Assume that the finishing times in a New York City 10-kilometer road race are normally distributed with a mean of 61minutes and a standard deviation of 9 minutes. let X be a randomly selected finishing time. Find (a) P(X72) (b.) P(52

Question

Assume that the finishing times in a New York City 10-kilometer road race are normally distributed with a mean of 61minutes and a standard deviation of 9 minutes. let X be a randomly selected finishing time. Find (a) P(X>72) (b.) P(52<X<70) (c.) Find the 95th percentile point

amistre64 · Answer

yay stats!!

amistre64 · Answer

gonna have to ti83 it for the p95 tho since there is no data

amistre64 · Answer

invnorm(.95,61,9) i think

amistre64 · Answer

75.8 is the p95 if i did it right

amistre64 · Answer

we got any choices to go by?

amistre64 · Answer

68% fall inbetween 52 and 70 since it is 1sd from the mean

amistre64 · Answer

1-.8888 might be the answer for A

amistre64 · Answer

61-72
------ = z score ; and zcore table says 1.2 qnd .02 = .8888 = 88%
   9

amistre64 · Answer

so above that is greater than 72

anonymous · Answer

how did you get 75.8 for the p95

amistre64 · Answer

i had to use my ti83 calculator; but I could probably get to it with a ztable

amistre64 · Answer

look for the number that is closest to .95, or in my book I gotta subtract .5 ......

amistre64 · Answer

z = (x-mean)/sd; so x = 

z(sd) +mean

amistre64 · Answer

my ztable says z = 1.64

amistre64 · Answer

1.64(9) + 61 = 75.76

amistre64 · Answer

on the calculator; i just go:

2nd -> vars -> invnorm(percentile,mean,sd) -> enter