# 12 the weights of adult men are approximately

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12.The weights of adult men are approximately normally distributed with a mean of 190 pounds and a standarddeviation of 30 pounds.(a) If you randomly select three men, what are the mean and standard deviation of the sum of their weights?(b) An elevator in a small apartment building has a maximum weight capacity of 700 pounds.If three(randomly selected) adult men get on the elevator, what is the probability that they exceed the maximumcapacity?
Starnes/Tabor,Updated Version of The Practice of Statistics, 6e© 2020 BFW Publishers, Inc.13.The Internal Revenue Serviceestimates that 8% of all taxpayers filling out forms make mistakes.(a) An IRS employee randomly selects and checks 10 forms for mistakes.What is the probability thatexactly one has an error?(b) The same IRS employee announces at lunch one day that she is going to check 25 randomly selectedforms after lunch. Calculate the mean and the standard deviation of the number of forms with mistakesshe should expect to find.(c) The IRS employee checked 25 randomly selected forms and didn’t find any mistakes. Does this provideconvincing evidence to cause her supervisor to believe that she is missing mistakes?Explain.
DO NOT POST THESE ANSWERS ONLINE© BFW Publishers 2020Starnes/Tabor,Updated Version of The Practice of Statistics, 6e© 2020 BFW Publishers, Inc.Chapter 6 SolutionsQuiz 6.1 Solutions1.(a) The histogram is given here.(b)P(X> 0) is the probability that Dave mowed at least 1 lawn on a randomly selected day.P(X> 0) = 0.4 + 0.3 + 0.1 = 0.8.(c) “Mows at most one lawn” is the eventX1.P(X1) = 0.2 + 0.4 = 0.6.2.(a) Dave charges \$32 for mowing and trimming with probability 0.85 and \$20 for just mowing with probability 0.15.Chargeci\$32\$20Probabilitypi0.850.15(b)The expected value ofCis(\$32)(0.85)(\$20)(0.15)\$30.20.Cµ=+=If many, many customers are randomly selected, the average charge would be about \$30.20.(c)222(3230.2) (0.85)(2030.2) (0.15)18.36.Cσ=+=So18.36\$4.28.Cσ==If many, many customers are randomly selected, the amount charged will typically vary about \$4.28 from the mean of\$30.20.Quiz 6.2 Solutions1.0.50.50.50.52.Tµ=+++=22220.50.50.50.51σ=+++=T2.Shape:Approximately NormalCenter:32373242TXµµ=++=++=minutes.Variability:3TXσσ=minutes.3.LetD= the difference in 1 mile race times andAbe Alyse’s time andJbe Jocelyn’s time. ThenD=AJThe mean is13.5121.5DAJµµµ===minutes.Because the random variables are independent, the standarddeviation is()()22222.51.52.92DAJσσσ=+=+=minutes.Because each individual’s time is approximatelyNormally distributed,Dis also approximately Normally distributed.ThusDfollows a Normal distribution with amean of 1.5 and a standard deviation of 2.92. We want to findP(AJ< 0) orP(D< 0).
DO NOT POST THESE ANSWERS ONLINE© BFW Publishers 2020Starnes/Tabor,Updated Version of The Practice of Statistics, 6e© 2020 BFW Publishers, Inc.1. Draw a Normal distribution.2. Perform calculations.(i)01.50.512.92z== −.Using Table A:P(Z <–0.51) = 0.3050.Using technology:P(Z< –0.51) = normalcdf(lower: –1000, upper: –0.51, mean: 0, SD: 1) = 0.3050.(ii)P(D< 0) = normalcdf(lower: –1000, upper: 0, mean: 1.5, SD: 2.92) = 0.3037.There is a 0.3037 probability that Alyse beats Jocelyn in a 1-mile race on a randomly selected day.Quiz 6.3 Solutions1.(a) This is a geometric setting.

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