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

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