Solution :
As given \(\bar{x}\) = 4, n = 5 and \({\sigma}^2\) = 5.2. If the remaining observations are \(x_1\), \(x_2\) then
\({\sigma}^2\) = \(\sum{(x_i – \bar{x})}^2\over n\) = 5.2
\(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2 + {(1-4)}^2 + {2-4)}^2 + {(6-4)}^2\over 5\) = 5.2
\(\implies\) \({(x_1-4)}^2 + {(x_2-4)}^2\) = 9 …..(1)
Also \(\bar{x}\) = 4 \(\implies\) \(x_1 + x_2 + 1 + 2 + 6\over 5\) = 4 \(\implies\) \(x_1 + x_2\) = 11 ….(2)
from eq.(1), (2) \(x_1\), \(x_2\) = 4, 7
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