AP Statistics help? Free response question

Question

Andrea and Ashley are middle-distance runners for their school's track team. Andrea's times in the 400-meter race are approximately Normally distributed with a mean of 62 seconds and a standard deviation of 0.8 seconds. Ashley's times are approximately Normally distributed with a mean of 62.8 seconds and a standard deviation of second.

(a) The school record for the 400-meter is 60.0 seconds. What is the probability that Andrea breaks the record in her next race?

(b) Assuming that Andrea and Ashley's times in any race are independent , what is the probability that Ashley beats Andrea in their next race?

Answer 1

Andrea has mean 62 and sd, so P(x <= 60) = P(z <= (60-62)/.8) = P(z <= 2.5) = norm.s.dist(-2.5,1) = .00621

Ashley has mean 62.8 and sd 1, so P(x <= 60) = P(z <= (60-62.8)/1) = P(z <= -2.8) = norm.s.dist(-2.8,1) = .002555

The mean difference for Ashley - Andrea = 62.8 - 62 = .8

Variance of the difference, given independence, is .8^2 + 1^2 = 1.64

Ashley is faster if the difference is less than 0.

Then, P(x <= 0) = P(z <= (0 - .8)/sqrt(1.64) ) = norm.s.dist(-.8/sqrt(1.64),1) = .2661