Central Limit Theorem

The Central Limit Theorem (CLT) is one of the most important results in statistics. It states that when we take sufficiently large random samples from any population with a finite mean and variance, the sampling distribution of the sample mean will be approximately normally distributed, regardless of the shape of the original population.

In simple words, even if your original data is skewed or irregular, the distribution of the sample means tends toward a bell curve as the sample size increases.

1.0Central Limit Theorem Statement

Statement:
If a random sample of size n is taken from a population with mean $\mu$ and standard deviation $\sigma$, then as n becomes large, the distribution of the sample mean $\bar{X}$ approaches a normal distribution with mean $\mu$ and standard deviation  $\frac{\sigma }{\sqrt{n}}$ .

2.0Central Limit Theorem Formula

The formula for the central limit theorem for the sample mean is:

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

Where:

• $\bar{X}$ = Sample mean
• $\mu$ = Population mean
• $\sigma$ = Population standard deviation
• n = Sample size

3.0Central Limit Theorem Equation

The probability density function for the sampling distribution of the mean (as n grows large) is given by:

$f(\bar{x})=\frac{1}{\sqrt{2\pi \left(\frac{{\sigma }^2}{n}\right)}}{e}^{-\frac{(\bar{x}-\mu {)}^2}{2\left({\sigma }^2/n\right)}}$

This is the normal distribution equation adapted for sample means.

4.0Central Limit Theorem Explanation

The CLT works because when independent random variables are added, their normalized sum tends to follow a normal distribution, even if the original variables themselves are not normally distributed. This is why normal distribution appears so often in real-world data analysis.

5.0Central Limit Theorem Example

Example:
Suppose the average height of students in a school is unknown, but the heights are skewed. You randomly select samples of 50 students at a time and record the average height for each sample. If you repeat this many times, the histogram of those sample averages will form an approximate normal curve, even though the original height distribution was skewed.

6.0Application of the Central Limit Theorem

The application of the central limit theorem is widespread in statistics and data science:

1. Confidence Intervals – Allows us to estimate population parameters from sample data.
2. Hypothesis Testing – Forms the foundation for many statistical tests like the Z-test and t-test.
3. Quality Control – Used in manufacturing to monitor processes.
4. Finance – Used in modeling returns and risk assessment.
5. Polling & Surveys – Helps predict population opinions from small samples.

7.0Solved Examples on Central Limit Theorem

Example 1: Population mean $\mu =50$, population standard deviation $\sigma =8$. A random sample of n = 36 is taken. Find $P(\bar{X}>52)$.

Solution.

1. Compute standard error: ${\sigma }_{\bar{x}}=\frac{\sigma }{\sqrt{n}}=\frac{8}{\sqrt{36}}=\frac{8}{6}=1.333333$.
2. Compute z-score:

$Z=\frac{52-50}{1.333333}=\frac{2}{1.333333}=\mathrm{1.5.}$

3. Use standard normal: $P(\bar{X}>52)=P(Z>1.5)$.

From normal table: $P(Z>1.5)\approx 0.0668072$.

Answer: 0.0668 (approx.) 

Example 2: True proportion p=0.40. Sample size n = 200. Find $P(0.35<\hat{p}<0.45)$ .

Solution.

1. Mean of $\hat{p}$ is 0.400.40. Standard error:

${\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.4\times 0.6}{200}}=\sqrt{\frac{0.24}{200}}=\sqrt{0.0012}\approx 0.034641$ 

2. Convert bounds to z-scores:

$\begin{array}{rl}& Z_{\text{low }}=\frac{0.35-0.40}{0.034641}\approx -1.4434,\\ & Z_{\text{high }}=\frac{0.45-0.40}{0.034641}\approx \mathrm{1.4434.}\end{array}$

3. Probability:

$P(-1.4434<Z<1.4434)=\Phi (1.4434)-\Phi (-1.4434)=2\Phi (1.4434)-1.$

$\Phi (1.4434)\approx 0.925543,soprobability\approx \mathrm{0.8511.}$

Answer: 0.8511 (approx.)

Example 3: Population standard deviation $\sigma =3$. How large a sample n is needed so that $P(|\bar{X}-\mu |<0.5)=0.95$ ?

Solution.

1. For two-sided 95% probability, critical z is  $z_{0.975}$ =1.96.
2. We want $P(|\bar{X}-\mu |<\mathrm{ME})=0.95$ with ME = 0.5. Use formula:

$\mathrm{ME}=z_{0.975}\frac{\sigma }{\sqrt{n}}\Rightarrow n={\left(\frac{z_{0.975}\sigma }{\mathrm{ME}}\right)}^2.$ 

3. Plug numbers:

$n={\left(\frac{1.96\times 3}{0.5}\right)}^2={\left(\frac{5.88}{0.5}\right)}^2=(11.76{)}^2\approx \mathrm{138.2976.}$ 

4. Round up to nearest whole person: n=139.

Answer: n = 139 

Example 4: Let $X_1,\dots ,X_{100}$ be good with mean 22 and variance 99. What is approximately $P\left(S_{100}>230\right)$ where $S_{100}={\sum }_{i=1}^{100}{X}_i$ ?

Solution.

1. Mean of sum: $E\left[S_{100}\right]=100\times 2=200.$
2. Variance of sum: $\mathrm{Var}\left(S_{100}\right)=100\times 9=900$ . Standard deviation: ${\sigma }_s=30$.
3. Convert: $Z=\frac{230-200}{30}=\frac{30}{30}=1$ . 
4. $P\left(S_{100}>230\right)=P(Z>1)=0.1587\text{ (approx) }$

Answer: 0.1587

8.0Practice Questions on Central Limit Theorem

1. Population mean $\mu =120,\sigma =15$. Sample n = 49. Find $P(118<\bar{X}<124)$.

Answer hint: .$S$$E=15/\sqrt{49}=15/7\approx 2.1429$

Compute z's, then probabilities.

Final answer: ≈ 0.8186.

2. A factory claims defect rate p = 0.05. Sample n = 500. What is probability that observed $\hat{p}>0.07$?

Answer hint: $SE=\sqrt{0.05\cdot 0.95/500}$

3. Compute z and tail.

Final answer: ≈ 0.067(approx).

4. For population sd $\sigma =10z_{0.995}\approx 2.5758$ , how large n for 99% confidence and margin 2?

Answer hint: $z_{0.995}\approx 2.5758$  Compute $n=(z\sigma /\mathrm{ME}{)}^2$

Final answer: $n\approx (2.5758\ast 10/2{)}^2\approx 165.9\Rightarrow 166$_.

5. IID mean 0.5, variance 0.25, n = 400. Approx $P\left(\sum X_i<210\right)$_.

Answer hint: Mean sum = 400 * 0.5 = 200, sd $=\sqrt{400^{\ast }0.25}=10$ . z = (210-200)/10 =1. So prob ≈ 0.1587 (upper tail) so P( <210 ) ≈ 0.8413.

6. The population is extremely skewed (exponential with mean 5). For sample size n = 10, is CLT safe to use for sample mean approx normal? For n = 100?

Answer: n = 10: approximation may be poor (skew shows). N = 100: CLT will give a good approximation.