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World class marathon runners are known to run that distance (26.2 miles) in an average of 143 minutes with a standard deviation of 13 minutes.If we sampled a group of world class runners from a particular race, find the probability of the following:**(use 4 decimal places)**a.) The probability that one runner chosen at random finishes the race in less than 137 minutes. b.) The probability that 10 runners chosen at random have an average finish time of less than 137 minutes. c.) The probability that 50 runners chosen at random have an average finish time of less than 137 minutes.

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EXPLANATION

We have μ = 144 and sd = 13

Computing the needed probabilities:

a) P(x < 137 ) =

[tex]=P(\frac{X-\mu}{\sigma}<\frac{137-\mu}{\sigma})[/tex][tex]=P({Z}\lt\frac{137-143}{13})[/tex][tex]=P(z<-0.4615[/tex][tex]=0.3228[/tex]

a) The probability is 0.3228

b) Number of runners ---> n=10

P(x < 137 ) =

[tex]=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{137-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex][tex]=P({Z}\lt\frac{137-143}{\frac{13}{\sqrt{10}}})[/tex][tex]=P(z<-1.4595)[/tex][tex]=0.0735[/tex]

b) The probability is 0.0735

c) Number of runners ---> n=50

P(x < 137 ) =

[tex]=P(\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{137-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex][tex]=P({Z}\lt\frac{137-143}{\frac{13}{\sqrt{50}}})[/tex][tex]=P(z<-3.2635)[/tex][tex]\approx0.0006[/tex]

b) The probability is approximately 0.0006

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