Here n = 20, Incorrect mean (x) = 10, Incorrect S.D. (s) = 2 Now, $\overset{-}{x}$ = $\frac{1}{n}\sum x_i$ ⇒ $\sum x_i$ = n × $\overset{-}{x}$ = 20 × 10 = 200 ∴ Incorrect $\sum x_i$ = 200 Also , $\frac{1}{n}\sum x_i^2 - \left(\overset{-}{x}\right)^2$ = 4 ⇒ $\frac{1}{20}\sum x_i^2 - (10)^2$ = 4 ⇒ $\sum x_i^2$ = 2080 ∴ Incorrect $\sum x_i^2$ = 2080 (i) When wrong item 8 is omitted from the data then we have 19 observations. ∴ Correct $\sum x_i$ = Incorrect $\sum x_i$ - 8 Correct $\sum x_i$ = 200 - 8 = 192 ∴ Correct mean = $\frac{192}{19}$ = 10.1 Also, correct $\sum x_i^2$ = Incorrect $\sum_i^2 - (8)^2$ ⇒ Correct $\sum x_i^2$ = 2080 - 64 = 2016 ∴ Correct variance = $\frac{1}{19}$ (correct $\sum x_i^2$) - $(\text{correct mean})^2$ = $\frac{1}{19}$ × 2016 - $\left(\frac{192}{19}\right)^2$ = $\frac{2016}{19} - \frac{36864}{361}$ = $\frac{38304-36864}{361}$ = $\frac{1440}{361}$ ∴ Correct S.D. = $\sqrt{\frac{1440}{361}}$ = $\sqrt{3.98}$ = 1.99 (ii) If wrong item 8 is replaced by 12 Correct $\sum x_i$ = Incorrect $\sum x_i$ – 8 + 12 = 200 – 8 + 12 = 204 ∴ Correct mean = $\frac{204}{20}$ = 10.2 Also correct $\sum x_i^2$ = Incorrect $\sum x_i^2$ - $(8)^2+(12)^2$ = 2080 - 64 + 144 = 2160 ∴ Correct variance = $\frac{1}{20}\left(\text{correct }\sum x_i^2\right) - (\text{correct mean})^2$ = $\frac{2160}{20} - \left(\frac{204}{20}\right)^2$ = $\frac{2160}{20} - \frac{41616}{400}$ = $\frac{43200-41616}{400}$ = $\frac{1584}{400}$ ∴ Correct S.D. = $\sqrt{\frac{1584}{400}}$ = $\sqrt{3.96}$ = 1.98