Prof. Miller ran a marathon. Given that the mean finishing time for all runners

Question

Prof. Miller ran a marathon. Given that the mean finishing time for all runners was 256 minutes with a standard deviation of 46 minutes and that finishing times were approximately normally distributed, find (a) the percent of finishers who ran faster than Prof. Miller’s time of 3 hr 30 min. (b) a time that was the 15th percentile (15% of finishers were under that time) (c) the percent of runners who finished between 3 hours and 4 hours (d) a random sample of 32 runners got a special vitamin supplement right before the race. Their mean time was 4 hr 2 min. What is the probability that a group of 32 runners chosen at random would have a mean time of 4 hr 2 minutes or less?

Explanation / Answer

a)percent of finishers who ran faster than Prof. Miller’s time of 3 hr 30 min(210 mins) =P(X>210)=1-P(X<210)

=1-P(Z<(210-256)/46)=1-P(Z<-1)=1-0.1587=0.8413 ~84.13%

b)

for 15th percentile ; z =-1.0364

therefore corresponding time =mean +z*Std deviaiton =256-1.0364*46=208.32 minute

c)

percent of runners who finished between 3 hours and 4 hours=P(180<X<240)=P((180-256)/46<Z<(240-256)/46)

=P(-1.6522<Z<-0.3478)=0.3640-0.0492 =0.3147

d)

here std error of mean =std deviation/(n)1/2 =46/(32)1/2 =8.1317

hence probability that a group of 32 runners chosen at random would have a mean time of 4 hr 2 minutes or less

=P(Xbar<242)=P(Z<(242-256)/8.1317)=P(Z<-1.7217)=0.0426

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