Question #4855

what is the probability that an athlete would clear a distance between 189-198 inches? if the average distance cleared by him during a large number of trail was 180 inches with standard deviation of 6 inches.

Expert's answer

P(x)=exp(-(x-mu)^2/2sigma^2)/(sigma*sqrt(2*mu))

Z=(x-mu)/sigma

mu=180 inches

sigma=6 inches

P(189<x<198)=P((189-180)/6<Z<(198-180)/6)=P(1.5<Z<3)=0.0655

Z=(x-mu)/sigma

mu=180 inches

sigma=6 inches

P(189<x<198)=P((189-180)/6<Z<(198-180)/6)=P(1.5<Z<3)=0.0655

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