The mean of the numbers \(1, 2, 4, 5, x, y\) is 5, and the variance is 10.
Step 1: Calculate the mean
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: \[ \frac{12 + x + y}{6} = 5 \implies 12 + x + y = 30 \implies x + y = 18. \]
Step 2: Use the formula for variance
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: \[ 46 + x^2 + y^2 = 210 \implies x^2 + y^2 = 164. \]
Step 3: Solve for \(x\) and \(y\)
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).
Step 4: Calculate the mean deviation
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: \[ \text{Mean Deviation} = \frac{4 + 3 + 1 + 0 + 3 + 5}{6} = \frac{16}{6} = \frac{8}{3}. \]
