0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Penguin_Knight Feb 19 '15 at 5:51
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – gung - Reinstate Monica Feb 23 '15 at 3:37
  • $\begingroup$ I actually had already gotten the answer...with help on homework from my teacher... and haven't had time to check back here. $\endgroup$ – Jeff Feb 23 '15 at 23:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.