The mean of the numbers \(1, 2, 4, 5, x, y\) is 5, and the variance is 10.
The formula for the mean is: \[ \text{Mean} = \frac{\text{Sum of all values}}{\text{Number of values}}. \] Substituting the given values: \[ \frac{1 + 2 + 4 + 5 + x + y}{6} = 5. \] Simplify: \[ \frac{12 + x + y}{6} = 5 \implies 12 + x + y = 30 \implies x + y = 18. \]
The formula for variance is: \[ \text{Variance} = \frac{\text{Sum of squares of values}}{\text{Number of values}} - (\text{Mean})^2. \] Substituting the given variance \(10\) and mean \(5\): \[ \frac{1^2 + 2^2 + 4^2 + 5^2 + x^2 + y^2}{6} - 5^2 = 10. \] Simplify: \[ \frac{1 + 4 + 16 + 25 + x^2 + y^2}{6} - 25 = 10. \] \[ \frac{46 + x^2 + y^2}{6} - 25 = 10 \implies \frac{46 + x^2 + y^2}{6} = 35. \] Multiply through by 6: \[ 46 + x^2 + y^2 = 210 \implies x^2 + y^2 = 164. \]
We are given: \[ x + y = 18, \quad x^2 + y^2 = 164. \] Use the identity: \[ (x + y)^2 = x^2 + y^2 + 2xy. \] Substituting: \[ 18^2 = 164 + 2xy \implies 324 = 164 + 2xy \implies 2xy = 160 \implies xy = 80. \] The quadratic equation for \(x\) and \(y\) is: \[ t^2 - (x + y)t + xy = 0 \implies t^2 - 18t + 80 = 0. \] Solve using the quadratic formula: \[ t = \frac{-(-18) \pm \sqrt{(-18)^2 - 4(1)(80)}}{2(1)} = \frac{18 \pm \sqrt{324 - 320}}{2}. \] \[ t = \frac{18 \pm 2}{2} \implies t = 10 \text{ or } t = 8. \] Thus, \(x = 8\) and \(y = 10\) (or vice versa).
The formula for mean deviation is: \[ \text{Mean Deviation} = \frac{\sum |x_i - \bar{x}|}{\text{Number of values}}, \] where \(\bar{x}\) is the mean. Substituting the values: \[ \text{Mean Deviation} = \frac{|1 - 5| + |2 - 5| + |4 - 5| + |5 - 5| + |8 - 5| + |10 - 5|}{6}. \] Simplify: \[ \text{Mean Deviation} = \frac{4 + 3 + 1 + 0 + 3 + 5}{6} = \frac{16}{6} = \frac{8}{3}. \]
The mean deviation is: \[ \frac{8}{3}. \]

A substance 'X' (1.5 g) dissolved in 150 g of a solvent 'Y' (molar mass = 300 g mol$^{-1}$) led to an elevation of the boiling point by 0.5 K. The relative lowering in the vapour pressure of the solvent 'Y' is $____________ \(\times 10^{-2}\). (nearest integer)
[Given : $K_{b}$ of the solvent = 5.0 K kg mol$^{-1}$]
Assume the solution to be dilute and no association or dissociation of X takes place in solution.
Inductance of a coil with \(10^4\) turns is \(10\,\text{mH}\) and it is connected to a DC source of \(10\,\text{V}\) with internal resistance \(10\,\Omega\). The energy density in the inductor when the current reaches \( \left(\frac{1}{e}\right) \) of its maximum value is \[ \alpha \pi \times \frac{1}{e^2}\ \text{J m}^{-3}. \] The value of \( \alpha \) is _________.
\[ (\mu_0 = 4\pi \times 10^{-7}\ \text{TmA}^{-1}) \]
A statistical measure that is used to calculate the average deviation from the mean value of the given data set is called the mean deviation.
The mean deviation for the given data set is calculated as:
Mean Deviation = [Σ |X – µ|]/N
Where,
Grouping of data is very much possible in two ways: