of the possible exam problems.

3 Problem 2. Ninety-eight percent of all babies survive delivery. However, 15 percent of all births involve Cesarean (C) sections, and when a C section is performed the baby survives 96 percent of the time. If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives? Solution. Let S be the event of survival, and C be the probability of Cesarian. We want to find P(S C ), and are given that P(S).98, P(S C).96, and P(C).15. The definition of conditional probability gives us P(S C ) P(S C ) P(C. ) We can find P(S C ) from the information given by using the identity P(S) P(S C) + P(S C ). In particular, rearranging and then using the conditional probability identity P(S C) P(S C)P(C) yields Since P(C ) 1 P(C).85, we have then P(S C ) P(S) P(S C) P(S) P(S C)P(C).98 (.96)(.15).836 P(S C )

If a randomly chosen pregnant woman does not have a C section, what is the probability that her baby survives? Solution. Let S be the event of survival, and C be the probability of Cesarian.

4 Problem 3. Urn A has 5 white and 7 black balls. Urn B has 3 white and 12 black balls. We flip a fair coin. If the outcome is heads, then a ball from urn A is selected, whereas if the outcome is tails, then a ball from urn B is selected. Suppose that a white ball is selected. What is the probability that the coin landed tails? Solution. P(T W ) P(T W ) P(W ) P(W H)P(H) P(W H)P(H) + P(W T )P(T )

from urn B is selected. Suppose that a white ball is selected.

5 Problem 4. Suppose that X has the distribution function Find E(X ). F(t) if t < 1 Solution. The density is given by differentiating F: f (t) 1 1 t 3 if t 1. if t < 1 3t 4 if t > 1. Thus, E(X ) t f (t)dt t(3t 4 2 3t )dt

6 Problem 5. Find the probability density function of R Asinθ, where A is a constant and θ is uniformly distributed on ( π/2, π/2). 1 Hint: The derivative of arcsin(t) is. 1 t 2 Solution. Instructor s Note: The derivative of arcsin was incorrectly given on the test. If you used this misinformation, you were not penalized! Notice that sinθ is one-to-one on ( π/2,π/2): Clearly then Asinθ is also one-to-one, but it ranges from A to A. For A t A We conclude that To get the pdf, we differentiate: -1 F R (t) PR t} PAsinθ t} Psinθ t/a} Pθ arcsin(t/a)} arcsin(t/a) + π/2 π t < π/2 arcsin(t/a)+π/2 F R (t) π π/2 t π/2 1 t > π/2. t < π/2 1 1 f R (t) π/2 t π/2 πa 1 (t/a) 2 t > π/2. 6

Notice that sinθ is one-to-one on ( π/2,π/2): 1.5-1.5-1 -.5.5 1 1.5 -.5 Clearly then Asinθ is also one-to-one, but it ranges from A to A.

7 Problem 6. Let X and Y have joint pdf f (s, t) se s(t+1) if s >, t >, otherwise. (a) Find the conditional probability density function of Y given X t. (b) Find the density function of Z X Y. Solution. f X (s) Consequently, if s > and t >, se s(t+1) dt e s(t+1) e s. f Y X (t s) f (s, t) f X (s) se s(t+1) e s se st. That is, given X s, the conditional distribution of Y is exponential with parameter s. We compute the density of Z in two ways. For u > F Z (u) PX Y u} u/s se s(t+1) dt ds e s u/s se st dt ds e s [ e st ] u/s ds e s [ 1 e u] ds [ 1 e u]. Differentiating, We can also compute as follows: f Z (u) e u if u, otherwise. PX Y u} P Y u/s X s} e s ds (1 e u )e s ds (1 e u ) e s ds (1 e u ). The second inequality follows from part (a). 7

That is, given X s, the conditional distribution of Y is exponential with parameter s. We compute the density of Z in two ways.

