parts a-f 2.18 Triathlon times, Part I. In triathlons, it is common for racers t
ID: 2929359 • Letter: P
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parts a-f
2.18 Triathlon times, Part I. 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 1:22:28 (4948 seconds), while Mary completed the race in 1:31:53 (5513 seconds). Obviously Leo finished faster, but they are curious about how they did within their respective groups. Can you help them? Here is some information on the performance of their groups: The finishing times of the Men, Ages 30 - 34 group has a mean of 4313 seconds with a standard deviation of 583 seconds. » The finishing times of the Women, Ages 25 - 29 group has a mean of 5261 seconds with a standard deviation of 807 seconds. » The distributions of finishing times for both groups are approximately Normal Remember: a better performance corresponds to a faster finish (a) Write down the short-hand for these two normal distributions (b) What are the Z scores for Leo's and Mary's finishing times? What do these Z scores tell you? (c) Did Leo or Mary rank better in their respective groups? Explain your reasoning. (d) What percent of the triathletes did Leo finish faster than in his group? (e) What percent of the triathletes did Mary finish faster than in her group? (f) If the distributions of finishing times are not nearly normal, would your answers to parts (b) (e) change? Explain your reasoningExplanation / Answer
(A) for Ages 30 - 34 group, normal distribution ~Norm(X, 4313 , 583)
for Ages 25-29 group, normal distribution ~ Norm(Y, 5261, 807)
(B) Z - score for Leo's running time
Z = (4948 - 4313)/ 583 = 1.09
Z - score for Mary's running time
Z= (5513 - 5261)/ 807 = 0.31
(C) No, Leo and Mary doesn't rank better in their respective group. as their Z value are positive in both cases.
(D) Pr(Athletes running time < 4948; 4313; 583) = Pr(Z < 1.09)
from Z- table
Pr(Z < 1.09) = 0.8621
so, 86.21 % of the triathletes leo finish faster in his group.
(e)
Pr(Athletes running time < 5513; 5261; 807) = Pr(Z < 0.31)
from Z- table
Pr(Z < 0.31) = 0.6217
so, 62.17 % of the triathletes finish faster than Mary in her group.
(f) If the distribution is not normal, answers to parts(b)- (e) will change but not that much as sample size is large enough
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