Need help with word problem.
In the life, inyears of a television set with a mean of 20 years and a standard deviation of 1.8 years, what should be the guarantee period if the sompany wants less than 1% of the tv sets to fail while under warranty?
should I take the .1 - 20/1.8=
no, you will need a table of values for the normal distribution or a webpage normal distribution calculator.
I find this one very good
http://davidmlane.com/hyperstat/z_table.html
in the second graph, enter
20 for mean, 1.8 for SD, and .99 as the shaded area.
click on "below" it should show 24.1874
To solve this problem, we can use the concept of standard deviation and the Z-score.
First, let's understand what the Z-score represents. The Z-score tells us how many standard deviations an observation is from the mean. In this case, we want to find the Z-score that corresponds to our desired guarantee period.
The formula to calculate the Z-score is:
Z = (X - μ) / σ
Where:
Z = Z-score
X = Observation (guarantee period in this case)
μ = Mean
σ = Standard deviation
We want less than 1% of the TV sets to fail while under warranty, which means we're looking for the Z-score that corresponds to the 99th percentile (100% - 1%). In a standard normal distribution (where the mean is 0 and the standard deviation is 1), the Z-score corresponding to the 99th percentile is approximately 2.33.
Now, let's calculate the Z-score for our problem:
Z = (X - μ) / σ
We know:
μ = 20 years
σ = 1.8 years
Z = 2.33 (corresponding to the 99th percentile)
Substituting the given values into the formula, we get:
2.33 = (X - 20) / 1.8
To solve for X (the guarantee period), we can rearrange the equation:
2.33 * 1.8 = X - 20
X - 20 = 4.194
X = 24.194
Therefore, the guarantee period should be approximately 24.194 years if the company wants less than 1% of the TV sets to fail while under warranty.