Chapter 4&5_Tutorial_W24_Solution

Department of Mathematics and Statistics STAT2910-01: Statistics for Sciences Faculty of Science University of Windsor Tutorial for Chapter 4 and 5 Questions for Chapter 4 Exercise 1. Of the delegates at a convention, 60% attended the breakfast forum, 70%attended the dinner speech, and 40% attended both events. Define events A and B as follows: A: attended the breakfast forum B: attended the dinner speech What is the probability that a randomly selected delegate either attended the breakfast forum, or attended the dinner speech, or attended both? Solution 1. The requested probability is given by P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) 0 . 60 + 0 . 70 − . 040 = 0 . 90 . Exercise 2. Heidi prepares for an exam by studying a list of 15 problems. She can solve 9 of them. For the exam, the instructor selects 7 questions at random from the list of 15. What is the probability that Heidi can solve all 7 problems on the exam? Solution 2. The instructor can select 7 out of the lsit of 15 questions in ( 15 7 ) ways. If Heidi is able to answer all 17 questions, the instructor must choose 7 questions out of the 9 that Heidi can answer, and none of the 6 questions that Heidi cannot answer. The number of ways in which this event can occur is ( 9 7 )( 6 0 ) . Hence, the probability that Heidi can answer all 17 questions is ( 9 7 )( 6 0 ) ( 15 7 ) = 36 6435 Exercise 3. a. How many different combinations of 5 students can be drawn from a class of 25 students? b. How many permutations of 3 colours can be drawn from a group of 20 colours? 1

Solution 3. a. There are ( 25 5 ) = 35130 combinations. b. There are P 3 20 = 6840 permutations. Exercise 4. A psychologist tests Grade 7 students on basic word association skills and number pattern recognition skills. Let W be the event a student does well on the word association test. Let N be the event a student does well on the number pattern recognition test. A student is selected at random, and the following probabilities are given: P ( W ∩ N ) = 0 . 25 , P W ∩ N ∁ = 0 . 15 , P W ∁ ∩ N = 0 . 10 , P W ∁ ∩ N ∁ = 0 . 50 a. What is the probability that the randomly selected student does well on the word association test? b. What is the probability that the randomly selected student does well on the number pattern recognition test? c. What is the probability that the randomly selected student does well on at least one of the tests? d. If the randomly selected student does well on the word association test, what is the probability he or she will also do well on the number pattern recognition test? e. If the randomly selected student does well on the number pattern recognition test, what is the proba- bility he or she will also do well on the word association test? f. Are the events W and N mutually exclusive? Justify your answer. g. Are the events W and N independent? Explain. Solution 4. a. The desired probability is given by P ( W ) = P ( W ∩ N ) + P W ∩ N ∁ = 0 . 25 + 0 . 15 = 0 . 40 b. The desired probability is given by P ( N ) = P ( W ∩ N ) + P W ∁ ∩ N = 0 . 25 + 0 . 10 = 0 . 35 c. The desired probability is given by P ( W ∪ N ) = P ( W ) + P ( N ) − P ( W ∩ N ) = 0 . 40 + 0 . 35 + − 0 . 25 = 0 . 5 Or P ( W ∪ N ) = 1 − P W ∁ ∩ N ∁ = 1 − 0 . 5 = 0 . 50 2

Solution 5. a. The requested probability is P ( A ) = 0 . 13 + 0 . 20 + 0 . 30 = 0 . 63 b. The requested probability is P ( B ) = 0 . 30 + 0 . 25 = 0 . 55 c. The requested probability is P ( A ∩ B ) = 0 . 30 d. The requested probability is P ( A ∪ B ) = P ( A ) + P ( B ) − P ( A ∩ B ) = 0 . 63 + 0 . 55 − 0 . 30 = 0 . 88 e. The requested probability is P ( B | A ) = P ( A ∩ B ) P ( A ) = 0 . 30 0 . 63 = 0 . 4762 f. This probability is given by P ( A | B ) = P ( A ∩ B ) P ( B ) = 0 . 30 0 . 55 = 0 . 5455 g. No. For example P ( A | B ) = 0 . 5455 ̸ = P ( A ) = 0 . 63 Exercise 6. A federal agency is trying to decide which of two waste management projects to investigate as the source of air pollution. In the past, projects of the first type were in violation of air quality standards with probability 0.3 on any given day, while projects of the second type were in violation of air quality standards with probability 0.25 on any given day. It is not possible for both projects to pollute the air in one day. Let A i , i = 1 , 2, denote that project of type i was in violation of air quality standards. a. Find the probability of an air pollution problem being caused by either the first project or the second project. b. If the first project is violating air quality standards, what is the probability the second project is also violating federal air quality standards? 4

Solution 6. a. Since A 1 and A 2 are mutually exclusive, the requested probability is given by P ( A 1 ∪ A 2 ) = P ( A 1 ) + P ( A 2 ) = 0 . 30 + 0 . 25 = 0 . 55 b. Since A 1 and A 2 are mutually exclusive, the requested probability is given by P ( A 1 | A 2 ) = P ( A 1 ∩ A 2 ) P ( A 2 ) = 0 0 . 30 = 0 Exercise 7. An experiment was conducted in which rats could choose to enter one of two corridors, A or B. A random sample of three rats is selected. Let X = number of rats that select corridor B. a. Assuming the rats select their favourite corridor independently of one another and that the two corri- dors are equally likely to be selected, find the probability distribution of X . b. What is the probability that, at most, one rat selects corridor B? c. What is the probability that at least one rat selects corridor B? Solution 7. a. The probability distribution of X is given by p (0) = P ( A ∩ A ∩ A ) = P ( A ) P ( A ) P ( A ) = 1 2 3 = 0 . 125 p (1) = P ( A ∩ A ∩ B ) + P ( A ∩ B ∩ A ) + P ( B ∩ A ∩ A ) = 3 1 2 3 = 0 . 375 p (2) = P ( B ∩ B ∩ A ) + P ( B ∩ A ∩ B ) + P ( A ∩ B ∩ B ) = 3 1 2 3 = 0 . 375 p (3) = P ( B ∩ B ∩ B ) = 0 . 125 As a table x 0 1 2 3 p ( x ) 0.125 0.375 0.375 0.125 Or graphically 5

Solution 8. a. The average can be calculated as µ = X x · p ( x ) = 0 × 0 . 1 + 5 × 0 . 5 + 10 × 0 . 3 + 15 × 0 . 1 = 7 b. Similarly, the variance is σ 2 = X ( x − µ ) 2 · p ( x ) = (0 − 7) 2 × 0 . 1 + (5 − 7) 2 × 0 . 5 + (10 − 7) 2 × 0 . 3 + (15 − 7) 2 × 0 . 1 = 16 c. P ( X ≥ 10) = P ( X = 10) + P ( X = 15) = 0 . 3 + 0 . 1 = 0 . 4 d. P (0 ≤ X ≤ 5) = P ( X = 0) + P ( X = 5) = 0 . 1 + 0 . 5 = 0 . 6 e. P (0 < X < 10) = P ( X = 5) = 0 . 5 Questions for Chapter 5 Exercise 9. A manufacturer of golf balls uses a production process that produces 10% defective balls. A quality inspector takes samples of a week’s output with replacement. Using the cumulative binomial probability table available in your text, which of the following probabilities can the inspector determine? a. If five units are inspected, the probability of at most three of these units being defective is 0.984. b. If 10 units are inspected, the probability of 5 or 6 of these units being defective is 0.002. c. If 15 units are inspected, the probability of at least 10 of these units being defective is 0.547. d. If 20 units are inspected, the probability of at least 19 of these units being defective is 0.0009. Solution 9. Let X be the number of defective balls, then X ∼ binomial( n, p = 0 . 1). The answer is b , which represents P ( X = 5) + P ( X = 6) = 0 . 002. 7

binom(10,0.1) Distribution: Shaded Area = 0.002 x p(x) 0.0 0.1 0.2 0.3 0.4 5 6 Exercise 10. The taste test for PTC (phenylthiourea) is a common class demonstration in the study of genetics. It is known that 70% of Canadians are “tasters” and 30% are “non-tasters.” Suppose a genetics class of size 20 does the test to see if they match the Canadian percentage of “tasters” and “non-tasters.” (Assume the assignment of students to classes constitutes a random process.) a. What is the probability distribution of the random variable X , the number of “non-tasters” in the class? b. Find P (3 < X < 9). c. Find the mean of X . d. Find the variance of X Solution 10. a. The random variable X follows a binomial distribution with parameters n = 20 and p = 0 . 3, namely P ( X = k ) = 20 k (0 . 3) k (0 . 7) 20 − k , k = 0 , 1 , 2 , . . . , 20 , where ( 20 k ) = 20! k !(20 − k )! . b. P (3 < X < 9) = P (4 ≤ X ≤ 8) = 0 . 78. 8

e. Find the probability that, at most, four children do not recognize their mother’s voice. Solution 12. a. Let X the number of one-year-old children can not distinguish their mother’s voice. Then, X ∼ Binomial(20 , 0 . 1). The requested probability is P ( X ≥ 3) = 1 − P ( X = 0) − P ( X = 1) − P ( X = 2) = 0 . 323 , (here p = 0 . 10) binom(20,0.1) Distribution: Shaded Area = 0.323 x p(x) 0.00 0.05 0.10 0.15 0.20 0.25 3 b. Let X the number of one-year-old children can distinguish their mother’s voice. Then, X ∼ Binomial(20 , 0 . 9). The requested probability is P ( X = 20) = 0 . 122 , (here p = 0 . 90) c. X ∼ Binomial(20 , 0 . 1), then µ = np = 20 × 0 . 1 = 2. d. X ∼ Binomial(20 , 0 . 1), then σ 2 = np (1 − p ) = 20 × 0 . 1 × 0 . 9 = 1 . 8. e. X ∼ Binomial(20 , 0 . 1), then P ( X ≤ 4) = P ( X = 0) + P ( X = 1) + P ( X = 2) + P ( X = 3) + P ( X = 4) = 0 . 957 10

