Question

This week, a very large running race (5K) occured in Denver. The times were normally distributed, with a mean of 24.03 minutes and a standard deviation of 3.04 minutes. a. What percent of runners took 26.614 minutes or less to complete the race? b. What time in minutes is the cutoff for the fastest 99.38 %? c. What percent of runners took more than 31.934 minutes to complete the race?

Answer 1

Solution :

Given that ,

mean = = 24.03

standard deviation = = 3.04

P(X<26.614 ) = P[(X- ) / < (26.614 -24.03) / 3.04]

= P(z <0.85 )

Using z table

= 0.8023

answer=80.23

(b)

Using standard normal table,

P(Z > z) = 99.38 %

= 1 - P(Z < z) = 0.9938

= P(Z < z ) = 1 - 0.9938

= P(Z < z ) = 0.0062

= P(Z < -2.50 ) = 0.0062

z =-2.50 (using standard normal (Z) table )

Using z-score formula  

x = z * +

x=- 2.50* 3.04+24.03

x= 16.43

x=16

(c)

P(x >31.934 ) = 1 - P(x<31.934 )

= 1 - P[(x -) / < (31.934-24.03) / 3.04]

= 1 - P(z <2.6 )

Using z table

= 1 - 0.9953

= 0.0047

answer=0.47%