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

Answer:

$a=200 +0.253*50=212.65$

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.  

Step-by-step 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:

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

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:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.6$  

$P(Z<\frac{a-\mu}{\sigma})=0.6$

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

$z=0.253<\frac{a-200}{50}$

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.