Question

The average student loan debt for college graduates is $25,850. Suppose that that distribution is normal and that the standard deviation is $13,750. 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 $28,750 and $46,650 in student loan debt.

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

Answer 1

Answer :- The average student loan debt for college graduates is $25,850. Suppose that that distribution is normal and that the standard deviation is $13,750. Let X = the student loan debt of a randomly selected college graduate.

Given that :-

• mean (μ) =  $25,850

• standard deviation (σ) = $13,750

a) The distribution of [ X ~ N = (μ, σ) ]

=> X ~ N = [ 25850, 13750 ]

b) probability that the college graduate has between $28,750 and $46,650 in student loan debt is :-

=> P($28,750 < X < $46,650)

=> P[(28750 - 25850/13750) < (X - μ/σ) < (46650 - 25850/13750)]

=> P[(2900/13750) < Z < (20800/13750)]

=> P(0.21 < Z < 1.51)

=> P(Z < 1.51) - P(Z < 0.21)

( Using Z table )

=> 0.9345 - 0.5832

=> 0.3513

[ Probability = 0.3513 ]

c) Middle 30℅ ,

=> Middle 30℅ the Z value are :

• α = 1 - 30℅

= 1 - 0.3

{ α = 0.7 }

• α/2 :- 0.7/2 => [ 0.35 ]

=> 1 - α/2

=> 1 - 0.35 => [ 0.65 ]

=> Z(0.35) = - 0.385

=> Z(0.65) = 0.385

Using 'Z' score formula :-

• X1 = Z × σ + μ

= -0.385 × 13750 + 25850

= -5293.75 + 25850

= 20556.25

[ X1 = 20556.25 ]

• X2 = Z × σ + μ

= 0.385 × 13750 + 25850

= 5293.75 + 25850

= 31143.75

[ X2 = 31143.75 ]

Hence,

Low : $20556.25

High : $ 31143.75