STA 2023 EXAM-2 Practice Problems From Chapters 4, 5, & Partly 6. With SOLUTIONS

STA 2023 EXAM-2 Practice Problems From Chapters 4, 5, & Partly 6 With SOLUTIONS Mudunuru Venkateswara Rao, Ph.D. STA 2023 Fall 2016 Venkat Mu

ALL THE CONTENT IN THESE SOLUTIONS PRESENTED IN BLUE AND BLACK SHOULD BE PRESENTED FOR FULL CREDIT. EXAM WILL BE GRADED BASED ON THESE STEPS 1. A committee of 5 persons is to be formed from 6 men and 4 women. What is the probability that the committee consists of exactly 2 women? For n = 10 and r = 5 we have n(5 member committee) = n(s) = ( 10 5 ) or 10C 5 = 10! = 252 ways 5! (10 5)! Number of committees consisting of exactly 2 women (so the rest of the five must be 3 men) are, n(2w 3M) = 4C 2 6C 3 = 120 ways n(2w 3M) P(2W 3M) = n(5 member committees) = 4C 2 6C 3 = 120 10C 5 252 = 0.4762 2. A coin is tossed 5 times. The probability of heads on any toss is 0.57. Let x denote the number of tails that comes up. a. Find the probability that tail shows up exactly twice. [Ans: 0.3424] NOTE: If probability of HEADS IS 0.57, PROBABILITY OF tails IS 0.43. Here in the questions, the success we are looking for is tails showing up. Hence p = 0.43 Here r = number of successes out of n trails = 2; n = number of trials = 5; p = probability of tail on each trail = 0.43; q = 1 p = probability of failure = 1 0.43 = 0.57; Here n = 5; r = 2; p = 0.43; q = 0.57 P(r) = ( n r ) pr q n r n! = r! (n r)! pr q n r P(r = 2) = P(2) = ( 5 2 ) (0.43)2 (0.57) 5 2 = 0.3424 b. Find the probability that tail shows up less than twice. [Ans: 0.2870] P(r < 2) = P(r 1) = P(0) + P(1) = ( 5 0 ) (0.43)0 (0.57) 5 0 + ( 5 1 ) (0.43)1 (0.57) 5 1 P(r < 2) = P(r 1) = 0.2870 c. Find the Probability that tail shows up more than twice and less than 5 times. [Ans: 0.3557] P(2 < r < 5) = P(3) + P(4) = ( 5 3 ) (0.43)3 (0.57) 5 3 + ( 5 4 ) (0.43)4 (0.57) 5 4 = 0.2583 + 0.0974 P(2 < r < 5) = 0.3557 1

3. Find the following a. In how many different three course meals can be selected from 2 salads, 6 main dishes, and 4 desserts? [Ans: 48 ways] Answer: 2 6 4 = 48 ways b. In how many different ways can I arrange 5 books in a book rack? [Ans: 120] Answer: 5! = 120 ways c. How many different 5-digit codes are possible without repetition of digits and if the first digit cannot be zero? [Ans: 27216] Answer: 9 9 8 7 6 = 27216 codes d. How many different 5-digit codes are possible with repetition of digits and if the first digit cannot be zero? [Ans: 90000] Answer: 9 10 10 10 10 = 90000 codes e. How many different 5-digit codes are possible with repetition of digits and if the first digit can be zero? [Ans: 100000] Answer: 10 10 10 10 10 = 100000 codes f. Given five women and three men, how many ways can a three-member committee be formed? [Ans: 56] n(3 member committee) = n(s) = ( 8 3 ) or 8C 8! 3 = = 56 ways 3! (8 3)! 4. What is the probability that when we roll a die the result is a number divisible by 1.5, given that the number is an even number? [Ans: 0.33] Solution: Here S = {1,2,3,4,5,6}; Let Event A be the number divisible by 1.5; A = {3,6}; Event B be an even number; B = {2,4,6}; A B = {6} P(A B) 1 P(A B) = = 6 P(B) 3 = 0.33 6 2

