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MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: the probability mass function is P ( X = 0) = P ( { a, b } ) = 1 6 + 1 6 = 1 3 , P ( X = 1 . 5) = P ( { c, d } ) = 1 3 , P ( X = 2) = P ( { e } ) = 1 6 , P ( X = 3) = P ( { f } ) = 1 6 . a) P ( X = 1 . 5) = 2 6 = 1 3 b) P (0 . 5 < X < 2 . 7) = P ( X = 1 . 5) + P ( X = 2) = 3 6 = 1 2 c) P ( X > 3) = 0 d) P (0 ≤ X < 2) = P ( X = 0) + P ( X = 1 . 5) = 4 6 = 2 3 e) P ( X = 0 or X = 2) = 3 6 = 1 2 University of Ottawa 2

MAT 2377 – Probability and Statistics for Engineers Practice Set Q33 . We say that X has uniform distribution on a set of values { X 1 , . . . , X k } if P ( X = X i ) = 1 k , i = 1 , . . . , k. The thickness measurements of a coating process are uniformly distributed with values 0 . 15 , 0 . 16 , 0 . 17 , 0 . 18 , 0 . 19 . Determine the mean and variance of the thickness measurements. Is this result compatible with a uniform distribution? University of Ottawa 5

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of non-mutated samples; then X has binomial distribution with n = 15 and p = 0 . 99 (no mutation is considered a success; note that if we chose mutation as a trial success, then p = 0 . 01 and we cannot use the Table): a) 0 mutated sample = 15 non-mutated samples, thus we need to evaluate P ( X = 15) = P ( X ≤ 15) − P ( X ≤ 14) = 1 − 0 . 1399 ≈ 0 . 8601; b) at most 1 mutated sample = at least 14 non-mutated samples, thus we need to evaluate P ( X ≥ 14) = 1 − P ( X < 14) = 1 − P ( X ≤ 13) = 1 − 0 . 0096 = 0 . 9904; c) more than half the samples are mutated = fewer than (or exactly) half the samples are non-mutated; P ( X ≤ 7 . 5) = P ( X ≤ 7) = 0 . 0000 . University of Ottawa 8

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) We have X ∼ B ( n, p ) , n = 20 , p = 0 . 01 (where a trial success = a part requires re-work). By definition of the binomial distribution, E[ X ] = np = 0 . 2 , Var[ X ] = np (1 − p ) = 0 . 2 × 0 . 99 = 0 . 198 , and SD[ X ] = p Var[ X ] ≈ 0 . 44 . What we want to compute is P ( X > E[ X ] + 3 · SD[ X ]) . Thus, P ( X > 0 . 2 + 3 · 0 . 44) = P ( X > 1 . 535) = P ( X ≥ 2) = 1 − P ( X ≤ 1) = 1 − ( P ( X = 0) + P ( X = 1)) = 1 − 20 0 0 . 01 0 · 0 . 99 20 − 20 1 0 . 01 1 · 0 . 99 19 (= 1 − pbinom ( 1 , 20 , 0 . 01 ) ) ≈ 0 . 017 . University of Ottawa 11

MAT 2377 – Probability and Statistics for Engineers Practice Set Q36 . In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a particular disease. The probability that the person carries a gene is 0 . 1 . a) What is the probability that 4 or more people will have to be tested in order to detect 1 person with the gene? b) How many people are expected to be tested in order to detect 1 person with the gene? c) How many people are expected to be tested before 2 with the gene are detected? University of Ottawa 14

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) Let X be the number of failures per 8 − hour shift. The failure rate per 8 hours is 8 × 0 . 02 = 0 . 16 . If X is Poisson random variable with λ = 0 . 16 , we want to evaluate P ( X = 0) = exp( − 0 . 16) = ppois ( 0 , 0 . 16 ) ≈ 0 . 85 . b) In this case, Y is the number of failures over a 24 − hour day. The failure rate over a full day is 24 × 0 . 02 = 0 . 48 . If Y is Poisson with λ = 0 . 48 , we want to evaluate P ( X ≥ 1) = 1 − P ( X = 0) = 1 − exp( − 0 . 48) = 1 − ppois ( 0 , 0 . 48 ) ≈ 0 . 38 . University of Ottawa 17

MAT 2377 – Probability and Statistics for Engineers Practice Set The samples and estimated values agree with the theoretical ones: X=rbinom(1000,20,0.2); mean(X); var(X); [1] 4.022 [1] 3.0745901 Y=rpois(1000,8.5); mean(Y); var(Y); [1] 8.388 [1] 9.090547 Agreement is a broad term to use. What does it mean, in this context? We’ll revisit this at a later stage. University of Ottawa 20

MAT 2377 – Probability and Statistics for Engineers Practice Set Q39 . A container of 100 light bulbs contains 5 bad bulbs. We draw 10 bulbs without replacement. Find the probability of drawing at least 1 defective bulb. 0 . 4164 a) 0 . 584 b) 0 . 1 c) 0 . 9 d) none of the preceding e) University of Ottawa 21

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: although this sounds like it could be a binomial experiment, it is in fact just a classical probability question; we are not working with independent trials because the bulbs are sampled WITHOUT replacement). Let X be the number of defective bulbs. We want to evaluate P ( X ≥ 1) = 1 − P ( X < 1) = 1 − P ( X = 0) . But P ( X = 0) = ( 95 10 )( 5 0 ) ( 100 10 ) = 0 . 584 , so P ( X ≥ 1) = 1 − 0 . 584 = 0 . 416 . University of Ottawa 22

MAT 2377 – Probability and Statistics for Engineers Practice Set Q40 . Let X be a disrete random variable with range { 0 , 1 , 2 } and probability mass function (p.m.f.) given by f (0) = 0 . 5 , f (1) = 0 . 3 , and f (2) = 0 . 2 . The expected value and variance of X are, respectively, 0 . 7 , 0 . 61 a) 0 . 7 , 1 . 1 b) 0 . 5 , 0 . 61 c) 0 . 5 , 1 . 1 d) none of the preceding e) University of Ottawa 23

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: by definition, E[ X ] = X i X i f ( X i ) = 0 · 1 . 5 + 1 · 0 . 3 + 2 · 0 . 2 = 0 . 7 and Var[ X ] = E[ X 2 ] − (E[ X ]) 2 = X i X 2 i f ( X i ) − (E[ X ]) 2 = 0 2 · 1 . 5 + 1 2 · 0 . 3 + 2 2 · 0 . 2 − 0 . 7 2 = 0 . 61 . University of Ottawa 24

MAT 2377 – Probability and Statistics for Engineers Practice Set Q41 . A factory employs several thousand workers, of whom 30% are not from an English-speaking background. If 15 members of the union executive committee were chosen from the workers at random, evaluate the probability that exactly 3 members of the committee are not from an English-speaking background. 0 . 17 a) 0 . 83 b) 0 . 98 c) 0 . 51 d) none of the preceding e) Use the following CDF table for the B ( n, p ) , with n = 15 and p = 0 . 30 if needed: University of Ottawa 25

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of executive committee members not from an English-speaking background. Because the number of employees is large, and the number of committee members is relatively small, we can treat the act of selecting committee members as (approximately) independent trials. Then X ∼ B (15 , 0 . 3) and we want to evaluate P ( X = 3) = 15 3 0 . 3 3 · 0 . 7 12 ≈ 0 . 17 . Alternatively, we could use the table and compute: P ( X = 3) = P ( X ≤ 3) − P ( X ≤ 2) = 0 . 2969 − 0 . 1268 = 0 . 1701 . University of Ottawa 26

MAT 2377 – Probability and Statistics for Engineers Practice Set Q42 . Assuming the context of Q11 , what is the probability that a majority of the committee members do not come from an English-speaking background? University of Ottawa 27

