Stat 118 - Chapter 6 Homework and Excel Lab

Choose the correct statements. (You may select more than one answer. Click the box with a check mark for the correct answer and click to empty the box for the wrong answer. Any boxes left with a question mark will be automatically graded as incorrect.) Check All That Apply A random variable is a quantitative or qualitative outcome which results from a chance experiment. A probability distribution includes the likelihood of each possible outcome or random variable. A probability distribution is the outcome of an experiment. A random variable represents the likelihood of an outcome.

A random variable is a quantitative or qualitative outcome which results from a chance experiment. A probability distribution includes the likelihood of each possible outcome or random variable.

Compute the mean and variance of the following probability distribution. (Round your answers to 2 decimal places.) X P(X) 4 0.10 8 0.30 12 0.25 16 0.35 MEAN: VARIANCE:

MEAN: 11.40 VARIANCE: 16.44

Keith's Florists has 16 delivery trucks, used mainly to deliver flowers and flower arrangements in the Greenville, South Carolina, area. Of these 16 trucks, seven have brake problems. A sample of six trucks is randomly selected. What is the probability that two of those tested have defective brakes?

Out of N = 16 trucks, K = 7 are defective and N-K=9 are fine. Let X = number of trucks that are defective in randomly selected n = 6 balls. P(X = a) = [ (a of K)*( n-a of N-K) ] / (n of N) P(2 of 7 defective brakes) ----> P(X = 2) P(X=2) = [(2 of 7)(6-2 of 9) )] / (6 of 16) Note: a = 2 K = 7 N = 16 n = 6 Use excel function HYPGEOM.DIST(a,n,K,N,cumulative) HYPGEOM.DIST(2,6,7,16,FALSE) P(X=2) = 0.3304 Use HYPEGEOMETRIC CALCULATOR ONLINE https://stattrek.com/online-calculator/hypergeometric.aspx Fill in: Population size = N Number of successes in population = K Sample size = n Number of successes in sample (x) = a

Assume that X is a Poisson random variable with μ = 20. Use the Excel's function options to find the following probabilities. μ = 20 in cell E8 P(X < 14) P(X >= 20) P(X = 25) P( 18 <= X <= 23)

P(X < 14) = P(X <= 13) =POISSON(13,E8, TRUE) 0.0661 P(X >= 20) =1 - POISSON(19,E8,TRUE) 0.5297 P(X = 25) =POISSON(25,E8,FALSE) 0.0446 P( 18 <= X <= 23) =POISSON(23,E8,TRUE) - POISSON(17,E8,TRUE) 0.4905

Assume that X is a Poisson random variable with μ = 15. Use the Excel's function options to find the following probabilities. μ = 15 in cell E8 P(X <= 10) P(X = 13) P(X > 15) P( 12 <= X <= 18)

P(X <= 10) =POISSON(10,E8,TRUE) 0.1185 P(X = 13) =POISSON(13,E8,FALSE) 0.0956 P(X > 15) = 1 - POISSON(15,E8,TRUE) 0.4319 P( 12 <= X <= 18) =POISSON(18,E8,TRUE) - POISSON(11,E8,TRUE) 0.6347

Let X represent a binomial random variable with n=200 and p=0.77. Use Excel's function options to find the following probabilities. n = 200 (cell E8) p = 0.77 (cell E10) P(X <= 150) P(X>160) P(155 <= X <= 165) P(X=160)

P(X <= 150) =BINOM.DIST(150,E8,E10,TRUE) 0.2750 P(X>160) = 1 - BINOM.DIST(160,E8,E10,TRUE) 0.1366 P(155 <= X <= 165) P(a <= X <= b) =BINOM.DIST(b,n,p,TRUE) - BINOM.DIST(a-1,n,p,TRUE) =BINOM.DIST(165,E8,E10,TRUE) - BINOM.DIST(154,E8,E10,TRUE) 0.4487 P(X=160) =BINOM.DIST(160,E8,E10,FALSE) 0.0416

