Question

normally distributed with mean 24.6 seconds and standard deviation 0.64 seconds;the lower finishing time, the better. If the qualifying time for this event for the regionalchampionship is set so that the top 15 percent of all runners qualify, then thatqualifying time in seconds is about

Answer 1

`z=1.04<(a-24.6)/(0.64)`

And if we solve for a we got

`a=24.6 +1.04*0.64=25.27`

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.

Explanation:

Previous concepts

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".

Solution to the problem

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

`X \sim N(24.6,0.64)`

Where
`\mu=24.6` and
`\sigma=0.64`

Solution to the problem

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.15` (a)

`P(X<a)=0.85` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.85 of the area on the left and 0.15 of the area on the right it's z=1.04. On this case P(Z<1.04)=0.85 and P(z>1.04)=0.15

If we use condition (b) from previous we have this:

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.85`

`P(z<(a-\mu)/(\sigma))=0.85`

But we know which value of z satisfy the previous equation so then we can do this:

`z=1.04<(a-24.6)/(0.64)`

And if we solve for a we got

`a=24.6 +1.04*0.64=25.27`

So the value of height that separates the bottom 85% of data from the top 15% is 25.27.