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be as in Q11 . We want to evaluate P ( X > 7 . 5) = P ( X ≥ 8) = 1 − P ( X ≤ 7) = 1 − 0 . 9500 = 0 . 0500 . University of Ottawa 28

MAT 2377 – Probability and Statistics for Engineers Practice Set Q43 . In a video game, a player is confronted with a series of opponents and has an 80% probability of defeating each one. Success with any opponent (that is, defeating the opponent) is independent of previous encounters. The player continues until defeated. What is the probability that the player encounters at least three opponents? 0 . 8 a) 0 . 64 b) 0 . 5 c) 0 . 36 d) none of the preceding e) You may need to use the following formula: ∞ X x = k q x = q k 1 − q , 0 < q < 1 . University of Ottawa 29

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of encountered opponents (including the last one, which is a loss by the player); X follows a geometric distribution, where a trial success is a loss against an opponent. The probability of success is p = 0 . 2 . Since X is geometric, and since the player faces at least 1 encounter, P ( X = x ) = (1 − p ) x − 1 p, x = 1 , 2 , . . . . We want to evaluate P ( X ≥ 3) = ∞ X x =3 (1 − p ) x − 1 p = p (1 − p ) 2 p = (1 − p ) 2 = 0 . 64 . University of Ottawa 30

MAT 2377 – Probability and Statistics for Engineers Practice Set Q44 . Assuming the context of Q13 , how many encounters is the player expected to have? 5 a) 4 b) 8 c) 10 d) none of the preceding e) University of Ottawa 31

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be as in Q13 . According to the expectation formula for a geometric distribution, E[ X ] = 1 /p = 1 / 0 . 2 = 5 . University of Ottawa 32

MAT 2377 – Probability and Statistics for Engineers Practice Set Q45 . From past experience it is known that 3% of accounts in a large accounting company are in error. The probability that exactly 5 accounts are audited before an account in error is found, is: 0 . 242 a) 0 . 011 b) 0 . 030 c) 0 . 026 d) none of the preceding e) University of Ottawa 33

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of accounts that need to be audited before an account in error is found. Then X follows a geometric distribution with a probability of success p = 0 . 03 . We want to evaluate P ( X = 5) = P ( First 4 are not in error ) P (5 th is in error ) = 0 . 97 4 · 0 . 03 ≈ 0 . 026 . University of Ottawa 34

MAT 2377 – Probability and Statistics for Engineers Practice Set Q46 . A receptionist receives on average 2 phone calls per minute. Assume that the number of calls can be modeled using a Poisson random variable. What is the probability that he does not receive a call within a 3 − minute interval? e − 2 a) e − 1 / 2 b) e − 6 c) e − 1 d) none of the preceding e) University of Ottawa 35

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of calls received over a 3 − minute interval. The call rate per 3 − minute interval is 3 × 2 = 6 . If X is a Poisson random variable with λ = 6 , we want to evaluate P ( X = 0) = exp( − 6) = e − 6 ≈ 0 . 0025 . University of Ottawa 36

MAT 2377 – Probability and Statistics for Engineers Practice Set Q47 . Consider a random variable X with probability density function (p.d.f.) given by f ( x ) =      0 if x ≤ − 1 0 . 75(1 − x 2 ) if − 1 ≤ x < 1 0 if x ≥ 1 What is the expected value and the standard deviation of X ? 0 , 3 a) 0 , 0 . 447 b) 1 , 0 . 2 c) 1 , 3 d) none of the preceding e) University of Ottawa 37

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: the expected value of X is given by E[ X ] = Z ∞ −∞ xf ( x ) dx = 0 . 75 Z 1 − 1 x (1 − x 2 ) dx = 0 . 75 Z 1 − 1 ( x − x 3 ) dx = 0 . 75 x 2 2 − x 4 4 1 − 1 = 0 . 75 1 2 − 1 4 − 1 2 − 1 4 = 0 The standard deviation is SD[ X ] = s Z ∞ −∞ ( x − E [ X ]) 2 f ( x ) dx = s 0 . 75 Z 1 − 1 ( x 2 − x 4 ) dx = s 0 . 75 x 3 3 − x 5 5 1 − 1 = s 2(0 . 75) 1 3 − 1 5 = √ 0 . 2 ≈ 0 . 447 . University of Ottawa 38

