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The average duration of Alzheimer’s disease is $8$ years and the standard deviation is $4$ years. For a clinical study $30$ patients, who have been determined to be at the very beginning stage of the disease, are randomly selected. (i) What is the probability that the average duration of the disease among the sampled patients will be less than $7$ years?

I believe if I let $Y$ denote the time it is normally distributed with parameters $8$ and $4^2$. I worked out $P(Y<7) = 0.4013$. However, I need to work out the probability of the average duration being less than $7$ years. How would I do this ? I know there are $30$ patients in total so the total sum must add to a number less than $210$ years.

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    $\begingroup$ Is the duration really modelled using the normal distribution? A normal random variable can have negative values which would be difficult to interpret here. Maybe some random variable with non-negative values would make more sense (perhaps the exponential distribution). $\endgroup$
    – Cm7F7Bb
    Mar 26 at 11:33
  • $\begingroup$ @Cm7F7Bb Yes exponential distribution that makes sense. $\endgroup$
    – Nikita Mazepin
    Mar 26 at 11:47
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    $\begingroup$ It looks like an exercise from a textbook so I would guess that at some point it should be mentioned what kind of a distribution should be used. It looks like you are required to use the normal distribution because you are given the expected value and the standard deviation but I am guessing here. $\endgroup$
    – Cm7F7Bb
    Mar 26 at 11:52

1 Answer 1

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The average also has the normal distribution. Suppose that $X_1,\ldots,X_n$ are independent and identically distributed $N(\mu,\sigma^2)$ random variables. Then $$ \bar X\sim N(\mu,\sigma^2/n), $$ where $$ \bar X=\frac1n\sum_{i=1}^nX_i. $$ I hope this helps.

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  • $\begingroup$ Shall I use the normal or exponential distribution then ? I would have thought normal because they gave the mean and standard deviation. How would I find out $\lambda$ for an exponential ? $\endgroup$
    – Nikita Mazepin
    Mar 26 at 11:48
  • $\begingroup$ I used what you said in this post and I obtained $0.08545$. Is this correct ? $\endgroup$
    – Nikita Mazepin
    Mar 26 at 11:52
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    $\begingroup$ @NikitaMazepin It is correct. $\endgroup$
    – Cm7F7Bb
    Mar 26 at 11:55
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    $\begingroup$ Okay thanks for your help on this question. I will accept your answer. $\endgroup$
    – Nikita Mazepin
    Mar 26 at 11:58

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