Question ID: #368
For a statistical data $x_{1},x_{2},…,x_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{i=1}^{10}x_{i}^{2}=371$. He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is
- (1) 7
- (2) 4
- (3) 9
- (4) 5
Solution:
Given $n=10$.
Incorrect Mean $\bar{x} = 5.5 \Rightarrow \sum x_{old} = 10 \times 5.5 = 55$.
Incorrect Sum of Squares $\sum x^2_{old} = 371$.
Correction of $\sum x$:
Sum is corrected by removing incorrect values (4, 5) and adding correct values (6, 8).
$$ \sum x_{new} = 55 – (4+5) + (6+8) = 55 – 9 + 14 = 60 $$
New Mean $\bar{x}_{new} = \frac{60}{10} = 6$.
Correction of $\sum x^2$:
$$ \sum x^2_{new} = 371 – (4^2 + 5^2) + (6^2 + 8^2) $$
$$ \sum x^2_{new} = 371 – (16 + 25) + (36 + 64) $$
$$ \sum x^2_{new} = 371 – 41 + 100 = 430 $$
Calculate New Variance $\sigma^2$:
$$ \sigma^2 = \frac{\sum x^2_{new}}{n} – (\bar{x}_{new})^2 $$
$$ \sigma^2 = \frac{430}{10} – (6)^2 $$
$$ \sigma^2 = 43 – 36 = 7 $$
Ans. (1)
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