# Variance and average

Of the 40 values were calculated average mx = 7.5 and variance sx = 2.25. After the control was found to lack the two items of the values of x41 = 3.8 and x42=7. Correct the above characteristics (mx and sx).

Result

m2 =  7.4
r2 =  2.455

#### Solution:

$m_{1}=7.5 \ \\ r_{1}=2.25 \ \\ m_{2}=(40 \cdot \ m_{1} + 3.8 + 7)/42=(40 \cdot \ 7.5 + 3.8 + 7)/42=\dfrac{ 37 }{ 5 }=7.4$
$r_{2}=(40 \cdot \ r_{1} + (3.8-m_{2})^2 + (7-m_{2})^2)/42=(40 \cdot \ 2.25 + (3.8-7.4)^2 + (7-7.4)^2)/42 \doteq \dfrac{ 1289 }{ 525 } \doteq 2.4552 \doteq 2.455$

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