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8 June, 08:29

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.

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Answers (1)
  1. 8 June, 09:43
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    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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