Random Variables and Probability Distributions Chapter 4

Transcription

1 Random Variables and Probability Distributions Chapter a. The closing price of a particular stock on the New York Stock Echange is discrete. It can take on only a countable number of values. b. The number of shares of a particular stock that are traded on a particular day is discrete. It can take on only a countable number of values. c. The quarterly earnings of a particular firm is discrete. It can take on only a countable number of values. d. The percentage change in yearly earnings between 2005 and 2006 for a particular firm is continuous. It can take on any value in an interval. e. The number of new products introduced per year by a firm is discrete. It can take on only a countable number of values. f. The time until a pharmaceutical company gains approval from the U.S. Food and Drug Administration to market a new drug is continuous. It can take on any value in an interval of time. 4.4 The number of customers,, waiting in line can take on values 0, 1, 2, 3,. Even though the list is never ending, we call this list countable. Thus, the random variable is discrete. 4.6 A banker might be interested in the number of new accounts opened in a month, or the number of mortgages it currently has, both of which are discrete random variables. 4.8 The manager of a hotel might be concerned with the number of employees on duty at a specific time, or the number of vacancies there are on a certain night A stockbroker might be interested in the length of time until the stockmarket is closed for the day a. The variable can take on values 1, 3, 5, 7, and 9. b. The value of that has the highest probability associated with it is 5. It has a probability of Chapter 4

2 c. Using MINITAB, the probability distribution of as a graph is: d. P( = 7) =.2 e. P( 5) = p(5) + p(7) + p(9) = =.7 f. P( > 2) = p(3) + p(5) + p(7) + p(9) = = a. This is not a valid distribution because p( ) =.9 1. b. This is a valid distribution because 0 p() 1 for all values of and p( ) = 1. c. This is not a valid distribution because p(4) =.3 < 0. d. The sum of the probabilities over all possible values of the random variable is p( ) = 1.1 > 1, so this is not a valid probability distribution a. μ = E() = p( ) = 10(.05) + 20(.20) + 30(.30) + 40(.25) + 50(.10) + 60(.10) = = 34.5 b. σ 2 = E( μ) 2 2 = ( μ) p ( ) = ( ) 2 (.05) + ( ) 2 (.20) + ( ) 2 (.30) + ( ) 2 (.25) + ( ) 2 (.10) + ( ) 2 (.10) = = σ = = Random Variables and Probability Distributions 83

3 c. μ ± 2σ 34.5 ± 2(13.219) 34.5 ± (8.062, ) P(8.062 < < ) = p(10) + p(20) + p(30) + p(40) + p(50) + p(60) = = a. It would seem that the mean of both would be 1 since they both are symmetric distributions centered at 1. b. P() seems more variable since there appears to be greater probability for the two etreme values of 0 and 2 than there is in the distribution of y. c. For : μ = E() = p( ) = 0(.3) + 1(.4) + 2(.3) = = 1 σ 2 = E[( μ) 2 2 ] = ( μ) p ( ) = (0 1) 2 (.3) + (1 1) 2 (.4) + (2 1) 2 (.3) = =.6 For y: μ = E(y) = yp( y) = 0(.1) + 1(.8) + 2(.1) = = 1 σ 2 = E[(y μ) 2 2 ] = ( y μ) p( y) = (0 1) 2 (.1) + (1 1) 2 (.8) + (2 1) 2 (.1) = =.2 The variance for is larger than that for y a. Yes. Relative frequencies are observed values from a sample. Relative frequencies are commonly used to estimate unknown probabilities. In addition, relative frequencies have the same properties as the probabilities in a probability distribution, namely 1. all relative frequencies are greater than or equal to zero 2. the sum of all the relative frequencies is 1 b. Using MINITAB, the graph of the probability distribution is: p(age) age c. Let = age of employee. Then P( > 30) = =.40. P( > 40) = 0 P( < 30) = =.51 d. P( = 25 or = 26) = = Chapter 4

4 4.22 a. The probability distribution for is: Grill Display Combination p() / 124 = / 124 = / 124 = / 124 = / 124 = / 124 =.274 b. P( > 10) = p(10) + p(11) = = a. First, we must find the probability distribution of. Define the following events: C: {Chicken is contaminated} N: {Chicken is not contaminated} If 3 slaughtered chickens are randomly selected, then the possible outcomes are: CCC, CCN, CNC, NCC, CNN, NCN, NNC, and NNN Each of these outcomes are NOT equally likely since P(C) = 1/100 =.01. P(N) = 1 P(C) = =.99. P(CCC) = P(C C C ) = P(C) P(C) P(C) =.01(.01)(.01) = P(CCN) = P(CNC) = P(NCC) = P(C C N ) = P(C) P(C) P(N) =.01(.01)(.99) = P(CNN) = P(NCN) = P(NNC) = P(C N N ) = P(C) P(N) P(N) =.01(.99)(.99) = P(NNN) = P(N N N ) = P(N) P(N) P(N) =.99(.99)(.99) = The variable is defined as the number of contaminated chickens in the sample. The value of for each of the outcomes is: Event p() CCC CCN CNC NCC CNN NCN NNC NNN Random Variables and Probability Distributions 85

5 The probability distribution of is: p() b. Using MINITAB, the probability graph for is: p() c. P( 1) = P( = 0) + P( = 1) = = To find the probability distribution of, we first list the possible values of. For this eercise, the possible values of are 3, 1, and 5. Net, we list the number of cases, f(), that result in the particular values of. To find the probability distribution of, we divide the number of cases for each value of, f(), by the total number of cases, 678. For = 3, the probability is p( 3) = 68 / 678 =.100. For = 1, the probability is p( 1) = 71 / 678 =.105. For = 5, the probability is p(5) = 539 / 678 =.795. The probability distribution of is: f() p() Total Chapter 4

6 Using MINITAB, the graph of the probability distribution is: p() a. E() = All p( ) Firm A: E() = 0(.01) + 500(.01) (.01) (.02) (.35) (.30) (.25) (.02) (.01) (.01) (.01) = = 2450 Firm B: E() = 0(.00) + 200(.01) + 700(.02) (.02) (.15) (.30) (.30) (.15) (.02) (.02) (.01) = = 2450 b. σ = 2 σ σ 2 = 2 ( μ) p ( ) All Firm A: σ 2 = (0 2450) 2 (.01) + ( ) 2 (.01) + + ( ) 2 (.01) = 60, , , , , , , , , ,025 = 437,500 σ = Firm B: σ 2 = (0 2450) 2 (.00) + ( ) 2 (.01) + + ( ) 2 (.01) = , , , , , , , , ,625 = 492,500 σ = Firm B faces greater risk of physical damage because it has a higher variance and standard deviation. Random Variables and Probability Distributions 87

