Two swimmers have been training for a big race in the 100 meter backstroke. Thei

Question

Two swimmers have been training for a big race in the 100 meter backstroke. Their times are known to be normally distributed with the means and standard deviations in seconds given below. We can also assume that the times of the two swimmers are independent.
Swimmer 1: mean = 76.76 standard deviation= 3.08
Swimmer 2: mean= 77.9 standard deviation= 2.49

a) the swimmers keep track of their progress over the season. What is the mean and standard deviation of the first swimmer's time (X) minus the second swimmer's time (Y) for a randomly selected race?

b) using your results in part (a) what is the probability that the second swimmer will beat the first swimmer on any given race? round your answer to the nearest hundredth. Please note that the lowest time wins.

c) Using your results in part (b) what is the probability that the second swimmer will beat the first swimmer in at least one of 10 races?

Explanation / Answer

Part a. since their times are independent,
    mean=76.76-77.9= -1.14
    SD=(3.08^2+2.49^2)^.5=3.96

Part b.P(X>Y)=P(X-Y>0)=P(Z>1.14/3.960618639)=P(Z>0.287833822)0.387

Part c. Prob=1 - P(first swimmer win all 10 races) = 1- (1-.386737)^10 = 0.992

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