# 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


Can somebody please help me understand why the correct answer is B??

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

## 1 Answer

We seek L such that P( < 8 + L) = 0.99, where = the average duration of the disease for the 30 patients. By the central limit theorem, has an approximately normal distribution with mean μ = 8 and standard deviation σ/√n = 4/√30 ≈ 0.73. Thus, P(< 8 + L) = P((- μ)/(σ/√n) < ((8 + L) - 8)/0.73) ≈ P(z < L/0.73), where z is the standard normal random variable. The area in the normal table closest to 0.99 is 0.9901, corresponding to an observation of z = 2.33. Setting L/0.73 = 2.33 and solving for L, we get L = (0.73)(2.33) ≈ 1.70.

• Please be cautious about providing complete answers to questions for coursework / homework (even if the test has presumably passed). Out policy is to provide hints only (see here). – gung Feb 23 '15 at 3:37
• I actually had already gotten the answer...with help on homework from my teacher... and haven't had time to check back here. – Jeff Feb 23 '15 at 23:50