7 4.30 a. If a large number of measurements are observed, then the relative frequencies should be very good estimators of the probabilities. b. E() = p( ) = 1(.01) + 2(.04) + 3(.04) + 4(.08) + 5(.10) + 6(.15) + 7(.25) + 8(.20) + 9(.08) + 10(.05) = = 6.50 c. σ 2 = The average number of checkout lanes per store is 6.5. = (1 6.5) 2 (.01) + (2 6.5) 2 (.04) + (3 6.5) 2 (.04) 2 ( μ) p ( ) All σ = 3.99 = (4 6.5) 2 (.08) + (5 6.5) 2 (.10) + (6 6.5) 2 (.15) + (7 6.5) 2 (.25) + (8 6.5) 2 (.20) + (9 6.5) 2 (.08) + (10 6.5) 2 (.05) = = 3.99 d. Chebyshev's Rule says that at least 0 of the observations should fall in the interval μ ± σ. Chebyshev's Rule says that at least 75% of the observations should fall in the interval μ ± 2σ. e. μ ± σ 6.5 ± (4.5025, ) P( ) = =.70 This is at least 0. μ ± 2σ 6.5 ± 2(1.9975) 6.5 ± (2.505, 10,495) P( ) = =.95 This is at least.75 or 75% Let = winnings in the Florida lottery. The probability distribution for is: p() $1 22,999,999/23,000,000 $6,999,999 1/23,000,000 The epected net winnings would be: μ = E() = ( 1)(22,999,999/23,000,000) + 6,999,999(1/23,000,000) = $.70 The average winnings of all those who play the lottery is $ Chapter 4

8 4.34 Each point in the system can have one of 2 status levels, free or obstacle. Define the following events: A F : {Point A is free} B F : {Point B is free} C F : {Point C is free} A O : {Point A is obstacle} B O : {Point B is obstacle} C O : {Point C is obstacle} Thus, the sample points for the space are: A F B F C F, A F B F C O, A F B O C F, A F B O C O, A O B F C F, A O B F C O, A O B O C F, A O B O C O Since it is stated that the probability of any point in the system having a free status is.5, the probability of any point having an obstacle status is also.5, Thus, the probability of each of the sample points above is P(A i B i C i ) =.5(.5)(.5) =.125. The values of Y, the number of free links in the system, for each sample point are listed below. A link is free if both the points are free. Thus, a link from A to B is free if A is free and B is free. A link from B to C is free if B is free and C is free. Sample point Y Probability A F B F C F A F B F C O A F B O C F A F B O C O A O B F C F A O B F C O A O B O C F A O B O C O The probability distribution for Y is: Y Probability Random Variables and Probability Distributions 89

9 4.36 a. is discrete. It can take on only si values. b. This is a binomial distribution. c. p(0) = p(1) = p(2) = p(3) = p(4) = p(5) = 5 0 (.7)0 (.3) 5-0 = 5 1 (.7)1 (.3) 5-1 = 5 2 (.7)2 (.3) 5-2 = 5 (.7) 3 (.3) 5-3 = (.7)4 (.3) 5-4 = 5 5 (.7)5 (.3) 5-5 = 5! 0!5! (.7)0 (.3) 5 = 5! 1!4! (.7)1 (.3) 4 = ! 2!3! (.7)2 (.3) 3 = ! 3!2! (.7)3 (.3) 2 = ! 4!1! (.7)4 (.3) 1 = ! 5!0! (.7)5 (.3) 0 = (1)(.00243) = d. μ = np = 5(.7) = 3.5 σ = npq = 5(.7)(.3) = e. μ ± 2σ = 3.5 ± 2(1.0247) (1.4506, ) 90 Chapter 4

10 4.38 a. p(0) = p(1) = p(2) = p(3) = 3 0 (.3)0 (.7) 3-0 = 3 1 (.3)1 (.7) 3-1 = 3 2 (.3)2 (.7) 3-2 = 3 (.3) 3 (.7) 3-3 = 3 3! 0!3! (.3)0 (.7) 3 = ! 1!2! (.3)1 (.7) 2 =.441 5! 2!1! (.3)2 (.7) 1 =.189 5! 3!0! (.3)3 (.7) 0 =.027 (1)(.7) 3 =.343 p() a. P( = 2) = P( 2) P( 1) = =.121 (from Table II, Appendi B) b. P( 5) =.034 c. P( > 1) = 1 P( 1) = =.081 d. P( < 10) = P( 9) = 0 e. P( 10) = 1 P( 9) = =.998 f. P( = 2) = P( 2) P( 1) = = a. We will check the 5 characteristics of a binomial random variable. 1. The eperiment consists of n = 200 identical trials. 2. There are only two possible outcomes on each trial. Let S = young adult owns a mobile phone with internet access and F = young adult does not own a mobile phone with internet access. 3. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) =.20. q = 1 p = 1.20 = The trials are independent. 5. The binomial random variable is the number of young adults in 200 trials that own a mobile phone with internet access. Thus, is a binomial random variable. b. From the eercise, p =.20. For any young adult, the probability that they own a mobile phone with internet access is.20. c. μ = E ( ) = np= 200(.20) = 40. On the average, for every 200 young people surveyed, 40 will own mobile phones with internet access. Random Variables and Probability Distributions 91

11 4.44 a. We will check the 5 characteristics of a binomial random variable. 1. The eperiment consists of n = 5 identical trials. We have to assume that the number of bottled water brands is large. 2. There are only 2 possible outcomes for each trial. Let S = brand of bottled water used tap water and F = brand of bottled water did not use tap water. 3. The probability of success (S) is the same from trial to trial. For each trial, p = P(S) =.25 and q = 1 p = = The trials are independent. 5. The binomial random variable is the number of brands in the 5 trials that used tap water. If the total number of brands of bottled water is large, then the above characteristics will be basically true. Thus, is a binomial random variable. b. The formula for the probability distribution for is for = 1, 2, 3, 4, 5. 5 p ( ) =.25 (.75) 5, c. d ! 2 3 P ( = 2) =.25 (.75) = = !3! 5 5 P ( 1) = P ( = 0) + P ( = 1) =.25 (.75) +.25 (.75) 0 1 5! 0 5 5! 1 4 = = = !5! 1!4! a. In order for to be a binomial random variable, the n trials must be identical. We can assume that the process of selecting of a worker is identical from trial to trial. There are two possible outcomes - a worker missed work due to a back injury or not. The probability of success must be the same from trial to trial. We can assume that the probability of missing work due to a back injury is constant. The trials must be independent of each other. We can assume that the outcome of one trials will not affect the outcome of any other. Thus, is a binomial random variable. b. From the information given in the problem, the estimate of p is.40. c. The mean is μ = E() = np = 10(.40) = 4. The standard deviation is σ = np(1 p) = 10(.40)(.60) = = d. Using Table II, Appendi B, with n = 10 and p =.40, P( = 1) = P( 1) P( 0) = =.040 P( > 1) = 1 P( 1) = = Chapter 4

