. During a recent track meet, the average time for all people who ran the 100-meter...

. During a recent track meet, the average time for all people who ran the 100-meter dash was 11.4 seconds with a standard deviation of 0.6 seconds. Assuming the times were approximately normally distributed:

a. What percentage of runners finished in less than 10.5 seconds?

b. What percentage of runners finished in greater than 12 seconds?

c. If a runner wanted to be in the top (fastest) 20% of times, what time would she or he need to beat?

d. What percentage of runners finished between 10.5 and 12 seconds?

Solutions

Expert Solution

P(z<Z) table :

z = (10.5-11.4)/0.6 = -1.5

P(z<-1.5) = 0.0668 {from table}

P(x<10.5) = 0.0668

z = (12-11.4)/0.6 = 1

P(z>1) = 1 - P(z<1)

= 1 - 0.8413

P(x>12) = 0.1587

P(z<Z) = 1 - 0.20 = 0.80

Z = 0.84

time = mean+ 0.84*SD

= 11.4 + 0.84*0.6

= 11.904 seconds

P(10.5<x<12) = 1 - P(x>12) - P(x<10.5)

= 1 - 0.1587 - 0.0668

= 0.7745

(please upvote)

orchestra answered 1 year ago

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