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

`P(170<X<220)=P((170-\mu)/(\sigma)<(X-\mu)/(\sigma)<(220-\mu)/(\sigma))=P((170-200)/(50)<Z<(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

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

Using the z score we got:

`P(170<X<220)=P((170-\mu)/(\sigma)<(X-\mu)/(\sigma)<(220-\mu)/(\sigma))=P((170-200)/(50)<Z<(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