12 4.48 Let = number of packets observed by a network sensor in 150 trials. Then has an approimate binomial distribution with n = 150 and p =.001. The virus will be detected if at least 1 packets is observed ! 150 P ( 1) = 1 P ( = 0) = (.999) = = = !150! 4.50 a. We must assume that the trials are identical, the probability of success is constant from trial to trial, and the trials are independent of each other. b. From the problem, we estimate p to be.20. Using Table II, Appendi B, with n = 25 and p =.20, P( 10) =.994 c. E() = np = 25(.20) = 5 σ = np (1 p ) = 25(.20)(.80) = 4 = 2 d. μ ± 2σ 5 ± 2(2) 5 ± 4 (1, 9) e. Using Table II, Appendi B, with n = 25 and p =.20, P(1 < < 9) = P( 8) P( 1) = = Assuming the supplier's claim is true, μ = np = 500(.001) =.5 σ = 500(.001)(.999) npq = = = If the supplier's claim is true, we would only epect to find.5 defective switches in a sample of size 500. Therefore, it is not likely we would find 4. Based on the sample, the guarantee is probably inaccurate. Note: μ 4.5 z = = = 4.95 σ.707 This is an unusually large z-score a. For this test, n = 20 and p =.10. Then is a binomial random variable with n = 20 and p =.10. Using Table II, Appendi, with n = 20 and p =.10, P( 1) =.392 Random Variables and Probability Distributions 93

13 b. For the eperiment in part a, the level of confidence is 1 P( 1) = =.608. Since this value is not close to 1, this would not be an acceptable level. c. Suppose we increased n from 20 to 25. Using Table II, Appendi B, with n = 25 and p =.10, P( 1) =.271. This value is smaller than the value found in part a. Now, suppose we keep n = 20, but change K to 0 instead of 1. Using Table II, Appendi B, with n = 20 and p =.10, P( 0) =.122. This value is again, smaller than the value found in part a. d. Suppose we let K = 0. Now, we need to find n such that the level of confidence.95, which means that P( = 0).05. n 0 n 0 P ( = 0) =.1 (.9).05 0 n! n !n! n.9.05 n ln(.9 ) ln(.05) nln(.9) ln(.05) ln(.05) n = = 28.4 ln(.9) Thus, if K = 0, then we need a sample size of 28 to get a level of confidence of at least.95. Now, suppose K = 1. Now, we need to find n such that the level of confidence is at least.95, which means that P( 1).05. n 0 n 0 n 1 n 1 P ( 1) = P ( = 0) + P ( = 1) =.1 (.9) +.1 (.9) n! n n! 1 n !n! 1(! n 1)! n n 1 1 n n ( n) ln(.05) From here, we will use trial and error. 94 Chapter 4

14 For n = 30, (.9+.1(30)) =.1837 n.9 n-1 (.9+.1n) (.9+.1(30)) = (.9+.1(40)) = (.9+.1(45)) = (.9+.1(46)) = μ = λ = 1.5 Thus, for K = 1, we would need a sample size of 46 to get a level of confidence of at least.95. Using Table III of Appendi B: a. P( 3) =.934 b. P( 3) = 1 P( 2) = =.191 c. P( = 3) = P( 3) P( 2) = =.125 d. P( = 0) =.223 e. P( > 0) = 1 P( = 0) = =.777 f. P( > 6) = 1 P( 6) = = a. To graph the Poisson probability distribution with λ = 5, we need to calculate p() for = 0 to 15. Using Table III, Appendi B, p(0) =.007 p(1) = P( 1) P( 0) = =.033 p(2) = P( 2) P( 1) = =.085 p(3) = P( 3) P( 2) = =.140 p(4) = P( 4) P( 3) = =.175 p(5) = P( 5) P( 4) = =.176 p(6) = P( 6) P( 5) = =.146 p(7) = P( 7) P( 6) = =.105 p(8) = P( 8) P( 7) = =.065 p(9) = P( 9) P( 8) = =.036 p(10) = P( 10) P( 9) = =.018 p(11) = P( 11) P( 10) = =.009 p(12) = P( 12) P( 11) = =.003 p(13) = P( 13) P( 12) = =.001 p(14) = P( 14) P( 13) = =.001 p(15) = P( 15) P( 14) = =.000 Random Variables and Probability Distributions 95

15 The graph is shown at right: b. μ = λ = 5 σ = λ = 5 = μ ± 2σ 5 ± 2(2.2361) 5 ± (.5278, ) c. P(.5278 < < ) = P(1 9) = P( 9) P( = 0) = = a. E() = μ = λ = 6 σ = λ = 6 = b. μ 1 6 z = = = σ c. Using Table III, Appendi B, with λ = 6, P( 10) = a. In the problem, it is stated that E() =.03. This is also the value of λ. σ 2 = λ =.03 b. The eperiment consists of counting the number of deaths or missing persons in a threeyear interval. We must assume that the probability of a death or missing person in a three-year period is the same for any three-year period. We must also assume that the number of deaths or missing persons in any three-year period is independent of the number of deaths or missing persons in any other three-year period. 96 Chapter 4