5. A box contains 16 colored mugs: 4 red, 6 yellow and 6 green. a. A mug is selected at random. What is the probability that the mug chosen is either red or green? Is this an independent event or dependent event or mutually exclusive event? [Ans: 0.625] P(Red Green) = P(Red) + P(Green) = 4 16 + 6 16 = 10 16 = 0.625; MutuallyExclusive b. A mug is selected at random. What is the probability that the mug chosen is either red or yellow? [Ans: 0.625] P(Red Yellow) = P(Red) + P(Yellow) = 4 16 + 6 16 = 10 16 = 0.625 c. A mug is selected at random. What is the probability that the mug chosen is both red and yellow? [Ans: 0] P(Red Yellow) = 0 d. Two mugs are drawn in succession with replacement. What is the probability that both mugs chosen are yellow? Is this an independent event or dependent event or mutually exclusive event? [Ans: 0.141] P(Y 1 Y 2 ) = P(Y 1 ) P(Y 2 ) = 6 16 6 = 0.141; Independent 16 e. Two mugs are drawn in succession without replacement. What is the probability that both mugs chosen are yellow? Is this an independent event or dependent event or mutually exclusive event? [Ans: 0.125] P(Y 1 Y 2 ) = P(Y 1 ) P(Y 2 Y 1 ) = 6 16 5 = 0.125; Dependent 15 6. Consider a pack of 52 playing cards. A card is selected at random. Also mention if these following events are independent or dependent or mutually exclusive? a. What is the probability that the card is either a diamond or an ace? [Ans: 0.308] P(D A) = P(D) + P(A) P(D A) = 13 52 + 4 52 1 52 = 0.308; Not Mutually Exclusive b. What is the probability that the card is both a diamond and a king? [Ans: 0.019] P(D K) = 1 52 = 0.019 c. What is the probability that the card is either a diamond or a spade? [Ans: 0.5] P(D S) = P(D) + P(S) = 13 52 + 13 = 0.5; Mutually Exclusive 52 3

d. What is the probability that the card is either a diamond or a face card?[ans: 0.423] P(D F) = P(D) + P(F) P(D F) = 13 52 + 12 52 3 52 = 0.423; Not Mutually Exclusive e. What is the probability that the card is both spade and a heart card? [Ans: 0] P(S H) = 0; 7. A box contains 10 defective and 18 non-defective bulbs. Two are drawn out together. One of them is tested and found to be non-defective bulb. What is the probability that the other one is also non-defective? [Ans: 0.63] P(D 2 D 1 ) = P(D 1 D 2 ) 18 = 28 17 27 = 0.63; P(D 1 ) 18 28 8. A student takes two core courses, statistics and biology. In any given semester the probability of failing statistics course is 2% and the probability of failing biology is 6%. In a given semester, what is the probability that a. Student will fail both the exams? [Ans: 0.0012] P(S B ) = P(S ) P(B ) = 0.02 0.06 = 0.0012; b. Student will fail one or the other (but not both) courses. [Ans: 0.078] P((S B)or(S B )) = P((S B) (S B )) = P(S ) P(B) + P(S) P(B ) = (0.02 0.94) + (0.98 0.06) = 0.078; OR P(S B) = 1 P(S B) = 1 [P(S) P(B)] = 1 [0.94 0.98] = 0.078 9. Answer the following - a. If the odds against an event occurring are 5:19, what is the probability of the event occurring? [Ans: 0.7916] We know that Odds against, Odds(A ) = n(a ): n(a) i. e., Odds(against) = (#against): (#favoring) and n(s) = n(a) + n(a ) Odds(against) = 5: 19 implying that n(s) = 5 + 19 = 24. Now P(Favoring), P(A) = n(a) n(s) = 19 24 = 0.7916 b. Given n (E) = 3, n(s) = 7. What is P (E )? [Ans: 0.5715] We know that, n(s) = n(e) + n(e ) 7 = 3 + n(e ) n(e ) = 4. Now P(E ) = n(e ) n(s) = P(E ) = 4 7 = 0.5715 4

c. Given n (E) = 3, n(s) = 7. What is Odds (E)? [Ans: 3:4] Here again n(s) = n(e) + n(e ) 7 = 3 + n(e ) n(e ) = 4. Hence, Odds(E) = n(e): n(e ) = 3: 4 d. Given n(e) = x, n(s) = x+y. What is P (E )? [Ans: y/(x+y)] Using n(s) = n(e) + n(e ) x + y = x + n(e ) n(e ) = y. Now P(E ) = n(e ) n(s) = y x + y e. Given n(e) = x, n(s) = x+y. What are Odds (E )? [Ans: y:x] Using n(s) = n(e) + n(e ) x + y = x + n(e ) n(e ) = y. Odds(E) = n(e ): n(e ) = y: x 10. The number of girls and boys enrolled for different majors in a community college are listed in the table below Major Girls Boys Total Biology 150 120 270 Chemistry 100 100 200 Stats 110 130 240 Medicine 175 150 325 Total 535 500 1035 a. If a student is chosen at random, what is the probability that the student is a boy? [Ans: 0.4830] P(Boy) = 500 1035 = 0.4830 b. If a student is chosen at random, what is the probability that the student is a boy enrolled in stats? [Ans: 0.1256] P(Boy Stats) = 130 1035 = 0.1256 c. What is the probability that the student is a girl given that she is biology major? [Ans: 0.5555] P(Girl Bio) 150 P(Girl Bio) = = 1035 = 0.5555 P(Bio) 270 1035 5

d. What is the probability that the student is biology major given that she is a girl? [Ans: 0.2804] P(Bio Girl) = P(Girl Bio) P(Girl) 150 = 1035 = 0.2804 535 1035 11. Below is a survey of 615 students regarding satisfaction with the choice their majors. Maths Biology Physics English Total Very satisfied 27 47 13 12 99 Satisfied 61 39 52 30 182 Not Satisfied 58 99 95 82 334 Total 146 185 160 124 615 Assume that the sample represents the entire population of students. Find the probability that a student selected at random is a. At least satisfied with her/his major. [Ans: 0.4569] 99 + 182 P(At least Satisfied) = = 0.4569 615 b. Satisfied, given that she/he is a Mathematics major. [Ans: 0.4178] P(Satisfied Math) 61 P(Satisfied Math) = = 615 = 0.4178 P(Math) 146 615 c. Not satisfied, given that she/he is a Biology major. [Ans: 0.5351] P(Not Satisfied Bio) 99 P(Not Satisfied Bio) = = 615 = 0.5351 P(Bio) 185 615 d. A Physics major. [Ans: 0.2602] P(Phy) = 160 615 = 0.2602 e. Not an English major. [Ans: 0.7984] P(Not Eng) = 1 P(Eng) = 1 124 615 = 0.7984 f. Satisfied and Not Satisfied. [Ans: 0] P(Satisfied Not Satisfied) = 0 615 = 0 6

12. A card is drawn from a deck of 52 cards. a. What is the probability that the card is diamond or a face card? [Ans: 0.4231] P(F D) = P(F) + P(D) P(F D) = 12 52 + 13 52 3 52 = 0.4231 b. What is the probability that the card is not a spade card? [Ans: 0.75] P(S ) = 1 13 52 = 0.75 c. What is the probability that the card is either a 6 or heart card? [Ans: 0.3077] P(6 heart) = P(6) + P(heart) P(6 heart) = 4 52 + 13 52 1 52 = 0.3077 d. What is the probability that the card is a king or queen or jack? [Ans: 0.2308] P(K Q J) = P(K) + P(Q) + P(J) = 4 52 + 4 52 + 4 52 = 0.2308 e. What are the odds in favor of selecting a face card? [Ans: 3:10] Odds(face card) = (#face cards): (#not face cards) = 12: 40 = 3: 10 13. A coin is tossed 5 times. The probability of tails on any toss is 0.52. Let x denote the number of tails that comes up. a. Find the probability that tail shows up at least twice. [Ans: 0.8365] P(r) = ( n r ) pr q n r Here: n = 5; p = 0.52; q = 1 p = 1 0.52 = 0.48; and r = 2; P(r 2) = 1 P(r 1) = 1 [P(0) + P(1)] = 1 [5C 0 (0.52) 0 (0.48) 5 0 + 5C 1 (0.52) 1 (0.48) 5 1 ] = 1 [0.1635] = 0.8365 b. Find the probability that tail shows up in all tosses. [Ans: 0.0380] P(r = 5) = 5C 5 (0.52) 5 (0.48) 5 5 = 0.038 c. Find the Probability that tail shows up more than twice and less than 5 times. [Ans: 0.4994] P(2 < r < 5) = P(3) + P(4) = [5C 3 (0.52) 3 (0.48) 5 3 + 5C 4 (0.52) 4 (0.48) 5 4 ] = 0.4994 7

14. The average number of injuries per week in an industry are recorded as 2.1. a. Find the probability that in a given week there can be no injuries? [Ans: 0.1224] Given average, λ = 2.1; We will use Poisson distribution. P(r) = e λ λ r r! P(r = 0) = e 2.1 (2.1) 0 = 0.1224 0! b. Find the probability that in a given week, there will be less than 3 injuries? [Ans: 0.6496] P(r < 3) = P(r 2) = P(0) + P(1) + P(2) = e 2.1 (2.1) 0 + e 2.1 (2.1) 1 + e 2.1 (2.1) 2 = 0.6496 0! 1! 2! c. Find the probability that in a given week, there will be at least 3 injuries? [Ans: 0.3504] P(r 3) = 1 P(r 2) = 1 [P(0) + P(1) + P(2)] P(r 3) = 1 0.6496 = 0.3504 {Using result of part b, above} d. Find the probability that in a given week, there will be at least 2 but at most 4 injuries? [Ans: 0.5582] P(2 r 4) = P(2) + P(3) + P(4) = e 2.1 (2.1) 2 + e 2.1 (2.1) 3 + e 2.1 (2.1) 4 = 0.5582 2! 3! 4! e. What is the mean and standard deviation of this problem? [Ans: 2.1; 1.449] Mean μ = λ = 2.1; Standard deviation σ = λ = 2.1 = 1.449 8

15. An insurance company has determined that on the average it receives 3 accident claims per day. Given average, λ = 3; We will use Poisson distribution. P(r) = e λ λ r r! a. Find the probability that the company receives at least 4 accident claims on a randomly selected day. [Ans: 0.3528] P(r 4) = 1 P(r 3) = 1 [P(0) + P(1) + P(2) + P(3)] = 1 [ e 3 (3) 0 + e 3 (3) 1 + e 3 (3) 2 + e 3 (3) 2 ] = 0.3528 0! 1! 2! 3! b. Find the probability that the company receives no claims on a randomly selected day. [Ans: 0.0498] P(r = 0) = e 3 (3) 0 = 0.0498 0! 16. Assuming that the heights of boys in a high-school basketball tournament are normally distributed with mean 80 inches and standard deviation 2 inches a. How many boys in a group of 40 are expected to be taller than 84 inches? [Ans: 1] Clearly, we have 2.5% of 40 boys who are taller than 84 height. Hence 2.5 40=1 There is 1 Boy in a group of 40 who is taller than 84" height. 100 9

