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.