7.2 The Central Limit Theorem for Sums

The Central Limit Theorem for sums tells you what happens when you add up a bunch of independent random variables: the total tends to follow a normal distribution, even if the individual variables aren't normal themselves.

This matters because it lets you estimate probabilities for real-world totals, like total sales over a month, combined test scores, or cumulative wait times, using the normal distribution and z-scores.

The Central Limit Theorem for Sums

Central Limit Theorem for Sums

The CLT for sums says that if you take a large number of independent and identically distributed (i.i.d.) random variables and add them together, the sum will be approximately normally distributed. This holds regardless of the shape of the original distribution, as long as the sample size is large enough (typically $n \geq 30$).

Three conditions need to be in place for the CLT for sums to apply:

• The random variables must be independent (the outcome of one doesn't affect the others)
• They should be identically distributed (drawn from the same population with the same mean and variance)
• The sample size must be sufficiently large ($n \geq 30$ is the common rule of thumb)

When these conditions are met:

• The mean of the sum equals the sum of the individual means: $\mu_{\sum X} = n \cdot \mu_X$
• The variance of the sum equals the sum of the individual variances: $\sigma^2_{\sum X} = n \cdot \sigma^2_X$
• The standard deviation of the sum is: $\sigma_{\sum X} = \sqrt{n} \cdot \sigma_X$

A related idea worth knowing: the law of large numbers says that as sample size increases, the sample mean converges to the true population mean. The CLT goes further by telling you the shape of the distribution of sums.

Mean and Standard Deviation of Sums

To find the mean and standard deviation of a sum, follow these steps:

1. Find the mean of the sum. Multiply the mean of a single variable by the number of variables: $\mu_{\sum X} = n \cdot \mu_X$

2. Find the variance of the sum. Multiply the variance of a single variable by the number of variables: $\sigma^2_{\sum X} = n \cdot \sigma^2_X$

3. Find the standard deviation of the sum. Take the square root of the variance: $\sigma_{\sum X} = \sqrt{n \cdot \sigma^2_X} = \sqrt{n} \cdot \sigma_X$

Example: Suppose individual shipping packages have a mean weight of 10 lbs and a standard deviation of 2 lbs. For a shipment of 36 packages:

• Mean of the total weight: $\mu_{\sum X} = 36 \times 10 = 360$ lbs
• Standard deviation of the total weight: $\sigma_{\sum X} = \sqrt{36} \times 2 = 12$ lbs

These formulas only work when the random variables are independent. If knowing the value of one variable changes what you'd expect for another, you can't simply add variances.

Z-Scores in Sum Analysis

Once you have the mean and standard deviation of the sum, you can convert any particular sum value to a z-score and use the standard normal distribution to find probabilities.

The z-score formula for sums:

$Z = \frac{\sum X - \mu_{\sum X}}{\sigma_{\sum X}}$

Here, $\sum X$ is the observed or target sum value, $\mu_{\sum X}$ is the mean of the sum, and $\sigma_{\sum X}$ is the standard deviation of the sum.

Step-by-step process for finding a probability:

1. Calculate $\mu_{\sum X}$ and $\sigma_{\sum X}$ using the formulas above
2. Plug your target sum value into the z-score formula
3. Look up the z-score in a z-table or use a calculator to find the probability

Example (continuing from above): What's the probability that 36 packages weigh less than 370 lbs total?

1. $\mu_{\sum X} = 360$ lbs, $\sigma_{\sum X} = 12$ lbs

2. $Z = \frac{370 - 360}{12} = \frac{10}{12} \approx 0.83$

3. From a z-table, $P(Z < 0.83) \approx 0.7967$

So there's about a 79.7% chance the total weight is under 370 lbs.

Statistical Inference and Sampling Distributions

The CLT for sums is a foundation of statistical inference, which is the process of drawing conclusions about a population based on sample data. Because the CLT guarantees the sum will be approximately normal for large samples, you can use normal-distribution tools (z-scores, confidence intervals) even when the underlying population isn't normal.

The sampling distribution of a statistic describes how that statistic varies across many different samples from the same population. As sample size increases, the sampling distribution of the sum becomes closer to normal, which makes probability calculations more reliable.