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

Answer:

$P(X>700)=P(\frac{X-\mu}{\sigma}>\frac{700-\mu}{\sigma})=P(Z>\frac{700-667}{65})=P(z>0.508)$

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

$P(z>0.508)=1-P(z$

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:

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

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

$P(X>700)=P(\frac{X-\mu}{\sigma}>\frac{700-\mu}{\sigma})=P(Z>\frac{700-667}{65})=P(z>0.508)$

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

$P(z>0.508)=1-P(z$