In: Statistics and Probability
. 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?
P(z<Z) table :
a.
z = (10.5-11.4)/0.6 = -1.5
P(z<-1.5) = 0.0668 {from table}
P(x<10.5) = 0.0668
b.
z = (12-11.4)/0.6 = 1
P(z>1) = 1 - P(z<1)
= 1 - 0.8413
P(x>12) = 0.1587
c.
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
d.
P(10.5<x<12) = 1 - P(x>12) - P(x<10.5)
= 1 - 0.1587 - 0.0668
= 0.7745
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