MATH302 WEEK 4 KNOWLEDGE CHECK

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School

American Military University *

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Course

302

Subject

Statistics

Date

Jan 9, 2024

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docx

Pages

9

Uploaded by carelessvicki

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1 / 1 point Find P(Z > -1.24). Round answer to 4 decimal places. Answer: ___.8925 ___ k 4,TRUE) 1 / 1 point Find P(Z ≥ 1.8). Round answer to 4 decimal places. Answer: ___.0359 ___ k TRUE) 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 70 inches with a standard deviation of 4.5 inches. Find length in inches of the 99th percentile for a man's arm span. Round answer to 2 decimal places. Answer: ___80.47 ___ k 5) 1 / 1 point Find the probability that�falls in the shaded area.
0.5 0.125 0.625 0.438 Hide question 4 feedback (1/8)*5 1 / 1 point The cost of unleaded gasoline in the Bay Area once followed a normal distribution with a mean of $4.59 and a standard deviation of $0.10. Thirty gas stations from the Bay area are randomly chosen. We are interested in the average cost of gasoline for the 30 gas stations. Find the exact probability that the average price for 30 gas stations is less than $4.55. 0 0.6554 0.0142 0.3446 Hide question 5 feedback New SD = .10/SQRT(30) = 0.018257419 P(x < 4.55) In Excel, =NORM.DIST(4.55,4.59,0.018257,TRUE)
1 / 1 point Suppose that the longevity of a light bulb is exponential with a mean lifetime of eight years. Find the probability that a light bulb lasts between six and ten years. 0.1175 0.1859 9.6318 0.3034 0.3682 Hide question 6 feedback P( 6 < x < 10) P(x < 10) - P( x < 6) In Excel, =EXPON.DIST(10,1/8,TRUE)-EXPON.DIST(6,1/8,TRUE) 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.) ___.4966 ___ k 0,TRUE) 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:
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___.9179 ___ k ,TRUE) 1 / 1 point The commute time for people in a city has an exponential distribution with an average of 0.5 hours. What is the probability that a randomly selected person in this city will have a commute time between 0.4 and 1 hours? Answer: (round to 3 decimal places) ___.314 ___ k TRUE)-EXPON.DIST(0.4,1/0.5,TRUE) 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 25 and 30 miles per gallon is: Answer: (Round to two decimal place) ___.50 ___ ck
x < 35 1(35−25) 1 / 1 point The waiting time in line at an ice cream shop has a uniform distribution between 0 and 9 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) _______ minutes Answer: (Round answer to one decimal places.) ___7.2 ___ ck < 9. 80th percentile use .80 1 / 1 point The waiting time for a table at a busy restaurant has a uniform distribution between 0 and 10 minutes. What is the 95th percentile of this distribution? (Recall: The 95th percentile divides the distribution into 2 parts so that 95% of area is to the left of 95th percentile) _______ minutes Answer: (Round answer to one decimal place.) ___9.5 ___ ck
< 10. 95th percentile use .95 1 / 1 point The waiting time for a bus has a uniform distribution between 0 and 10 minutes. What is the probability that the waiting time for this bus is less than 5 minutes on a given day? Answer: (Round to two decimal place.) ___.50 ___ ck < 10 0)) 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) ___.50 ___ ck x ≤ 35 (1(35−25)) 1 / 1 point
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A local pizza restaurant delivery time has a uniform distribution over 0 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.) ___.58 ___ ck < 60 0−0) 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 0.3 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 16.2 ounces of beverage? Answer: (round to 4 decimal places) ___.9962 ___ ck 1,0.0375,TRUE) 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)
___.7907 ___ ck ) 2333,TRUE)-NORM.DIST(75,76,1.2333,TRUE) 1 / 1 point The average amount of water in randomly selected 16-ounce bottles of water is 15.9 ounces with a standard deviation of 0.6 ounces. If a random sample of sixty-four 16-ounce bottles of water are selected, what is the probability that the mean of this sample is less than 15.7 ounces of water? Answer: (round to 4 decimal places) ___.0038 ___ ck 9,0.075,TRUE) 1 / 1 point The average credit card debt for college seniors is $16,601 with a standard deviation of $4100. What is the probability that a sample of 35 seniors owes a mean of more than $18,000? Round answer to 4 decimal places. Answer: ___.0218 ___ ck 35)
el ,16601,693.0265,TRUE) 1 / 1 point 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) ___.9772 ___ Hide question 20 feedback New SD = .4/SQRT(64) P(x < 16.1), in Excel =NORM.DIST(16.1,16,0.05,TRUE)
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