Question

A running shoe company wants to sponsor the fastest 5% of runners. You know that in this race, the running times are normally distributed with a mean of 7.2 minutes and a standard deviation of 0.56 minutes. How fast would you need to run to be sponsored by the company?

Answer 1

Answer:

We need a time of 8.118 minutes or higher in order to be sponsored by the company

Explanation:

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

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

$X \sim N(7.2,0.56)$  

Where $\mu=7.2$ and $\sigma=0.56$ minutes.  

And we want the fastest 5% of runners, so we need a value a such that:

$P(X>a)=0.05$ or $P(X$

We need on the right tail of the distribution a value a that gives to us 95% of the area below and 5% of the area above. Both conditions are equivalent.

Let's use the condition $P(X$, the best way to solve this problem is using the z score with the following formula:

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

So we need a value from the normal standard distribution that accumulates 0.95 of the area on the left and 0.05 on the right. This value on this case is 1.64 and we can founded with the following code in excel:

"=NORM.INV(0.95,0,1)"

If we apply the z score formula to our case we have this:

$P(X$

So then based on the equalities we have this:

$\frac{a-7.2}{0.56}=1.64$

And if we solve for a we got:

$a=(0.56*1.64) +7.2=8.118$

So then we need a time of 8.118 minutes or higher in order to be sponsored by the company