3) High school track students running a race have times which are approximately normally distributed with mean 14.3s and variance 2.2s². In a large competition, the top 15% of athletes qualify for the next round. Estimate the required qualifying time. Solution? The variance is 2.2s? → the standard deviation is 2.2 s. Since we want the best 15%, that means the area to the left of the value will be 85% or 0.85 We use the inverse normal function with an area of 0.85, a mean of 14.3, and a standard deviation of v2.2, to get a time of approximately 15.8 seconds.

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Author:Amos Gilat
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3) High school track students running a race have times which are approximately normally distributed with mean
14.3s and variance 2.2s². In a large competition, the top 15% of athletes qualify for the next round. Estimate the
required qualifying time.
Solution?
The variance is 2.2s² → the standard deviation is 2.2 s.
Since we want the best 15%, that means the area to the left of the value will be 85% or 0.85
We use the inverse normal function with an area of 0.85, a mean of 14.3, and a standard deviation of v2.2, to get a
time of approximately 15.8 seconds.
Transcribed Image Text:3) High school track students running a race have times which are approximately normally distributed with mean 14.3s and variance 2.2s². In a large competition, the top 15% of athletes qualify for the next round. Estimate the required qualifying time. Solution? The variance is 2.2s² → the standard deviation is 2.2 s. Since we want the best 15%, that means the area to the left of the value will be 85% or 0.85 We use the inverse normal function with an area of 0.85, a mean of 14.3, and a standard deviation of v2.2, to get a time of approximately 15.8 seconds.
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