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 - 34group while Mary competed in the Women, Ages 25 - 29 group. Leo completed the race in 4989 seconds, while Mary completed the race in 5790 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 4302 seconds with a standard deviation of 598 seconds.
The finishing times of the Women, Ages 25 - 29 group has a mean of 5210 seconds with a standard deviation of 846 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 (4302 , 598).
The Women, Ages 25 - 29 group has a distribution of N(5210 , 846).
2. What is the Z score for Leo's finishing time? z = 1.1488
3. What is the Z score for Mary's finishing time? z = 0.6856
4. Did Leo or Mary rank better in their respective groups?
A. They ranked the same
B. Mary ranked better
C. Leo ranked better
B.
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 28% of athletes in the men's group, i.e. those who took the shortest 28% of time to finish?
__?___ seconds
8. What is the cutoff time for the slowest 15% of athletes in the women's group?
__?___seconds
1)
Short hand for men group N(4302, 598)
For women group N(5210, 846)
2)
Z = (4989-4302)/598 = 1.1488
3)
Z = (5790-5210)/846 = 0.6856 or 0.69
4)
As Z score is less for mary, So mary ranked better than the LEO
5)
P(X > 4989) = 1 - P(X < 4989)
= 1 - (z < (4989 - 4302)/598)
= 1 - P(z < 0.15)
= 0.4404
from Z -table we can find p-value
P(X > 4989) = 0.4404
6)
P(X < 5708) = P( < (5708 - 5210)/846)
At Z = 0.59 we found p-value
P(X < 5708) = 0.7224
7)
p = 0.28 so Z = -0.5828
-0.5828 = (X - 4302)/598
X = 3953.4856
8)
p= 0.85, so Z score = 1.04
1.04 = (X-5210)/846
X = 5210+879.84 = 6089.84
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