Question

While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and the variance.
[NCERT EXEMPLAR]

Answer 1

Given:

Number of observations, n = 10

Mean, $\overline{x}$ = 45

Variance, ${\sigma }^2$ = 16

Now,

Incorrect mean, $\overline{x}$ = 45

$$\begin{gathered} \Rightarrow \frac{\mathrm{Incorrect}{\displaystyle \sum _{}}{x}_i}{10}=45 \\ \Rightarrow \mathrm{Incorrect}\sum _{}{x}_i=450 \end{gathered}$$

∴ Correct $\sum _{}{x}_i=450-52+25=423$

⇒ Correct mean = $\frac{\mathrm{Correct}\sum _{}{x}_i}{10}=\frac{423}{10}=42.3$

Incorrect variance, ${\sigma }^2$ = 16

$$\begin{gathered} \Rightarrow 16=\frac{\mathrm{Incorrect}{\displaystyle \sum _{}}{x}_i^2}{10}-{\left(45\right)}^2 \\ \Rightarrow \mathrm{Incorrect}\sum _{}{x}_i^2=10\left(16+2025\right)=20410 \end{gathered}$$

∴ Correct $\sum _{}{x}_i^2=20410-{\left(52\right)}^2+{\left(25\right)}^2=20410-2704+625=18331$

Now,

Correct variance = $\frac{18331}{10}-{\left(42.3\right)}^2=1833.1-1789.29=43.81$