MAT 2377 – Probability and Statistics for Engineers Practice Set Q48 . A random variable X has a cumulative distribution function (c.d.f.) F ( x ) =      0 if x ≤ 0 x/ 2 if 0 < x < 2 1 if x ≥ 2 What is the mean value of X ? 1 a) 2 b) 0 c) 0 . 5 d) none of the preceding e) University of Ottawa 39

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: the corresponding p.d.f. is f ( x ) = F ′ ( x ) =      0 if x ≤ 0 1 / 2 if 0 < x < 2 0 if x ≥ 2 Therefore, E[ X ] = Z ∞ −∞ xf ( x ) dx = 0 . 5 Z 2 0 x dx = 0 . 5 x 2 2 2 0 = 0 . 5 2 2 2 − 0 2 2 = 1 . University of Ottawa 40

MAT 2377 – Probability and Statistics for Engineers Practice Set Q49 . Let X be a random variable with p.d.f. given by f ( x ) = 3 2 x 2 for − 1 ≤ x ≤ 1 , and f ( x ) = 0 otherwise. Find P ( X 2 ≤ 0 . 25) . 0 . 250 a) 0 . 125 b) 0 . 500 c) 0 . 061 d) none of the preceding e) University of Ottawa 41

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: the corresponding distribution function is F ( x ) = Z x − 1 f ( t ) dt =      0 if x ≤ − 1 x 3 2 + 1 2 if − 1 ≤ x ≤ 1 1 if x ≥ 1 (You should verify that F ( x ) is continuous.) In that case, P ( X 2 ≤ 0 . 25) = P ( − 0 . 5 ≤ X ≤ 0 . 5) = F (0 . 5) − F ( − 0 . 5) = 0 . 5 3 2 + 1 2 − ( − 0 . 5) 3 2 + 1 2 = 0 . 125 . University of Ottawa 42

MAT 2377 – Probability and Statistics for Engineers Practice Set Q50 . In the inspection of tin plate produced by a continuous electrolytic process, 0 . 2 imperfections are spotted per minute, on average. Find the probability of spotting at least 2 imperfections in 5 minutes. Assume that we can model the occurrences of imperfections as a Poisson process. 0 . 736 a) 0 . 264 b) 0 . 632 c) 0 . 368 d) none of the preceding e) University of Ottawa 43

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of imperfections found in 5 minutes. The imperfection spotting rate per 5 minutes is 0 . 2 × 5 = 1 . If X is a Poisson random variable with λ = 1 , we want to evaluate P ( X ≥ 2) = 1 − P ( X ≤ 1) = 1 − ( P ( X = 0) + P ( X = 1)) = 1 − P ( X = 0) − P ( X = 1) = 1 − exp( − λ ) − λ exp( − λ ) = 1 − 2 exp( − 1) ≈ 0 . 264 . University of Ottawa 44

MAT 2377 – Probability and Statistics for Engineers Practice Set Q51 . If X ∼ N (0 , 4) , the value of P ( | X | ≥ 2 . 2) is (using the normal table): 0 . 2321 a) 0 . 8438 b) 0 . 2527 c) 0 . 2713 d) 0 . 7286 e) none of the preceding f) University of Ottawa 45

