Question 969359: A company knows that replacement times for the DVD players it produces are Normally distributed with a mean of 4.3 years and a standard deviation of 2.4 years.
If the company wants to provide a warranty so that only 2.9% of the DVD players will be replaced before the warranty expires, what is the time length of the warranty?
Enter your answer as a number accurate to 1 decimal place. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! With 2 SDs, it would be 9.1 days, and 97.1% working is very close to 97.5%, so I would expect the number to be close to 9.1 days, but a littleness.
Using a z-table or a calculator, 0.971 on the z-table (0.029) is with a z-score of 1.895 (at 1.89, the z-score is 0.9706 and at 1.90 it is 0.9713).
z=(Value-mean)/SD
1.895=Value-4.3/2.4 Multiply by 2.4
4.548=Value-4.3 Add 4.3
8.848=Value or 8.8 days.
Check
(8.8-4.3)/2.4 = z= (4.5/2.4)=1.875. This rounds to 0.970.
(8.9-4.3)/2.4=z=1.917 This rounds to 0.972.
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