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

Answer:

$z=1.036$

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.  

Step-by-step 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)

$P(X$   (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:

$P(X$  

$P(z$

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

$z=1.036$

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.