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The Central Limit Theorem explains why normal curves appear so often in statistics, even when the original population is not normal. This cheat sheet helps students recognize when sample means or sample proportions can be modeled with an approximately normal distribution. It is useful for solving probability problems, checking conditions, and preparing for inference topics like confidence intervals and hypothesis tests.

Key Facts

  • For sample means, the Central Limit Theorem says that if nn is large enough, the sampling distribution of xˉ\bar{x} is approximately normal with mean μxˉ=μ\mu_{\bar{x}} = \mu and standard deviation σxˉ=σn\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}.
  • The standard error of the sample mean is SExˉ=σnSE_{\bar{x}} = \frac{\sigma}{\sqrt{n}}, which measures the typical distance between xˉ\bar{x} and μ\mu.
  • A common rule of thumb is that the sampling distribution of xˉ\bar{x} is approximately normal when n30n \ge 30, especially if the population is not strongly skewed.
  • If the original population is normal, then the sampling distribution of xˉ\bar{x} is normal for any sample size nn.
  • For sample proportions, the sampling distribution of p^\hat{p} is approximately normal with mean μp^=p\mu_{\hat{p}} = p and standard error SEp^=p(1p)nSE_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} when np10np \ge 10 and n(1p)10n(1-p) \ge 10.
  • To standardize a sample mean, use z=xˉμσ/nz = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} when the population standard deviation σ\sigma is known.
  • Larger sample sizes make the standard error smaller because σn\frac{\sigma}{\sqrt{n}} decreases as nn increases.
  • The Central Limit Theorem describes the distribution of sample statistics, not the shape of the original population data.

Vocabulary

Central Limit Theorem
A theorem stating that the sampling distribution of a mean or proportion becomes approximately normal as the sample size becomes large enough.
Sampling Distribution
The probability distribution of a statistic, such as xˉ\bar{x} or p^\hat{p}, calculated from many samples of the same size.
Sample Mean
The average value from a sample, written as xˉ\bar{x}.
Standard Error
The standard deviation of a sampling distribution, such as SExˉ=σnSE_{\bar{x}} = \frac{\sigma}{\sqrt{n}} for sample means.
Sample Proportion
The fraction of a sample with a certain characteristic, written as p^\hat{p}.
Normal Approximation
The use of a normal distribution to estimate probabilities for a sampling distribution when the required conditions are met.

Common Mistakes to Avoid

  • Confusing the population distribution with the sampling distribution is wrong because the Central Limit Theorem describes the behavior of statistics like xˉ\bar{x}, not the original data values.
  • Using σ\sigma instead of σn\frac{\sigma}{\sqrt{n}} for sample mean problems is wrong because sample means vary less than individual observations.
  • Assuming n30n \ge 30 always guarantees accuracy is wrong because strong skewness or extreme outliers may require a larger sample size.
  • Forgetting the success-failure condition for proportions is wrong because p^\hat{p} is not safely normal unless np10np \ge 10 and n(1p)10n(1-p) \ge 10.
  • Thinking a larger sample size changes the mean of the sampling distribution is wrong because μxˉ=μ\mu_{\bar{x}} = \mu stays the same while SExˉSE_{\bar{x}} gets smaller.

Practice Questions

  1. 1 A population has mean μ=80\mu = 80 and standard deviation σ=12\sigma = 12. For samples of size n=36n = 36, find μxˉ\mu_{\bar{x}} and SExˉSE_{\bar{x}}.
  2. 2 A population has p=0.40p = 0.40 and sample size n=100n = 100. Check whether the normal approximation for p^\hat{p} is reasonable, then find SEp^SE_{\hat{p}}.
  3. 3 A population has mean μ=50\mu = 50 and standard deviation σ=10\sigma = 10. For n=25n = 25, calculate the z-score for a sample mean of xˉ=54\bar{x} = 54 using z=xˉμσ/nz = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}}.
  4. 4 Explain why increasing the sample size makes the sampling distribution of xˉ\bar{x} narrower but does not change its center.