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                  

Answer 2

A type of payment card issued by banks to their customers for purchasing goods and services on credit is called the credit card. The accumulation of interests and amounts of purchase on this type of card is called credit card debt.

The probabilities are a. 0.98, b. 0.25 and c. 0.34 approximately.

The probabilities can be calculated as:

Given,

• The standard deviation of debt distribution (β) = $1100

• Average debt for college seniors (α) = $3262

• Random variable depicting debt of college senior = D

a. Probability (P) of senior owning $1000:

$\rm P ( X \geq 1000) = P ( D \geq \frac{1000 - \alpha} {\beta})$  

Replacing values in the equation:

$\rm P ( D \geq \frac{1000 - \3262} {\1100})$

$\therefore \rm P (D \geq - 2.06)$

Solving further:

$\rm 1 - p \;( D \leq - 2.06)$

Therefore, the probability of college senior's debt = 1 - 0.0199

Probability = 0.9801

b. Probability of senior owning more than $4000:

$\rm P ( X \geq 4000) = P ( D \geq \frac{4000 - \alpha} {\beta})\\\\P = ( D \geq \frac{4000 - 3262} {1100})\\\\1- p (D \leq 0.67)$

The probability of senior owning more than $4000 = 1 - 0.7468

Probability = 0.2514

c. Probability of debt between $3000- $4000:

$\rm P ( 3000\leq X \leq 4000) = P ( \frac{3000 - \alpha }{\beta } \leq D \leq \frac{4000 - \alpha }{\beta } )\\\\P ( \frac{3000 - 3262}{1100} \leq D \leq \frac{4000 - 3262}{1100} )$

Solving further:

$\rm P ( -.024 \leq D \leq 0.67) \\\\p (D \leq 0.67 ) - p (D \leq -0.24) \\\\1-p (D \leq 0.67 ) -1-p (D \leq -0.24)$

Replacing with values:

Probability = 0.7846 - 0.4052

Probability = 0.3434      

Thus, approximate probabilities are a. 0.98, b. 0.25 and c. 0.34.

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