8 Problem people get on an elevator on the ground floor of a 1 story building. Each person selects a floor; assume that each person selects independently and each person picks one of the 1 possible floors uniformly at random. No new people get on the elevator after the ground floor. Compute the expected number of stops the elevator makes. Solution. Let X be the number of stops the elevator makes. We can write X 1 i1 I i, where I i Since expectation is linear, we have Now E(X ) E 1 if the elevator stops at floor i, ( 1 otherwise. i1 I i ) 1 i1 E(I i ) 1 i1 Pstop at floor i}. Pstop at floor i} 1 Pno-one picks floor i} ( ) So [ ( ) 9 12 ] E(X )

No new people get on the elevator after the ground floor. Compute the expected number of stops the elevator makes. Solution.

9 Problem 8. Suppose two roommates, Bill and Bob, share a phone line. The number of calls received during the evening is a Poisson random variable with parameter 1. Bill is more popular, so each call independently has probability.7 of being for Bill. (a) Find the conditional pmf for X, the number of calls received by Bill, given that the total number of calls is n. (b) Find the pmf for X. (c) Given that Bill receives 5 calls, what is the probability that Bob receives exactly one call. Solution. Let X be the number of calls for Bill, and Y the number of calls for Bob. Let Z X + Y. Given that Z n, each of the n call is for Bill with probability.7, and not for Bill with probability.3. The calls are independent of each other. Thus the conditional distribution of X is Binomial(n, p.7). Then ( ) n p X (k) p X Z (k n)p Z (n) (.7) k (.3) n k 1 (1)n e nk nk k n! e 1 (.7) k (1) k (.3) n k (1) n k k! (n k)! nk e 1 7 k k! Thus X has a Poisson(7) distribution. Similarly, Y has a Poisson(3) distribution. We want to find PY 1 X 5}. PX 5,Y 1} PY 1 X 5} PX 5} PX 5 Z 6}PZ 6} PX 5} ( 6 5) (.7) 5 (.3) e 1 (1) 6 3e 3. e ! 6! e 3 e 7 7 k. k! 9

(a) Find the conditional pmf for X, the number of calls received by Bill, given that the total number of calls is n. (b) Find the pmf for X.

10 Problem 9. A column bet in roulette wins 2$ with probability 12/38, and losses 1$ with probability 26/38. (a) Compute the mean and standard deviation of your winnings on a single game. (b) You place this bet 25 times. Estimate the probability that you have won a positive amount. (c) You place this bet 1 times. Estimate the a probability that you have won a positive amount. Solution. Suppose that X i is the amount won on the ith game. X 1,X 2,... are independent and all have the same distribution. Then E(X 1 ) and Thus and so SD(X 1 ) Let S n n E ( X1 2 ) V (X 1 ) E ( X1 2 ) [E(X1 )] i1 X i. Write µ for E(X 1 ) and σ for SD(X 1 ). S25 25µ PS 25 > } P > 25(.526) } 5σ 5(1.394) S25 25µ P > } 5σ 6.97 } S25 25µ P >.189 5σ 1 Φ(.189).425 S1 1µ PS 1 > } P > } 1σ 44.8 } S25 1µ P > σ 1 Φ(1.194).116 1

Suppose that X i is the amount won on the ith game. X 1,X 2,... are independent and all have the same distribution. Then E(X 1 ) 2 12 38 26 38 1 19.5263. and Thus and so SD(X 1 ) 1.394.

11 Problem 1. Let X have a Gamma(α,λ) distribution, and let Y be an independent Gamma(β,λ) random variable. Let Z X + Y. (a) Find the MGF of Z. You can use Table 7.2. (b) What is the distribution of Z? Solution. M Z (t) M X (t)m Y (t) ( ) λ α ( λ λ t λ t ( ) λ α+β λ t Using Table 7.2, we see that Z has a Gamma(α + β,λ) distribution. ) β 11

Binomial random variables

Binomial and Poisson Random Variables Solutions STAT-UB.0103 Statistics for Business Control and Regression Models Binomial random variables 1. A certain coin has a 5% of landing heads, and a 75% chance