MCQ
Choose the correct answer. Let a, b, c, d, e be the observations with mean m and standard deviation s. The standard deviation of the observations a + k, b + k, c + k, d + k, e + k is:
  • s
  • B
    ks
  • C
    s + k
  • D
    $\frac{\text{s}}{\text{k}}$

Answer

Correct option: A.
s
Given observations are a, b, c d and e.
$\text{Mean}=\text{m}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}$
$\sum\text{x}_\text{i}={\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}={5}\text{m}$
Now, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$
$=\frac{(\text{a}+\text{b}+\text{c}+\text{d}+\text{e})+5\text{k}}{5}=\text{m}+\text{k}$
$\therefore\ \text{SD}=\sqrt{\frac{{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{k}\text{x}_\text{i})}}{\text{n}}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$
$=\text{s}$

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