Let α, β ∈ R. Let the mean and the variance of 6 observations -3, 4, 7, -6, α, β be 2 and 23, respec

Q: Wait, why do we use the variance formula $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$ instead of $\frac{\sum (x_i - \bar{x})^2}{n}$?

Both formulas are mathematically equivalent! However, in JEE problems involving unknown variables, the first form is often much easier to work with algebraically. It allows you to isolate the sum of squares $\sum x_i^2$ directly, which is usually the missing piece of the puzzle. Using the second form would require you to expand $(\alpha - \bar{x})^2$, which is perfectly valid but often leads to more tedious algebraic steps.

Q: Does the order of $\alpha$ and $\beta$ matter when we find them to be 4 and 6?

Great question! In this specific problem, the set of observations is $\{-3, 4, 7, -6, \alpha, \beta\}$. Since the mean and variance are symmetric functions of $\alpha$ and $\beta$, swapping them does not change the set of observations. Therefore, the mean deviation about the mean will remain the same regardless of which variable you assign to 4 and which to 6. The set of values is what matters, not the labels.

Q: What exactly is 'mean deviation about the mean'?

Think of it as the 'average distance' of all your data points from the center (the mean). While variance squares the distances to penalize outliers, mean deviation simply takes the absolute distance $|x_i - \bar{x}|$. It tells you, on average, how far each observation is from the mean line. It is a very intuitive measure of dispersion!

Question

Let $\alpha, \beta \in \mathbb{R}$. Let the mean and the variance of 6 observations $-3, 4, 7, -6, \alpha, \beta$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

Options:

(A) $\frac{13}{3}$

(B) $\frac{16}{3}$

(C) $\frac{11}{3}$

(D) $\frac{14}{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Statistics is the language of uncertainty, and in this problem, we are tasked with playing the role of a detective. We have a set of six observations: $-3, 4, 7, -6, \alpha, \beta$.

We know the mean is $\bar{x} = 2$ and the variance is $\sigma^2 = 23$. Our mission is to find the mean deviation about the mean.

### The First Constraint: The Balance Point

The mean is defined as the sum of all observations divided by the total count. Mathematically, this is expressed as:

$$\bar{x} = \frac{\sum x_i}{n}$$

Substituting our knowns and unknowns, we get:

$$\frac{-3 + 4 + 7 - 6 + \alpha + \beta}{6} = 2$$

Simplifying the knowns, $-3 + 4 + 7 - 6$ equals $2$. Thus, the equation becomes:

$$\frac{2 + \alpha + \beta}{6} = 2$$

Multiplying by $6$ gives $2 + \alpha + \beta = 12$, which simplifies beautifully to:

$$\alpha + \beta = 10$$

This is our first vital clue.

### The Second Constraint: The Measure of Spread

Now, we turn to the variance, $\sigma^2 = 23$. The variance measures how spread out the data is from the mean. We use the computational formula:

$$\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2$$

Substituting our values, we have:

$$\frac{(-3)^2 + 4^2 + 7^2 + (-6)^2 + \alpha^2 + \beta^2}{6} - 2^2 = 23$$

Calculating the squares of the knowns: $9 + 16 + 49 + 36 = 110$. The equation becomes:

$$\frac{110 + \alpha^2 + \beta^2}{6} - 4 = 23$$

Adding $4$ to both sides gives:

$$\frac{110 + \alpha^2 + \beta^2}{6} = 27$$

Multiplying by $6$ yields $110 + \alpha^2 + \beta^2 = 162$, so:

$$\alpha^2 + \beta^2 = 52$$

### The Algebraic Bridge

We now have two equations: $\alpha + \beta = 10$ and $\alpha^2 + \beta^2 = 52$. To find the individual values, we use the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$.

Substituting our known sums, we get $10^2 = 52 + 2\alpha\beta$, which means $100 = 52 + 2\alpha\beta$. Solving for the product:

$$2\alpha\beta = 48 \Rightarrow \alpha\beta = 24$$

We need two numbers that add to $10$ and multiply to $24$. A quick mental check reveals these numbers are $4$ and $6$. Thus, our missing observations are $4$ and $6$.

### The Final Journey: Mean Deviation

With our full set of observations $\{-3, 4, 7, -6, 4, 6\}$, we can calculate the mean deviation about the mean. This is defined as:

$$MD = \frac{\sum |x_i - \bar{x}|}{n}$$

We calculate the absolute distance of each point from the mean $2$:

$|-3 - 2| = 5$
$|4 - 2| = 2$
$|7 - 2| = 5$
$|-6 - 2| = 8$
$|4 - 2| = 2$
$|6 - 2| = 4$

Summing these distances: $5 + 2 + 5 + 8 + 2 + 4 = 26$. Finally, we divide by the total number of observations, $6$:

$$MD = \frac{26}{6} = \frac{13}{3}$$

We have successfully navigated the complexity of the problem. The final result is **$\frac{13}{3}$**.

Animated Solution for Mathematics - Statistics: Let $α, β \in R$ . Let the mean and the variance of 6 observations $- 3, 4, 7, - 6, α, β$ be 2 and 23, respectively. The mean deviation about the mean of these 6 observations is :

Select Answer:

Visualized Solution

Given Data

Mean Equation Setup

Simplifying for $α + β$

Variance Equation Setup

Simplifying for $α^{2} + β^{2}$

Finding the Product $α β$

Solving for $α$ and $β$

Absolute Deviations

Sum of Absolute Deviations

Final Mean Deviation

The Sigma Insight: Mean Deviation

Analyzing the Setup

The First Constraint

The Second Constraint

The Algebraic Bridge

The Final Journey

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