The mean of five observations is 4 and their variance is 5.2 . If three of them are 1, 2, 6, then other two are
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The mean of five observations is 4 and their variance is 5.2 . If three of them are 1, 2, 6, then other two are
2, 9
4, 7
5, 6
2, 10
Mean = (1+2+6+x+y)/5 = 4 => x+y = 11. Variance = (sum(x_i^2)/5) - mean^2 = 5.2. (1+4+36+x^2+y^2)/5 - 16 = 5.2 => (41+x^2+y^2)/5 = 21.2 => x^2+y^2 = 65. Solving x+y=11 and x^2+y^2=65 gives {4, 7}.
Let the two unknown observations be a and b, so their sum a plus b equals five times the mean minus the sum of the known items, giving 20 minus 9 equal to 11. Using the variance formula, the sum of all squared observations equals 5 multiplied by (variance plus mean squared), which is 5 times (5.2 plus 16) equal to 106. Subtracting the sum of the squares of the known items leaves a squared plus b squared equal to 106 minus 41, which is 65. Solving the system a plus b equals 11 and a squared plus b squared equals 65 gives the two values 4 and 7.