statistics
posted by Holly .
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

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 = (scoremean)/SD
2.33 = (score75)/8
Multiply both sides by 8
18.64 = score  75
ADD 75 to both sides.
93.64 = score 
The Z score should be 2.33 making the score 56.36
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