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let Z = X − 0 √ 4 . Then Z ∼ N (0 , 1) and we have P ( | X | ≥ 2 . 2) = 1 − P ( | X | ≤ 2 . 2) = 1 − P ( − 2 . 2 ≤ X ≤ 2 . 2) = 1 − P − 2 . 2 − 0 √ 4 ≤ X − 0 √ 4 ≤ 2 . 2 − 0 √ 4 = 1 − P ( − 1 . 1 ≤ Z ≤ 1 . 1) = 1 − (Φ(1 . 1) − Φ( − 1 . 1)) = 1 − ( pnorm ( 1 . 1 , 0 , 1 ) − pnorm ( − 1 . 1 , 0 , 1 )) ≈ 0 . 2713 . This could have been computed directly as 1 − P ( − 2 . 2 ≤ X ≤ 2 . 2) = 1 − ( pnorm ( 2 . 2 , 0 , 2 ) − pnorm ( − 2 . 2 , 0 , 2 )) . University of Ottawa 46

MAT 2377 – Probability and Statistics for Engineers Practice Set Q52 . If X ∼ N (10 , 1) , the value of k such that P ( X ≤ k ) = 0 . 701944 is closest to 0 . 59 a) 0 . 30 b) 0 . 53 c) 10 . 53 d) 10 . 30 e) 10 . 59 f) University of Ottawa 47

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let Z = X − 10 √ 1 = X − 10 . Then Z ∼ N (0 , 1) and we have P ( X ≤ k ) = P X − 10 1 ≤ k − 10 1 = P ( Z ≤ k − 10) = 0 . 701944 . According to the table (or R ), k − 10 = Φ − 1 (0 . 701944) = qnorm ( 0 . 701944 , 0 , 1 ) ≈ 0 . 53 , whence k ≈ 10 . 53 . University of Ottawa 48

MAT 2377 – Probability and Statistics for Engineers Practice Set Q53 . The time it takes a supercomputer to perform a task is normally distributed with mean 10 milliseconds and standard deviation 4 milliseconds. What is the probability that it takes more than 18 . 2 milliseconds to perform the task? (use the normal table or R ). 0 . 9798 a) 0 . 8456 b) 0 . 0202 c) 0 . 2236 d) 0 . 5456 e) none of the preceding f) University of Ottawa 49

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X ∼ N (10 , 4 2 ) and Z = X − 10 4 . Then Z ∼ N (0 , 1) and P ( X ≥ 18 . 2) = 1 − P ( X ≤ 18 . 2) = 1 − P X − 10 4 ≤ 18 . 2 − 10 4 = 1 − P ( Z ≤ 2 . 05) ≈ 0 . 0202 . University of Ottawa 50

MAT 2377 – Probability and Statistics for Engineers Practice Set Q54 . Roll a fair 4 − sided die twice, and let X equal the larger of the two outcomes if they are different and the common value if they are the same. Find the p.m.f. and the c.d.f. of X . University of Ottawa 51

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: The outcome space for this experiment is S = { ( d 1 , d 2 ) : d 1 = 1 , 2 , 3 , 4; d 2 = 1 , 2 , 3 , 4 } . It is assumed that each of these 16 outcomes has equal probability 1 / 16 . By definition, X ( d 1 , d 2 ) = max { d 1 , d 2 } for any ( d 1 , d 2 ) ∈ S . Then P ( X = 1) = P [(1 , 1)] = 1 16 , P ( X = 2) = P [ { ((1 , 2) , (2 , 1) , (2 , 2) } ] = 3 16 , P ( X = 3) = P [ { (1 , 3) , (2 , 3) , (3 , 1) , (3 , 2) , (3 , 3) } ] = 5 16 P ( X = 4) = P [ { (1 , 4) , (2 , 4) , (3 , 4) , (4 , 1) , (4 , 2) , (4 , 3) , (4 , 4) } ] = 7 16 , and the p.d.f. can be simply re-written as f ( x ) = P ( X = x ) = 1 16 (2 x − 1) , for x = 1 , 2 , 3 , 4 , and f ( x ) = 0 otherwise . University of Ottawa 52

MAT 2377 – Probability and Statistics for Engineers Practice Set We can verify that 4 X x =1 f ( x ) = 1 16 4 X x =1 (2 x − 1) = 2 16 4 X x =1 x − 1 16 4 X x =1 1 = 2 16 · 4(5) 2 − 1 16 · 4 = 5 4 − 1 4 = 1 . Since 2 x > 1 for x = 1 , 2 , 3 , 4 , f ( x ) ≥ 0 for all x and f is indeed a p.m.f. The c.d.f. is F ( x ) = P ( X ≤ x ) =      0 if x < 1 ⌊ x ⌋ 16 if 1 ≤ x < 4 1 if x ≥ 4 University of Ottawa 53

MAT 2377 – Probability and Statistics for Engineers Practice Set University of Ottawa 54

MAT 2377 – Probability and Statistics for Engineers Practice Set The graph on the preceding slide was produced with the following (not commented) R code: X <- 1:4 P <- c(1/16,3/16,5/16,7/16) require(graphics) par(mfrow=c(2,1)) plot(X,P,type="h",col=2,main="PMF",xlim=c(0,5),ylim=c(0,0.5),xlab="x", ylab="f(x)") points(X,P,col=2) abline(h=0,col=4) F <- cumsum(P) plot(c(1,X),c(0,F),type="s",main="CMF",xlim=c(0,5),ylim=c(0,1),col=2,xlab="x",ylab="F(x)") abline(h=0:1,col=4) University of Ottawa 55

MAT 2377 – Probability and Statistics for Engineers Practice Set Q55 . Compute the mean and the variance of X as defined in Q24 , as well as E[ X (5 − X )] . University of Ottawa 56

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be as in question Q24 . Then E[ X ] = 4 X x =1 xP ( X = x ) = 1 · 1 16 + 2 · 3 16 + 3 · 5 16 + 4 · 7 16 = 25 8 = 3 . 125 Var[ X ] = E[ X 2 ] − E 2 [ X ] = 4 X x =1 x 2 P ( X = x ) − 25 8 2 = ( 1 2 · 1 16 + 2 2 · 3 16 + 3 2 · 5 16 + 4 2 · 7 16 ) − 625 64 = 233 16 − 625 64 = 307 64 ≈ 4 . 797 , whereas E[ X (5 − X )] = E[5 X − X 2 ] = 5E[ X ] − E[ X 2 ] . The first and second moments were computed above: E[ X ] = 25 8 and E[ X 2 ] = 233 16 , so E[ X (5 − X )] = 5 · 25 8 − 233 16 = 17 16 ≈ 1 . 0625 . University of Ottawa 57

MAT 2377 – Probability and Statistics for Engineers Practice Set Q56 . In 80% of cases when a basketball player attempts a free throw, they are successful. Assume that each of the free throw attempts are independent. Let X be the minimum number of attempts in order to succeed 10 times. Find the p.m.f. of X and the probability that X = 12 . University of Ottawa 58

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: the player requires at least x = 10 throws to be successful on 10 separate attempts. There must be 10 successes in total (with probability 0 . 8 10 ), and x − 10 failures (with probability 0 . 2 10 − x ). Furthermore, the player has to be successful on the x th throw (otherwise x would not be the minimum number of attempts in order to succeed 10 times); within the first x − 1 throws, exactly 9 must be successful, and there are x − 1 C 9 = ( x − 1 9 ) ways for these to be ordered. Thus, the p.m.f. of X is f ( x ) = P ( X = x ) = x − 1 9 (0 . 8) 10 (0 . 2) x − 10 , x = 10 , 11 , 12 , .... and P ( X = 12) = f (12) = ( 12 9 ) (0 . 80) 10 (0 . 20) 2 ≈ 0 . 2362 . University of Ottawa 59

MAT 2377 – Probability and Statistics for Engineers Practice Set Q57 . Let X be the minimum number of independent trials (each with probability of success p ) that are needed to observe r successes. The p.m.f. of X is f ( x ) = P ( X = x ) = x − 1 r − 1 p r (1 − p ) x − 1 , x = r, r + 1 , . . . The mean and variance of X are E[ X ] = r p and Var[ X ] = r (1 − p ) p 2 . Compute the mean minimum number of independent free throw attempts required to observe 10 successful free throws if the probability of success at the free thrown line is 80% . What about the standard deviation of X ? University of Ottawa 60

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be as in Q26 . We have E[ X ] = r p = 10 0 . 80 = 12 . 5 and Var[ X ] = r (1 − p ) p 2 = 10(0 . 20) 0 . 80 2 ≈ 3 . 125 , from which we conclude that SD[ X ] ≈ √ 3 . 125 ≈ 1 . 768 . University of Ottawa 61

MAT 2377 – Probability and Statistics for Engineers Practice Set Q58 . If n ≥ 20 and p ≤ 0 . 05 , it can be shown that the binomial distribution with n trials and an independent probability of success p can be approximated by a Poisson distribution with parameter λ = np . This is called the Poisson approximation : ( np ) x e − np x ! ≈ n x p x (1 − p ) n − x . A manufacturer of light bulbs knows that 2% of its bulbs are defective. What is the probability that a box of 100 bulbs contains exactly at most 3 defective bulbs? Use the Poisson approximation to estimate the probability. University of Ottawa 62

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X be the number of defective bulbs in the box of 100 . Since X ∼ B (100 , 0 . 02) , we have P ( X = x ) = 100 x (0 . 02) x (0 . 98) 100 − x and P ( X ≤ 3) = 3 X x =0 100 x (0 . 02) x (0 . 98) 100 − x . But n = 100 ≥ 20 and p = 0 . 02 ≤ 0 . 05 and the Poisson approximation applies. If λ = np = 2 , then 2 x e − 2 x ! ≈ 100 x (0 . 02) x (0 . 98) 100 − x , University of Ottawa 63

MAT 2377 – Probability and Statistics for Engineers Practice Set so P ( X ≤ 3) ≈ 3 X x =0 2 x e − 2 x ! = e − 2 2 0 0! + 2 1 1! + 2 2 2! + 2 3 3! ≈ 0 . 857 . University of Ottawa 64

MAT 2377 – Probability and Statistics for Engineers Practice Set Q59 . Consider a discrete random variable X which has a uniform distribution over the first positive m integers, i.e. f ( x ) = P ( X = x ) = 1 m , x = 1 , . . . , m, and f ( x ) = 0 otherwise. Compute the mean and the variance of X . For what values of m is E[ X ] > Var[ X ] ? University of Ottawa 65

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: we have E[ X ] = m X x =1 xf ( x ) = m X x =1 x · 1 m = 1 m m X x =1 x = 1 m · m ( m + 1) 2 = m + 1 2 , Var[ X ] = E [ X 2 ] − E 2 [ X ] = m X x =1 x 2 f ( x ) − ( m + 1) 2 4 = 1 m m X x =1 x 2 − ( m + 1) 2 4 = 1 m · m ( m + 1)(2 m + 1) 6 − ( m + 1) 2 4 = m 2 − 1 12 . The mean is greater than the variance when m + 1 2 > m 2 − 1 12 ↔ 6( m + 1) > m 2 − 1 ↔ m 2 − 6 m − 7 < 0 ↔ 1 ≤ m < 7 . University of Ottawa 66

MAT 2377 – Probability and Statistics for Engineers Practice Set Q60 . Let X be a random variable. What is the value of b (where b is not a function of X ) which minimizes E[( X − b ) 2 ] ? University of Ottawa 67

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: write g ( b ) = E[( X − b ) 2 ] = E[ X 2 − 2 bX + b 2 ] = E[ X 2 ] − 2 b E[ X ] + E[ b 2 ] = E[ X 2 ] − 2 b E[ X ] + b 2 . To find the minimum of g ( b ) with respect to b , set g ′ ( b ) = 0 and solve for b : g ′ ( b ) = − 2E[ X ] = 2 b = 0 b = E[ X ] . Since g ′′ ( b ) = 2 > 0 , E[ X ] is the value of b that minimizes E[( X − b ) 2 ] . University of Ottawa 68