Suppose the time to complete a 200-meter backstroke swim for female competitive swimmers is normally distributed with a mean μ =
141 seconds and a standard deviation σ = 7 seconds. Suppose that to qualify for the Nationals, a woman must complete the 200-meter backstroke in less than 128 seconds. What proportion of competitive female swimmers will qualify for the Nationals? Give your answer to four (4) decimal places.
Answer: The proportion that will qualify is 0.0314 or 3.14%.
First, we need to find the z-score of a time of 128 seconds. To do this, we find the difference of the mean and score and divide by the standard deviation.
(128 - 141) / 7 = -1.86
Now, use a standard normal distribution table to determine the percent below a z-score of -1.86. That value is 0.0314 or 3.14%.
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