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 [tex]P_{60}[/tex] , the score which separates the lower 60% from the top 40%.

Please explain!

Respuesta :

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.

Step-by-Step Explanation:

The normal distribution is a "probability distribution that is symmetric around the mean, demonstrating that data near the mean occur more frequently than data distant from the mean."

The Z-score is defined as "a numerical measurement used in statistics of a value's relationship to the mean (average) of a set of values, expressed in standard deviations from the mean."

Let X be the random variable that represents how a loan officer scores credit applications from a population, and we know that the distribution for X is provided by:

X ~ N (200,50)

Where u = 200 and o = 50

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

Z = x - u/o

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)

In this example, both conditions are equal.

The z value that meets the criterion with 0.60 of the area on the left and 0.40 of the area on the right is z=0.253, as shown in the attached figure. P(Z0.253)=0.60 and P(z>0.253)=0.4 in this scenario.

Using the prior condition (b), we get:

P (X < a) = P (X-u/a < a - u/o) = 0.6

P (Z < a-u/o) = 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.