STAT/MA 416 Midterm Exam 2 Thursday, October 18, Circle the section you are enrolled in:

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1 STAT/MA 46 Midterm Exam 2 Thursday, October 8, 27 Name Purdue student ID ( digits) Circle the section you are enrolled in: STAT/MA 46-- STAT/MA : AM :5 AM 3: PM 4:5 PM REC 4 UNIV 23. The testing booklet contains 2 questions. 2. Permitted Texas Instruments calculators: BA-35 BA II Plus BA II Plus Professional TI-3Xa TI-3X IIS (solar) TI-3X IIB (battery) The memory of the calculator should be cleared at the start of the exam, if possible. 3. Mark your answers on the Scantron sheet using a #2 pencil. 4. Make sure to supply answers to all of the questions. There is no penalty for guessing. 5. No partial credit will be given. 6. Show all of your work in the exam booklet. 7. Extra sheets of paper are available from the instructor.

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3 . How many rolls of a die are needed so that the probability of 5 appearing at least once is at least 9/? A.) 2 B.) 6 C.) 9 D.) 3 E.) 6 Answer. Let E n denote the event that at least one of the first n rolls show 5 ; thus E c n is the event that none of the first n rolls show 5, and P (E c n) = (5/6) n. We compute 9 P (E n) = P (E c n) = (5/6) n Thus we neet (5/6) n /, i.e., n ln(5/6) ln(/). Dividing by ln(5/6) (which is a negative number) reverses the inequality, and we obtain n ln(/)/ ln(5/6) So the correct answer is D, namely, 3.

4 2. Suppose that a die is rolled twice. Let X denote the maximum value to appear in the two rolls. Find the expected value of X. A.) 2.53 B.) 3. C.) 3.5 D.) 4.47 E.) 5.6 Answer. Considering the various outcomes, we notice that /36 i = Thus 3/36 i = 2 5/36 i = 3 P (X = i) = 7/36 i = 4 9/36 i = 5 /36 i = 6 E[X] = ()(/36)+(2)(3/36)+(3)(5/36)+(4)(7/36)+(5)(9/36)+(6)(/36) = 6/ So the correct answer is D, namely,

5 3. Consider an exponential random variable X with parameter λ = 3. Find E[e X ]. A.) /3 B.) /2 C.) 2/3 D.) E.) 3/2 Answer. One method is to compute directly E[e X ] = e x f X (x) dx = So the correct answer is E, namely, 3/2. (e x )(3e 3x ) dx = 3 e 2x dx = 3 e 2x 2 x= = 3/2 Another method is to write Y = e X and the proceed as follows. Since X, then Y. So F Y (a) = P (Y a) = for a <. For a, we have F Y (a) = P (Y a) = P (e X a) = P (X ln a) = F X (ln a) = e 3 ln a = a 3. So x 3 x F Y (x) = otherwise Differentiating with result to X yields f Y (x) = 3x 4 for x, and f Y (x) = otherwise. So E[Y ] = (x)(3x 4 ) dx = 3x 3 dx = 3 x 2 2 = 3/2 x= 3

6 4. Let X be an exponential random variable with parameter λ. Define Y = 3X. Find the density of Y. 3 A.) f Y (y) = λe λy/3 y y < λe λy/3 y B.) f Y (y) = y < λe 3λy y C.) f Y (y) = y < 3λe 3λy y D.) f Y (y) = y < λe λy y E.) f Y (y) = y < Answer. We first find the cumulative distribution function Y. Since X then Y, so F Y (a) = P (Y a) = for a <. For a >, we have F Y (a) = P (Y a) = P (3X a) = P (X a/3) = F X (a/3) = e λa/3. Thus e λx/3 x F Y (x) = x < Differentiating with respect to x yields f Y (x) = So the correct answer is A, namely, f Y (y) = 3 λe λx/3 x x < 3 λe λy/3 y y < 4

7 5. Suppose that, in a given population, the weight X of a randomly-selected person is approximately normal with mean µ = 6 and variance σ 2 =. Find the probability that a randomly-selected person weighs less than 5 pounds. A.).587 B.).462 C.).5398 D.).843 E.).9452 Answer. We have ( X 6 P (X < 5) = P < Since Φ().843, then P (X < 5).843 =.587. So the correct answer is A, namely, ) = P (Z < ) = P (Z > ) = P (Z ) = Φ() 5

8 6. The average visitor s purchase on an internet web site contains two books. (All purchases are independent of each other.) What is the approximate probability that a visitor s purchase contains at least 3 books? A.) e 2 B.) 2e 2 C.) 3e 2 D.) 4e 2 E.) 5e 2 Answer. Let X denote the number of books in a visitor s purchase. Then X is approximately Poisson with average λ = 2, so an approximation of the probabilty that a purchase contains at least 3 books is P (X 3) = P (X = ) P (X = ) P (X = 2) = e 2 2! + e 2 2! + e ! = 5e 2 So the correct answer is E, namely, 5e 2. 6

9 7. If X is uniformly distributed over the interval (, 2), compute E[X n ]. A.) 2/n B.) (2 n ) n+ C.) 2n D.) 2 n 2 E.) n+ 2 n+ n+ n+ Answer. There are several methods: Method #: We use Proposition 2.: E[X n ] = x n f X (x) dx = 2 So the correct answer is C, namely, 2 n n+. x n 2 dx = x n+ 2 2 n + = 2 n+ x= 2 n + = 2n n + Method #2: We write Y = X n. Since < X < 2, then < Y < 2 n. For < a < 2 n, we have F Y (a) = P (Y a) = P (X n a) = P (X a /n ) = a/n, and thus 2 2 F Y (x) = x/n < x < 2 n otherwise Differentiating with respect to x yields f Y (x) = x n < x < 2 n 2n otherwise Thus E[Y ] = xf Y (x)dx = 2 n x 2 n 2n x n dx = 2n x/n dx = 2n x n + n + 2 n x= = 2n 2+n n + = 2n n + Method #3: We write Y = X n and we use Lemma 2.; for y, we have P (Y > y) = if y 2 n (since Y = X n 2 n always). On the other hand, if y 2 n, then P (Y > y) = P (X n > y) = P (X n y) = P (X y /n ) = y/n 2 since X is uniform on (, 2). So E[Y ] = P (Y > y) dy = 2 n Simplifying yields E[Y ] = 2n 2( n +) 2+n ) ( ) ( y/n dy = y y n + 2 2( + ) n 2 n y= = 2n+ 2( n +2n+ 2 +n = 2n+ n +) 2( n 2n = n +) 2( +) n+ n = 2 n 2+n 2( n + ) 7

10 8. Buses arrive at a specified stop at -minute intervals starting at 7 AM. That is, they arrive at 7:, 7:, 7:2, 7:3, 7:4, 7:5, 8:, and so on. If a passenger arrives at the stop at a time that is uniformly distributed between 7: and 7:3, find the probability that he waits more than 6 minutes for a bus. A.) 2/5 B.) /5 C.) 4/5 D.) 2/5 E.) 3/5 Answer. The desired times are between 7: and 7:4, or between 7: and 7:4, or between 7:2 and 7:24. So the probability is ( )/3 = 2/3 = 2/5. So the correct answer is D, namely, 2/5. Another possibility is to use integration: dx dx dx = = 2 5 8

11 9. The density function of X is given by a + bx 3 x f(x) = otherwise If E[X] = 3/5, find a and b. A.) a = 4/5, b = 4/5 B.) a = 2/3, b = 4/3 C.) a = 7/, b = 3/2 D.) a = 2/5, b = 6/5 E.) a = 3/5, b = 6/5 Answer. We have Also = f(x) dx = 3/5 = E[X] = (a + bx 3 ) dx = a + b/4 (x)(a + bx 3 ) dx = a 2 + b 5 Solving the equations = a + b/4 and 3/5 = a/2 + b/5 yields a = 2/3 and b = 4/3. So the correct answer is B, namely, a = 2/3, b = 4/3. 9

12 . An inspector of widgets buys them in lots of size. His policy is to inspect 2 components from a lot and to accept the lot only if both components are functioning. Suppose: 6 percent of the lots have functioning components; 3 percent of the lots have 9 functioning components ( is defective); percent of the lots have 8 functioning components (2 are defective). What proportion of lots does the inspector reject? A.).6222 B.).9778 C.).2222 D.).5778 E.) Answer. Let E denote the event that he accepts a lot he is inspecting. Let F, F, F 2 denote (respectively) the events that the lot he is inspecting actually contains zero, one, or two defective widgets. Then P (E) = P (E F )P (F ) + P (E F )P (F ) + P (E F 2 )P (F 2 ) ( 9 )( ( 8 )( 2 ) ( = ()(.6) + ( ) + 8 )( 2 ) (.3) + ( 2) ) (.) So the correct answer is B, namely,.9778.

13 . Fruit flies have white eyes with proability 9/ and red eyes with probability /. We assume that the eye colors of different fruit flies are independent. When checking a large number of fruit flies, find the probability that the 8th white-eyed fruit fly is seen before the 3rd red-eyed fruit fly. A.).7 B.).937 C.).6242 D.).8368 E.).9298 Answer. We let X denote the number of fruit flies examined in order to see that 8th whiteeyed fruit fly. Then X is a negative binomial random variable with p = 9/ and r = 8. The desired probability is P (X ), i.e., we need the 8th white-eyed fruit fly within the first flies. Of course, X 8, since at least 8 flies are required to see 8 white-eyed flies. So the desired probability is P (8 X ). We know P (X = n) = ( ) n (9/) 8 (/) n 8. The probability that the first 8 flies have white eyes is: 7 P (X = 8) = ( ) 7 (9/) 8 (/) = (9/) The probability that exactly 7 out of the first 8 flies have white eyes, and then the 9th fly is white-eyed too, is: ( ) 8 P (X = 9) = (9/) 8 (/) = 8(9/) 8 (/) The probability that exactly 8 out of the first 9 flies have white eyes, and then the th fly is white-eyed too, is: ( ) 9 P (X = ) = (9/) 8 (/) 2 = (36)(9/) 8 (/) So the desired probability is P (8 X ).9298 [Using the notation of the negative binomial X is not crucial in this problem, but it just provides more insight. The problem can easily be solved without using a negative binomial random variable, by just focusing on the arrival of the 8th white-eyed fly.] So the correct answer is E, namely,.9298.

14 2. Two percent of the population is left-handed. Use a normal random variable to approximate the probability that there are at least five lefthanders in a school of 2 students. Hint: Remember to use the continuity correction, since we are using a normal random variable to approximate a binomial random variable on this problem. [As usual, if Z denotes a normal random variable (i.e., µ = and σ 2 = ), then Φ(a) = P (Z a) = a e x2 /2 dx is the cumulative distribution function of Z.] A.) Φ() B.) Φ(.25) C.) Φ(.5) D.) Φ(.6) E.) Φ(.76) Answer. Let X denote the number of lefthanders in the school of 2 students. Then X X np is binomial with n = 2 and p =.2. So is approximately a standard normal random variable. We compute P (X 5) = P (X 4.5) ( = P np( p) X 2(.2) (2)(.2)(.98) ) 4.5 (2)(.2) (2)(.2)(.98) P (Z.2525) = P (Z.2525) = Φ(.2525) () So the correct answer is B, namely, Φ(.25). 2