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 an applicant is randomly selected, find the probability of a rating that is between 170 and 220. Group of answer choices 0.2257 0.1554 0.0703 0.3811

Answer 1

Answer:

$P(170<X<220)=P(\frac{170-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{220-\mu}{\sigma})=P(\frac{170-200}{50}<Z<\frac{220-200}{50})=P(-0.6<z<0.4)$

And we can find the probability with this difference:

$P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)$

And using the normal standard table or excel we got:

$P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.6552-0.2742=0.3811$

And the best answer would be:

0.3811

Step-by-step explanation:

Let X the random variable that represent the ratings 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 $\sigma=50$

We are interested on this probability

$P(170<X<220)$

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

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

Using the z score we got:

$P(170<X<220)=P(\frac{170-\mu}{\sigma}<\frac{X-\mu}{\sigma}<\frac{220-\mu}{\sigma})=P(\frac{170-200}{50}<Z<\frac{220-200}{50})=P(-0.6<z<0.4)$

And we can find the probability with this difference:

$P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)$

And using the normal standard table or excel we got:

$P(-0.6<z<0.4)=P(z<0.4)-P(z<-0.6)=0.6552-0.2742=0.3811$

And the best answer would be:

0.3811