Question

g Credit card applicants have an average credit rating score of 667. Assume the distribution of credit scores is Normal with a standard deviationof 65. Use this information to answer the problems below. Write probability statementsand show all of your work.70. What is the probability that a single applicant for a credit card will have a credit rating score above 700?

Answer 1

`P(X>700)=P((X-\mu)/(\sigma)>(700-\mu)/(\sigma))=P(Z>(700-667)/(65))=P(z>0.508)`

And we can find this probability using the complement rule and excel or a calculator and we got:

`P(z>0.508)=1-P(z<0.508)=1-0.694=0.306`

Step-by-step explanation:

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".

Solution to the problem

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

`X \sim N(667,65)`

Where
`\mu=667` and
`\sigma=65`

We are interested on this probability

`P(X>700)`

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

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

If we apply this formula to our probability we got this:

`P(X>700)=P((X-\mu)/(\sigma)>(700-\mu)/(\sigma))=P(Z>(700-667)/(65))=P(z>0.508)`

And we can find this probability using the complement rule and excel or a calculator and we got:

`P(z>0.508)=1-P(z<0.508)=1-0.694=0.306`