3.33 Theorems
\[\sum_{i=0}^n (x_i-\bar x)=0\]
\[\sum_{i=0}^n f_i(x_i-\bar x)=0\]
\[\sum_{i=1}^n (x_i-\bar x)^2 \lt \sum_{i=1}^n (x_i-a)^2; a\ne\bar x\]
\[\sum_{i=1}^n f_i(x_i-\bar x)^2 \lt \sum_{i=1}^n f_i(x_i-a)^2; a\ne\bar x\]
(AM~origin & scale)
Prove that AM depends on origin and scale Use frequency as well i.e,
- \[\bar x=\frac {\sum_{i=1}^nx_i}{n}\]
- \[\bar x=\frac {\sum_{i=1}^n f_ix_i}{n}\]
If the GM of \(n_1\) va;ues is \(G_1\), and of \(n_2\) values is \(G_2\), show GM of \(n_1+n_2\) values is \(G=\sqrt{G_1G_2}\)
For two non-zero positive numbers, prove \(AM \ge GM \ge HM\)
Let, the numbers be a, b
\(\therefore AM = \frac{a+b}{2}\)
\(GM = \sqrt{ab}\)
\(HM = \frac{2}{\frac 1 a +\frac 1 b}\)We know, \[\begin{eqnarray} & &(a-b)^2\ge 0 \nonumber \\ & \Rightarrow & (a+b)^2-4ab \ge 0 \nonumber \\ & \Rightarrow & (a+b)^2 \ge 4ab \nonumber \\ & \Rightarrow & (a+b) \ge 2 \sqrt{ab} \nonumber \\ & \Rightarrow & \frac{a+b} 2 \ge \sqrt{ab} \nonumber \\ & \Rightarrow & AM \ge GM \nonumber \\ \end{eqnarray}\]Similarly, \[\begin{eqnarray} & &(\frac{1}{a}-\frac{1}{b})^2\ge 0 \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b})^2 -4 \cdot \frac 1 a \cdot \frac 1 b\ge 0 \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b})^2\ge \frac 4 {ab} \nonumber \\ & \Rightarrow & (\frac{1}{a}+\frac{1}{b}) \ge \frac 2 {\sqrt{ab}} \nonumber \\ & \Rightarrow & \sqrt{ab}(\frac{1}{a}+\frac{1}{b}) \ge 2 \nonumber \\ & \Rightarrow & \sqrt{ab} \ge \frac{2}{(\frac{1}{a}+\frac{1}{b})} \nonumber \\ & \Rightarrow & GM \ge HM \nonumber \\ \end{eqnarray}\]For two non-zero positive numbers, \(AM \times HM =(GM)^2\)
Mean and Median of first n natural numbers are \(\frac {n+1} 2\)
If \(\bar x_1\) and \(\bar x_2\) are means of 2 data sets of sizes \(n_1\) and \(n_2\), respectively, the combined mean is \(\bar x_c=\frac{n_1 \bar x_1+n_2 \bar x_2}{n_1+n_2}\)
If \(u=x+y, \bar u=\bar x + \bar y\); if \(n_1=n_2=n\)
Given \(u=x+y\)
\[\begin{eqnarray} \bar u &=& \frac{\sum u}{n} \nonumber \\ &=& \frac{\sum (x+y)}{n} \nonumber \\ &=& \frac{\sum x}{n}+ \frac{\sum y}{n} \nonumber \\ &=& \bar x + \bar y \nonumber \\ \end{eqnarray}\]
For equal number of observations, GM of two variables is equal to the product of their individual means.
\(GM=Antilog(\frac{\sum \log x_i}{n})\) or \(Antilog(\frac{\sum f_i \log x_i}{\sum f_i})\) 14.If \(y = a + bx, \bar y = a + b \bar x\)
If \(z_i=ax_i+by_i, \bar z=a \bar x + b \bar y\)