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February 28, 2015

February 28, 2015

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

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

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**PsyDAG**, Saturday, February 12, 2011 at 1:00pmFind 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

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**Anonymous**, Thursday, February 19, 2015 at 11:20pm1

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