The average credit card debt for college seniors is $16,601 with a standard deviation of $4100. What is the probability that a sample of 35 seniors owes a mean of more than $18,000? Round answer to 4 decimal places.

0.0218

Explanation:

Data provided in the question:

Average credit card debt, µ = $16,601

Standard deviation, σ = $4,100

Sample size, n = 35

To find : P (X > 18000)

Now,

P (X > $18,000) = 1 - P (X < $18,000)

For Standardizing the value, we have

Z = [ X - µ ] : [ σ : √n ]

Z = [ $18,000 - $16,601 ] : [ $4,100 : √35 ]

or

Z = 2.02

Thus,

P ([ X - µ ] : [ σ : √n ] > [ $18,000 - $16,601 ] : [ 4100 : √35 ]

or

P (Z > 2.02)

or

P (X > 18000) = 1 - P (Z < 2.02)

[ from standard Z-value table P (Z < 2.02) = 0.9782 ]

therefore,

P (X > 18000) = 1 - 0.9782

or

P (X > 18000) = 0.0218