Week 4 Practice: Probability and Distribution Problems

Week 4 Knowledge Check Homework Practice Questio... X Attempt 1 of 4 Written Sep 24, 2023 3:51 PM - Sep 24, 2023 4:50 PM Attempt Score 20/ 20 -100 % Overall Grade (Highest Attempt) 20 /20 - 100 % Question 1 1/ 1 point The length of a human pregnancy is normally distributed with a mean of 272 days with a standard deviation of 9 days (Bhat & Kushtagi, 2006). How many days would a pregnancy last for the shortest 20%? Round answer to 2 decimal places. Answer: ___264.43 v w Hide question 1 feedback In Excel, =NORM.INV(0.2,272,9) Question 2 1/ 1 point Find P(1.31 < Z < 2.15). Round answer to 4 decimal places. Answer: 0.0793 v w Hide question 2 feedback In Excel, =NORM.S.DIST(2.15,TRUE)-NORM.S.DIST(1.31,TRUE)

Question 3 Find P(Z = 1.8). Round answer to 4 decimal places. Answer: w Hide question 3 feedback In Excel, =1-NORM.S.DIST(1.8,TRUE) Question 4 Find the probability that falls in the shaded area. 1/ 1 point 1 8 | ) ) | 0O 1 2 3 4 (00> ¢<> 0.625 0.0359 v 1/ 1 point — X 10

w Hide question 4 feedback (1/8)*5 Question 5 1/ 1 point The manufacturer of a new compact car claims the miles per gallon (mpg) for the gasoline consumption is mound-shaped and symmetric with a mean of 24.6 mpg and a standard deviation of 9.5 mpg. If 30 such cars are tested, what is the probability the average mpg achieved by these 30 cars will be greater than 28? Answer: ___ Round your answer to 4 decimal places as necessary. For example, 0.1357 would be a legitimate entry. Make sure you include the O before the decimal. Answer: 0.0250 v w Hide question 5 feedback This is a sampling distribution problem with pu = 24.6. o = 9.5, and sample size n = 30. New SD = 9.5/SQRT(30) = 1.734454765 P(x > 28) = 1 - NORM.DIST(28, 24.6,1.734454765 , TRUE) Question 6 1/ 1 point The life of an electric component has an exponential distribution with a mean of 10 years. What is the probability that a randomly selected one such component has a life more than 7 years? Answer: (Round to 4 decimal places.) 0.4966 v

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¥ Hide question 6 feedback P(x > 7) 1-P(x < 7) =1-EXPON.DIST(7,1/10,TRUE) Question 7 1/ 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. What is the median lifetime of these phones (in years)? 55452 () 1.3863 ) 0.1941 v 20794 ¥ Hide question 7 feedback Median Lifetime is the 50th percentile. Use .50 in the equation and the rate of decay is 1/3

Question 8 1/ 1 point The caller times at a customer service center has an exponential distribution with an average of 10 seconds. Find the probability that a randomly selected call time will be less than 25 seconds? (Round to 4 decimal places.) Answer: 0.9179 v w Hide question 8 feedback P(x < 25) In Excel, =EXPON.DIST(25,1/10,TRUE) Question 9 1/ 1 point The average lifetime of a certain new cell phone is three years. The manufacturer will replace any cell phone failing within two years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. The decay rate is: ~0.1666 ) 0.6666 \/C> 0.3333 w Hide question 9 feedback 1/3 =.3333 Question 10 1/ 1 point

A local grocery delivery time has a uniform distribution over O to 60 minutes. What is the probability that the grocery delivery time is more than 15 minutes on a given day? Answer: (Round to 2 decimal places.) 0.75 v w Hide question 10 feedback Interval goes from O < x < 60 P(x > 15) = Question 11 1/ 1 point Miles per gallon of a vehicle is a random variable with a uniform distribution from 25 to 35. The probability that a random vehicle gets between 27 and 32 miles per gallon is: Answer: (Round to two decimal place) ___ 0.50 « w Hide question 11 feedback Interval goes from 25 < x < 35 1 P27 < z < 32) (35— 25) (32 — 27) * Question 12 1/ 1 point Suppose the time it takes a barber to complete a haircuts is uniformly distributed between 5 and 17 minutes, inclusive. Let X = the time, in minutes,

