Theorem 8.2.1.
In short, the Central Limit Theorem says:
For any population distribution with a sufficiently large sample, the sample means are normally distributed.
Section 8.2 The Central Limit Theorem
There are basically three scenarios that come up that involve the use of the Central Limit Theorem:
- If the samples are drawn from a normal population, then the sampling distribution of the sample means will also be normally distributed regardless of the size of the samples.
- The sampling distribution of the sample means will not be normally distributed when samples of size \(n\geq 30\) are drawn from a population that is not necessarily normal.
- If samples of size \(n\lt 30\) are drawn from a population that is not necessarily normal, then the Central Limit Theorem does not apply and we cannot draw conclusions about the sampling distribution of the sample means.
Bottom line: If the sample size is large enough, then we can use the Central Limit Theorem to answer the questions about sample means using the same techniques that we used to answer questions about individual values of a normally distributed population. All we need to do is adjust the \(z\)-score calculation to account for the new standard deviation.
\begin{equation*}
\text{Generic $z$-score formula: } Z=\frac{X-\text{mean}}{\text{standard deviation}}
\end{equation*}
Exercise 8.2.2.
(Donnelly 7.8)
For a normal population with a mean equal to 87 and a standard deviation equal to 16, determine the probability of observing a sample mean of 90 or less from a sample of size 15.
Answer.
\begin{equation*}
P(\overline{X}\leq 90) = P\left(Z\leq \frac{90-87}{(16/\sqrt{15})}\right)\approx 0.7661
\end{equation*}
Exercise 8.2.3.
(Donnelly 7.9)
For a population that is left-skewed with a mean of 21 and a standard deviation equal to 15, determine the probability of observing a sample mean of 18 or more from a sample of size 33.
Answer.
\begin{equation*}
P(\overline{X}\geq 18)=P\left(Z\geq \frac{18-21}{15/\sqrt{33}}\right)\approx 1-P(Z\lt -1.15)=1-NORM.S.DIST(-1.15,1)\approx 0.8747
\end{equation*}
Exercise 8.2.4.
(Donnelly 7.21)
According to the Organization for Economic Cooperation and Development (OECD), adults worked an average of 1771 hours in 2016. Assume the population standard deviation is 390 hours and that a random sample of 50 adults was selected.
(a)
Calculate the standard error of the mean.
Answer.
\begin{equation*}
\sigma_{\overline{x}}=\frac{390}{\sqrt{50}}\approx 55.1543
\end{equation*}
(b)
What is the probability that the sample mean will be more than 1790 hours?
Answer.
\begin{equation*}
P(\overline{x}\gt 1790)=P\left( Z\gt \frac{1790-1771}{\sigma_{\overline{x}}}\right)\approx 1-P(Z\lt 0.3444879)\approx 0.3652
\end{equation*}
(c)
What is the probability that the sample mean will be between 1720 and 1760 hours?
Answer.
\begin{equation*}
P(1720\lt \overline{x}\lt 1760) \approx P(-0.92\lt z\lt -0.20)\approx 0.242
\end{equation*}
(d)
Would a sample mean of 1799 hours support the claim made by the organization?
Answer.
\begin{equation*}
P(\overline{x}\gt 1799)\approx 0.3050
\end{equation*}
There is a \(30\%\) chance of observing a sample mean as high as 1799, so 1799 is not unusual. Therefore, the claim is reasonable.
(e)
Identify the symmetrical interval that includes \(95\%\) of the sample means if the true population mean is 1771 hours.
Answer.
\(\overline{x}\pm z\cdot\sigma_{\overline{x}}\rightarrow \boxed{(1663,1879)}\)
