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?

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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Answer :

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 ≈ 1.645z = (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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