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it takes a barber to complete a haircut. Then X ~ U (5, 17). Find the probability that a randomly selected barber needs at least seven minutes to complete the haircut, P(x > 7) (round to 4 decimal places) Answer: 0.8333 v w Hide question 12 feedback Interval goes from 5 <x <17 P(x > 7) = Question 13 1/ 1 point The waiting time for a taxi has a uniform distribution between 0 and 10 minutes. What is the probability that the waiting time for this taxi is less than 7 minutes on a given day? Answer: (Round to two decimal place.) ___0.70 __ v W Hide question 13 feedback Interval goes from O < x < 10 P(x < 7) = Question 14 1/ 1 point

The waiting time in line at an ice cream shop has a uniform distribution between 0 and 92 minutes. What is the 80th percentile of this distribution? (Recall: The 80th percentile divides the distribution into 2 parts so that 80% of area is to the left of 80th percentile) ___ answer to one decimal places.) 7.2 minutes Answer: (Round W Hide question 14 feedback Interval goes from 0 < x < 9. 80th percentile use .80 P(X < x) =.80 x — ( 80 = 9—-0 .80 *9 = x 7.2 =X Question 15 1/ 1 point A local pizza restaurant delivery time has a uniform distribution over O to 60 minutes. What is the probability that the pizza delivery time is more than 25 minutes on a given day? Answer: (Round to 2 decimal places.) ___0.58 v Ww Hide question 15 feedback Interval goes from O < x < 60 P(x > 25) =

Question 16 1/ 1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 15.8 ounces with a standard deviation of 0.5 ounces. If a random sample of thirty-six 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 15.7 ounces of beverage? Answer: (round to 4 decimal places) 0.1151 v w Hide question 16 feedback New SD =.5/SQRT(36) P(x <15.7), in Excel =NORM.DIST(15.7,15.8,0.08333,TRUE) Question 17 1/ 1 point The final exam grade of a statistics class has a skewed distribution with mean of 76 and standard deviation of 7.6. If a random sample of 32 students selected from this class, then what is the probability that average final exam grade of this sample is between 75 and 80?7 Answer: (round to 4 decimal places) ___0.7702 v W Hide question 17 feedback New SD = 7.6/SQRT(32) P(75 < x < 80), in Excel =NORM.DIST(80,76,1.3435,TRUE)-NORM.DIST(75,76,1.3435,TRUE) Question 18 1/ 1 point

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The average amount of a beverage in randomly selected 16-ounce beverage can is 16 ounces with a standard deviation of 0.4 ounces. If a random sample of sixty-four 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 16.1 ounces of beverage? Answer: (round to 4 decimal places) __ 0.9772 v w Hide question 18 feedback New SD = .4/SQRT(64) P(x < 16.1), in Excel =NORM.DIST(16.1,16,0.05,TRUE) Question 19 1/ 1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 16.1 ounces with a standard deviation of 0.4 ounces. If a random sample of eighty-one 16-ounce beverage cans are selected, what is the probability that mean of this sample is less than 16.2 ounces of beverage? Answer: (round to 4 decimal places) 0.9878 v w Hide question 19 feedback New SD = .4/SQRT(81) P(x < 16.2), in Excel =NORM.DIST(16.2,16.1,0.0444,TRUE) Question 20 1/ 1 point The final exam grade of a statistics class has a skewed distribution with mean of 76 and standard deviation of 7.4. If a random sample of 36 students selected from this class, then what is the probability that the average final

exam grade of this sample is between 75 and 80? Answer: (round to 4 decimal places) __ 0.7907 w Hide question 20 feedback New SD = 7.4/SQRT(36) P(75 < x < 80), in Excel =NORM.DIST(80,76,1.2333,TRUE)-NORM.DIST(75,76,1.2333,TRUE) Done

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