binom(20,0.1) Distribution: Shaded Area = 0.957 x p(x) 0.00 0.05 0.10 0.15 0.20 0.25 4 Exercise 13. Hotels, like airlines, often overbook, counting on the fact that some people with reservations will cancel at the last minute. A certain hotel chain has found that 20% of the reservations will not be used. a. If we randomly selected 15 reservations, what is the probability more than 8 but fewer than 12 reser- vations will be used? b. If four reservations are made, what is the chance fewer than two will cancel? Solution 13. a. Let X be the number used. Then, X ∼ Binomial(15 , 0 . 8). The requested probability is P (8 < X < 12) = P ( X = 9) + P ( X = 10) + P ( X = 11) = 0 . 334. 11

binom(4,0.2) Distribution: Shaded Area = 0.819 x p(x) 0.0 0.1 0.2 0.3 0.4 1 Exercise 14. Let X be a binomial random variable with n = 15 and p = 0 . 20. a. Calculate P ( X ≤ 4) using the binomial formula. b. Calculate P ( X ≤ 4) using the cumulative binomial probabilities table in Appendix I. c. Compare the results of (a) and (b). d. Calculate the mean and standard deviation of the random variable X . e. Use the results of (d) to calculate the intervals µ ± σ , µ ± 2 σ and µ ± 3 σ . Find the probability that an observation will fall into each of these intervals. f. Are the results of part (e) consistent with Tchebysheff’s Theorem? With the Empirical Rule? Solution 14. a. Given that X ∼ binomial( n = 15 , p = 0 . 2), just plug the numbers into the binomial probability formula (omitted here), we will have P ( X ≤ 4) = 0 . 835767. b. P ( X ≤ 4) = 0 . 836. c. The results are almost identical. d. µ = np = 15 × 0 . 2 = 3. σ = √ npq = √ 15 × 0 . 2 × 0 . 8 = 1 . 549. 13

e. µ ± σ is [1 . 451 , 4 . 549], so P (1 . 451 ≤ X ≤ 4 . 549) = 0 . 669. µ ± 2 σ is [ − 0 . 098 , 6 . 098], so P ( − 0 . 098 ≤ X ≤ 6 . 098) = 0 . 982. µ ± 3 σ is [ − 1 . 647 , 7 . 647], so P ( − 1 . 647 ≤ X ≤ 7 . 647) = 0 . 996. f. The results are consistent with Tchebysheff’s Theorem and the Empirical Rule. See an illustration for the empirical rule bellow Empirical Rule: mean = 3 , SD = 1.54919333848297 x density 3 ~68% Empirical Rule: mean = 3 , SD = 1.54919333848297 x density 3 ~95% Empirical Rule: mean = 3 , SD = 1.54919333848297 x density 3 ~99.7% binom(15,0.2) Distribution: Shaded Area = 0.669 x p(x) 0.00 0.05 0.10 0.15 0.20 0.25 2 4 binom(15,0.2) Distribution: Shaded Area = 0.982 x p(x) 0.00 0.05 0.10 0.15 0.20 0.25 0 6 binom(15,0.2) Distribution: Shaded Area = 0.996 x p(x) 0.00 0.05 0.10 0.15 0.20 0.25 0 7 Exercise 15. The number of telephone calls coming into a business’s switchboard averages four calls per minute. Let X be the number of calls received. a. Find P ( X = 0). b. What is the probability there will be at least one call in a given one-minute period? c. What is the probability at least one call will be received in a given two-minute period? Solution 15. a. X ∼ Poisson(4). The requested probability is P ( X = 0) = e − 4 = 0 . 018. b. P ( X ≥ 1) = 1 − P ( X = 0) = 0 . 982. c. Y ∼ Poisson(8) .The requested probability is P ( Y ≥ 1) = 1 − P ( Y = 0) = 1 − e − 8 = 0 . 99966. 14

Solution 18. a. The random variable X = number of misfiring spark plugs has a Hypergeometric distribution with n = 4 , M = 2 , N = 8. Then, P ( x = 2) = C 2 2 C 6 2 C 8 4 = 0 . 214 . b. µ = n M N = 1 . c. σ 2 = n M N N − M N N − n N − 1 = 0 . 4286 . 16