Let X represent a binomial random variable with n=150 and p=0.36. Use Excel's function options to find the following probabilities. P(X <= 50) P(X = 40) P(X > 60) P(X>=55)

P(X <= 50) =BINOM.DIST(50,E8,E10,TRUE) 0.2776 P(X = 40) =BINOM.DIST(40,E8,E10,FALSE) 0.0038 P(X > 60) = 1 - BINOM.DIST(60,E8,E10,TRUE) 0.1348 P(X>=55) = 1 - BINOM.DIST(54,E8,E10,TRUE) 0.4630

Assume that X is a hypergeometric random variable with N=50, S = 20, and n=5. Use Excel's function options to find the following probabilities. N = 50 (cell C9) S = 20 (cell C10) n = 5 (cell C11) P(X = 2) P(X >= 2) P(X <=3)

P(X = 2) =HYPGEOM.DIST(2,C11,C10,C9,FALSE) =HYPGEOM.DIST(2,5,20,50,FALSE) 0.3641 P(X >= 2) =1-HYPGEOM.DIST(1,C11,C10,C9,TRUE) 0.6741 P(X <=3) =HYPGEOM.DIST(3,C11,C10,C9,TRUE) 0.9241

Industry standards suggest that 9% of new vehicles require warranty service within the first year. Jones Nissan in Sumter, South Carolina, sold 11 Nissans yesterday. (Round your mean answer to 2 decimal places and the other answers to 4 decimal places.) a. What is the probability that none of these vehicles requires warranty service? b. What is the probability exactly one of these vehicles requires warranty service? c. Determine the probability that exactly two of these vehicles require warranty service. d. Compute the mean and standard deviation of this probability distribution.

P(X = x) = nCx * p^x * (1 - p)^(n - x) n = 11 p = 0.09 a) 0.3544 P(X=0) = 11C0 * 0.09^0 * (1-.09)^(11-0) b) 0.3855 P(X=1) = 11C1 * 0.09^1 * (1-.09)^(11-1) P(X=1) = 11 * 0.09 * 0.91^10 c) 0.1906 P(X=2) 11C2 * 0.09^2 * (1-.09)^(11-2) P(X=2) = 55 * 0.0081 * 0.91^9 d) MEAN: E(X) = np 11 * 0.09 = 0.99 VARIANCE: V(X) = np(1-p) 11 * 0.09 * (1-0.09) 11*0.09*0.91 = 0.9009 STANDARD DEVIATION: σ = square root of V(X) = 0.9492

Refer to the Baseball 2018 data. Compute the mean number of home runs per game. To do this, first find the mean number of home runs per team for 2018. Next, divide this value by 165 (a season comprises 165 games). Then multiply by 2 because there are two teams in each game. Use the Poisson distribution to estimate the number of home runs that will be hit in a game. (Round your answers to 4 decimal places.) a. Find the probability that there are no home runs in a game. b. Find the probability that there are two home runs in a game. c. Find the probability that there are at least four home runs in a game.

STEP 1: Find the MEAN Mean of HR per Team = Total of HR Column / Number of teams Using Excel, calculate Total of HR by =SUM(D4:D33) = 5,585 There are 30 teams total 5,585 / 30 = 186.1667 Mean of HR per Game = [Mean of HR per Team / number of Games] * 2 [186.1667 / 165 games] * 2 = 2.2566 STEP 2: Use Excel =POISSON Funtion Note: μ = 2.2566 P(X=x) = [ e^-x * μ^x ] / x! P(X=x) =POISSON(x, μ,FALSE) a) 0.1047 P(X=0) =POISSON(0, 2.2566, FALSE) b) 0.2666 P(X=2) =POISSON(2, 2.2566, FALSE) c) 0.1919 P(at least 4) P(X>=4) =1-POISSON(3, 2.2566, TRUE) Use POISSON DISTRIBUTION online Calculator https://stattrek.com/onlinecalculator/poisson.aspx Poisson random variable (x) Average rate of success: mean NO ONE: Poisson Probability: P(X = 0) TWO: Poisson Probability: P(X = 2) AT LEAST 4: Cumulative Probability: P(X > 4)

In a binomial situation, n = 4 and π = 0.45. Find the probabilities for all possible values of the random variable, x. (Round your answers to 4 decimal places.)

