Chapter: 2021· JEE
Answer: the other two numbers are 10 and 11
Let the other two numbers be \(a\) and \(b\).
Mean: sum of all six \(= 6\times6.5 = 39\). The given four add to \(2+4+5+7 = 18\), so \(a + b = 21\).
Variance: \(\sigma^{2} = \dfrac{\sum x^{2}}{n} - \bar x^{2}\), so
\[\frac{\sum x^{2}}{6} = 10.25 + 6.5^{2} = 52.5\]
\[\sum x^{2} = 315\]
The given four give \(4 + 16 + 25 + 49 = 94\), so \(a^{2} + b^{2} = 221\).
Then \(2ab = (a+b)^{2} - (a^{2}+b^{2}) = 441 - 221 = 220\), so \(ab = 110\).
\(a\) and \(b\) are roots of \(t^{2} - 21t + 110 = 0\), i.e. \((t-10)(t-11) = 0\). So the numbers are 10 and 11.