Question

We saw two distributions for triathlon times: N(µ = 4313, Ï = 583) for Men, Aged 30 - 34 and N(µ = 5261, Ï = 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 menâs 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 womenâs group.

Answer 1

Answer:

a) 3354 seconds

b) 6294 seconds

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

(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.

$\mu = 4313, \sigma = 583$

The cutoff for the shortest 5% is the 5th percentile, which is X when Z has a pvalue of 0.05. So X when Z = -1.645.

$Z = \frac{X - \mu}{\sigma}$

$-1.645 = \frac{X - 4313}{583}$

$X - 4313 = -1.645*583$

$X = 3354$

Cutoff of 3354 seconds.

(b) The cutoff time for the slowest 10% of athletes in the women's group.

$\mu = 5261, \sigma = 807$

The slowest 10% is the 10% that takes more time, so the cutoff is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9. So X when Z = 1.28.

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 5261}{807}$

$X - 5261 = 1.28*807$

$X = 6294$

Cutoff time of 6294 seconds