JEE Main 2023 — Statistics Question with Solution

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, $\frac{8}{3}$.