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: \[ \frac{8}{3}. \]
The figure shows a pipe with cross-section area 10 \( cm^2 \). Water flows from one end with velocity 20 cm/s. The other end of the pipe is closed and consists of 10 holes each of area 30 \( mm^2 \). Find the velocity of water coming out from each hole: 
Which of the following best represents the temperature versus heat supplied graph for water, in the range of \(-20^\circ\text{C}\) to \(120^\circ\text{C}\)? 
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: