Central Limit Theorem

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PAS

Cegep/2

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281 words

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2 minutes

Consider a random sample of $n$ observations selected from any population with mean $\mu$ and standard deviation $\sigma$. Then, when $n$ is sufficiently large (typically >= 30), the sampling distribution of sample mean will be approximately normal:

$$\begin{array}{rl}\bar{X}& \sim N(\mu ,\frac{{\sigma }^2}{n})\\ E(\bar{X})& =\mu \\ \sigma (\bar{X})& =\frac{\sigma }{\sqrt{n}}\end{array}$$

Explicitly assume the population is normal when $n < 30$

Consequences

z-score applied to sample mean:

$$\begin{array}{r}\\ Z& \sim \frac{\bar{X}-\mu }{\frac{\sigma }{\sqrt{n}}}\end{array}$$

t-statistic applied to sample mean:

$$\begin{array}{rl}{T}_{n-1}& \sim \frac{\bar{X}-\mu }{\frac{s}{\sqrt{n}}}\\ df& =n-1\end{array}$$

Examples

Suppose we have selected a random sample of 36 observations from a population with mean 80 and standard deviation 6. What is the probability that the sample mean will be greater than 82?

$$\begin{array}{rl}\bar{X}& \sim N({\mu }_{\bar{X}}=80,{\sigma }_{\bar{X}}=\frac{6}{\sqrt{36}}=1)\\ \\ P(\bar{X}>82)& =P(\frac{\bar{X}-{\mu }_{\bar{X}}}{{\sigma }_{\bar{X}}}>\frac{82-80}{1})\\ & =P(Z>2)\\ & =P(Z<-2)\\ & =0.0228\end{array}$$

The distribution of the lengths of life of a certain brand’s car battery has a mean of 54 months and a standard deviation of 6 months. Suppose you purchase a sample of 50 of these batteries and subject them to tests that estimate the battery’s life. What is the probability that your sample has a mean life of 52 or fewer months?

$$\begin{array}{rl}\bar{X}& \sim N({\mu }_{\bar{X}}=52,{\sigma }_{\bar{X}}=\frac{\sigma }{\sqrt{n}}=\frac{6}{\sqrt{50}})\\ \\ P(\bar{X}\le 52)& =P(\frac{\bar{X}-{\mu }_{\bar{X}}}{{\sigma }_{\bar{X}}}\le \frac{52-54}{\frac{6}{\sqrt{50}}})\\ & =P(Z\le -2.36)\\ & =0.0091\end{array}$$

Contributors

Ajitani Hifumi

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