MCQ
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
The given observations are a, b, c, d, e.
$\text{Mean}=\text{m}=\frac{\text{a+b+c+d+e}}{5}$
$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$
Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
Now, consider the observations a + k, b + k, c + k, d + k, e + k.
New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$
$=\frac{\text{a+b+c+d+e+5k}}{5}$
$=\frac{5\text{m}+5\text{k}}{5}$
$=\text{m}+\text{k}$
$\therefore$ New standard deviation
$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$
$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}}$ [Using (1)]
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
$=\text{s}$
Hence, the correct answer is option (a).

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