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The meanand S.D of 20 items were found to be 10 and 2. If an item 12 was misread as 8 the correct mean and S.D are

Updated 5 months ago

Given mean = 10 = \(\bar x\)

Given standard Deviation = 2 = \(\sigma\)

No. of observations = 20 = n

**Step (1)**

\({\sum_{i=1}^{20} x_i}\over n \) = \(\bar x\)

\(\therefore\) \({\sum_{i=1}^{20} x_i}\over 20 \) = 10 \(\therefore\) \({\sum_{i=1}^{20} x_i} \) = 200

But observation 8 was wrong & is replaced by 12.

\(\therefore\) \({\sum_{i=1}^{20} x_i} \) = \(200-8 + 12 = 204\)

\(\therefore\) New mean = \(204\over 20\) = 10.2

Step (2)

Given \(\sigma\) = 2

Variance = \(\sigma^2\) = \(2^2\) = 4

\(\therefore\) \({{\sum_{i=1}^{20} x_i^2}\over n} - \left({\sum_{i=1}^{20} x_i}\over n\right)^2 \) = 4

\({\sum_{i=1}^{20} x_i^2}\over 20 \) \(- 10^2 = 4\)

\(\therefore\) \({\sum_{i=1}^{20} x_i^2} = 104 \times 20 = 2080 \)

But observation 8 is replaced by 12 \(\therefore\)

\({\sum_{i=1}^{20} x_i^2} = 2080 - 8^2 + 12^2 = 2160 \)

\(\therefore\) New variance = \({{\sum_{i=1}^{20} x_i^2}\over 20} - \left({\sum {x_i}}\over 20\right)^2 \)

= \({2160 \over 20} - (10.2)^2\)

= 3.96

\(\therefore\) New S. D. = \(\sqrt{3.96} = 1.99\)

\(\therefore\) New S. D. = 1.99

Updated 5 months ago

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