QJEEM23Ma0512: JEE Main 2023 Mathematics - Class 11 JEEM23Ma0512 Solution: Let the mean of 6 observations and be 5 and their variance be…

Given that the mean of the observations ${x, y, 1, 2, 4, 5}$ is 5, we get the equation:

$x + y + 1 + 2 + 4 + 5 = 6 \cdot 5 \Rightarrow x + y = 18$ .......... $(1)$

We are also given that the variance of the observations is 10. Using the formula for variance, we have:

$V = \frac{\Sigma x_i^2}{n} - \bar{x}^2$ , where $n$ is the number of observations, and $\bar{x}$ is the mean.

Substituting the given values in the variance formula, we get:

$10 = \frac{\Sigma x_i^2}{6} - 5^2 = \frac{\Sigma x_i^2}{6} - 25 \Rightarrow \Sigma x_i^2 = 210$ .

Here, $\Sigma x_i^2$ is the sum of the squares of all the observations, thus:

$x^2 + y^2 + 1 + 4 + 16 + 25 = 210 \Rightarrow x^2 + y^2 = 164$ ......... $(2)$

By solving the system of equations $(1)$ and $(2)$ , we get $x = 8$ and $y = 10$ .

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:

$MD(5) = \frac{1}{6} [|1-5| + |2-5| + |4-5| + |5-5| + |8-5| + |10-5|] = \frac{16}{6} = \frac{8}{3}$ .

So, the mean deviation about the mean is $\frac{8}{3}$ , and the correct answer is Option B, $\frac{8}{3}$ .