16 c. P( = 1) = P( = 0) = 1 -λ λ e.03 e = = ! 1! 0 -λ λ e.03 e = = ! 0! 4.64 a. Using Table III and λ = 6.2, P( = 2) = P( 2) P( 1) = =.039 P( = 6) = P( 6) P( 5) = =.160 P( = 10) = P( 10) P( 9) = =.047 b. The plot of the distribution is: c. μ = λ = 6.2, σ = λ = 6.2 = μ ± σ 6.2 ± 2.49 (3.71, 8.69) μ ± 2σ 6.2 ± 2(2.49) 6.2 ± 4.98 (1.22, 11.18) μ ± 3σ 6.2 ± 3(2.49) 6.2 ± 7.47 ( 1.27, 13.67) See the plot in part b. d. First, we need to find the mean number of customers per hour. If the mean number of customers per 10 minutes is 6.2, then the mean number of customers per hour is 6.2(6) = 37.2 = λ. μ = λ = 37.2 and σ = λ = 37.2 = μ ± 3σ 37.2 ± 3(6.099) 37.2 ± (18,903, ) Using Chebyshev's Rule, we know at least 8/9 or 88.9% of the observations will fall within 3 standard deviations of the mean. The number 75 is way beyond the 3 standard deviation limit. Thus, it would be very unlikely that more than 75 customers entered the store per hour on Saturdays. Random Variables and Probability Distributions 97

17 4.66 Let = number of minor flaws in one square foot of a door's surface. Then has a Poisson distribution with λ =.5. μ= λ =.5, using Table III, Appendi B: P(fail inspection) = P(2 or more minor flaws in the square foot inspected) = P( 2) = 1 P( 1) = =.090 P(pass inspection) = P( < 2) = P( 1) = If it takes eactly 5 minutes to wash a car and there are 5 cars in line, it will take 5(5) = 25 minutes to wash these 5 cars. Thus, for anyone to be in line at closing time, more than 1 car must arrive in the final ½ hour. In addition, if on average 10 cars arrive per hour, then an average of 5 cars will arrive per ½ hour (30 minutes). If we let = number of cars to arrive in ½ hour, then is a Poisson random variable with λ = 5. P( > 1) = 1 P( 1) = 1.04 =.96 (Using Table III, Appendi B) Since this probability is so big, it is very likely that someone will be in line at closing time..04 (20 45) 4.70 From Eercise 4.69, f() = 0 otherwise a. P(20 30) = (30 20)(.04) =.4 b. P(20 < < 30) = (30 20)(.04) =.4 c. P( 30) = (45 30)(.04) =.6 d. P( 45) = (45 45)(.04) = 0 e. P( 40) = (40 20)(.04) =.8 f. P( < 40) = (40 20)(.04) =.8 g. P(15 35) = (35 20)(.04) =.6 h. P( ) = ( )(.04) =.4 1 (3 7) 4.72 From Eercise 4.71, f() = 4 0 otherwise 1 a. P( a) =.6 (7 a) 4 =.6 7 a = 2.4 a = Chapter 4

18 1 b. P( a) =.25 (a 3) 4 =.25 a 3 = 1 a = 4 1 c. P( a) = 1 (a 3) 4 = 1 a 3 = 4 a = 7 For any value of a 7, P( a) = 1. Thus, a 7. 1 d. P(4 a) =.5 (a 4) 4 =.5 a 4 = 2 a = μ = σ = c + d = 10 c + d = 20 c = 20 - d 2 d - c = 1 d c = Substituting, d (20 d) = 12 2d 20 = 12 2d = d = 2 d = Since c + d = 20 c = 20 c = f() = (c d) d c = = =.289 d c ( ) Therefore, f() = 0 otherwise The graph of the probability distribution for is given here. Random Variables and Probability Distributions 99

19 4.76 a. For this problem, c = 0 and d = = (0 1) f() = d c otherwise c + d μ = = = σ 2 ( d c) (1 0) = = = 1 12 =.0833 b. P(.2 < <.4) = (.4.2)(1) =.2 c. P( >.995) = (1.995)(1) =.005. Since the probability of observing a trajectory greater than.995 is so small, we would not epect to see a trajectory eceeding a. For layer 2, let = amount loss. Since the amount of loss is random between.01 and.05 million dollars, the uniform distribution for is: f() = 1 d c (c d) = = d c = 25 Therefore, f() = 25 (.01.05) 0 otherwise A graph of the distribution looks like the following: μ = c + d = 2 2 =.03 σ = d c = =.0115, σ 2 = (.0115) 2 = The mean loss for layer 2 is.03 million dollars and the variance of the loss for layer 2 is million dollars squared. 100 Chapter 4

20 b. For layer 6, let = amount loss. Since the amount of loss is random between.50 and 1.00 million dollars, the uniform distribution for is: f() = d 1 c (c d) = = d c = 2 Therefore, f() = 2 ( ) 0 otherwise A graph of the distribution looks like the following: μ = c + d = = σ = d c = =.1443, σ 2 = (.1443) 2 = The mean loss for layer 6 is.75 million dollars and the variance of the loss for layer 6 is.0208 million dollars squared. c. A loss of $10,000 corresponds to =.01. P( >.01) = 1 A loss of $25,000 corresponds to = P( <.025) = (Base)(Height) = ( c) d c = ( ) =.015(25) =.375 Random Variables and Probability Distributions 101

21 d. A loss of $750,000 corresponds to =.75. A loss of $1,000,000 corresponds to = 1. 1 P(.75 < < 1) = (Base)(Height) = (d - ) d c = ( ) =.25(2) =.5 A loss of $900,000 corresponds to = P( >.9) = (Base)(Height) = (d ) d c = ( ) =.10(2) =.20 P( =.9) = Let = cycle availability, where has a uniform distribution on the interval from 0 to a. c+ d 0+ 1 Mean = μ = = = d c 1 0 Standard deviation = σ = = = The 10 th percentile is that value of such that 10% of all observations are below it. Let K 1 = 10 th percentile. P( K 1 ) = (K 1 0)(1 0) = K 1 =.10 The lower quartile is that value of such that 25% of all observations are below it. Let K 2 = 25 th percentile. P( K 2 ) = (K 2 0)(1 0) = K 2 =.25 The UPPER quartile is that value of such that 75% of all observations are below it. Let K 3 = 75 th percentile. P( K 3 ) = (K 3 0)(1 0) = K 3 = Chapter 4

22 b. μ = σ = c + d 0+ 1 = = d c 1 0 = =.289 σ 2 = = c. P(p >.95) = (1.95)(1) =.05 P(p <.95) = (.95 0)(1) =.95 d. The analyst should use a uniform probability distribution with c =.90 and d = = = = 20 (.90 p.95) f(p) = d c otherwise 4.84 Table IV in the tet gives the area between z = 0 and z = z 0. In this eercise, the answers may thus be read directly from the table by looking up the appropriate z. a. P(0 < z < 2.0) =.4772 b. P(0 < z < 3.0) =.4987 c. P(0 < z < 1.5) =.4332 d. P(0 < z <.80) = a. P( 1 z 1) = A 1 + A 2 = =.6826 b. P( 2 z 2) = A 1 + A 2 = =.9544 c. P( 2.16 < z 0.55) = A 1 + A 2 = =.6934 Random Variables and Probability Distributions 103

23 d. P(.42 < z < 1.96) = P(.42 z 0) + P(0 z 1.96) = A 1 + A 2 = =.6378 e. P(z 2.33) = P( 2.33 z 0) + P(z 0) = A 1 + A 2 = =.9901 f. P(z < 2.33) = P(z 0) + P(0 z 2.33) = A 1 + A 2 = = a. P(z = 1) = 0, since a single point does not have an area. b. P(z 1) = P(z 0) + P(0 < z 1) = A 1 + A 2 = =.8413 (Table IV, Appendi B) c. P(z < 1) = P(z 1) =.8413 (Refer to part b.) d. P(z > 1) = 1 P(z 1) = =.1587 (Refer to part b.) 4.90 Using Table IV, Appendi B: a. P(z z 0 ) =.05 A 1 =.5.05 =.4500 Looking up the area.4500 in Table IV gives z 0 = b. P(z z 0 ) =.025 A 1 = =.4750 Looking up the area.4750 in Table IV gives z 0 = c. P(z z 0 ) =.025 A 1 = =.4750 Looking up the area.4750 in Table IV gives z = Since z 0 is to the left of 0, z 0 = Chapter 4

24 d. P(z z 0 ) =.10 A 1 =.5.1 =.4 Looking up the area.4000 in Table IV gives z 0 = e. P(z > z 0 ) =.10 A 1 =.5.1 =.4 z 0 = 1.28 (same as in d) 4.92 a. z = 1 b. z = 1 c. z = 0 d. z = 2.5 e. z = Using Table IV of Appendi B: a. To find the probability that assumes a value more than 2 standard deviations from μ: P( < μ 2σ) + P( > μ + 2σ) = P(z < 2) + P(z > 2) = 2P(z > 2) = 2( ) = 2(.0228) =.0456 To find the probability that assumes a value more than 3 standard deviations from μ: P( < μ 3σ) + P( > μ + 3σ) = P(z < 3) + P(z > 3) = 2P(z > 3) = 2( ) = 2(.0013) =.0026 b. To find the probability that assumes a value within 1 standard deviation of its mean: P(μ σ < < μ + σ) = P( 1 < z < 1) = 2P(0 < z < 1) = 2(.3413) =.6826 Random Variables and Probability Distributions 105

25 To find the probability that assumes a value within 2 standard deviations of μ: P(μ 2σ < < μ + 2σ) = P( 2 < z < 2) = 2P(0 < z < 2) = 2(.4772) =.9544 c. To find the value of that represents the 80th percentile, we must first find the value of z that corresponds to the 80th percentile. P(z < z 0 ) =.80. Thus, A 1 + A 2 =.80. Since A 1 =.50, A 2 = =.30. Using the body of Table IV, z 0 =.84. To find, we substitute the values into the z-score formula: z = μ σ = 10 =.84(10) = To find the value of that represents the 10th percentile, we must first find the value of z that corresponds to the 10th percentile. P(z < z 0 ) =.10. Thus, A 1 = =.40. Using the body of Table IV, z 0 = To find, we substitute the values into the z-score formula: z = μ σ = 10 = 1.28(10) = The random variable has a normal distribution with μ = 50 and σ = 3. a. P( 0 ) =.8413 So, A 1 + A 2 =.8413 Since A 1 =.5, A 2 = = Look up the area.3413 in the body of Table IV, Appendi B; z 0 = Chapter 4

26 To find 0, substitute all the values into the z-score formula: z = μ σ = 3 0 = (1.0) = 53 b. P( > 0 ) =.025 So, A = =.4750 Look up the area.4750 in the body of Table IV, Appendi B; z 0 = To find 0, substitute all the values into the z-score formula: c. P( > 0 ) =.95 z = μ σ = 3 0 = (1.96) = So, A 1 + A 2 =.95. Since A 2 =.5, A 1 =.95.5 = Look up the area.4500 in the body of Table IV, Appendi B; (since it is eactly between two values, average the z-scores). z To find 0, substitute into the z-score formula: z = μ σ = 3 0 = 50 3(1.645) = d. P(41 < 0 ) =.8630 z = μ = = 3 σ 3 A 1 = P(41 μ) = P( 3 z 0) = P(0 z 3) =.4987 A 1 + A 2 =.8630, since A 1 =.4987, A 2 = = Look up.3643 in the body of Table IV, Appendi B; z 0 = 1.1. Random Variables and Probability Distributions 107

27 To find 0, substitute into the z-score formula: e. P( < 0 ) =.10 z = μ σ = 3 0 = (1.1) = 53.3 So A = =.4000 Look up area.4000 in the body of Table IV, Appendi B; z 0 = Since z 0 is to the left of 0, z 0 = To find 0, substitute all the values into the z-score formula: f. P( > 0 ) =.01 z = μ σ = 3 0 = (3) = So A = =.4900 Look up area.4900 in the body of Table IV, Appendi B; z 0 = To find 0, substitute all the values into the z-score formula: z = μ σ = 3 0 = (3) = a. Using Table IV, Appendi B, P ( > 0) = P z> = Pz ( > 0.526) 10 =.5 + P( 0.53 < z< 0) = =.7019 b P(5 < < 15) = P < z< = P( < z< 0.974) = P(.03 < z< 0) + P(0 < z<.97) = = Chapter 4

28 c. d P ( < 1) = P z< = Pz ( < 0.426) 10 =.5 P( 0.43 < z< 0) = = P ( 25) = P z = Pz ( 3.026) 10 =.5 P( 3.03 z< 0) = =.0012 Since the probability of seeing a win percentage of -25% or anything more unusual is so small (p =.0012), we would conclude that the average casino win percentage is not 5.26% Let = driver s head injury rating. The random variable has a normal distribution with μ = 605 and σ = 185. Using Table IV, Appendi B, a. b. c. d P(500 < < 700) = P < z< = P( 0.57 < z< 0.51) = P( 0.57 < z < 0) + P(0 < z < 0.51) = = P(400 < < 500) = P < z< = P( 1.11< z< 0.57) = P( 1.11< z< 0) P( 0.57 < z< 0) = = P ( < 850) = P z< = Pz ( < 1.32) =.5 + P(0 < z< 1.32) 185 = = , P ( > 1,000) = P z> = Pz ( > 2.14) =.5 P(0 < z< 2.14) 185 = = a. Let = crop yield. The random variable has a normal distribution with μ = 1,500 and σ = ,600-1,500 P( < 1,600) = P z < 250 (Using Table IV) = P(z <.4) = =.6554 Random Variables and Probability Distributions 109