Using EXCEL =ROUND(BINOM.DIST(1,4,0.45,FALSE),4) use ROUND function to round to 4 decimal places Here, n = 4, p = 0.45, (1 - p) = 0.55 and x = 0 As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x) We need to calculate P(X = 0) P(X = 0) = 4C0 * 0.45^0 * 0.55^4 P(X = 0) = 0.0915 P(X = 1) = 4C1 * 0.45^1 * 0.55^3 P(X = 1) = 0.2995

A recent study conducted by Penn, Shone, and Borland, on behalf of LastMinute.com, revealed that 52% of business travelers plan their trips less than two weeks before departure. The study is to be replicated in the tri-state area with a sample of 12 frequent business travelers. a. Develop a probability distribution for the number of travelers who plan their trips within two weeks of departure. b. Find the mean and the standard deviation of this distribution. c. What is the probability exactly 5 of the 12 selected business travelers plan their trips within two weeks of departure? d. Probability of 5 or fewer of the 12 selected business travelers plan their trips within two weeks of departure.

a) Using Excel, open Excel data on file Input n = 12 in cell C2 and p = 0.52 in cell C5 Binomial Distribution function =BINOM.DIST(number_s, trials, probability_s, cumulative) in cell B2: =BINOM.DIST(A2,$C$2,$C$5,FALSE) b) MEAN= SUM of (x times P(x)) 6.24 Variance = SUM(x^2 times P(x)) - mean^2 Standard Deviation = Variance^0.5 1.7307 c) n = 12 p = 0.52 EXACTLY 5 of the 12 P(X=5): Excel function: =BINOM.DIST(5, 12, 0.52, FALSE) d) P(X <= 5) =BINOM.DIST(5,12,0.52,TRUE)

The game called Lotto sponsored by the Louisiana Lottery Commission pays its largest prize when a contestant matches all four of the 29 possible numbers. Assume there are 29 ping-pong balls each with a single number between 1 and 29. Any number appears only once, and the winning balls are selected without replacement. a. The commission reports that the probability of matching all the numbers are 1 in 23,751. What is this in terms of probability? (Round your answer to 8 decimal places.) b. Use the hypergeometric formula to find the probability of matching all 4 winning numbers. The lottery commission also pays if a contestant matches two or three of the four winning numbers. Hint: Divide the 29 numbers into two groups, winning numbers and nonwinning numbers. (Round your answer to 8 decimal places.) c. Find the probability, again using the hypergeometric formula, for matching two of the four winning numbers. (Round your answer to 8 decimal places.) d. Find the probability of matching three of the four winning numbers. (Round your answer to 8 decimal places.)

a) 0.00004210 1 / 23,751 = 4.210349038e^-5 b) 0.00004210 Out of 29 ping-pong balls, 4 are winning and remaining 25 are losing. Let X = number of winning among randomly selected 4 balls. All 4 winning numbers P(X=4) = [ (Choose 4 of 4 winning) * (Choose 0 of 25 losing) ] / ( select 4 from 29 total) Choose 4 of 4 winning: 4C4 Choose 0 of 25 losing: 25C0 Select 4 from 29 total: 29C4 [ (4C4) * (25C0) ] / (29C4) = 4.210349038x^-5 c) 0.07578628 two of the four winning numbers P(X=2) = [ (Choose 2 of 4 winning) * (Choose 2 of 25 losing) ] / ( select 4 from 29 total) [ (4C2) * (25C2) ] / (29C4) = d) 0.00421035 three of the four winning numbers P(X=3) = [ (Choose 3 of 4 winning) * (Choose 1 of 25 losing) ] / ( select 4 from 29 total) [ (4C3) * (25C1) ] / (29C4) = 0.0757862827

n a Poisson distribution, μ=3.90 . (Round your answers to 4 decimal places.) a. What is the probability that x=0? b. What is the probability that x>0 ?

