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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 :

JEE · Math · previous-year question

  1. A.$$\frac{10}{3}$$
  2. B.$$\frac{8}{3}$$correct
  3. C.$$\frac{7}{3}$$
  4. D.3

Answer

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

Explanation

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

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