Assume that finishing times in a marathon have a distribution that is mound-shap

Question

Assume that finishing times in a marathon have a distribution that is mound-shaped and symmetric with a mean of 4.72 hours and a standard deviation of 1.42 hours Draw and label a picture of the distribution, showing the mean time and times one standard deviation below and one standard deviation above the mean. Determine the z-score of a runner who finishes the marathon in 3.4 hours. Determine the proportion of runners who finish the marathon faster than (i.e., in less than) 3.4 hours. If a runner is 1.6 standard deviations slower than the mean, what is her finishing time? Determine the time for which 75% of runners are slower.

Explanation / Answer

b)zscore =(X-mean)/std deviation =(3.4-4.72)/1.42=-0.9296

c)P(X<3.4)=P(Z<-0.9296) =0.1763

d)as z score is 1.6; hence time =4.72+1.6*1.42=6.992

e)for 75% runners to be slower, at 25% z=-0.6745

hence corresponding time =4.72-0.6745*1.42=3.7622

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