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 13.00 and a standard deviation equal to 0.2 seconds. What percentage of players on the team run the 100-meter dash in 13.28 seconds or faster

Answer 1

`P(X<13.28)=P((X-\mu)/(\sigma)<(13.28-\mu)/(\sigma))=P(Z<(13.28-13)/(0.2))=P(z<1.4)`

And we can find this probability using the normal standard distribution table and we got:

`P(z<1.4) =0.919`

And the percentage faster than 13.28 seconds would be 91.9%

Explanation:

Let X the random variable that represent the runtimes of a population, and for this case we know the distribution for X is given by:

`X \sim N(13,0.2)`

Where
`\mu=13` and
`\sigma=0.2`

We want to find this probability:

`P(X<13.28)`

And we can use the z score formula given by:

`z=(x-\mu)/(\sigma)`

Using this formula we have:

`P(X<13.28)=P((X-\mu)/(\sigma)<(13.28-\mu)/(\sigma))=P(Z<(13.28-13)/(0.2))=P(z<1.4)`

And we can find this probability using the normal standard distribution table and we got:

`P(z<1.4) =0.919`

And the percentage faster than 13.28 seconds would be 91.9%