a) 0.0202 P(X=x) = [ e^-x * μ^x ] / x! P(X = 0) = [ e^-3.90 * 3.90^0 ] / 0! = 0.0202 b) 0.9798 P(X > x) = 1 - P(X <= x) 1 - [ P(X=x) + P(X=x....) P(X > 0) = 1 - P(X<=0) 1 - P(X=0) P(X = 0) = [ e^-3.90 * 3.90^0 ] / 0! = 0.020 1 - 0.0202 = 0.9798 USING EXCEL see attached photo

The 1989 U.S. Open golf tournament was played on the East Course of the Oak Hills Country Club in Rochester, New York. During the second round, four golfers scored a hole in one on the par 3 sixth hole. The odds of a professional golfer making a hole in one are estimated to be 3,708 to 1, so the probability is 1/3,709. There were 170 golfers participating in the second round that day. a. What is the probability that no one gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.) b. What is the probability that exactly one golfer gets a hole in one on the sixth hole? (Round your answer to 5 decimal places.) c. What is the probability that four golfers score a hole in one on the sixth hole? (Round your answer to 3 decimal places.)

a) 0.95519 NO ONE P(X = 0) b) 0.04379 EXACTLY ONE P(X=1) c) 0.000 FOUR GOLFERS P(X=4) n = 170 p = 1 / 3,709 P(X=x) Excel Function =BINOM.DIST(x,n,p,FALSE) Binomial Distribution Calculator Online https://stattrek.com/online-calculator/binomial.aspx Probability of success on a single trial (p): 1/3,709 = 0.000269614 Number of trials (n): 170 Number of successes (x): Binomial probability: P(X = x)

The director of admissions at Kinzua University in Nova Scotia estimated the distribution of student admissions for the fall semester on the basis of past experience. Admissions Probability 1,200 0.5 1,300 0.4 1,000 0.1 a. What is the expected number of admissions for the fall semester? b. Compute the variance and the standard deviation of the number of admissions. (Round your standard deviation to 2 decimal places.)

a) 1220 x = Admissions P(x) = Probability find the sum of Mean Mean: x times P(x) b) Variance = SUM(x^2 times P(x)) minus Mean^2 7600 Standard Deviation = Variance ^ 0.5 87.18

[PART 1] The Internal Revenue Service is studying the category of charitable contributions. A sample of 30 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 30 returns, 4 had charitable contributions of more than $1,000. Suppose 3 of these returns are selected for a comprehensive audit. a. Explain why the hypergeometric distribution is appropriate. Because - you are sampling a small population without replacement. - you are sampling a small population with replacement. - you are sampling a large population without replacement. b. What is the probability exactly one of the three audited had a charitable deduction of more than $1,000? (Round your answer to 4 decimal places.)

a) you are sampling a small population without replacement. A hypergeometric distribution is used when we are sampling from a finite population without replacement. b) 0.3202 the number of successes is M = 4 the number of failures is L = 26 (Total - Successes) total samples N = 30 The n of elements that are selected at random without replacement is n = 3 P(X=x) = [ (Choose x of M) * (Choose (n - x) of L) ] / ( select n of N) Probability at exactly one of 3 : X=1 P(X=1) = [ (Choose 1 of 4 winning) * (Choose (3-1) of 26 losing) ] / ( select 3 from 30 total) P(X=1) = [ (4C1) * (26C2) ] / (30C3) = 0.3201970443

The Downtown Parking Authority of Tampa, Florida, reported the following information for a sample of 233 customers on the number of hours cars are parked. a-1. Convert the information on the number of hours parked to a probability distribution. (Round your answers to 3 decimal places.) a-2. Is this a discrete or a continuous probability distribution? b-1. Find the mean and the standard deviation of the number of hours parked. (Do not round the intermediate calculations. Round your final answers to 3 decimal places.) b-2. How long is a typical customer parked? (Do not round the intermediate calculations. Round your final answer to 3 decimal places.) c. What is the probability that a car would be parked for 6 hours or more? What is the probability that a car would be parked for 3 hours or less? (Round your answers to 3 decimal places.)