29 b. Let 1 = crop yield in first year and 2 = crop yield in second year. If 1 and 2 are independent, then the probability that the farm will lose money for two straight years is: 1,600 1,500 P( 1 < 1,600) P( 2 < 1,600) = P z 1 < 250 P z 2 1,600 1,500 < 250 = P(z 1 <.4) P(z 2 <.4) = ( )( ) =.6554(.6554) =.4295 (Using Table IV) [1,500 2 σ] 1,500 [1,500 2 σ] 1,500 c. P(1,500 2σ 1, σ) = P z + σ σ = P( 2 z 2) = 2P(0 z 2) = 2(.4772) =.9544 (Using Table IV) Let = wage rate. The random variable is normally distributed with μ = 16 and σ = Using Table IV, Appendi B, a. b P ( > 17.30) = P z> = Pz ( > 1.04) 1.25 =.5 P(0 < z< 1.04) = = P ( > 17.30) = P z> = Pz ( > 1.04) 1.25 =.5 P(0 < z< 1.04) = =.1492 c. P( η) = P( η) =.5 Thus, μ = η = 16. (Recall from section 2.4 that in a symmetric distribution, the mean equals the median.) a. The contract will be profitable if total cost,, is less than $1,000,000. P( < 1,000,000) = 1,000, ,000 P z < 170,000 = P(z <.88) = =.8106 b. The contract will result in a loss if total cost,, eceeds 1,000,000. P( > 1,000,000) = 1 P( < 1,000,000) = = Chapter 4

30 c. P( < R) =.99. Find R. P( < R) = R 850,000 P z < 170,000 = P(z < z 0) =.99 A 1 =.99.5 =.4900 Looking up the area.4900 in Table IV, z 0 = 2.33 R 850,000 R 850, 000 z 0 = 2.33 = 170, , 000 R = 2.33(170,000) + 850,000 = $1,246, a. Let = quantity injected per container. The random variable has a normal distribution with μ = 10 and σ =.2. P( < 10) = P( 10) = P z < P z.2 = P(z < 0.0) =.5 = P(z 0.0) =.5 b. Since the container needed to be reprocessed, it cost $10. Upon refilling, it contained units with a cost of 10.60($20) = $212. Thus, the total cost for filling this container is $10 + $212 = $222. Since the container sells for $230, the profit is $230 $222 = $8. c. Let = quantity injected per container. The random variable has a normal distribution with μ = and σ =.2. The epected value of is E() = μ = The cost of a container with units is 10.10($20) = $202. Thus, the epected profit would be the selling price minus the cost or $230 $202 = $ a. If z is a standard normal random variable, Q L = z L is the value of the standard normal distribution which has 25% of the data to the left and 75% to the right. Find z L such that P(z < z L ) =.25 A 1 = =.25. Look up the area A 1 =.25 in the body of Table IV of Appendi B; z L =.67 (taking the closest value). If interpolation is used,.675 would be obtained. Random Variables and Probability Distributions 111

31 Q U = z U is the value of the standard normal distribution which has 75% of the data to the left and 25% to the right. Find z U such that P(z < z U ) =.75 A 1 + A 2 = P(z 0) + P(0 z z U ) =.5 + P(0 z z U ) =.75 Therefore, P(0 z z U ) =.25. Look up the area.25 in the body of Table IV of Appendi B; z U =.67 (taking the closest value). b. Recall that the inner fences of a bo plot are located 1.5(Q U - Q L ) outside the hinges (Q L and Q U ). To find the lower inner fence, Q L 1.5(Q U Q L ) = (.67 (.67)) = (1.34) = ( 2.70 if z L =.675 and z U = +.675) The upper inner fence is: Q U + 1.5(Q U - Q L ) = (.67 (.67)) = (1.34) = 2.68 (+2.70 if z L =.675 and z U = +.675) c. Recall that the outer fences of a bo plot are located 3(Q U Q L ) outside the hinges (Q L and Q U ). To find the lower outer fence, Q L 3(Q U Q L ) =.67 3(.67 (.67)) =.67 3(1.34) = ( if z L =.675 and z U = +.675) The upper outer fence is: Q U + 3(Q U Q L ) = (.67 (.67)) = (1.34) = 4.69 (4.725 if z L =.675 and z U = +.675) 112 Chapter 4

32 d. P(z < 2.68) + P(z > 2.68) = 2P(z > 2.68) = 2( ) (Table IV, Appendi B) = 2(.0037) =.0074 (or 2( ) =.0070 if 2.70 and 2.70 are used) P(z < 4.69) + P(z > 4.69) = 2P(z > 4.69) 2( ) 0 e. In a normal probability distribution, the probability of an observation being beyond the inner fences is only.0074 and the probability of an observation being beyond the outer fences is approimately zero. Since the probability is so small, there should not be any observations beyond the inner and outer fences. Therefore, they are probably outliers a. IQR = Q U Q L = = 123 b. IQR/s = 123/95 = c. Yes. Since IQR is approimately 1.3, this implies that the data are approimately normal a. Using MINITAB, the stem-and-leaf display is: Stem-and-leaf of X N = 28 Leaf Unit = Since the data do not form a mound-shape, it indicates that the data may not be normally distributed. b. Using MINITAB, the descriptive statistics are: Variable N Mean Median TrMean StDev SE Mean X Variable Minimum Maimum Q1 Q3 X The standard deviation is Random Variables and Probability Distributions 113

33 c. Using the printout from MINITAB in part b, Q L = 3.35, and Q U = The IQR = Q U Q L = = 4.7. If the data are normally distributed, then IQR/s 1.3. For this data, IQR/s = 4.7/2.765 = This is a fair amount larger than 1.3, which indicates that the data may not be normally distributed. d. Using MINITAB, the normal probability plot is: The data at the etremes are not particularly on a straight line. This indicates that the data are not normally distributed From the normal probability plot, it appears that the data may not be normal. The points with small observed values and the points with large observed values do not fall on the straight line. This implies that the data may not be from a normal distribution a. We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the fish weights is: Frequency Weight From the histogram, the data appear to be fairly mound-shaped. This indicates that the data may be normal. 114 Chapter 4

