Thread: Normal Distribution Problem

If the life, in years, of a television set is normally distributed with a mean of 35 years and a standard deviation of 8 years, what should be the guarantee period if the company wants less than 3% of the television sets to fail while under warranty?

I cannot figure this out.

Thanks

Sorry if I'm wrong, not much of an expert either but I hope I help!

1. Decide if this 3% (0.03) should be on the extreme left or extreme right.
(I concluded it should be extreme left of the normally distributed curve.)

2. Let X be max number of years for warranty. Which means you want to find, P(X<0.03)

3. Now find the area under the curve starting from the center (i.e. mean)

3.1. Area under the curve is 1. Area under curve starting from center is 0.5.
Therefore, 0.5 - 0.03 = 0.47

3.2. Referring to the Z-table: The corresponding Z value for an area of 0.47 is 1.88. However you need to factor in this is on the left side of the standard normal curve, and need to add a negative sign to it. i.e. -1.88

4. Given, Z=-1.88, mean= 35 and sigma(std dev)=8 substitute them into the formula X=mean+(Z * sigma)

Ans: X = 35 + (-1.88 * 8) = 35 -15.04 = 19.96 ~ 20 years

The warranty period should be 20 years.

P.S. I'm so so sorry if I'm wrong. I just picked up stats again after a 2 year hiatus having to serve in the army so I'm kind of rusty!