Question

The mean and standard deviation of a series of 20 items are 20 and 5 respectively. While calculating these measures, an item of 13 was wrongly read as 30. Find out the correct mean and standard deviation.

Answer 1

Calculating correct Mean, by using the following observations:

$$\begin{gathered} {\overline{X}}_{wrong}=\frac{\Sigma X_{wrong}}{N} \\ \Sigma X_{wrong}={\overline{X}}_{wrong}\times N \\ \Sigma X_{wrong}=20\times 20 \\ \Rightarrow \Sigma X_{wrong}=400 \\ {\mathrm{\Sigma X}}_{correct}=\Sigma X_{wrong}-\mathrm{incorrect}\mathrm{item}+\mathrm{correct}\mathrm{item} \\ {\mathrm{\Sigma X}}_{correct}=400-30+13 \\ \Rightarrow {\mathrm{\Sigma X}}_{correct}=383 \\ {\overline{\mathrm{X}}}_{\mathrm{correct}}=\frac{{\mathrm{\Sigma X}}_{correct}}{N}=\frac{383}{20}=19.15 \\ \mathrm{Calculation}\mathrm{of}\mathrm{Correct}\mathrm{S}.\mathrm{D}. \\ \sigma =\sqrt{\frac{\Sigma x^2}{N}-{\left(\overline{X}\right)}^2} \\ 5=\sqrt{\frac{\Sigma x^2}{20}-{\left(20\right)}^2} \end{gathered}$$

Squaring both sides

$25=\frac{\mathrm{\Sigma }{x}^2}{20}-400$

425 × 20 = Σx²

$\Sigma {x^2}_{wrong}$= 8500

$\Sigma {x^2}_{correct}$ = $\Sigma {x^2}_{wrong}$ − (Incorrect item)² + (Correct item)²

$\Sigma {x^2}_{correct}$= 8500 − (30)² + (13)²
$\Sigma {x^2}_{correct}$= 8500 − 900 + 169 = 7769

$$\begin{gathered} {\sigma }_{\mathrm{Correct}}=\sqrt{\frac{\Sigma {x^2}_{Correct}}{N}-{\left({\overline{X}}_{Correct}\right)}^2} \\ or,{\sigma }_{\mathrm{Correct}}=\sqrt{\frac{7769}{20}-{\left(19.15\right)}^2} \\ or,{\sigma }_{\mathrm{Correct}}=\sqrt{388.45-366.72} \\ or,{\sigma }_{\mathrm{Correct}}=\sqrt{21.73} \\ {\sigma }_{\mathrm{Correct}}=4.66 \end{gathered}$$

Hence, correct mean and standard deviation are 19.15 and 4.66 respectively.