MTH4451Test#2-Solutions Spring 2009

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1 Pat Rossi Instructions. MTH4451Test#2-Solutions Spring 2009 Name Show CLEARLY how you arrive at your answers. 1. A large jar contains US coins. In this jar, there are 350 pennies ($0.01), 300 nickels ($0.05), 250 dimes ($0.10), and 300 quarters ($0.25) - each type of coin is evenly distributed throughout the jar. I take one coin from the jar. Let the random variable Y be the value of the coin that I take. Then Y = $0.01 when the coin is a penny $0.05 when the coin is a nickel $0.10 when the coin is a dime $0.25 when the coin is a quarter Furthermore, the probability function for Y is P (Y = y) = = when y =$0.01 =0.25 when y =$0.05 = when y =$0.10 =0.25 when y =$0.25 (a) What is the expected value of the coin? E (Y )= P Y =y y P (Y = y) =($0.01) ( )+($0.05) (0.25)+($0.10) ( ) +($0.25) (0.25) = $ i.e., E (Y )=$ (b) If I perform this experiment a number of times, what is the variance? V (Y )=E (Y 2 ) [E (Y )] 2 We need to find the value of E (Y 2 ) E (Y 2 )= P Y =y y 2 P (Y = y) =($0.01) 2 ( )+($0.05) 2 (0.25)+($0.10) 2 ( ) +($0.25) 2 (0.25) = $ i.e., E (Y 2 )=$ Thusthevarianceis: V (Y )=E (Y 2 ) [E (Y )] 2 =$ ( ) 2 =$ i.e., V (Y )=$

2 2. A group of 6 software packages available to solve a linear programming problem has been ranked from 1 to 6 (best to worst). An engineering team, unaware of the rankings, randomly selected and then purchased two of the packages. Let Y denote the number of packages purchased (by the team) that are ranked 3, 4, 5, 6. (a) Give the probability distribution for Y. In this experiment, we have two identical trials, and each trial has exactly two outcomes - Success (Package is ranked 3, 4, 5, 6) or Failure (Package is ranked 1, 2). Because the sample size (n =2)is large compared to the population size (N =6), the trials are NOT independent. This is a hypergeometric distribution N = 6 n = 2 M = 4 P (Y = k) = (M K)( N M n k ) ( N n) = k)( 2 k) (4 2 ( 6 2) for k =0, 1, 2 (b) Compute P (1) (i.e., compute P (Y =1)) P (Y =1)= 1)( 2 1) (4 2 = ( 6 2) 8 = A recent survey indicated that a record low number (41%) of adult American citizens expressed a great deal of confidence in the US Supreme Court. If you conduct your own random poll on the same issue, (a) find the probability distribution for Y, the number people who must be polled in order to find a person who does express a great deal of confidence in the US Supreme Court. Each time a person is polled, there is a 41% probability that they will express a great deal of confidence in the US Supreme Court. So each time a person is polled, it constitutes a trail in which there are two outcomes - Success (a person DOES express a great deal of confidence) or Failure (a person does NOTexpress a great deal of confidence). Furthermore each trial is independent of all other trials. Since we are looking for the number of trials necessary to acheive the first success, the experiment is geometric. p =0.41 q =0.59 Y = the number of the trial on which the first success is acheived. 2

3 P (Y = y) =q y 1 p =(0.59) y 1 (0.41) i.e., P (Y = y) =(0.59) y 1 (0.41) (b) Compute P (5) (i.e., compute P (Y =5)) P (Y =5)=(0.59) 5 1 (0.41) = i.e., P (Y =5)= A multiple choice examination has 15 questions, each with 5 possible answers, of which only one answer is correct. Suppose that one of the students who takes the examination answers each of the questions with an independent, random guess. What is the probability that he answers at least 10 and no more than 12 questions correctly? Each question constitutes a trial which has exactly two outcomes - Success (Correct) or Failure (incorrect). The probability of getting a success is constant from one problem to the next and each trial is independent of all others. This is binomial. Y = the number of successes (number of questions answered correctly) n =15 p =0.2 q =0.8 We want P [(Y =10) (Y = 11) (Y =12)]=P (Y =10)+P (Y =11)+P (Y =12) because the trials are independent. Therefore, P [(Y =10) (Y =11) (Y =12)]=P (Y =10)+P (Y =11)+P (Y =12) = ³ = ³ p10 q ³ 15 p11 q ³ p12 q (0.2) 10 (0.8) ³ 15 (0.2) 11 (0.8) ³ 15 (0.2) 12 (0.8) = i.e., P [(Y =10) (Y =11) (Y =12)]=

4 5. Themeannumberofcarsenteringanarrowmountaintunnelduringa5minuteperiod is 6. (a) Compute the probability that during the next 5 minutes exactly 4 cars enter the tunnel. This is a Poisson Distribution with µ =6 (i.e., µ = 6 cars ) 5 minutes Y = number of cars entering the tunnel We want P (Y =4)= µy e µ y! = 64 e 6 4! = i.e., P (Y =4)= (b) Compute the probability that during the next 15 minutes at least 18 and no more than 20 cars enter the tunnel. We must re-parameterize µ, since the time period is different. µ = 6 cars = 6 cars 3 18 cars = 5 minutes 5 minutes 3 15 minutes i.e., µ =18 (i.e., µ = 18 cars ) 15 minutes We want P [(Y =18) (Y =19) (Y =20)]=P (Y =18)+P (Y =19)+P (Y =20) because the trials are independent. Therefore, P [(Y =18) (Y =19) (Y =20)]=P (Y =18)+P (Y =19)+P (Y =20) = µ18 e µ 18! + µ19 e µ 19! + µ20 e µ 20! = 1818 e 18 18! e 18 19! e 18 20! = i.e., P [(Y =18) (Y =19) (Y =20)]=

5 6. A student answers a multiple choice question that offers five possible answers. Suppose the probability that the student knows the answer to the question is 0.7 and the probability that the student will guess is 0.3. Assume that if the student guesses, the probability of guessing the correct answer is 0.2. If the student correctly answers the question, what is the probability that the student really knew the correct answer? Let s name the events as follows: B 1 Student knows the correct answer B 2 Student guesses the answer (does not know the correct answer) A 1 Student correctly answers the question A 2 Student answers the question incorrectly Then: P (B 1 )=0.7 P (B 2 )=0.3 P (A 1 B 2 )=0.2 Also, note that we can justify assuming that the following must be true: P (A 1 B 1 )=1.0 Note that: (a) B 1 B 2 = (mutually exclusive) and (b) B 1 B 2 = S (collectively exhaustive) Thus, Bayes Rule applies. We want P (B 1 A 1 )= i.e., P (B 1 A 1 )= P (B 1 )P (A 1 B 1 ) = (0.7)(1.0) = P (B 1 )P (A 1 B 1 )+P (B 2 )P (A 1 B 2 ) (0.7)(1.0)+(0.3)(0.2) 5