Question

The average student loan debt for college graduates is $25,300. Suppose that that distribution is normal and that the standard deviation is $11,250. 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 $7,500 and $18,100 in student loan debt.

c. The middle 30% of college graduates' loan debt lies between what two numbers?
     Low: $
     High: $

Answer 1

Solution :

Given that ,

mean = = $25300

standard deviation = = $11250

a.

X N (25300 , 11250)

b.

P($7500 < x < $18100) = P[(7500 - 25300)/ 11250) < (x - ) /  < (18100 - 25300) / 11250) ]

= P(-1.58 < z < -0.64)

= P(z < -0.64) - P(z < -1.58)

= 0.2611 - 0.0571

= 0.2040

Probability = 0.2040

c.

Middle 30%

Middle 30% thr z value are

= 1 - 30% = 1 - 0.3 = 0.7

/ 2 = 0.7 / 2 = 0.35

1 - / 2 = 1 - 0.35 = 0.65

z 0.35 = -0.3853

z 0.65 = 0.3853

Using z-score formula,

x = z * +

x = -0.3853 * 11250 + 25300 = 20965

and

x = 0.3853 * 11250 + 25300 = 29635

Two numbers are Low: $20965 High: $29635