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: \[ \frac{8}{3}. \]
| Class | 0 – 15 | 15 – 30 | 30 – 45 | 45 – 60 | 60 – 75 | 75 – 90 |
|---|---|---|---|---|---|---|
| Frequency | 11 | 8 | 15 | 7 | 10 | 9 |
Let the Mean and Variance of five observations $ x_i $, $ i = 1, 2, 3, 4, 5 $ be 5 and 10 respectively. If three observations are $ x_1 = 1, x_2 = 3, x_3 = a $ and $ x_4 = 7, x_5 = b $ with $ a>b $, then the Variance of the observations $ n + x_n $ for $ n = 1, 2, 3, 4, 5 $ is
Find the mean of the following distribution:
\[\begin{array}{|c|c|c|c|c|c|c|c|} \hline \textbf{Class-interval} & 11-13 & 13-15 & 15-17 & 17-19 & 19-21 & 21-23 & 23-25 \\ \hline \text{Frequency} & \text{7} & \text{6} & \text{9} & \text{13} & \text{20} & \text{5} & \text{4} \\ \hline \end{array}\]
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is:
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: