Question 357731: As reported in Runner’s World magazine, the times of the finishers in the New York City 10K Run are normally distributed with a mean of 61.2 minutes and a standard deviation of 9.3 minutes.
a. A runner is selected at random. Find the probability that the runner’s finishing time was more than 75 minutes.
b. A runner is selected at random. Find the probability that the runner’s finishing time was between 60 and 63 minutes
c. Runners whose finishing times are in the top 5% are automatically invited to participate in next year’s race. What finishing time separates the top 5% from the others? Round answer to 1 decimal place.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! As reported in Runner’s World magazine, the times of the finishers in the New York City 10K Run are normally distributed with a mean of 61.2 minutes and a standard deviation of 9.3 minutes.
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a. A runner is selected at random. Find the probability that the runner’s finishing time was more than 75 minutes.
z(75) = (75-61.2)/9.3 = 1.4839
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P(x > 75) = P(z > 1.4839) = normalcdf(1.4839,100) = 0.0689
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b. A runner is selected at random. Find the probability that the runner’s finishing time was between 60 and 63 minutes
Find the z-value of 60 and of 63.
Find the probability that z is between those z-values.
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c. Runners whose finishing times are in the top 5% are automatically invited to participate in next year’s race. What finishing time separates the top 5% from the others? Round answer to 1 decimal place.
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find the z-value with a left tail of 5%.
invNorm(0.95) = -1.645
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Find the x-value coressponding to that z-value:
x = zs + u
x = -1.645*9.3+61.2 = 45.90 seconds
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Cheers,
stan H.
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