At least one of the answers above is NOT correct. As reported in \"Runner\'s wor

Question

At least one of the answers above is NOT correct. As reported in "Runner's world' magazine, the times of the finishers in the New York Cty10 km run are normally distributed with mu = 61 minutes and sigma = 9 minutes. Let X denote finishing time for the finishers. The percentage of finishers with times between 60 and 90 minutes is equal to the area under the standard normal curve between and The percentage of finishers with times exceeding 84 minutes is equal to the area under the standard normal curve that lies to the of

Explanation / Answer

a) P(60<X<90 ) =P((60-61)/9<Z<(90-61)/9)= P(-0.1111<Z<3.2222) =b/w 0.4558 and 0.9994 or 45.58% to 99.94%

b)P(X>85)=P(Z>(85-61)/9)=P(Z>2.6667) hence to the right of 2.6667

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