Question

5.5 (1 point) Triathlon times. In triathlons, it is common for racers to be placed into age and gender groups. Friends Leo and Mary both completed the Hermosa Beach Triathlon, where Leo competed in the Men, Ages 30 - 34 group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 4841 seconds, while Mary completed the race in 5724 seconds. Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Round all calculated answers to four decimal places.

Here is some information on the performance of their groups:

The finishing times of the Men, Ages 30 - 34 group has a mean of 4345 seconds with a standard deviation of 580 seconds.

The finishing times of the Women, Ages 25 - 29 group has a mean of 5285 seconds with a standard deviation of 831 seconds.

The distributions of finishing times for both groups are approximately Normal.

Remember: a better performance corresponds to a faster finish.

1. Write the short-hand for these two normal distributions.

The Men, Ages 30 - 34 group has a distribution of N(__, __).

The Women, Ages 25 - 29 group has a distribution of N(__, __).

2. What is the Z score for Leo's finishing time? z = __

3. What is the Z score for Mary's finishing time? z = __

5. What percent of the triathletes did Leo finish slower than in his group? __%.

6. What percent of the triathletes did Mary finish faster than in her group? __ %.

7. What is the cutoff time for the fastest 7% of athletes in the men's group, i.e. those who took the shortest 7% of time to finish?

__ seconds

8. What is the cutoff time for the slowest 13% of athletes in the women's group?

__ seconds

Answer 1

The Men, Ages 30 - 34 group has a distribution of N(4345,580²).

The Women, Ages 25 - 29 group has a distribution of N(5285,831²).

Z score for Leo's finishing time

z_Leo = (4841-4345)/580 0.8552

Z score for Mary's finishing time

z_Mary = (5724-5285)/831 0.5283

Let X be the finishing time of a triathlete in the Men, Ages 30 - 34 group.

Percentage of the triathletes Leo finished slower than in his group = P(X<4841)100% = P(Z<z_Leo)100% ; where ZN(0,1)

P(Z<z_Leo)100% = P(Z<0.8552)100% 0.803772100% = 80.3772%

[ The above probability is obtained is obtained from the Standard Normal distribution table. ]

Let Y be the finishing time of a triathlete in the Women, Ages 25 - 29 group.

Percentage of the triathletes Mary finished faster than in his group = P(Y>5724)100% = P(Z>z_Mary)100% ; where ZN(0,1)

P(Z>z_Mary)100% = [1 - P(Z<z_Mary)]100% = [1 - P(Z<0.5283)]100% [1-0.701347]100% = 0.298653100% = 29.8653%

[ The above probability is obtained is obtained from the Standard Normal distribution table. ]

Let X be the finishing time of a triathlete in the Men, Ages 30 - 34 group.

Let cutoff time for the fastest 7% of athletes in the men's group be X_c.

P(X<X_m) = 7% P[(X-4345)/580 < (X_c-4345)/580] = 0.07

X N(4345,580) (X-4345)/580 N(0,1) distribution

From the Standard Normal Probability table it is obtained that :- P[Z<-1.47579 | Z N(0,1)] 0.07

  (X_c-4345)/580 -1.47579   X_c -1.47579580 + 4345 = 3489.0418

Let Y be the finishing time of a triathlete in the Women, Ages 30 - 34 group.

Let cutoff time for the slowest 13% of athletes in the men's group be Y_c.

P(Y>Y_c) = 13% P(Y<Y_c) = 87% P[(Y-5285)/831 < (Y_c-5285)/831] = 0.87

Y N(5285,831) (X-5285)/831 N(0,1) distribution

From the Standard Normal Probability table it is obtained that :- P[Z<1.12639 | Z N(0,1)] 0.87

  (Y_c-5285)/831 1.12639   Y_c 1.12639831 + 5285 = 3489.0301