Let X = the average of the 49 races.

Question

Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 144 minutes with a standard deviation of 13 minutes. Consider 49 of the races.

Let X

 = the average of the 49 races.

Part (a)

Give the distribution of X. (Round your standard deviation to two decimal places.)

1. Xbar  ~ _____ (_____, _____) 

  ,  

Part (b)

Find the probability that the average of the sample will be between 143 and 147 minutes in these 49 marathons. (Round your answer to four decimal places.)_____

Part (c)

Find the 70th percentile for the average of these 49 marathons. (Round your answer to two decimal places.)

1. _____ min

Part (d)

Find the median of the average running times.

1. _____ min

Answer 1

a) xbar has a mean of 144 and a SD of 13/sqrt(49) = 1.86

b) use transformation to std normal distribution: z = (xbar - pop mean)/SD xbar = (xbar - 144)/1.86

P(143 < xbar < 147) = P((143-144)/1.86 < z < (147-144)/1.86) =

P(-0.54 < z < 1.61) =

P(z < 1.61) - P(z < -0.54) =

0.9463 - 0.2946 = 0.6517

c) z for 70th percentile is z associated with 0.7 probability = 0.525.

using above z transformation:

0.525 = (xbar - 144)/1.86 ---> xbar = 144.98

d) For the standard normal distribution mean = median = mode = 144.