# sampling distributions / Probability

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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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.
@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