TutWeek8_F23

EC285 A,B Tutorial Questions Week8 Random Variables and Probability Distributions Edda Claus EC285 A,B - Fall 2023 10 November 2023 1. A new sales gimmick has 30% of the M&M’s covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for the speckles. Note that this situation involves Bernoulli trials. There are two outcomes: success = speckles, failure = ordinary. The probability of success, based on the information from the candy company, is 30%. Trials can be assumed independent – there is no reason to believe that finding a speckled candy reveals anything about whether the next one out of the bag will have speckles. a. What’s the probability that the first speckled one we see is the fourth candy we get? Answer P (First speckled is fourth candy) = 0 . 7 9 · 0 . 3 ≈ 0 . 0121 b. What’s the probability that the first speckled one is the tenth one? Write a general formula. What’s the probability the first speckled one is one of the first three we look at? How many do we expect to have to check, on average, to find a speckled one? Answer P (First speckled is tenth candy) = 0 . 7 3 · 0 . 3 = 0 . 1029 1

General formula: P ( X = x ) = q x − 1 p P (First speckled is among first three) = 0 . 3 + 0 . 7 · 0 . 3 + 0 . 7 2 · 0 . 3 ≈ 0 . 657 How many do we expect to check, on average to find a speckled one? E ( X ) = 1 p = 1 0 . 3 = 3 1 3 c. What’s the probability that we’ll find two speckled ones in a handful of five candies? Answer P (Two speckled candies in first five) = 5 2 ! 0 . 3 2 · 0 . 7 3 = 0 . 3087 2. Consider a dice game: no points for rolling a 1, 2, or 3; 5 points for a 4 or 5; 50 points for a 6. a. Find the expected value and the standard deviation. Answer We want to find the mean and expected value of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 5 points, and 6 is worth 50 points. Let Y = the number of points. Roll 1, 2, 3 4, 5 6 Y 0 5 50 P ( Y ) 1 6 2 6 1 6 µ = E ( Y ) = 0 ( 3 6 ) + 5 ( 2 6 ) + 50 ( 1 6 ) = 10 σ 2 = V ar ( Y ) = (0 − 10) 2 ( 3 6 ) (5 − 10) 2 ( 2 6 ) + (50 − 10) 2 ( 1 6 ) = 325 σ = SD ( Y ) = p V ar ( Y ) = √ 325 ≈ 18 . 03 I expect to win 10 points per roll in the long run. The standard deviation is 18 . 03 points, suggesting a highly variable game. b. Imagine doubling the points awarded. What are the new mean and standard devi- ation? Answer I want to find the mean and expected value of a dice game where 1, 2, and 3 are worth 0 points, 4 or 5 are worth 10 points, and 6 is worth 100 points. This is double the payout of the original dice game. E (2 Y ) = 2 E ( Y ) = 20 and V ar (2 Y ) = 2 2 V ar ( Y ) = 4 · 325 = 1300, so SD (2 Y ) = 36 . 06 c. Now imagine just playing the game twice. What are the mean and the standard deviation of your total points? 2

sell. The mechanical checkup costs $ 50 per hour and takes an average of 90 minutes, with a standard deviation of 15 minutes. The appearance prep (wash, polish, etc.) costs $ 6 per hour and takes an average of 60 minutes, with a standard deviation of 5 minutes. a. What are the mean and standard deviation of the total time spent preparing a car? (Note that we cannot find the standard deviation if we do not believe that the two phases of the process are independent, an important assumption to check.) Answer We want to know the mean and standard deviation of the total time spent preparing a car. We believe that the two phases of the process are independent, so variances add. Let M = mechanical checkup time, A = appearance prep time, T = total time T = M + A E ( T ) = E ( M ) + E ( A ) = 90 + 60 = 150 minutes V ar ( T ) = V ar ( M ) + V ar ( A ) = 15 2 + 5 2 = 250 SD ( T ) = √ 250 = 15 . 8 minutes We expect the total preparation time to take an average of 150 minutes (2.5 hours), with a standard deviation of 15.8 minutes. b. What are the mean and standard deviation of the total expense to prepare a car? Answer We want to know the mean and standard deviation of the total expense of preparing a car. We can find mean and standard deviation of expenses for each part of the preparation by multiplying the mean and standard deviation of the times (in hours) by the rate of $ 50/hr. We believe that the two phases of the process are independent, so variances add. Let M = mechanical checkup hours E ( M ) = 1 . 5 SD ( M ) = 0 . 25 A = appearance prep hours E ( A ) = 1 . 0 SD ( A ) = 5 60 = 1 12 = 0 . 0833 TE = total expenses TE = 50 M + 6 A E ( TE ) = E (50 M +6 A ) = E (50 M )+ E (6 A ) = 50 E ( M )+6 E ( A ) = 50(1 . 5)+6(1 . 0) = $81 V ar ( TE ) = V ar (50 M + 6 A ) = V ar (50 M ) + V ar (6 A ) = 50 2 V ar ( M ) + 6 2 V ar ( A ) = 2500 · 0 . 25 2 + 36 · ( 1 12 ) 2 = 156 . 50 SD ( T ) = √ 156 . 50 = $12 . 51 We expect the total preparation expense to average $ 81, with a standard deviation of $ 12.51. 4

c. What are the mean and standard deviation of the difference in costs for the two phases of the operation? Answer We want to know the mean and standard deviation of the difference in expenses of preparing a car. We can find mean and standard deviation of expenses for each part of the preparation by multiplying the mean and standard deviation of the times (in hours) by the rate of $ 50/hr. We believe that the two phases of the process are independent, so variances add. D = 50 M − 6 A E ( D ) = E (50 M − 6 A ) = E (50 M ) − E (6 A ) = 50 E ( M ) − 6 E ( A ) = 50 · 1 . 5 − 6 · 1 . 0 = $69 V ar ( D ) = V ar (50 M − 6 A ) = V ar (50 M ) − V ar (6 A ) = 50 2 V ar ( M ) − 6 2 V ar ( A ) = 2500 · 0 . 25 2 + 36 · ( 1 12 ) 2 = 156 . 50 SD ( D ) = √ 156 . 50 = $12 . 51 We expect the difference in preparation expenses to average $ 69, with a standard deviation of $ 12.51. d. What is the probability that it will take longer to do the appearance prep than the mechanical checkup? (Note that we cannot answer this question unless we believe each phase of the process can be described by a Normal model.) Answer To calculate the probability that it will take longer to do the appearance prep than the mechanical checkup, we need to know the mean and standard deviation of the difference in time spent on each. If the difference (mechanical checkup time – appearance prep time) is less than 0, then the appearance prep takes longer. The two phases of the process are independent, so variances add. We will assume that each of the processes varies according to a Normal model. Let D = M − A E ( D ) = E ( M − A ) = E ( M ) − E ( A ) = 90 − 60 = 30 minutes SD ( D ) = SD ( M − A ) = p V ar ( M − A ) = p V ar ( M + A ) = p V ar ( M ) + V ar ( A ) = √ 15 2 + 5 2 = √ 250 ≈ 15 . 8 minutes The difference in times varies according to the Normal model with µ = 30 and σ = 15 . 8. The next step is to find the z -score associated with y = 0. Using software, we find Area ( y < 0) = Area ( z < − 1 . 899) = 0 . 028782 ≈ 0 . 0288 which means that 2.88% of the scores fall below − 1 . 899. So, the Normal model estimates that about 2.88% of differences fall below zero. 5

a. What is P ( X = 0)? b. What is P ( X = 2)? c. What is P ( X ≤ 2)? d. Find E ( X ) and SD ( X ). Answer a. 0.1074 b. 0.3020 c. 0.6778 d. 2 and 1.265 7. Suppose that X follows the Poisson probability model with λ = 2 for some specified time interval. a. What is P ( X = 0)? b. What is P ( X = 2)? c. What is P ( X ≤ 2)? d. Find E ( X ) and SD ( X ). Answer a. 0.1353 b. 0.2707 c. 0.6767 d. 2 and 1.414 8. Suppose that X is uniformly distributed between a = 1 and b = 9. a. What is P ( X = 1)? b. What is P (4 ≤ X ≤ 8)? c. What is P ( X ≥ 4)? d. Find E ( X ) and SD ( X ). Answer 7

a. 0 b. 0.50 c. 0.625 d. 05 and 2.309 9. Suppose that X is normally distributed with µ = 50 and σ = 2 a. What is P ( X ≥ 50)? b. What is P ( X ≥ 55)? c. What is P ( X ≤ 55)? d. What is P (45 ≤ X ≤ 55)? Answer a. 0.50 b. 0.0062 c. 0.9938 d. 0.9876 10. Suppose that X follows the Binomial distribution with n = 500 and p = 0 . 60. a. Can the Normal probability model be used to approximate this distribution? Ex- plain. b. Find E ( X ) and SD ( X ). c. What is P ( X ≥ 320)? d. What is P ( X ≥ 280)? Answer a. Yes, the success/failure condition is satisfied. b. 300 and 10.95 c. 0.0375 d. 0.9693 11. Suppose that X is the time between arrivals and that arrivals follow the Poisson dis- tribution with λ = 6 per hour. 8