b. What percentage of boys falls between 78 and 84 inches height? [Ans: 0.815 or 81.5%] There are 68% of boys who are falling between 78 to 82 & 13.5% from 82 to 84, so, There are a total of 68% + 13.5% = 81.5% of boys who are between 78 and 84 inches c. What percentage of boys falls below 82 inches? [Ans: 84%] There are 50% of boys who are falling below 80 inches & 34% from 80 to 82 inches. so There are a total of 50% + 34% = 84% of boys who are below 82 inches 17. Given the experiment: a fair die is rolled; if a multiple of three appears uppermost, then the die is rolled again, otherwise a fair coin is tossed. a. Represent this scenario using a tree diagram with the respective probabilities on each branch mentioned. [Ans: ] P(3)=P(6)=1/6 Start P(1)= =P(6)=1/6 P(1)=P(2)=P(4)=P(5)=1/6 3,6 1,2,4,5 b. What is the sample space of this experiment? 0.5 S = {(3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6), (1, H), (1, T), (2, H), (2, T), (4, H), (4, T), (5, H), (5, T)} c. What is the associated probability distribution in this experiment? (3,1) = (3,2) = (3,3) = (3,4) = (3,5) = (3,6) = (6,1) = (6,2) = (6,3) = (6,4) = (6,5) = (6,6) = 0.02778 (1, H) = (1, T) = (2, H) = (2, T) = (4, H) = (4, T) = (5, H) = (5, T) = 0.08333 1 2 3 4 5 6 H T 10

18. Given the experiment: a fair die is rolled followed by tossing a fair coin. a. Represent this scenario using a tree diagram with the respective probabilities on each branch mentioned. [Ans: try or guess ] b. What is the sample space of this experiment? Ans: S = {1H, 1T, 2H, 2T, 6H, 6T} c. What is the associated probability distribution in this experiment? Ans: 1H = (1/6) (1/2) = 0.0833 = 1T = 2H = 2T = = 6H = 6T 19. A TV company when purchasing thousands electronic components apply this sampling plan: randomly select 15 of them and then accept the whole batch if there are at most 3 defective. a. If a particular lot has a 3% of defective components, what is the probability that the whole lot is accepted? [Ans: 0.9991] Solution: n = 15; p = 0.03; q = 1 p = 1 0.03 = 0.97; and Using: P(r) = ( n r ) pr q n r P(r 3) = P(0) + P(1) + P(2) + P(3) = 15C 0 (0.03) 0 (0.97) 15 0 + 15C 1 (0.03) 1 (0.97) 15 1 + 15C 2 (0.03) 2 (0.97) 15 2 + 15C 3 (0.03) 3 (0.97) 15 3 = 0.9991 b. In a sample of 2000 electronic components, what will be the expected number of the defective components and what is the standard deviation? [Ans: 60; 7.6288] Also since given a sample of size n = 2000; and p = 0.03; The average number of defective components are E(X) = μ = np = 2000 0.03 = 60 The standard deviation is given by σ = npq σ = 2000 0.03 0.97 = 7.6288 11

20. Draw a probabilistic tree diagram for the following experiment: a fair coin is tossed; if a head appears uppermost, then the fair coin is tossed again, otherwise an unfair coin with P (Head) = 0.55 is tossed. What is the sample space of this experiment? What is the associated probability distribution in this experiment? Start Head Tail Sample space, S = {HH, HT, TH, TT} Probability Distribution: HH = HT = 0.5 0.5 = 0.25; TH = 0.5 0.55 = 0.275; TT = 0.5 0.45 = 0.225 Head Tail Head Tail 21. Consider the following experiment. One card is selected from a deck of cards. Find the probability the card selected is: a. A six number card b. A seven & a king card c. A diamond or a club card d. A king card e. A five & a diamond card f. A face or an ace card g. A red card h. A ten or an ace card i. A face or a heart card j. A four and a black card k. A queen and a king card l. A red and a black card 12

