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

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]