34 Net, we look at the intervals ± s, ± 2 s, ± 3s. If the proportions of observations falling in each interval are approimately.68,.95, and 1.00, then the data are approimately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: Weight Variable N Mean Median TrMean StDev SE Mean Weight Variable Minimum Maimum Q1 Q3 Weight ± s ± (673.2, 1,426.2) 98 of the 144 values fall in this interval. The proportion is.68. This is eactly the.68 we would epect if the data were normal. ± 2s ± 2(376.5) ± 753 (296.7, ) 140 of the 144 values fall in this interval. The proportion is.97. This is somewhat larger than the.95 we would epect if the data were normal. ± 3s ± 3(376.5) ± ( 79.8, ) 143 of the 144 values fall in this interval. The proportion is.993. This is close to the 1.00 we would epect if the data were normal. From this method, it appears that the data are normal. Net, we look at the ratio of the IQR to s. IQR = Q U Q L = = IQR = = 1.22 This is close to the 1.3 we would epect if the data were normal. s This method indicates the data are normal. Finally, using MINITAB, the normal probability plot is: Normal Probability Plot for Weight ML Estimates - 95% CI ML Estimates Percent Mean StDev Goodness of Fit AD* Data Since the data form a fairly straight line, the data appear to be normal. Random Variables and Probability Distributions 115

35 From the 4 different methods, all indications are that the fish weight data are approimately normal. b. We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the fish DDT levels is: Frequency DDT 1000 From the histogram, the data appear to be skewed to the right. This indicates that the data may not be normal. Net, we look at the intervals ± s, ± 2 s, ± 3s. If the proportions of observations falling in each interval are approimately.68,.95, and 1.00, then the data are approimately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: DDT Variable N Mean Median TrMean StDev SE Mean DDT Variable Minimum Maimum Q1 Q3 DDT ± s ± ( 74.03, ) 138 of the 144 values fall in this interval. The proportion is.96. This is much greater than the.68 we would epect if the data were normal. ± 2s ± 2(98.38) ± ( , ) 142 of the 144 values fall in this interval. The proportion is.986 This is much larger than the.95 we would epect if the data were normal. ± 3s ± 3(98.38) ± ( , ) 142 of the 144 values fall in this interval. The proportion is.986. This is somewhat lower than the 1.00 we would epect if the data were normal. From this method, it appears that the data are not normal. Net, we look at the ratio of the IQR to s. IQR = Q U Q L = = Chapter 4

36 IQR 9.67 = = This is much smaller than the 1.3 we would epect if the data were s normal. This method indicates the data are not normal. Finally, using MINITAB, the normal probability plot is: Normal Probability Plot for DDT ML Estimates - 95% CI Percent ML Estimates Mean StDev Goodness of Fit AD* Data Since the data do not form a straight line, the data are not normal. From the 4 different methods, all indications are that the fish DDT level data are not normal We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the sanitation scores is: Histogram of SCORE Frequency SCORE Random Variables and Probability Distributions 117

37 From the histogram, the data appear to be skewed to the left. This indicates that the data are not normal. Net, we look at the intervals ± s, ± 2 s, ± 3s. If the proportions of observations falling in each interval are approimately.68,.95, and 1.00, then the data are approimately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: DDT Variable N Mean Median TrMean StDev SE Mean DDT Variable Minimum Maimum Q1 Q3 DDT ± s ± (90.086, ) 137 of the 169 values fall in this interval. The proportion is.81. This is much larger than the.68 we would epect if the data were normal. ± 2s ± 2(4.825) ± 9.65 (85.261, ) 165 of the 169 values fall in this interval. The proportion is.98. This is somewhat larger than the.95 we would epect if the data were normal. ± 3s ± 3(4.825) ± (80.436, ) 166 of the 169 values fall in this interval. The proportion is.982. This is somewhat smaller than the 1.00 we would epect if the data were normal. From this method, it appears that the data are not normal. Net, we look at the ratio of the IQR to s. IQR = Q U Q L = = 5. IQR 5 = = This is much smaller than the 1.3 we would epect if the data were s normal. This method indicates the data are not normal. Finally, using MINITAB, the normal probability plot is: Probability Plot of SCORE Normal - 95% C I Percent Mean StDev N 169 AD P-Value < SCORE Chapter 4

38 Since the data do not form a straight line, the data are not normal. From the 4 different methods, all indications are that the sanitation scores data are not normal We will look at the 4 methods for determining if the data are normal. First, we will look at a histogram of the data. Using MINITAB, the histogram of the tensile strength values is: Histogram of Strength Frequency Strength From the histogram, the data appear to be somewhat skewed to the left. This might indicate that the data are not normal. Net, we look at the intervals ± s, ± 2s, ± 3s. If the proportions of observations falling in each interval are approimately.68,.95, and 1.00, then the data are approimately normal. Using MINITAB, the summary statistics are: Descriptive Statistics: Strength Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 Strength Variable Maimum Strength ± s ± 7.91 (334.22, ) 8 of the 11 values fall in this interval. The proportion is.73. This is somewhat larger than the.68 we would epect if the data were normal. ± 2s ± 2(7.91) ± 9.65 (326.34, ) All 11 of the 11 values fall in this interval. The proportion is This is somewhat larger than the.95 we would epect if the data were normal. ± 3s ± 3(7.91) ± (318.43, ) Again, all 11 of the 11 values fall in this interval. The proportion is This is equal to the 1.00 we would epect if the data were normal. Random Variables and Probability Distributions 119

39 From this method, it appears that the data are quite normal. Net, we look at the ratio of the IQR to s. IQR = Q U Q L = = IQR 13.1 = = This is much larger than the 1.3 we would epect if the data were normal. s 7.91 This method indicates the data are not normal. Finally, using MINITAB, the normal probability plot is: Probability Plot of Strength Normal - 95% CI Percent Mean StDev N 11 AD P-Value Strength Since the data do form a fairly straight line, the data could be normal. From the 4 different methods, three of the four indicate that the data probably are not from a normal distribution a. In order to approimate the binomial distribution with the normal distribution, the interval μ ± 3σ np ± 3 npq should lie in the range 0 to n. When n = 25 and p =.4, np ± 3 npq 25(.4) ± 3 25(.4)(1.4) 10 ± ± (2.6515, ) Since the interval calculated does lie in the range 0 to 25, we can use the normal approimation. b. μ = np = 25(.4) = 10 σ 2 = npq = 25(.4)(.6) = 6 c. P( 9) = 1 P( 8) = =.726 (Table II, Appendi B) (9.5) 10 d. P( 9) P z 6 = P(z.61) = =.7291 (Using Table IV in Appendi B.) 120 Chapter 4

