Question

Suppose that finishing times of runners in a local 10K race follow an approximate normal distribution with mean 61 minutes and standard deviation 9 minutes.

(a) Melissa finished the race in 52 minutes. How many standard deviations below average is Melissa's time?

(b) What is the probability that a randomly selected runner completed the race in between 43 and 52 minutes?

(c) Jennifer's finishing time was 79 minutes. What percentage of runners had faster times than Jennifer?

(d) The fastest 16% of runners finished the race in less than how many minutes?

Answer 1

Solution :

Given that ,

mean = = 61

standard deviation = = 9

(b) P(43< x < 52) =P[(43 - 61) / 9< (x - ) / < (52 - 61) / 9)]

= P( -2< Z <-1 )

= P(Z <-1 ) - P(Z <-2 )

Using z table,  

= 0.1587 - 0.0228  

=0.1359

(c)P(x >79 ) = 1 - P(x < 79)

= 1 - P[(x - ) / < (79 - 61) /9 ]

= 1 - P(z <2 )

Using z table,

= 1 -0.9772

=0.0228

(d)Using standard normal table,

P(Z > z) = 16%

= 1 - P(Z < z) = 0.16

= P(Z < z) = 1 - 0.16

= P(Z < z ) = 0.84

= P(Z < 0.99 ) = 0.84  

z =0.99

Using z-score formula,

x = z * +

x = 0.99 * 9+61

x = 69.91