a. A six number card P(6) = 4 52 = 0.0769 d. A king card P(K) = 4 52 = 0.0769 g. A red card P(R) = 26 52 = 0.5 j. A four & a black card P(4 B) = 2 52 = 0.0384 b. A seven & a king card P(7 K) = 0 52 = 0 e. A five & a diamond card P(5 D) = 1 52 = 0.0192 h. A ten or an ace card P(10 A) = 4 52 + 4 52 = 0.1538 c. A diamond or a club card P(D C) = 13 52 + 13 52 = 0.5 f. A face or an ace card P (F A) = 12 52 + 4 52 = 0.3077 i. A face or a heart card P(F H) = 12 52 + 13 52 3 52 = 0.4231 k. A queen and a king card P(Q K) = 0 l. A red and a black card 52 = 0 P(R B) = 0 52 = 0 22. A quality control engineer is in charge of testing whether or not 90% of the DVD players produced by his company conform to specifications. To do this, the engineer randomly selects a batch of 12 DVD players from each day s production. The day s production is acceptable provided no more than 1 DVD player fails to meet specifications. Otherwise, the entire day s production has to be tested. What is the probability that the engineer incorrectly passes a day s production as acceptable if only 80% of the day s DVD players actually conform to specification? Using: n = 12; p = 0.2; q = 1 p = 1 0.2 = 0.8; and P(r) = ( n r ) pr q n r P(r 1) = P(0) + P(1) = 12C 0 (0.2) 0 (0.8) 12 0 + 12C 1 (0.2) 1 (0.8) 12 1 = 0.275 23. Are the following experiments follow binomial distribution or not? Give your answer with Yes/No followed by (fixed number of trials, possible outcomes). a. Surveying 84 people to determine if they like dessert after dinner. Yes (84, like or dislike) b. Rolling a die 70 times to find a prime number occur. Yes (70, Prime or Not a prime) c. Drawing a card from a deck and getting a Jack card Yes (1, jack or not a jack) d. Surveying 100 people which cuisine they prefer the most. No (100, more than 2 possible cuisines) e. Testing 8 different brands of phones to see which brand is best to purchase No (NA, 8 brands) 13

f. Testing 1 brand of Tylenol by using 10 people to determine whether it is effective Yes (10, effective or not) g. Surveying 143 high school students to see what would they prefer as their major for a college degree duties in a row. No (143, more than 2 majors) 24. A die is rolled. Find the Here let S = set of all posisble outcomes when we roll a die. Therefore S = {1,2,3,4,5,6} n(s) = 6 a. Odds in favor of getting a number less than 2 [Ans: 1:5] Here let E = set of numbers less than 2. Therefore E = {1} n(e) = 1 and therefore, n(e ) = n(s) n(e) = 6 1 = 5. Odds(E) = n(e): n(e ) = 1: 5 b. Odds in favor of getting a number less than 4 [Ans: 1:1] Here let E = set of numbers less than 4. Therefore E = {1,2,3} n(e) = 3 and therefore, n(e ) = n(s) n(e) = 6 3 = 3. Odds(E) = n(e): n(e ) = 3: 3 = 1 1 c. Odds against getting a number more than 5 [Ans: 5:1 Here let E = set of numbers more than 5. Therefore E = {6} n(e) = 1 and therefore, n(e ) = n(s) n(e) = 6 1 = 5. Odds(E ) = n(e ): n(e) = 5: 1 d. Probability of getting an even prime [Ans: 0.1667] Here let E = set of even prime numbers. Therefore E = {2} n(e) = 1 P(E) = 1 6 = 0.1667 e. Probability of getting a multiple of 4 [Ans: 0.1667] Here let E = set of multiples of 4. Therefore E = {4} n(e) = 1 P(E) = 1 6 = 0.1667 f. Probability of getting a number divisible by 1.75 [Ans: ] Here let E = set of numbers divisible by 1.75. Therefore E = { } n(e) = 0 P(E) = 0 6 = 0 14

25. Let exam scores out of 150 points follow normal distribution data with μ = 112 and σ = 12. Consider a class of 100 students. a. What is the percentage of students falling in the interval between 112 & 124 points? We have 34% of total students in the interval 112 and 124 b. What is the probability that a student selected at random has a score more than 136? P(x > 136) = 2.5% = 0.025 c. How many students in the class of 100 scored points between 100 & 124? We have 68% of data falling between 100 and 124. 68 Hence 100=68 There are 68 students in a group of 100 who 100 scored points between 100 and 124. d. How many students in the class of 100 scored points between 112 & 136? We have 47.5% of data falling between 112 and 136. Hence 47.5 100=47.5~48 There are 48 students in a group of 100 who 100 scored points between 112 and 136. e. How many students in the class of 100 scored points less than 88 points? We have 2.5% of data falling below 88 points. Hence 2.5 100=2.5~3 There are 3 students in a group of 100 who 100 scored points below 88. 15