40 4.126 μ = np = 1000(.5) = 500, σ = npq = 1000(.5)(.5) = a. Using the normal approimation, P( > 500) ( ) 500 P z > = P(z >.03) = =.4880 (from Table IV, Appendi B) b. P(490 < 500) (490.5) 500 (500.5) 500 P z < = P(.66 z <.03) = =.2334 (from Table IV, Appendi B) c. P( > 550) ( ) 500 P z > = P(z > 3.19).5.5 = 0 (from Table IV, Appendi B) a. E() = μ = np = 350(.27) = b. c. σ 2 = σ = npq = = = μ z = = = 0.60 σ (.27)(.73) d. To see if the normal approimation is appropriate, we use: μ ± 3σ 94.5 ± 3(8.306) 94.5 ± (69.582, ) Since the interval lies in the range of 0 to 350, the normal approimation is appropriate. P ( 100) Pz ( 0.60) = =.2743 (Using Table IV, Appendi B) Let = number of white-collar employees in good shape who will develop stress related illnesses in a sample of 400. Then is a binomial random variable with n = 400 and p =.10. To see if the normal approimation is appropriate for this problem: np ± 3 npq 400(.1) ± 3 400(.1)(.9) 40 ± 18 (22, 58) Since this interval is contained in the interval 0, n = 400, the normal approimation is appropriate. (60 +.5) 40 P( > 60) P z > 6 = P(z > 3.42) = 0 Random Variables and Probability Distributions 121

41 4.132 a. For n = 100 and p =.01: μ ± 3σ np ± 3 npq 100(.01) ± 3 100(.01)(.99) 1 ± 3(.995) 1 ± ( 1.985, 3.985) Since the interval does not lie in the range 0 to 100, we cannot use the normal approimation to approimate the probabilities. b. For n = 100 and p =.5: μ ± 3σ np ± 3 npq 100(.5) ± 3 100(.5)(.5) 50 ± 3(5) 50 ± 15 (35, 65) Since the interval lies in the range 0 to 100, we can use the normal approimation to approimate the probabilities. c. For n = 100 and p =.9: μ ± 3σ np ± 3 npq 100(.9) ± 3 100(.9)(.1) 90 ± 3(3) 90 ± 9 (81, 99) Since the interval lies in the range 0 to 100, we can use the normal approimation to approimate the probabilities b. Let v = number of credit card users out of 100 who carry Visa. Then v is a binomial random variable with n = 100 and p v =.539. E(v) = np v = 100(.539) = Let d = number of credit card users out of 100 who carry Discover. Then d is a binomial random variable with n = 100 and p d =.040. E(d) = np d = 100(.040) = 4.0. c. To see if the normal approimation is valid, we use: μ ± 3σ npv ± 3 npvqv 100(.539) ± 3 100(.539)(.461) 53.9 ± 3(4.9848) 53.9 ± (38.946, ) Since the interval lies in the range 0 to 100, we can use the normal approimation to approimate the probability. (50.5) 53.9 Pv ( 50) P z = Pz (.88) = = Chapter 4

42 Let a = number of credit card users out of 100 who carry American Epress. Then a is a binomial random variable with n = 100 and p a =.132. To see if the normal approimation is valid, we use: μ ± 3σ npa ± 3 npaqa 100(.132) ± 3 100(.132)(.868) 13.2 ± 3(3.385) 13.2 ± (3.045, ) Since the interval lies in the range 0 to 100, we can use the normal approimation to approimate the probability. (50.5) 13.2 Pa ( 50) P z = Pz ( 10.72) = d. In order for the normal approimation to be valid, μ ± 3σ must lie in the interval (0, n). This check was done in part c for both portions of the question. In both cases, the normal approimation was justified a. If 80% of the passengers pass through without their luggage being inspected, then 20% will be detained for luggage inspection. The epected number of passengers detained will be: E() = np = 1,500(.2) = 300 b. For n = 4,000, E() = np = 4,000(.2) = 800 c. P( > 600) ( ) 800 P z > 4000(.2)(.8) = P(z > 7.89) = = E() = μ = p( ) = 1(.2) + 2(.3) + 3(.2) + 4(.2) + 5(.1) = = 2.7 E( ) = p( ) = 1.0(.04) + 1.5(.12) + 2.0(.17) + 2.5(.20) + 3.0(.20) + 3.5(.14) + 4.0(.08) + 4.5(.04) + 5.0(.01) = = The sampling distribution is approimately normal only if the sample size is sufficiently large or if the population being sampled from is normal a. μ = μ = 10, b. μ = μ = 100, c. μ = μ = 20, d. μ = μ = 10, σ = σ / n = 3/ 25 = 0.6 σ = σ / n = 25/ 25 = 5 σ = σ / n = 40/ 25 = 8 σ = σ / n = 100/ 25 = 20 Random Variables and Probability Distributions 123

43 4.148 a. μ = μ = 20, σ = σ / = 16 / 64 = 2 n b. By the Central Limit Theorem, the distribution of is approimately normal. In order for the Central Limit Theorem to apply, n must be sufficiently large. For this problem, n = 64 is sufficiently large. c. z = d. z = μ σ μ σ = = = = For this population and sample size, E( ) = μ = 100, σ = σ / n = 10/ 900 = 1/3 a. Approimately 95% of the time, will be within two standard deviations of the mean, i.e., 1 μ ± 2σ 100 ± ± 2 (99.33, ). Almost all of the time, the 3 sample mean will be within three standard deviations of the mean, i.e., μ ± 3σ 100 ± ± 1 (99, 101). 3 b. No more than three standard deviations, i.e., = 1 c. No, the previous answer only depended on the standard deviation of the sampling distribution of the sample mean, not the mean itself a. μ = μ = 98,500 b. σ 30,000 σ = = = 4, n 50 c. By the Central Limit Theorem, the sampling distribution of is approimately normal. d. μ 89,500 98,500 z = = = 2.12 σ 4, e. P ( > 89,500) = Pz ( > 2.12) = =.9830 (Using Table IV, Appendi B) 124 Chapter 4