Answer:
A) 0.3531%
B) 45.6598%
C) $4696.90
Step-by-step explanation:
The credit card debt of graduating seniors at a certain university are normally distributed with a mean (μ) of $3400 and a standard deviation (σ) of $1225. Therefore:
[tex]\rm X \sim N(\mu,\sigma^2)\implies \boxed{\rm X \sim N(3400,1225^2)}[/tex]
where X is the credit card debt in dollars.
[tex]\dotfill[/tex]
Part A
To find what percent of seniors graduating from this university have less than $100 in credit card debt, we need to find P(X ≤ 100).
Calculator input for "normal cumulative distribution function (cdf)":
- Upper bound: x = 100
- Lower bound: x = -10000
- μ = 3400
- σ = 1225
This gives the percent of seniors graduating from this university that have less than $100 in credit card debt as:
P(X ≤ 100) = 0.00353130...
P(X ≤ 100) = 0.3531% (4 d.p.)
[tex]\dotfill[/tex]
Part B
To find what percent of seniors graduating from this university have have between $2500 and $4000 in credit card debt, we need to find P(2500 ≤ X ≤ 4000).
Calculator input for "normal cumulative distribution function (cdf)":
- Upper bound: x = 4000
- Lower bound: x = 2500
- μ = 3400
- σ = 1225
This gives the percent of seniors graduating from this university that have between $2500 and $4000 in credit card debt as:
P(2500 ≤ X ≤ 4000) = 0.456597865...
P(2500 ≤ X ≤ 4000) = 45.6598% (4 d.p.)
[tex]\dotfill[/tex]
Part C
To find how much credit card debt is owed for a graduating senior from this university if he has more credit card debt than 90% of his classmates, we need to find the value of 'a' for P(X ≤ a) = 0.9.
Calculator input for "inverse normal":
- Area: 0.9
- μ = 3400
- σ = 1225
This gives the value of 'a' as 4969.900757. So, if a graduating senior from this university has more credit card debit than 90% of his classmates, his debt will be $4696.90.
