Question
While calculating the mean and variance of 10 readings, a student wrongly uses the reading 52 for the correct reading 25 . He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and variance.
✓
Answer
Number of observations, $n=10$
Mean,
$\bar{x}=45$ Variance,
$\sigma^{2}=16$
Now,
Incorrect mean,
$\bar{x}=45$
$\Rightarrow \frac{\text { Incorrect } \sum x_{i}}{10}=45$
$\Rightarrow$ Incorrect $\sum \mathrm{x}_{\mathrm{i}}=450$
$\therefore$ Correct $\sum \mathrm{x}_{\mathrm{i}}=450-52+25=423$
$\Rightarrow$ Correct mean $=\frac{\text { Correct } \sum x_{i}}{10}=\frac{423}{10}=42.3$
Incorrect variance,
$\sigma^{2}=16$
$\Rightarrow 16=\frac{\text { Incorrect } \sum x_{i}^{2}}{10}-(45)^{2}$
$\Rightarrow$ Incorrect $\sum x_{i}^{2}=10(16+2025)=20410$
$\therefore$ Correct $\sum x_{i}^{2}=20410-(52)^{2}+(25)^{2}=20410-2704+625=18331$
Now,
Correct variance $=\frac{18331}{10}-(42.3)^{2}=1833.1-1789.29=43.81$
Mean,
$\bar{x}=45$ Variance,
$\sigma^{2}=16$
Now,
Incorrect mean,
$\bar{x}=45$
$\Rightarrow \frac{\text { Incorrect } \sum x_{i}}{10}=45$
$\Rightarrow$ Incorrect $\sum \mathrm{x}_{\mathrm{i}}=450$
$\therefore$ Correct $\sum \mathrm{x}_{\mathrm{i}}=450-52+25=423$
$\Rightarrow$ Correct mean $=\frac{\text { Correct } \sum x_{i}}{10}=\frac{423}{10}=42.3$
Incorrect variance,
$\sigma^{2}=16$
$\Rightarrow 16=\frac{\text { Incorrect } \sum x_{i}^{2}}{10}-(45)^{2}$
$\Rightarrow$ Incorrect $\sum x_{i}^{2}=10(16+2025)=20410$
$\therefore$ Correct $\sum x_{i}^{2}=20410-(52)^{2}+(25)^{2}=20410-2704+625=18331$
Now,
Correct variance $=\frac{18331}{10}-(42.3)^{2}=1833.1-1789.29=43.81$
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