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In an experiment with 8 observations X_{i},i=1,2,3,...8, the following results were available:

\sum_{i=1}^{8}x_{i}^{2}=510,\sum_{i=1}^{8}x_{i}=58

If at the time of calculation the observation 21 was wrongly taken as 12, then the corrected variance is

Option: 1

\frac{1967}{16}


Option: 2

100


Option: 3

\frac{1967}{8}

 


Option: 4

\frac{1967}{64}


Answers (1)

best_answer

Corrected \sum x^{2}=510-144+441=807

Corrected \sum x=58-12+21=67

\therefore Corrected Variance = \frac{807}{8}-\left ( \frac{67}{8} \right )^{2}

=\frac{807}{8}-\frac{4489}{8\times8}

=\frac{807\times8-4489}{64}=\frac{6456-4489}{64}

=\frac{1967}{64}

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