9.2: The Central Limit Theorem for Sample Sums

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Theorem $\PageIndex{1}$ Central Limit Theorem for Sample Sums

For a relatively large sample size $n$ ($n\geq30$) from a population $X$ with parameters $\mu_X$ and $\sigma_X$, the sample sum, $S$, is approximately normally distributed with parameters $\mu_{S}=n\cdot\mu_X$ and $\sigma_{\bar{X}}=\sigma_X\cdot\sqrt{n}$, regardless of the distribution of $X$. For normally distributed $X$, the sample doesn’t have to be large.

Symbolically, the statement can be written in the following way:

For $X\sim?(\mu, \sigma)$ and $n\geq30$ or for $X\sim N(\mu, \sigma)$ and any $n$

$S\sim N\left(\mu_{S}=n\cdot\mu_X, \sigma_{S}=\sigma\cdot\sqrt{n}\right)$

Example $\PageIndex{1}$

According to a TSA study, the weights of passengers' carry-ons are distributed with a mean of 25 lbs. and a standard deviation of 4.5 lbs. Find the probability that the total weight of 200 randomly selected carry-ons is less than 5200 lbs. - the maximum recommended weight of carry-ons on a plane of such size.

$Image of checked-in luggage at an airport.$ Solution

Let $X$ be the weight of the carry-on of a randomly selected passenger, then

$X\sim ?(\mu_X=25, \sigma_X=4.5)$

Let $S$ be the total weight of the carry-ons of 200 randomly selected passengers, then, by CLT,

$S\sim N\left(\mu_{S}=n\cdot\mu_X=200\cdot25=5000, \sigma_{S}=\sigma_X\cdot\sqrt{n}={4.5}\cdot\sqrt{200}=63.64\right)$

Then the probability that the total weight of 200 randomly selected carry-ons is less than 5200 lbs. can be expressed as

$P(S<5200)=P\left(\dfrac{S-\mu_{S}}{\sigma_{S}}<\dfrac{5200-5000}{63.64}\right)=P(Z<3.14)=0.9992=99.92\%$

Or using technology,

$P(S<5200)=0.9992=99.92\%$

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