If the life in years, of a television set is normally distributed with a mean of 46 years and a standard deviation of 4 years, what should be the guarantee period if the company wants less than 3% of the television sets to fail while under warranty?

That would be about two sigma less than the mean, but did you really mean 46 years mean life? No one keeps a TV set that long, nor is there accurate lifetime data available.

I know but that's what the question asked.

Thanks

In that case, 97% will last longer than 38 years.

To determine the guarantee period, we need to find the value that corresponds to the lower 3% of the distribution.

First, we need to convert the guarantee period to a standardized score (also known as a z-score). The formula for the z-score is:

z = (x - μ) / σ

Where:
z = z-score
x = value we want to convert (in this case, the guarantee period)
μ = mean of the distribution
σ = standard deviation of the distribution

In this case, we want to find the value (x) that corresponds to the lower 3% of the distribution. So we need to find the z-score that corresponds to the cumulative probability of 0.03 (3%).

Using a standard normal distribution table or a z-score calculator, we can find that the z-score for a cumulative probability of 0.03 is approximately -1.8808.

Now we can use the z-score formula to solve for x:

-1.8808 = (x - 46) / 4

Rearranging the formula to solve for x:

x - 46 = -1.8808 * 4

x - 46 = -7.5232

x = 46 - 7.5232

x ≈ 38.48

Therefore, to ensure that less than 3% of the television sets fail while under warranty, the guarantee period should be approximately 38.48 years.