sarahelizabethoz61jk
Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Get quick and reliable solutions to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

The average student loan debt for college graduates is $26,000. Suppose that that distribution is normal and that the standard deviation is $12,300. Let X = the student loan debt of a randomly selected college graduate. Round all probabilities to 4 decimal places and all dollar answers to the nearest dollar.

a.  What is the distribution of X? X ~ N(

b   Find the probability that the college graduate has between $24,900 and $30,950 in student loan debt.

c.  The middle 20% of college graduates' loan debt lies between what two numbers?
     Low: $
     High: $​

Sagot :

Using the normal distribution, it is found that:

a. X ~ N(26000, 12300).

b. 0.1913 = 19.13% probability that the college graduate has between $24,900 and $30,950 in student loan debt.

c. Low: $22,888, High: $29,112.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It measures how many standard deviations the measure is from the mean.
  • After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

Item a:

  • The mean is of $26,000, hence [tex]\mu = 26000[/tex].
  • The standard deviation is of $12,300, hence [tex]\sigma = 12300[/tex].

Then, the distribution is:

X ~ N(26000, 12300).

Item b:

The probability is the p-value of Z when X = 30950 subtracted by the p-value of Z when X = 24900, hence:

X = 30950:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{30950 - 26000}{12300}[/tex]

[tex]Z = 0.4[/tex]

[tex]Z = 0.4[/tex] has a p-value of 0.6554.

X = 24900:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{24900 - 26000}{12300}[/tex]

[tex]Z = -0.09[/tex]

[tex]Z = -0.09[/tex] has a p-value of 0.4641.

0.6554 - 0.4641 = 0.1913

0.1913 = 19.13% probability that the college graduate has between $24,900 and $30,950 in student loan debt.

Item c:

  • Between the 40th percentile(low) and the 60th percentile(high).
  • The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.
  • The 60th percentile is X when Z has a p-value of 0.6, so X when Z = 0.253.

Then, the 40th percentile is found as follows.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.253 = \frac{X - 26000}{12300}[/tex]

[tex]X - 26000 = -0.253(12300)[/tex]

[tex]X = 22888[/tex]

For the 60th percentile:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.253 = \frac{X - 26000}{12300}[/tex]

[tex]X - 26000 = 0.253(12300)[/tex]

[tex]X = 29112[/tex]

Hence:

Low: $22,888, High: $29,112.

You can learn more about the normal distribution at https://brainly.com/question/24663213

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.