A set of numbers has a mean of 22 and a standard deviation of 6. If 3 is added to each number of the set, and each resulting number is then doubled, find the mean(50) and standard deviation(12) of the new set. Ans in brackets.
For the mean;
Ex=22n+3n
Ex=25n*2
Ex=50n
X(mean)=50
But i can't figured out for the S.d, do i need to use y=ax+b or other method ??
On
The expectation is linear, so you have $$ \mathbb{E}(aX + b) = \mathbb{E}(aX) + \mathbb{E}(b) = a\mathbb{E}(X) + b $$
For the variance, you have (using that $\mathbb{E}(\mathbb{E}(Y)) = Y$) $$ \begin{eqnarray} \mathbb{V}(X) &=& \mathbb{E}((X-\mathbb{E}(X))^2) \\ &=& \mathbb{E}(X^2 -2X\mathbb{E}(X) + (\mathbb{E}(X))^2) \\ &=& \mathbb{E}(X^2) - \underbrace{\mathbb{E}(2X\mathbb{E}(X))}_{=2\mathbb{E}(X)\mathbb{E}(\mathbb{E}(X))} + (\mathbb{E}(X))^2) \\ &=& \mathbb{E}(X^2) - (\mathbb{E}(X))^2 \end{eqnarray} $$ and thus $$ \begin{eqnarray} \mathbb{V}(aX + b) &=& \mathbb{E}((aX+b)^2) - (\mathbb{E}(aX+b))^2 \\ &=& \mathbb{E}(a^2X^2 + 2abX + b^2) - (a\mathbb{E}(X) + b)^2 \\ &=& a^2\mathbb{E}(X)^2 + 2ab\mathbb{E}(X) + b^2 - (a^2(\mathbb{E}(X))^2 + 2ab\mathbb{E}(X) + b^2) \\ &=& a^2\mathbb{E}(X)^2 - a^2(\mathbb{E}(X))^2 \\ &=& a^2\mathbb{V}(X) \end{eqnarray} $$ Since the standard deviation $\sigma_X$ is $\sqrt{\mathbb{V}(X)}$, $$ \sigma_{aX + b} = \sqrt{\mathbb{V}(aX+b)} = \sqrt{a^2\mathbb{V}(X)} = a\sqrt{\mathbb{V}(X)} = a\sigma_X. $$
You are considering a set of numbers $x_1,\ldots,x_n$ with the property that $$ \bar{x}=\frac1n\sum_{i=1}^n x_i=22,\quad \sigma_x=\sqrt{\frac1n\sum_{i=1}^n(x_i-\bar{x})^2}=6. $$ Now we construct a new set of numbers $y_1,\ldots,y_n$ by letting $y_i=2(x_i+3)=2x_i+6$ for all $i$. Then $\bar{y}=2\bar{x}+6$ and hence $$ y_i-\bar{y}=2x_i+6-(2\bar{x}+6)=2(x_i-\bar{x}). $$ Thus the standard deviation of $y_1,\ldots,y_n$ is $$ \sigma_y=\sqrt{\frac1n\sum_{i=1}^n(y_i-\bar{y})^2}=\sqrt{4\frac1n\sum_{i=1}^n(x_i-\bar{x})^2}=2\sigma_x. $$