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Posted by on Friday, February 11, 2011 at 10:35pm.

Please let me know if I solved this right
The lifetime of television produced by the Hishobi Company are normally distributed with a mean of 75 months and a standard deviation of 8 months. If the manufacturer wants to have to replace only 1% of its televisions, what should its warranty be?




P (Z01<0) = 0.5 – 0.01 =.4900 2.33 × 8 =18.64 75 - 18.64 = 56.36 X=56.36

  • statistics - , Saturday, February 12, 2011 at 1:00pm

    Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score related to that proportion (.01).

    Z = 2.33 (approx. for .01)

    Z = (score-mean)/SD

    2.33 = (score-75)/8

    Multiply both sides by 8

    18.64 = score - 75

    ADD 75 to both sides.

    93.64 = score

  • statistics - , Monday, January 23, 2017 at 3:45pm

    The Z score should be -2.33 making the score 56.36

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