More for Exam-2: 1) The probability of the sample space is ONE. 2) The probability of any event is between 0 and 1 3) Given P(E) = 0. 38, then P(E C ) = 0. 62 4) Let n(e) be the number of outcomes in an event E. Given n(e) = 21 and P(E) = 0. 7, then n(e C ) = 9 5) Given n(e) = 21 and P(E) = 0. 7, then O(E) = 21: 9 or 7: 3. 6) Given A and B are dependent then P(A B) = P(A)P(B A) or P(B)P(A B) 7) Given A and B are independent then P(A B) = P(A) 8) Given A and B are independent then P(A B) = P(A)P(B) 9) Given A and B are mutually exclusive then P(A B) = 0. 10) Given A and B are mutually exclusive then P(A B) = 0 11) Given A and B are mutually exclusive then A B = φ 12) Given two independent random variables with E(x 1 ) = μ 1, Var(x 1 ) = σ 1 2, E(x 2 ) = μ 2, and Var(x 2 ) = σ 2 2 ; if we transform the data using the linear combination y = 3x 1 5x 2, then E(y) = 3μ 1 5μ 2 and Var(y) = 9σ 1 2 + 25σ 2 2 13) In a normal distribution, the empirical rule states approximately 68% of the observed data fall within one standard deviation of the mean. 14) In a normal distribution, the empirical rule states approximately 95% of the observed data fall within two standard deviation of the mean. 15) In a normal distribution, the empirical rule states approximately 99.7% of the observed data fall within three standard deviation of the mean. 16) Given two independent random variables with E(x 1 ) = 5, Var(x 1 ) = 4, E(x 2 ) = 6, and Var(x 2 ) = 9; if we transform the data using the linear combination y = x 1 4x 2, then E(y) = 19 and Var(y) = 148 17) Given two independent random variables with E(x 1 ) = 5, Var(x 1 ) = 16,; if we transform the data using the linear combination y = 2 3x 1, then E(y) = 13 and Var(y) = 144 16

18) According to normal control charts, out of control signal-1 indicates that there is at least ONE point beyond three standard deviations of the mean. 19) According to normal control charts, a run of nine consecutive points on one side of the center line indicates out of control signal-2. 20) According to normal control charts, at least two out of three points beyond the two standard deviations of center line indicates out of control signal-3. Helpful Rules: Exactly at least or no less than at most or no more than Less than More than k successes k successes k successes k successes k successes r = k r k r k r < k r > k Conditional Probability P(A B) P(A B) = P(B) Odds(E) O(E) = n(e): n(e ) Odds(E ) O(E ) = n(e ): n(e) Probability of an event P(E) = n(e) n(s) Probability of a compliment P(E ) = n(e ) or 1 P(E) n(s) Factorial n! = n (n 1) (n 2) 2 1 Combinations ( n r ) or nc n! r = r! (n r)! Permutations n! np r = (n r)! Binomial P(r) = nc r p r q n r ; q = 1 p Poisson n(s) = n(e) + n(e ) n(e ) = n(s) n(e) n(s) = 1 P(r) = e λ λ r r! 17

Few quick examples of Sample Spaces Tossing a coin S = {H, T} Tossing two coins S = {HH, HT, TH, TT} Tossing three coins S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT} Rolling a die S = {1,2,3,4,5,6} Rolling two dice S = {(1,1), (1,2), (1,3), (6,4), (6,5), (6,6)} Rule Example E(a) = a E(2) = 2 E(aX) = ae(x) E( 5X) = 5E(X) V(a) = 0 V(4) = 0 V(aX) = a 2 V(X) V( 3X) = ( 3) 2 V(X) = 9V(X) 18

Venkat Mudun 19