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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

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