What the central limit theorem actually claims, which distribution becomes normal, and why the spread shrinks like the square root of n.
The top histogram is the population: heavily skewed, nothing like a bell. The bottom one is the distribution of the sample mean. Increase the sample size and watch only one of them change.
The population never changes. Only the thing you are averaging does.
If X has mean μ and variance σ², then for large n the sample mean is approximately normal: X̅ ~ N(μ, σ²n), whatever the shape of X itself.
It is a claim about the mean, not about the data. Household incomes stay skewed however many people you survey. What becomes normal is the distribution of the average of a sample, which is a different thing and the one every confidence interval and test is built on.
Quadrupling n halves the standard error. The square root is why precision is expensive: to be twice as precise you need four times the data, and to be ten times as precise you need a hundred times.
It depends on how odd the parent distribution is: the more skewed, the larger n has to be. In examinations n > 30 is taken as sufficient. If the population is already normal, the sample mean is exactly normal for any n at all.
1. σ = 6 and n = 36. Find the standard error.
2. To halve the standard error, you must multiply n by:
3. Thai household incomes are strongly skewed. A sample of 100 is taken. Which is approximately normal?
Writing X̅ ~ N(μ, σ²/n) with the numbers. Using σ instead of σ/√n is the error that quietly ruins every probability after it.
Saying why normality applies: either the population is normal, or n is large enough for the central limit theorem.
Keeping variance and standard error apart. One is σ²/n, the other is its square root, and questions use both.
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