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

Answer:

$P(X<13.28)=P(\frac{X-\mu}{\sigma}<\frac{13.28-\mu}{\sigma})=P(Z<\frac{13.28-13}{0.2})=P(z<1.4)$

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%

Step-by-step 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:

$z=\frac{x-\mu}{\sigma}$

Using this formula we have:

$P(X<13.28)=P(\frac{X-\mu}{\sigma}<\frac{13.28-\mu}{\sigma})=P(Z<\frac{13.28-13}{0.2})=P(z<1.4)$

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%