math exam 302 course hero

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1 1 / 1 point The mean yearly rainfall in Sydney, Australia, is about 134 mm and the standard deviation is about 66 mm ("Annual maximums of," 2013). Assume rainfall is normally distributed. How many yearly mm of rainfall would there be in the top 15%? Round answer to 2 decimal places. Answer: ___202.40 ___ estion 1 feedback %, is in the bottom 85% percentile. , .INV(.85,134,66) on 2 1 / 1 point Find the area under the standard normal distribution to the left of z = -1.35. Round answer to 4 decimal places. Answer: ___0.0885 ___ estion 2 feedback , .S.DIST(-1.35,TRUE) on 3 1 / 1 point Find P(Z ≤ 2.76). Round answer to 4 decimal places. Answer: ___0.9971 ___ estion 3 feedback , .S.DIST(2.76,TRUE) on 4 0 / 1 point
Which type of distribution does the graph illustrate? Poisson Distribution Right skewed Distribution Uniform Distribution Normal Distribution Question 5 1 / 1 point The cost of unleaded gasoline in the Bay Area once followed a normal distribution with a mean of $4.74 and a standard deviation of $0.16. Fifteen gas stations from the Bay area are randomly chosen. We are interested in the average cost of gasoline for the 15 gas stations. What is the approximate probability that the average price for 15 gas stations is over $4.99? .0256 0.0000 0.1587 0.0943 Hide question 5 feedback New SD = .16/SQRT(15) = .0413 P(x > 4.99) = 1 - P(x < 4.99) In Excel, =1-NORM.DIST(4.99,4.74,.0413,TRUE) You might get an answer with an "E" in it. The "E"; means scientific notation. 7.1E-10 decimal answer is, . 00000000071 n 6 1 / 1 point The average lifetime of a certain new cell phone is 3.4 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. 68% of these phone last how long (in years)?
4.5686 3.8741 4.7862 2.0794 Hide question 6 feedback For 68% use .68 in the equation and the rate of decay is 1/3.4 − �� (1−.68)13.4 n 7 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 0.3261 8.5634 0.6739 0.3907 Hide question 7 feedback P(x < 3) In Excel, =EXPON.DIST(3,1/7.6,TRUE) n 8 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: 0.2381 0.7619 0.3333 0.6667
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Hide question 8 feedback 1/4.2 n 9 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 ___ estion 9 feedback , =EXPON.DIST(4,1/7.2,TRUE) on 10 1 / 1 point The waiting time for a table at a busy restaurant has a uniform distribution between 15 and 45 minutes. What is the 90th percentile of this distribution? (Recall: The 90th percentile divides the distribution into 2 parts so that 90% of area is to the left of 90th percentile) _______ minutes Answer: (Round answer to one decimal place.) ___42 ___ estion 10 feedback goes from 15 < x < 45. 90th percentile use .90 = .90 −1545−15 = x - 15 15 on 11 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 28 and 36 miles per gallon is: Answer: (Round to four decimal places) ___0.3333 ___ estion 11 feedback goes from 23 < x < 47 < 36) = (36 - 28)*147−23 on 12 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.) ___.25 ___ estion 12 feedback goes from 5 < x < 37 ) = (13 - 5) *137−5 on 13 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 30 minutes on a given day? Answer: (Round to four decimal places.) ___.6923 ___ estion 13 feedback goes from 10 < x < 75 ) = (75 - 30) *175−10
on 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 30 miles per gallon is: Answer: (Round to four decimal places) ___0.2083 ___ estion 14 feedback goes from 23 < x < 47 < 30) = (30 - 25) *147−23 on 15 1 / 1 point The mail arrival time to a department has a uniform distribution over 5 to 45 minutes. What is the probability that the mail arrival time is more than 25 minutes on a given day? Answer: (Round to 2 decimal places.) ___0.50 ___ estion 15 feedback goes from 5 < x < 45 ) = (45 - 25) *145−5 on 16 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 ___ estion 16 feedback
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=.3/SQRT(35) = 0.050709 .7), in Excel .DIST(15.7,15.85,0.050709,TRUE) on 17 1 / 1 point The average amount of a beverage in randomly selected 16-ounce beverage can is 16.18 ounces with a standard deviation of 0.4 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 16.1 ounces of beverage? Answer: (round to 4 decimal places) ___0.0534 ___ estion 17 feedback = .4/SQRT(65) = 0.049614 .1), in Excel .DIST(16.1,16.18,0.049614,TRUE) on 18 1 / 1 point The final exam grade of a statistics class has a skewed distribution with mean of 79.8 and standard deviation of 8.2. If a random sample of 45 students selected from this class, then what is the probability that the average final exam grade of this sample is between 80 and 83? Answer: (round to 4 decimal places) ___0.4306 ___ estion 18 feedback = 8.2/SQRT(45) = 1.222384 < 83), in Excel .DIST(83,79.8,1.222384,TRUE)-NORM.DIST(80,79.8,1.222384,TRUE) on 19 0 / 1 point
The time a student sleeps per night has a distribution with mean 6.15 hours and a standard deviation of 0.5 hours. Find the probability that average sleeping time for a randomly selected sample of 40 students is more than 6.29 hours per night. Answer: (round to 4 decimal places) ___0.0382 ___ (0.0383, .0383) estion 19 feedback = .5/SQRT(40) = 0.079057 29), in Excel M.DIST(6.29,6.15,0.079057,TRUE) on 20 1 / 1 point The final exam grade of a statistics class has a skewed distribution with mean of 79 and standard deviation of 8.2. If a random sample of 35 students selected from this class, then what is the probability that average final exam grade of this sample is between 76 and 82? Answer: (round to 4 decimal places) ___0.9696 ___ Hide question 20 feedback New SD = 8.2/SQRT(35) = 1.386053 P( 76 < x < 82), in Excel =NORM.DIST(82,79,1.386053,TRUE)-NORM.DIST(76,79,1.386053,TRUE)
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