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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 64 min and a standard deviation of 5 min. To the nearest minute, what is the qualifying time for the race?

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OZARKIUS
it is within 5 minutes ahead or behind 5 minutes of 64 minutes because deviantion means a different face value than the original hope this helps
if i had some answer choices i could give you the exact answer but here is my best answer for you.

The qualifying time for the race is 59 minutes if the running times are normally distributed, with a mean of 64 min and a standard deviation of 5 min.

What is a normal distribution?

It is defined as the continuous distribution probability curve which is most likely symmetric around the mean. At Z=0, the probability is 50-50% on the Z curve. It is also called a bell-shaped curve.

We have:
Mean = 64 min

Standard deviation = 5 min

As it is given:

[tex]\rm P(z)=16\% \ where \ z =\dfrac{(t-63)}{4}[/tex]

As we know from the Normal Distribution table, a left tail of 16% implies

z=-0.9945

Now,

t=-4 x 0.9945 + 63

t =59.022 ≈ 59 minutes

Thus, the qualifying time for the race is 59 minutes if the running times are normally distributed, with a mean of 64 min and a standard deviation of 5 min.

Learn more about the normal distribution here:

brainly.com/question/12421652

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