The Leadville 100, also called the "Race across the sky," is a 100-mile race through the Colorado Rockies. The race director reports that approximately 5% of runners finish the race in less than 24 hours. Let X be the number of runners who finish the race in less than 24 hours. If 900 runners sign up for the race, what is the standard deviation of X? Assume that the runners’ times are independent of one another.

For each runner, there are only two possible outcomes. Either they finish the race is less than 24 hours, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, with p probability and X can only have two outcomes.

Has a standard deviation of:

[tex]\sqrt{Var(X)} = \sqrt{np(1-p)}[/tex]

In this problem, we have that:

5% of runners finish the race in less than 24 hours. This means that [tex]p = 0.05[/tex].

If 900 runners sign up for the race, what is the standard deviation of X?

We have [tex]n = 900[/tex].

[tex]\sqrt{Var(X)} = \sqrt{np(1-p)} = \sqrt{900*0.05*0.95} = 6.5383[/tex]

The standard deviation of X is 6.5383.