a-1) see attached photo Pr(x) a-2) Discrete b-1) Mean: 4.060 Std. Dev: 2.196 b-2) That is the mean of b-1 c) P(time parked >= 6) : 0.245 P(time parked <= 3) : 0.476

For each of the following indicate whether the random variable is discrete or continuous. a. The length of time to get a haircut. b. The number of cars a jogger passes each morning while running. c. The number of hits for a team in a high school girls' softball game. d. The number of patients treated at the South Strand Medical Center between 6 and 10 P.M. each night. e. The distance your car traveled on the last fill-up. f. The number of customers at the Oak Street Wendy's who used the drive-through facility. g. The distance between Gainesville, Florida, and all Florida cities with a population of at least 50,000.

a. continuous b. discrete c. discrete d. discrete e. continuous f. discrete g. continuous

Which of these variables are discrete and which are continuous random variables? a. The number of new accounts established by a salesperson in a year. b. The time between customer arrivals to a bank ATM. c. The number of customers in Big Nick's barber shop. d. The amount of fuel in your car's gas tank. e. The number of minorities on a jury. f. The outside temperature today

a. discrete b. continuous c. discrete d. continuous e. discrete f. continuous

[PART 2] The Internal Revenue Service is studying the category of charitable contributions. A sample of 30 returns is selected from young couples between the ages of 20 and 35 who had an adjusted gross income of more than $100,000. Of these 30 returns, 4 had charitable contributions of more than $1,000. Suppose 3 of these returns are selected for a comprehensive audit. c. What is the probability at least one of the audited returns had a charitable contribution of more than $1,000? (Round your answer to 4 decimal places.)

c) 0.3596 the number of successes is M = 4 the number of failures is L = 26 (Total - Successes) total samples N = 30 The n of elements that are selected at random without replacement is n = 3 P(X=x) = [ (Choose x of M) * (Choose (n - x) of L) ] / ( select n of N) Probability AT LEAST ONE of 3 : X>=1 P(X>=1) = 1 - P(X=0) P(X=0) = [ (Choose 0 of 4 winning) * (Choose (3-0) of 26 losing) ] / ( select 3 from 30 total) P(X=0) = [ (4C0) * (26C3) ] / (30C3) = 0.6403940887 P(X>=1) = 1 - 0.6403940887 = 0.3596

[PART TWO] In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that: c. More than five have large-screen TVs? (Round your answer to 3 decimal places.) d. At least seven homes have large-screen TVs? (Round your answer to 3 decimal places.)

c) 0.992 Using the complement rule, P(A) = 1 - P(~A), P(X>5) = 1 - P(X <= 5) 1 - [(P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)] 1 - [0.00089092 + P(X=5) P(X = 5) = 9C5 x 0.90^5 x (1 - 0.90)^9-5 = 0.007440174 = 0.00089092 + 0.007440174 =0.008331094 1 - 0.008331094 = 0.992 d) 0.947 P(X>=7) = P(X=7) + P(X=8) + P(X=9) (9C7 x 0.90^7 x 0.10^2) + (9C8 x 0.90^8 x 0.10^1) + (9C9 x 0.90^9 x 0.10^0) = 0.947

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70% of the cases. Suppose the 15 cases reported today are representative of all complaints. a-1. How many of the problems would you expect to be resolved today? a-2. What is the standard deviation? b. What is the probability 10 of the problems can be resolved today? c. What is the probability 10 or 11 of the problems can be resolved today? d. What is the probability more than 10 of the problems can be resolved today?

n = 15 p = 0.70 a) 10.5 15 x 0.70 = 10.5 a-2) 1.7748 σ = square root of (n x p(1-p)) square root of (15 x 0.70 x 0.30) = 1.7748 b) 0.2061 P(X=10) = 15C10 x (0.70^10) x (1 - 0.70)^(15-10) P(X=10) = 3003 x 0.7^10 x 0.30^5 = 0.2061 c) 0.4247 P(X=10) + P(X=11) P(X=10) = 0.2061 P(X=11) = 15C11 x (0.70^11) x (1 - 0.70)^(15-11) P(X=11) = 1365 x 0.7^11 x 0.30^4 = 0.2186 0.2061 + 0.2186 = 0.4247 d) 0.5155 P(X > 10) = P(X=11) + P(X=12) + P(X=13) + P(X=14) + P(X=15) [15C11 x (0.70^11) x (1 - 0.70)^(15-11)] + [15C12 x (0.70^12) x (1 - 0.70)^(15-12)] + [15C13 x (0.70^13) x (1 - 0.70)^(15-13)] + [15C14 x (0.70^14) x (1 - 0.70)^(15-14)] + [15C15 x (0.70^15) x (1 - 0.70)^(15-15)]

The U.S. Postal Service reports 95% of first-class mail within the same city is delivered within 2 days of the time of mailing. Six letters are randomly sent to different locations. a. What is the probability that all six arrive within 2 days? (Round your answer to 3 decimal places.) b. What is the probability that exactly five arrive within 2 days? (Round your answer to 3 decimal places.) c. Find the mean number of letters that will arrive within 2 days. (Round your answer to 1 decimal place.) d-1. Compute the variance of the number that will arrive within 2 days. (Round your answer to 3 decimal places.) d-2. Compute the standard deviation of the number that will arrive within 2 days. (Round your answer to 3 decimal places.)

n = 6 p = 0.95 X = number of mails arrive within two days a. 0.735 P(X = 6) = (0.95)^6 = 0.7351 b. 0.232 P(X = 5) = 6 x (0.95)^5 x (1-0.95)^(6-5) = 6 x 0.7737809375 x 0.05^1 = 0.232 c. 5.7 MEAN: E(X) = np 6 x 0.95 = 5.7 d-1. 0.285 VARIANCE: V(X) = np(1-p) = 6 x 0.95 x (1-0.95) = 6 x 0.95 x 0.05 = 0.285 d-2. 0.534 STANDARD DEVIATION: σ = square root of V(X) Square root of 0.285 = 0.5338

[PART ONE] In a recent study, 90% of the homes in the United States were found to have large-screen TVs. In a sample of nine homes, what is the probability that: a. All nine have large-screen TVs? (Round your answer to 3 decimal places.) b. Less than five have large-screen TVs? (Round your answer to 3 decimal places.)

n = 9 p = 0.90 P(X) = (combination formula) x (p^n) x ((n-p)^n-x) a) 0.387 P(X = 9) = 0.90 ^ 9 = 0.387 b) 0.001 nCr = MATH --> PRB --> 3:nCr P(X = 0) = 9C0 x 0.90^0 x (1 - 0.90)^9-0 P(0) = 1 x 0.90^0 x 0.10^9 = 0.000000001 P(X = 1) = 9C1 x 0.90^1 x (1 - 0.90)^9-1 P(1) = 9 x 0.90^1 x 0.10^8 = 0.000000081 P(X = 2) = 9C2 x 0.90^2 x (1 - 0.90)^9-2 P(2) = 36 x 0.90^2 x 0.10^7 = 0.000002916 P(X = 3) = 9C3 x 0.90^3 x (1 - 0.90)^9-3 P(3) = 84 x 0.90^3 x 0.10^6 = 0.000061236 P(X = 4) = 9C4 x 0.90^4 x (1 - 0.90)^9-4 P(4) = 126 x 0.6561 x 0.10^5 = 0.000826686 P(X < 5 ) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) 0.000000001 + 0.000000081 + 0.000002916 + 0.000061236 + 0.000826686 = 0.00089092

In a binomial distribution, n = 7 and π=0.31 . Find the probabilities of the following events. (Round your answers to 4 decimal places.) a. x = 4. b. x ≤ 4. c. x ≥ 5.

see attached photo

Croissant Bakery Inc. offers special decorated cakes for birthdays, weddings, and other occasions. It also has regular cakes available in its bakery. The following table gives the total number of cakes sold per day and the corresponding probability. Number of Cakes Sold in a Day Probability 12 0.25 13 0.40 14 0.25 15 0.10 a. Compute the mean of the number of cakes sold per day. b. Compute the variance of the number of cakes sold per day. c. Compute the standard deviation of the number of cakes sold per day.

x = number of cakes P(x) = Probability a) 13.2 Mean = x times P(x) 12*0.25 + 13*0.40 + 14*0.25 +15*0.10 = 13.2 Using Excel File. Add a new column C for Mean multiply x in column A and P(x) in Column B =sum(C2:C5) b.) 0.86 Variance= SUM(x^2 times P(x)) - mean^2 Excel: =E6-F6 c) 0.9274 Standard Deviation = Variance ^ 0.5 0.86 ^ 0.5 = 0.92736185 Excel: =Power(F9,0.5)

An internal study by the Technology Services department at Lahey Electronics revealed company employees receive an average of 4.2 non-work-related e-mails per hour. Assume the arrival of these e-mails is approximated by the Poisson distribution. a. What is the probability Linda Lahey, company president, received exactly 3 non-work-related e-mails between 4 p.m. and 5 p.m. yesterday? b. What is the probability she received 6 or more non-work-related e-mails during the same period? c. What is the probability she received two or less non-work-related e-mails during the period?

μ = 4.2 a) P(X=x) = [ e^-x * μ^x ] / x! By Calculator P(X = 3) = [ e^-4.2 * 4.2^3 ] / 3! = 0.1851 Excel Function =POISSON(3,4.2,FALSE) b) 0.2469 P(X>=6) = 1 - P(X < 5) 1 - [ P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) ] Excel Function =1 - POISSON(5,4.2,TRUE) c) 0.2102 P(X<=2) Excel Function =POISSON(2,4.2,TRUE) Use POISSON CALCULATOR ONLINE https://stattrek.com/online-calculator/poisson.aspx a) exactly 3 Poisson random Variable (x) : 3 Average rate of success: 4.2 Poisson Probability: P(X = 3) = 0.18517 b) 6 or more Poisson random Variable (x) : 6 Average rate of success: 4.2 Cumulative Probability: P(X > 6) = 0.24686 c) 2 or less Poisson random Variable (x) : 2 Average rate of success: 4.2 Cumulative Probability: P(X < 2) = 0.21024

Recent crime reports indicate that 5.5 motor vehicle thefts occur each minute in the United States. Assume that the distribution of thefts per minute can be approximated by the Poisson probability distribution. a. Calculate the probability exactly two thefts occur in a minute. (Round your probability to 3 decimal places.) b. What is the probability there are no thefts in a minute? (Round your probability to 3 decimal places.) c. What is the probability there is three or less thefts in a minute?

μ = 5.5 a) 0.062 P(X=x) = [ e^-x * μ^x ] / x! By Calculator P(X = 2) = [ e^-5.5 * 5.5^2 ] / 2! = 0.06181 Excel Function =POISSON(2,5.5,FALSE) b) 0.004 P(X=0) Excel Function =POISSON(0,5.5,FALSE) c) 0.202 P(X<=2) Excel Function =POISSON(2,5.5,TRUE) Use POISSON CALCULATOR ONLINE https://stattrek.com/online-calculator/poisson.aspx a) exactly 2 thefts Poisson random Variable (x) : 2 Average rate of success: 5.5 Poisson Probability: P(X = 2) = 0.06181 b) no thefts Poisson random Variable (x) : 0 Average rate of success: 5.5 Cumulative Probability: P(X = 0) = 0.00409 c) 3 or less Poisson random Variable (x) : 3 Average rate of success: 5.5 Cumulative Probability: P(X < 3) = 0.2017