A physical fitness association is including the mile run in its secondary school

Question

A physical fitness association is including the mile run in its secondary school fitness test. The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 440 seconds and a variance of 50 seconds. a. Find the probability that a randomly selected boy in secondary school completes the mile run in less than 325 seconds? b. What proportion of boys in secondary school completes the mile run in between 325 seconds and 500 seconds? c. Calculate the 90^th percentile of the time that a boy in secondary school completes the mile run.

Explanation / Answer

mean = 440 seconds

variance = 50 seconds2

standard deviation = sqrt(50) = 7.071

a) Z = (325 - 440)/7.071 = -16.264

P(X < 325) = P(Z < -16.264) = 0

b) P(325 < X < 500) = P(X < 500) - P(X < 325)

Z value for X = 500 is (500 - 440)/7.071 = 8.485

P(325 < X < 500) = P(X < 500) - P(X < 325)  = P(Z < 8.485) - P(Z < -16.264) = 1 - 0 = 1

So 100% of boys complete the mile run between 325 sec and 500 sec.

c) 90th percentile means that 90% of the time are smaller

that is we need to find, P(X > x) = P(Z > z) = 0.9

z-value for 0.9 is 1.282

so, (X- 440)/7.071 = 1.282

gives, X = 449.055

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