Question

Credit card debt. The average credit card debt for college seniors in $3262. If the debt is normally distributed with a standard deviation of $1100, find these probabilities (6 points). Please show your steps. a.That the senior owes at least $1000 b.That the senior owes more than $4000 c.That the senior owes between $3000 and $4000

Answer 1

Answer:

• 0.98 approximately
• 0.25 approximately
• 0.34 approximately

Explanation:

standard deviation by which debt is distributed  = $1100

average credit card debt for college seniors = $3262

A)  probability that senior owes at least $1000

P (x ≥ 1000 ) = P ( Z ≥ $\frac{1000- \alpha }{\beta }$  )

where $\alpha = average credit card debt$ = $3262

$\beta = standard deviation$ = $1100

Z = random variable representing credit card debt for college seniors

back to equation P ( Z ≥ $\frac{1000- 3262}{1100}$ )

therefore  P ( z ≥ -2.06 )

                 1 - p ( z ≤ -2.06 )

therefore probability of the senior owing $1000 = 1 - 0.0199 = 0.9801

B) probability that senior owes more than $4000

p ( x ≥ 4000) = P ( Z ≥ $\frac{4000 - \alpha }{\beta }$ )

                     = P ( Z ≥ $\frac{4000 - 3262}{1100}$ )

                     = 1 - p ( Z ≤ 0.67 ) therefore probability that senior owes more than $4000 = 1 - 0.7468 = 0.2514

C ) Probability that the senior owes between $300 and $4000

P ( 3000 ≤ x ≤ 4000) = P ( $\frac{3000 - \alpha }{\beta }$ ≤ Z ≤ $\frac{4000 - \alpha }{\beta }$ )

                                  = P ( $\frac{3000 - 3262}{1100}$ ≤ z ≤ $\frac{4000 - 3262}{1100}$ )

                                  = P ( - 0.24 ≤ z ≤ 0.67 )  

                                 = p ( z ≤ 0.67 ) - p ( z ≤ - 0.24 )        

                                 = 1 - p ( z ≥ 0.67 ) - 1 - p ( z ≥ -0.24 )

                                 = 0.7846 - 0.4052 = 0.3434