The times for completing one circuit of a bicycle course are normally distributed with a mean of 72.5 minutes and a standard deviation of 6.5 minutes. An association
wants to sponsor a race but will eliminate the bottom 75% of riders. In a trial run,
what should be the cutoff time?

The answer is 76.9 rounded or rounds.

Step-by-step explanation:

Solution

Given that:

Mean = μ = 72.5

Standard deviation = σ = 6.5

Now

By applying the standard normal table we have the following:

P(Z < z) = 75%

= P(Z < z)

= 0.75

Thus

z =0.67 by using the standard normal z table

Applying the  z-score formula we have:

x = z * σ + μ

x = 0.67*6.5+72.5

x= 76.855 ≈ 76.9

Therefore the cutoff time is 76.9 rounds.

ou would have 12 of each

your answer is the frist one

hop i