Question

A bank's loan officer rates applicants for credit.

The ratings are normally distributed with a mean of 200 and a standard deviation of 50.

Find P60, the score which separates the lower 60% from the top 40%.

Round your answer to the nearest tenth.

Answer 1

`a=200 +0.253*50=212.65`

So the value of height that separates the bottom 60% of data from the top 40% is 212.65. And rounded would be 212.7.

Explanation:

1) 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".

2) Solution to the problem

Let X the random variable that represent the loan officer rates applicants for a credit of a population, and for this case we know the distribution for X is given by:

`X \sim N(200,50)`

Where
`\mu=200` and
`\sigma=50`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

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

`P(X>a)=0.40` (a)

`P(X<a)=0.60` (b)

Both conditions are equivalent on this case.

As we can see on the figure attached the z value that satisfy the condition with 0.60 of the area on the left and 0.40 of the area on the right it's z=0.253. On this case P(Z<0.253)=0.60 and P(z>0.253)=0.4

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.6`

`P(Z<(a-\mu)/(\sigma))=0.6`

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

`z=0.253<(a-200)/(50)`

And if we solve for a we got

`a=200 +0.253*50=212.65`

So the value of height that separates the bottom 60% of data from the top 40% is 212.65. And rounded would be 212.7.