To qualify as a contestant in a race, a runner has to be in the top 16% of all entrants. the running times are normally distributed, with a mean of 63 min and a standard deviation of 4 min. to the nearest minute, what is the qualifying time for the race

Find P(z)=16% where z =(t-63)/4

And from the Normal Distribution tables you find that a left tail of 16% implies z=-0,9945 then:

t=-4 x 0.9945 + 63=59.22 and the solution would be 59 minutes

according to egenuity the answer is

step-by-step explanation:

d. 344.22

answer by yourhope:

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