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Suppose life expectancy is normally distributed with mean 60 and variance 9. (Actually this must be an approximation but assume it is exact, just for simplicity.)
(a) For a randomly selected person, what is the probability of a life span greater than 62 years?

(b) For a group of 4 randomly selected people, what is the probability of an average life span greater than 62 years?

(c) For a group of 16 randomly selected people, what is the probability of an average life span greater than 62 years?

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What have you tried? –  Henry Mar 7 '12 at 18:45
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1 Answer 1

Hint: If $X$ has normal distribution with mean $\mu$ and variance $\sigma^2$, and $\bar X_n$ is the mean of a random sample of size $n$ from $X$, then $\bar X_n$ has normal distribution with mean $\mu$ and variance $\sigma^2/n$. So, for example, in part b), if $\bar X_4$ is the average lifespan of 4 randomly chosen people, then $\bar X_4$ is normally distributed with mean $60$ and variance $9/4$.

To calculate probabilities for a normal variable, convert to the standard normal: If $X$ has normal distribution with mean $\mu$ and variance $\sigma^2$, then $$ P[X\ge a]= P\Bigl[ Z\ge {a-\mu\over \sigma}\Bigr], $$ where $Z$ is the standard normal variable. Values of $P[Z\ge a]=1-P[Z\le a]$ can be found from tables, such as those found here.

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should I use σ^2/n or σ/n^(1/2)? for b) if i use σ^2/n, its 9/4, if I use σ/n^(1/2), the answer is 3/2 –  Forest Mar 7 '12 at 19:29
    
@Forest In the "conversion to the standard normal" formula, you divide by the standard deviation $\sigma$; so by $\sqrt{9/4}=3/2$ for b). –  David Mitra Mar 7 '12 at 19:49
    
thanks for the help. Able to solve the problem now :) –  Forest Mar 7 '12 at 19:59
    
@Forest You're welcome. Glad to help :) –  David Mitra Mar 7 '12 at 20:12
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