Question

The average student loan debt for college graduates is $25,300. Suppose that that distribution is normal and that the standard deviation is $11,500. 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 $22,600 and $32,750 in student loan debt. c. The middle 20% of college graduates' loan debt lies between what two numbers? Low: $ High: $

Answer 1

Part a.

From the given information, we have that

$\begin{gathered} \mu=25300 \\ \sigma=11500 \end{gathered}$

Then, the distribution of X

$N(\mu,\sigma)$

is given by:

$N(25300,11500)$

Part b.

In this case, we need to find the following probability:

$P(22600then we need to find the corresponding z values for there values, that is, [tex]z=\frac{22600-25300}{11500}=-0.23478$

and

$z\frac{32750-25300}{11500}=0.64782$

So we need to find on the z-table the following probability:

$P(-0.23478which gives 0.33426

Therefore, by rounding to 4 decimal places, the answer for part b is: 0.3343.

Part c.

The middle 20% of college graduate loans debt lies within the interval 10% below the mean and 10% over the mean. Then, the z-value for this interval is z=+/- 0.253

Then, we can find the lower and upper bound for this interval as

[tex]\begin{gathered} Lower=\mu-z\times\sigma=25300-0.253\times11500 \\ Upper=\mu+z\sigma=25,300+0.253\times11,500 \end{gathered}$

which gives

$\begin{gathered} Lower=22390.5 \\ Upper=28209.5 \end{gathered}$

Therefore, by rounding up to the neares dollar, the answers for part c are:

$\begin{gathered} Low:\text{ \ 22,391} \\ High:\text{ \ 28,210} \end{gathered}$