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Math Help - Standard Normal Distribution Problem...

  1. #1
    Newbie MrRedNapoleon's Avatar
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    Unhappy Standard Normal Distribution Problem...

    Hi all,

    I have a problem that I can't get my head around - maybe one of you bright sparks can help me out!? The question goes:
    "A hard drive has a lifetime which is Normally distributed with a mean of 30,000 hours and a standard deviation of 8,000 hours. The warranty is for 2 years and is included in the sale at $100. An extended warranty to cover an extra year is available at a cost of $39.

    Q1: Using variable 't' (in hours) to represent the lifetime of the HD, sketch a graph of the Normal distribution and then calculate to probability that any HD dies during its 3rd year of operation.

    Q2: Using the answer to Q1 calculate the expected (or average) cost of replacing a HD when it dies during its 3rd year of operation."
    I have attempted it and calculated an answer to Q1 (have yet to draw the bell graph), however I am not sure whether this is correct or not :S ...

    P(t<=17,520) = P[(t-30,000)/8,000 <= (17520-30,000)/8,000]


    = P(Z < -2.60)
    = P(Z > 2.60)


    = 0.5 - P(0 <= Z <= 2.60)
    = 0.5 - 0.495
    = 0.005
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by MrRedNapoleon View Post
    Hi all,

    I have a problem that I can't get my head around - maybe one of you bright sparks can help me out!? The question goes:
    I have attempted it and calculated an answer to Q1 (have yet to draw the bell graph), however I am not sure whether this is correct or not :S ...

    P(t<=17,520) = P[(t-30,000)/8,000 <= (17520-30,000)/8,000]

    = P(Z < -2.60)
    = P(Z > 2.60)

    = 0.5 - P(0 <= Z <= 2.60)
    = 0.5 - 0.495
    = 0.005
    What you have calculated is failure within 3 years rather than failure during the third year ....
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