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. If 40 different applicants are randomly selected, find the probability that their mean is above 215.

Answer 1

Answer:

$P(\bar X>215)$

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

$z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this: for the value of 215

$z = \frac{215-200}{\frac{50}{\sqrt{40}}}= 1.897$

And we can find this probability using the complement rule and with the normal standard distribution or excel we got:

$P(z >1.897) = 1-P(z<1.897) =1- 0.971= 0.029$

Step-by-step explanation:

Let X the random variable that represent the ratings of applicants from a population, and for this case we know the distribution for X is given by:

$X \sim N(200,50)$  

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

We select a sample size of n =40. We are interested on this probability

$P(\bar X>215)$

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

$z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}$

If we apply this formula to our probability we got this: for the value of 215

$z = \frac{215-200}{\frac{50}{\sqrt{40}}}= 1.897$

And we can find this probability using the complement rule and with the normal standard distribution or excel we got:

$P(z >1.897) = 1-P(z<1.897) =1- 0.971= 0.029$