2011SP-II-45

$\begin{array}{cl}
= & (32+3)\times 2 \\
= & 70
\end{array}$

設 $Q_1$ 及 $Q_3$ 分別為原來數據的下四分位數及上四分位數。則新的四分位數間距

$\begin{array}{cl}
= & (Q_3+3)\times 2 – (Q_1+3)\times 2 \\
= & 2(Q_3-Q_1)+6-6 \\
= & 2(27) \\
= & 54
\end{array}$

設 $x_1$、$x_2$、…、$x_n$ 為原來的數據，且 $\overline{x}$ 為原來的平均值。

新的平均值

$\begin{array}{cl}
= & \dfrac{2(x_1+3)+2(x_2+3)+\cdots+2(x_n+3)}{n} \\
= & \dfrac{2(x_1+x_2+\cdots +x_n)+6n}{n} \\
= & 2\times \dfrac{x_1+x_2+\cdots + x_n}{n} + 6 \\
= & 2\overline{x}+6
\end{array}$

由此，新的標準差

$\begin{array}{cl}
= & \displaystyle \sum_{i=1}^n \dfrac{[2(x_i+3)-(2\overline{x}+6)]^2}{n} \\
= & \displaystyle \sum_{i=1}^n \dfrac{4(x_i-\overline{x})^2}{n} \\
= & \displaystyle 4\sum_{i=1}^n \dfrac{(x_i-\overline{x})^2}{n} \\
= & 4(25) \\
= & 100
\end{array}$