# Question about probability of 0.99 that an average lies less than L years above overall mean

The duration of Alzheimer’s disease, from the onset of symptoms until death, ranges from 3 to 20 years, with a mean of 8 years and a standard deviation of 4 years. The administrator of a large medical center randomly selects the medical records of 30 deceased Alzheimer’s patients and records the duration of the disease for each one. Find the value L such that there is a probability of 0.99 that the average duration of the disease for the 30 patients lies less than L years above the overall mean of 8 years.

A. 0.72
B. 1.70
C. 2.33


I understand how to get this I just can't get the right answer.

If the probability is 0.99, the z-score is 2.33. 2.33 = $\frac{L - mean}{standard-deviation}$

Would the standard deviation be 4? Or would it be $\frac{sigma}{square-root-n}$ = $\frac{4}{square-root-30}$?

I need to know this for my test tomorrow.

• You confused standard deviation and standard error. SD is 4, SE is $4/\sqrt{30}$. What you want is 2.33 * SE. Although I would opt for t distribution rather than normal distribution, aka I may replace 2.33 with 2.46. However that answer is not available. – Penguin_Knight Feb 19 '15 at 5:51