Question

An athletics coach states that the distribution of player run times (in seconds) for a 100-meter dash is normally distributed with a mean equal to 15.00 and a standard deviation equal to 0.2 seconds. What percentage of players on the team run the 100-meter dash in 15.28 seconds or faster? (Round your answer to two decimal places.)
%

Answer 1

Solution :

Given that ,

mean = $\mu$ = 15.00

standard deviation = $\sigma$ = 0.2

P(x > 15.28) = 1 - p( x< 15.28 )

=1- p P[(x - $\mu$ ) / $\sigma$ < (15.28 - 15.00) / 0.2]

=1- P(z < 1.4 )

Using z table,

= 1 - 0.9192

=0.0808

The percentage = 8.08%