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Assuming a normal distribution:

For the fastest 5% we need to find the z value which gives 100% - 5% = 95% of the area under the normal curve (from -∞).

Using single tailed tables, we need the z value which gives 95% - 50% = 0.45 (above the mean); this is found to be z ≈ 1.645

z = (value - mean)/standard deviation

→ value = mean + z × standard deviation

≈ 5 min 17 sec + 1.645 × 12 sec

≈ 5 min 17 sec + 20 s

= 5 min 37 sec

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Q: What is the qualifying time for the fastest 5 percent of all runners if the average finishing time is 5 minutes 17 seconds with standard deviation of 12 seconds?
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