Question

We saw two distributions for triathlon times: N (μ = 4313, σ = 583) for Men, Ages 30 - 34 and N (μ = 5261, σ = 807) for the Women, Ages 25 - 29 group. Times are listed in seconds. Use this information to compute each of the following:
a. The cutoff time for the fastest 5% of athletes in the men's group, i.e. those who took the shortest 5% of time to finish.
b. The cutoff time for the slowest 10% of athletes in the women's group.

Answer 1

Answer:

Step-by-step explanation:

Hello!

Using the information of the triathlon times of men and women you have to find:

a. The cutoff time for the fastest 5% men, if the times of the male triathletes have a normal distribution, this value will separate the top 5% from the rest of the distribution, symbolically:

P(X≥x₀)= 0.05

Where x₀ represents the cutoff time that you need to find.

Now if this value has 5% of the distribution above it, then 95% of the distribution will be below it, symbolically:

P(X<x₀)= 0.95

Next is to use the standard normal distribution, you have to find the Z value that accumulates 0.95 of probability: P(Z<z₀)= 0.95

z₀= 1.648

Using the formula: Z= (X-μ)/δ what is left to do is to reverse the standardization and reach the value of the variable:

z₀= (x₀-μ)/δ

x₀= (z₀*δ)+μ

x₀= (1.648*583)+4313

x₀= 5273.784 seconds

The cutoff time for the fastest 5% is 5273.784 seconds.

b. The cutoff time for the slowest 10% women, if you are taking the "slowest" into account, then this is the lowes 10% of the distribution. Meaning that the value of time separates the "lowest" 10% of the distribution from the rest, symbolically:

P(X≤x₀)=0.10

Same as before, you have to use the standard normal distribution:

P(Z≤z₀)= 0.10

z₀= - 1.283

Now using this value you have to reverse the standardization:

z₀= (x₀-μ)/δ

x₀= (z₀*δ)+μ

x₀= (-1.283*807)+5261

x₀= 4225.619 seconds

The cutoff time for the slowest 10% is 4225.619 seconds.

I hope it helps!