Question

6. The length of life in months of a certain product is approximately normally distributed with a mean of 92 and a standard deviation of 17. For each question below, draw a picture indicating what you know you need to find. (3 points)

a. the manufacturer decides to guarantee the product for five years. What percentage of items will fail before the warranty expires?

b. If the manufacturer wanted to replace only 1% of the product due to failure under warranty, what should the length in months of the warranty be?

Answer 1

Using Zscore =( X - mean )/ standard deviation

From Normal distribution find the probability

A) Required percentage=3.01℅

B) we get length in month is 52.46= 52 months

Solution file is attached go through it

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