WK 4 Test

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American Military University *

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302

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Statistics

Date

Jun 12, 2024

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pdf

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Uploaded by HighnessAtom2424

Week 4 Test Your work has been saved and submitted Written May 30, 2024 1:56 PM - May 30, 2024 3:30 PMAttempt 1 of 2 Attempt Score 19/ 20-95% Overall Grade (Highest Attempt) 19 /20-95% Question 1 1 /1 point Find P(Z > -.98). Round answer to 4 decimal places. Answer: ___0.8365___ v W Hide question 1 feedback In Excel, =1- NORM.S.DIST(-.98, TRUE)

Question 2 1 /1 point Arm span is the physical measurement of the length of an individual's arms from fingertip to fingertip. A man's arm span is approximately normally distributed with mean of 68 inches with a standard deviation of 3.7 inches. Find length in inches of the 97th percentile for a man's arm span. Round answer to 2 decimal places. Answer: 74.96 v W Hide question 2 feedback In Excel, =NORM.INV(0.97,68,3.7) Question 3 1 /1 point Find P(Z = .42). Round answer to 4 decimal places. Answer: ___0.3372 __ v W Hide question 3 feedback In Excel, =1-NORM.S.DIST(0.42,TRUE) Question 4 1 /1 point Which type of distribution does the graph illustrate?

() Poisson Distribution v ) Right skewed Distribution O Uniform Distribution () Normal Distribution Question 5 0/ 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 27.4 mpg and a standard deviation of 12.3 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 0 before the decimal. Answer: 0.3949 3 (0.3947) W Hide question 5 feedback This is a sampling distribution problem with y = 27.4. 0 = 12.3, and sample size n = 30.

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New SD = 12.3/SQRT(30) = 2.245662 P(x > 28) = 1 - NORM.DIST(28, 27.4, 2.245662, TRUE) Question 6 1 /1 point Suppose that the longevity of a light bulb is exponential with a mean lifetime of 7.6 years. Find the probability that a light bulb lasts between seven and eleven years. ) 0.1325 ) 0.1859 v 0.1629 ) 0.8371 W Hide question 6 feedback P(7<x<11) Px < 11)-P(x<7) In Excel, =EXPON.DIST(11,1/7.6,TRUE)-EXPON.DIST(7,1/7.6,TRUE) Question 7 1 /1 point

The average lifetime of a certain new cell phone is 4.2 years. The manufacturer will replace any cell phone failing within three years of the date of purchase. The lifetime of these cell phones is known to follow an exponential distribution. The decay rate is: v ) 0.238] ) 0.7619 () 0.3333 ) 0.6667 W Hide question 7 feedback 1/4.2 Question 8 1 /1 point The life of an electric component has an exponential distribution with a mean of 7.2 years. What is the probability that a randomly selected one such component has a life less than 4 years? Answer: (round to 4 decimal places) 0.4262 v W Hide question 8 feedback P(x < 4) In Excel, =EXPON.DIST(4,1/7.2,TRUE) Question 9 1 /1 point

Suppose that the longevity of a light bulb is exponential with a mean lifetime of 7.6 years. Find the probability that a light bulb lasts less than three years. ~ ) 0.1175 v ) 0.3261 () 8.5634 . 0.6739 () 0.3907 W Hide question 9 feedback P(x < 3) In Excel, =EXPON.DIST(3,1/7.6,TRUE) Question 10 1 /1 point The waiting time for an Uber has a uniform distribution between 5 and 37 minutes. What is the probability that the waiting time for this Uber is less than 13 minutes on a given day? Answer: (Round to two decimal places.) ___0.25 v W Hide question 10 feedback Interval goes from 5 < x < 37 P(x < 13) = (13 - 5) *

