Question 1137371
the population mean is 3.1 years and the population standard deviation is .6 years.


your sample size is 24.


your standard error is therefore population standard deviation / square root of sample size = .6 / sqrt(24) = .1224744871.


you are looking to find the the z-score where the area to the left of it is 3% or .03.


you can use a z-score calculator to determine that the z-score associated with an area of .03 to the left of it is equal to -1.88079361.


you will want to find the raw score associated with that z-score.


the z-score formula is z = (x - m) / s


z is the z-score.
x is the raw score.
m is the mean.
s is the standard error.


the formula becomes:


-1.88079361 = (x - 3.1) / .1224744871


solve for x to get x = .1224744871 * -1.88079361 + 3.1 = 2.869650767 years.


this can be seen visually using the following online normal distribution calculator.


<a href = "http://davidmlane.com/hyperstat/z_table.html" target = "_blank">http://davidmlane.com/hyperstat/z_table.html</a>


here's what it looks like.


<img src = "http://theo.x10hosting.com/2019/032601.jpg" alt="$$$" >


what this is showing you is that, 3% of the time, the mean life of the sample of size 24 will be less than 2.869650767 years.


the calculator limits the number of digits that can be entered into it, therefore some discrepancies with my more detailed values will be noted.


i rounded my values to the maximum number of decimal digits allowed by the online calculator.


the answer wasn't affected by this rounding.


it said 3% as well.


the interpretation that you can take from this is that, if you took a very large number of samples of size 24, approximately 3% of the time the mean of the sample will be less than 2.869650767


the rest of the time,it will be more than 2.869650767 years.