Question

In order to quality for police academy training, recruits are tested for stress tolerance. The scores are normally distributed, with a mean of 50 and SD of 12. If only the top 15% of recruits are selected, find the cutoff score. Note: make a picture of a bell-shaped distribution, show all information on it

Answer 1

`z=1.036<(a-50)/(12)`

And if we solve for a we got

`a=50 +1.036*12=62.432`

So the value of height that separates the bottom 85% of data from the top 15% is 62.432.

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 scores of a population, and for this case we know the distribution for X is given by:

`X \sim N(50,12)`

Where
`\mu=50` and
`\sigma=12`

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=1.036. On this case P(Z<1.036)=0.85 and P(z>1.036)=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=1.036<(a-50)/(12)`

And if we solve for a we got

`a=50 +1.036*12=62.432`

So the value of height that separates the bottom 85% of data from the top 15% is 62.432.