WK 3 Test 2

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American Military University *

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302

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Statistics

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Jun 12, 2024

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Week 3 Test Your work has been saved and submitted Written May 22, 2024 3:25 PM - May 22, 2024 4:59 PMAttempt 2 of 2 Attempt Score 20 /20-100% Overall Grade (Highest Attempt) 20 /20-100% Question 1 1 /1 point It is known that 50% of adult workers have a high school diploma. If a random sample of 8 adult workers is selected, what is the probability that less than 6 of them have a high school diploma? (That is, what is P(X < 6) (round to 4 decimal places) Answer: .8555 v W Hide question 1 feedback P(x < 6) = P(x < 5), in Excel =BINOM.DIST(5,8,0.5,TRUE) Question 2 1 /1 point If random variable X has a binomial distribution with n =8 and P(success) = p =0.5, find the probability that X is at most 3. (That is, find P(X < 3)) (round to 4 decimal places) Answer: 0.3633 v W Hide question 2 feedback

In Excel, =BINOM.DIST(3,8,0.5,TRUE) Question 3 1 /1 point Suppose a random variable, x, arises from a binomial experiment. If n = 25, and p = 0.85, find the P(X = 15) using Excel. Round answer to 4 decimal places. Answer: 0.9995 v W Hide question 3 feedback P(x > 15)=1 - P(x < 14), in Excel =1-BINOM.DIST(14,25,0.85,TRUE) Question 4 1 /1 point Approximately 10% of all people are left-handed. Out of a random sample of 15 people, what is the probability that 4 of them are left-handed? Round answer to 4 decimal places. Answer: 0.0428 v W Hide question 4 feedback P(x = 4), In Excel, =BINOM.DIST(4,15,0.1,FALSE) Question 5 1 /1 point Suppose a random variable, x, arises from a binomial experiment. If n = 14, and p = 0.13, find the P(X = 3) using Excel. Round answer to 4 decimal places. Answer: 0.1728 v W Hide question 5 feedback In Excel, =BINOM.DIST(3,14,0.13,FALSE) Question 6 1 /1 point The table shows a random sample of musicians and how they learned to play their instruments.

Gender | Self-Taught | Studied in School | Private Instruction | Total Female |12 38 22 72 Male 19 24 15 58 Total 31 62 37 130 Find P(musician is a male AND had private instruction). D 0.71 v 0.12 fj) 0.88 :) 0.29 W Hide question 6 feedback Good job, 15/130 Question 7 1 /1 point You have a club of fifteen people. You need to pick a president, treasurer, and secretary from the fifteen. How many different ways can you do this? Leave answer as whole number, do not include decimals or commas. Answer: 2730 v W Hide question 7 feedback Order matters. In Excel, =PERMUT(15,3) Question 8 1 /1 point How many ways can you choose 4 cookies from a cookie jar containing 25 cookies of all the same type? Leave answer a whole number, do not include decimals or commas. Answer: 12650 v W Hide question 8 feedback Order does not matter. In Excel, =COMBIN(25,4)

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Question 9 1 /1 point You are going to a benefit dinner, and need to decide before the dinner what you want for salad, main dish, and dessert. You have 2 different salads to choose from, 3 main dishes, and 5 desserts. How many different meals are available? Answer: v 30 O 25 O 15 ()10 W Hide question 9 feedback 2%3*5 Question 10 1 /1 point Does the following table represent a valid discrete probability distribution? X 1 2 3 4 5 P(X=x) 0.11]0.06]0.18]0.06]0.59 v yes /) no W Hide question 10 feedback Yes, because the probabilities add up to 1. Question 11 1 /1 point The random variable X = the number of vehicles owned. Find the probability that a person owns at least 2 vehicles. Round to two decimal places.

P(X=x)|0.1]0.35]0.25]0.2]0.1 Answer: 0.55 v W Hide question 11 feedback At least 2 vehicles is the probability of P(x = 2) + P(x = 3) + P(4) 25 +.2 + .1 Question 12 1 /1 point The random variable X = the number of vehicles owned. Find the standard deviation of the number of vehicles owned. Round answer to 4 decimal places. X |0l 1]12]3]4 P(X)|0.1]0.35]0.25{0.2]0.1 Answer: 1.1522 v W Hide question 12 feedback Expected value = 0*.1 + 1*.35 + 2*.25 + 3*.2 + 4*.1 = 1.85 SD = Question 13 1 /1 point Does the following table represent a valid discrete probability distribution? X 1 12131 4 |5

P(X=x)]0.16]0.11]0.06|- 0.36]0.21 D yes ,/Q no W Hide question 13 feedback No, since the probabilities do not add up to 1 and all probabilities need to be between 0 and 1. Probabilities cannot be negative. Question 14 1 /1 point If random variable X has a Poisson distribution with mean = 4.5 find the probability that X is more than 4. (That is, find P(X>4) (round to 4 decimal places) Answer: 0.4679 v W Hide question 14 feedback Px >4)= 1-P(x <4), in Excel =1-POISSON.DIST(4,4.5,TRUE) Question 15 1 /1 point If random variable X has a Poisson distribution with mean = 10, find the probability that X is more than 8. (That is, find P(X>8) (round to 4 decimal places) Answer: 0.6672 v W Hide question 15 feedback Px > 8)=1-P(x < 8), in Excel =1-POISSON.DIST(8,10,TRUE) Question 16 1 /1 point A bank gets an average of 12 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that there will be 4 or more customers at this bank in one hour. (That is, find P(X>=4)) (round to 4 decimal places) Answer: 0.9977 v W Hide question 16 feedback

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PXx >4) =1-P(x < 3), in Excel =1-POISSON.DIST(3,12,TRUE) Question 17 1 /1 point A bank gets an average of 8.5 customers per hour. Assume the variable follows a Poisson distribution. Find the probability that there will be 4 or more customers at this bank in one hour. (That is, find P(X>=4)) (round to 4 decimal places) Answer: 0.9699 v W Hide question 17 feedback Px>4) =1 -P(x < 3), in Excel =1-POISSON.DIST(3,8.5,TRUE) Question 18 1 /1 point The mean number of visitors at a national park in one weekend is 55. Assume the variable follows a Poisson distribution. Find the probability that there will be at most 71 visitors at this park in one weekend. (That is, find P(X<71) (round to 4 decimal places) Answer: 0.9841 v W Hide question 18 feedback In Excel, =POISSON.DIST(71,55,TRUE) Question 19 1 /1 point A furniture manufacturer offers bookcases in 6 different sizes and 3 different colors. If every color is available in every size, then the total number of different bookcases is: 9 j) 15 ()6 \/O 18

W Hide question 19 feedback 6*3 Question 20 1 /1 point On a baseball team, there are infielders and outfielders. Some players are great hitters, and some players are not great hitters. Let | = the event that a player in an infielder. Let O = the event that a player is an outfielder. Let H = the event that a player is a great hitter. Let N = the event that a player is not a great hitter. Write the symbols for the probability that a player is an outfielder or is a great hitter. . )P(N or O) ¢Q P(O or H) ~ ) P(HIO) () P(Oand H) W Hide question 20 feedback You need to use the word "or" in the probability. O is Outfield and H is hitter. P(O or H) Done

WK 3 Test 2

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