To qualify as a contestant in a race a runner has to be in the fastest 16% of all applicants. The running times are normally distributed, with a mean of 61 min and a standard deviation of 5 min. To the nearest minute, what is the qualifying time for the race?



Answer :

Answer:

66 minutes

Step-by-step explanation:

Let X be the running time of a contestant in the race.

Given that the running times are normally distributed, with a mean (μ) of 61 min and a standard deviation (σ) of 5 min, then:

X ~ N(61, 5²)

To find the qualifying time for the race, we need to determine the running time that corresponds to the top 16% of all applicants, so we need to find the value of a for the following:

P(X ≥ a) = 0.16

P(X ≤ a) = 1 - 0.16 = 0.84

Input the following in the inverse normal function on a calculator:

  • Area = 0.84
  • σ = 5
  • μ = 61

Therefore:

a = 65.9722855... = 66

So:

P(X ≥ 66) ≈ 0.16

Therefore, the qualifying time for the race to the nearest minute is approximately 66 minutes.