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Question 11 1 /1 point The waiting time for a train has a uniform distribution between 3 and 22 minutes. What is the probability that the waiting time for this train is more than 5 minutes on a given day? Answer: (Round to four decimal places.) 0.8947 v W Hide question 11 feedback Interval goes from 3 < x < 22 P(x > 5) =(22 - 5) * Question 12 1 /1 point A local pizza restaurant delivery time has a uniform distribution over 10 to 75 minutes. What is the probability that the pizza delivery time is more than 45 minutes on a given day? Answer: (Round to four decimal places.) 0.4615 v

W Hide question 12 feedback Interval goes from 10 < x < 75 P(x > 45) = (75 - 45) * Question 13 1 /1 point The waiting time for a bus has a uniform distribution between 2 and 13 minutes. What is the probability that the waiting time for this bus is less than 4.5 minutes on a given day? Answer: (Round to four decimal places.) 0.2273 v W Hide question 13 feedback Interval goes from 2 < x < 13 P(x <4.5)=(4.5-2)* Question 14 1 /1 point

Miles per gallon of a vehicle is a random variable with a uniform distribution from 23 to 47. The probability that a random vehicle gets between 25 and 34 miles per gallon is: Answer: (Round to three decimal places) 0.375 v W Hide question 14 feedback Interval goes from 23 < x < 47 P(25 <x<34) =34 -25)* Question 15 1 /1 point Suppose the time it takes a barber to complete a haircuts is uniformly distributed between 8 and 22 minutes, inclusive. Let X = the time, in minutes, it takes a barber to complete a haircut. Then X ~U (8, 22). Find the probability that a randomly selected barber needs at least 14 minutes to complete the haircut, P(x > 14) (round answer to 4 decimal places) Answer: 0.5714 v W Hide question 15 feedback Interval goes from 8 < x < 22 P(x > 14) = (22-14) *

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Question 16 1 /1 point The final exam grade of a statistics class has a skewed distribution with mean of 81.2 and standard deviation of 6.95. If a random sample of 42 students selected from this class, then what is the probability that the average final exam grade of this sample is between 75 and 807 Answer: (round to 4 decimal places) 0.1316 v W Hide question 16 feedback New SD = 6.95/SQRT(42) = 1.072408 P(75 < x < 80), in Excel =NORM.DIST(80,81.2,1.072408,TRUE)-NORM.DIST(75,81.2,1.072408,TRUE) Question 17 1 /1 point The final exam grade of a mathematics class has a normal distribution with mean of 81 and standard deviation of 6.6. If a random sample of 40 students selected from this class, then what is the probability that the average final exam grade of this sample is between 77 and 827 Answer: (round to 4 decimal places) 0.8310 v W Hide question 17 feedback New SD = 6.6/SQRT(40) = 1.043552

P(77 < x < 82), in Excel =NORM.DIST(82,81,1.043552,TRUE)-NORM.DIST(77,81,1.043552,TRUE) Question 18 1 / 1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 15.85 ounces with a standard deviation of 0.3 ounces. If a random sample of thirty-five 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) 0.0015 v W Hide question 18 feedback New SD =.3/SQRT(35) = 0.050709 P(x <15.7), in Excel =NORM.DIST(15.7,15.85,0.050709,TRUE) Question 19 1 / 1 point The time a student sleeps per night has a distribution with mean 6.06 hours and a standard deviation of 0.55 hours. Find the probability that average sleeping time for a randomly selected sample of 35 students is more than 6.15 hours per night. Answer: (round to 4 decimal W Hide question 19 feedback

New SD = .55/SQRT(35) = 0.092967 P(x > 6.15), in Excel =1-NORM.DIST(6.15,6.06,0.092967,TRUE) Question 20 1 /1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 15.96 ounces with a standard deviation of 0.5 ounces. If a random sample of sixty-five 16-ounce beverage cans are selected, what is the probability that the mean of this sample is less than 16.05 ounces of beverage? Answer: (round to 4 decimal places) 0.9266 v W Hide question 20 feedback New SD = .5/SQRT(65) = 0.062017 P(x < 16.05), in Excel =NORM.DIST(16.05,15.96,0.062017,TRUE) Done

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WK 4 Test

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