Let the mean of 6 observations $1,2,4,5, \mathrm{x}$ and $\mathrm{y}$ be 5 and their variance be 10 . Then their mean deviation about the mean is equal to :

<p>Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:</p> <p>$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ ..........$(1)$</p> <p>We are also given that the variance of the observations is 10. Using the formula for variance, we have:</p> <p>$V = \frac{\Sigma x_i^2}{n} - \bar{x}^2$, where $n$ is the number of observations, and $\bar{x}$ is the mean.</p> <p>Substituting the given values in the variance formula, we get:</p> <p>$10 = \frac{\Sigma x_i^2}{6} - 5^2 = \frac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$.</p> <p>Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus:</p> <p>$x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ .........$(2)$</p> <p>By solving the system of equations $(1)$ and $(2)$, we get $x = 8$ and $y = 10$.</p> <p>Now, the mean deviation about the mean ($\bar{x} = 5$) is calculated by taking the average of the absolute differences of each observation from the mean:</p> <p>$MD(5) = \frac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \frac{16}{6} = \frac{8}{3}$.</p> <p>So, the mean deviation about the mean is $\frac{8}{3}$, and the correct answer is Option B, $\frac{8}{3}$.</p>