Central limit theorem

Central limit theorem

The central limit theorem says that even if the population distribution isn’t normal, the distribution of sample means becomes approximately normal when the sample size is large enough. For example, the number of hours of television people watch each day might be skewed (not normal). But if you randomly survey $100$ people each day and record the average number of hours watched, those daily averages will have an approximately normal distribution.

Definitions

Central limit theorem

If one starts with a population with any distribution and picks random samples of at least $30$, the means of each of these samples will be approximately normally distributed.

In order to use the central limit theorem, it is crucial to check that:

1. The data is collected using a random sampling process.
2. Observations must be independent of one another.
3. The sample size is at least $30$.

Sidenote

Important to know!

• Means vary less than individual values.
• Means of higher sample size vary less than means of lower sample size.
• The larger the sample size, the closer the averages are to a normal distribution.

Sampling distribution

When you take many random samples of the same size $n$ and compute the sample mean $\bar{x}$ each time, the collection of those means forms the sampling distribution of $\bar{x}$. If $X$ is a population variable with mean $\mu$ and standard deviation $\sigma$, then the sampling distribution of $\bar{x}$ for samples of size $n$ is:

$$\bar{x}\sim N\left(\mu ,\frac{\sigma }{\sqrt{n}}\right)$$

Where:

${\mu }_{\bar{x}}=\mu$ (mean of the sampling distribution)

$${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}\text{ (standard error)}$$

The z-score for sample means is:

$$z=\frac{\bar{x}-\mu }{\frac{\sigma }{\sqrt{n}}}$$

Practice problem

Example 1:

The mean SAT Math score is $\mu =520$ with a standard deviation of $\sigma =100$.

Part a.

If $36$ students are randomly selected, what is the probability that their average score is greater than $540$?

First, identify the given values:

$\mu =520$, $\sigma =100$, $n=36$

Next, find the standard error:

$${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}=\frac{100}{\sqrt{36}}=\frac{100}{6}\approx 16.67$$

Now compute the z-score for $\bar{x}=540$:

$$z=\frac{\bar{x}-{\mu }_{\bar{x}}}{{\sigma }_{\bar{x}}}=\frac{540-520}{16.67}\approx \frac{20}{16.67}\approx 1.20$$

Finally, use the z-score table or technology to find the right-tail probability:

$$\begin{array}{rl}P(\bar{x}>540)& =P(z>1.20)\\ & =1-P(z\le 1.20)\\ & =1-0.8849\\ & =0.1151\end{array}$$

Solution:

Part b.

What is the value of $\bar{x}$ such that only $5\%$ of samples of size $36$ have a mean above that value?

“Only $5\%$ above” means we want the cutoff at the $95$th percentile of the sampling distribution. The z-score for the $95$th percentile is $z=1.645$ (or use invNorm$(0.95)$).

Start with the z-score formula and solve for $\bar{x}$:

$$\begin{array}{rl}z& =\frac{\bar{x}-\mu }{{\sigma }_{\bar{x}}}\\ \bar{x}& =\mu +z\cdot {\sigma }_{\bar{x}}\\ & =520+1.645\cdot 16.67\\ & \approx 520+27.45\\ & =547.42\end{array}$$

Solution:

Part c.

Why is it appropriate to use the normal distribution in this context?

The sample size is $36$, which meets the central limit theorem’s minimum threshold of $30$. This means the distribution of sample means is approximately normal even if the population distribution is not. Additionally, since the population standard deviation is known, the z-score model applies directly.

Solution:

Definitions

Standard error

How much a sample statistic is expected to vary from sample to sample. It measures the typical distance between a sample statistic (like $\bar{x}$) and the population parameter (like $\mu$).

If the population standard deviation is known, use:

$$\text{SE}={\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}$$

If only the sample standard deviation is known, use:

$$\text{SE}=\frac{s}{\sqrt{n}}$$

Where $\sigma$ is the population standard deviation, $s$ is the sample standard deviation, $n$ is the sample size, and ${\sigma }_{\bar{x}}$ is the standard error of the sample mean.

Practice problem

Example 2:

One Pizza Pizza location in California knows that the population standard deviation of delivery times is $12$ minutes. A random sample of $36$ deliveries is selected.

Part a.

Can the central limit theorem be applied to this situation? Why or why not?

Check each condition:

1. The deliveries are randomly selected. ✓
2. It’s reasonable to treat delivery times as independent from one delivery to the next. ✓
3. The sample size is $36$, which is at least $30$. ✓

Solution:

Part b.

What is the standard error and what does it mean in this context? Use the standard error formula with known $\sigma$:

$$\begin{array}{rl}\text{SE}& =\frac{\sigma }{\sqrt{n}}\\ & =\frac{12}{\sqrt{36}}\\ & =\frac{12}{6}\\ & =2\end{array}$$

Solution: