Statistics for Everyday Life: Understanding Distributions and Probabilities

n 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 ___ estion 1 feedback , .INV(0.2,272,9) on 2 1 / 1 point The mean yearly rainfall in Sydney, Australia, is about 137 mm and the standard deviation is about 69 mm ("Annual maximums of," 2013). Assume rainfall is normally distributed. How many yearly mm of rainfall would there be in the top 25%? Round answer to 2 decimal places. Answer: ___183.54 ___ estion 2 feedback % is the bottom 75%. 1 - .25 = .75. , .INV(0.75,137,69) on 3 1 / 1 point Find P(Z ≤ 3). Round answer to 4 decimal places. Answer: ___0.9987 ___ estion 3 feedback , .S.DIST(3,TRUE)

on 4 1 / 1 point Which type of distribution does the graph illustrate? Right Skewed Distribution Left skewed Distribution Uniform Distribution Normal Distribution 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 26.3 mpg and a standard deviation of 11.2 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 zero before the decimal. ___

Answer: 0.2029 Hide question 5 feedback This is a sampling distribution problem with μ = 26.3. σ = 11.2, and sample size n = 30. New SD = 11.2/SQRT(30) = 2.044831 P(x > 28) = 1 - NORM.DIST(28, 26.3, 2.044831, TRUE) n 6 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 less than one year. 0.1175 0.1859 9.6318 0.3034 0.3682 Hide question 6 feedback P(x < 1) In Excel, =EXPON.DIST(1,1/8,TRUE) n 7 1 / 1 point The life of an electric component has an exponential distribution with a mean of 8 years. What is the probability that a randomly selected one such component has a life less than 5 years? Answer: (round to 4 decimal places) ___0.4647 ___ estion 7 feedback , =EXPON.DIST(5,1/8,TRUE)

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on 8 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)? 5.5452 1.3863 0.1941 2.0794 Hide question 8 feedback Median Lifetime is the 50th percentile. Use .50 in the equation and the rate of decay is 1/3 − �� (1−.5)13 n 9 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 9 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) n 10 1 / 1 point The waiting time for a train has a uniform distribution between 0 and 10 minutes. What is the probability that the waiting time for this train is more than 4 minutes on a given day? Answer: (Round to two decimal place.)

___0.60 ___ estion 10 feedback goes from 0 < x < 10 = (10−4) ∗ 1(10−0) on 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 ___ estion 11 feedback goes from 25 ≤ x ≤ 35 � <32)=(32−27) ∗ (1(35−25)) on 12 1 / 1 point 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 30 minutes on a given day? Answer: (Round to 2 decimal place.) ___0.50 ___ estion 12 feedback goes from 0 < x < 60 ) = (60−30) ∗ 1(60−0) on 13 1 / 1 point

The mail arrival time to a department has a uniform distribution over 0 to 60 minutes. What is the probability that the mail arrival time is more than 20 minutes on a given day? Answer: (Round to 2 decimal places.) ___0.67 ___ estion 13 feedback goes from 0 < x < 60 x) = (60−20) ∗ 1(60−0) on 14 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 ___ estion 14 feedback goes from 0 < x < 9. 80th percentile use .80 = .80 −09−0 x on 15 1 / 1 point 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 ___

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estion 15 feedback goes from 0 < x < 60 ) =(60−25) ∗ 1(60−0) on 16 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) ___0.0038 ___ estion 16 feedback = .6/SQRT(64) .7), in Excel .DIST(15.7,15.9,0.075,TRUE) on 17 1 / 1 point The final exam grade of a mathematics class has a skewed distribution with mean of 78 and standard deviation of 7.2. If a random sample of 38 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) ___0.9515 ___ estion 17 feedback = 7.2/SQRT(38) < 80), in Excel .DIST(80,78,1.1679943,TRUE)-NORM.DIST(75,78,1.1679943,TRUE) on 18 1 / 1 point

The time a student sleeps per night has a distribution with mean 6.3 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 42 students is more than 6.5 hours per night. Answer: (round to 4 decimal places) ___0.0154 ___ estion 18 feedback = .6/SQRT(42) 5), in Excel M.DIST(6.5,6.3,0.092582,TRUE) on 19 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? Answer: (round to 4 decimal places) ___0.7702 ___ estion 19 feedback = 7.6/SQRT(32) < 80), in Excel .DIST(80,76,1.3435,TRUE)-NORM.DIST(75,76,1.3435,TRUE) on 20 1 / 1 point The time a student sleeps per night has a distribution with mean 6.1 hours and a standard deviation of 0.6 hours. Find the probability that average sleeping time for a randomly selected sample of 36 students is more than 6 hours per night. Answer: (round to 4 decimal places) ___0.8413 ___ Hide question 20 feedback New SD = .6/SQRT(36)

P(x > 6), in Excel =1-NORM.DIST(6,6.1,0.1,TRUE)

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