For a statistical data of 10 values, a student obtained the mean at 5.5 and . 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

Question

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 [2025]

Smart Achievers · Accepted Answer

(1)

$\sum_{i = 1}^{10} x_{i}^{} = 5.5 \times 10 = 55$

Correct mean $= \frac{\sum_{i = 1}^{10} x_{i}^{} - 4 - 5 + 6 + 8}{10} = \frac{55 - 9 + 14}{10} = 6$

Corrected $\sum_{i = 1}^{10} x_{i}^{2} = 371 - 4^{2} - 5^{2} + 6^{2} + 8^{2} = 430$

Corrected variance $= \frac{Corrected \sum_{i = 1}^{10} x_{i}^{2}}{10} - {(Corrected \bar{x})}^{2}$

$= \frac{430}{10} - {(6)}^{2} = 43 - 36 = 7$

Solution

1 7

2 5

3 9

4 4

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Solution

Q. For a statistical data x1,x2,...,x10 of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=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 [2025]

1 7

2 5

3 9

4 4

Ans. (1) ∑i=110xi=5.5×10=55 Correct mean =∑i=110xi–4–5+6+810=55–9+1410=6 Corrected ∑i=110xi2=371–42–52+62+82=430 Corrected variance =Corrected∑i=110xi210–(Corrected x–)2 =43010–(6)2=43–36=7

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