SOLUTION: Finishing times in the annual cross country meet hosted by Altoona High School have a symmetrical distribution with a mean of 22.7 minutes and a standard deviation of 3.22 minutes.
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Question 1196427: Finishing times in the annual cross country meet hosted by Altoona High School have a symmetrical distribution with a mean of 22.7 minutes and a standard deviation of 3.22 minutes.
What number would correctly complete the sentence below?
Approximately 16% of the runners in the Altoona High School cross country meet have finishing times below _____ minutes. Round your answer to one decimal place.
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According to the Empirical rule, roughly 68% of the population from a normal distribution is between z = -1 and z = 1
I.e. about 68% of the population is within 1 standard deviation of the mean.
P(-1 < z < 1) = 0.68 approximately
This leaves 1-0.68 = 0.32 as the area of the two tails combined
P(z < -1 or z > 1) = 0.32 approximately
One tail is 0.32/2 = 0.16
About 16% of the area is to the left of z = -1
P(z < -1) = 0.16 approximately
mu = 22.7 = mean
sigma = 3.22 = standard deviation
Let's find the raw score x based on the z score z = -1
So,
z = (x - mu)/sigma
-1 = (x - 22.7)/(3.22)
-1*3.22 = x - 22.7
-3.22 = x - 22.7
-3.22 + 22.7 = x
19.48 = x
x = 19.48
x = 19.5
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