If both the mean and the standard deviation of 50 observations x1, x2, ..., x50 are equal to 16, the

Q: Wait, why can't I just subtract 4 from the mean and square it? Isn't the mean of $(x-4)^2$ just $(\bar{x}-4)^2$?

That is a classic trap! The mean of a function is not the function of the mean. You must expand the quadratic term first and apply the linearity of summation. Jensen's Inequality tells us that $E[f(X)] \ge f(E[X])$ for convex functions, so your intuition is actually mathematically impossible here!

Q: Why do we need to calculate $\sum x_i^2$ separately? Can't we just use the variance of the new set?

You could, but that is a longer path. The question asks for the *mean* of the squared terms, not the variance of the new set. Calculating $\sum x_i^2$ directly is the most efficient route to finding the mean of the squares.

Q: Is the sum of a constant $c$ over $n$ observations always $n \times c$?

Exactly! When you sum a constant $c$ for $n$ times, you get $n \times c$. When you divide by $n$ to find the mean, the $n$ cancels out, leaving you with just the constant $c$. It is a beautiful simplification that often appears in JEE problems.

Question

If both the mean and the standard deviation of 50 observations $x_1, x_2, ..., x_{50}$ are equal to 16, then the mean of $(x_1 - 4)^2, (x_2 - 4)^2, ..., (x_{50} - 4)^2$ is :

Options:

(A) 525

(B) 380

(C) 480

(D) 400

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are going to dismantle a problem that often trips up even the brightest minds in the JEE arena. Statistics is not just about crunching numbers; it is about pattern recognition and understanding the architecture of data.

We are given $50$ observations, $x_1, x_2, \dots, x_{50}$. We know the mean $\bar{x} = 16$ and the standard deviation $\sigma = 16$.

In the world of JEE, the first step is always to translate the problem statement into the language of mathematics. We have $n = 50$, $\bar{x} = 16$, and $\sigma = 16$. These are our building blocks.

### The Variance Key: Bridging the Gap

Whenever you see standard deviation and you need to find something related to squares, your mind should immediately jump to the variance formula. Variance is simply the square of the standard deviation, so $\sigma^2 = 16^2 = 256$.

The variance formula acts as the bridge between the mean and the sum of squares:

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

Let us plug in our known values:

$$ 256 = \frac{\sum x_i^2}{50} - (16)^2 $$

Since $16^2 = 256$, our equation becomes $256 = \frac{\sum x_i^2}{50} - 256$. By rearranging this, we find:

$$ \frac{\sum x_i^2}{50} = 256 + 256 = 512 $$

This value, $512$, is the mean of the squares of our original observations. Keep this safe; it is the missing piece of our puzzle.

### The Transformation: Expanding the Target

Now, let us shift our focus to the target. We need to calculate the mean of a new set of observations: $(x_1 - 4)^2, (x_2 - 4)^2, \dots, (x_{50} - 4)^2$.

Let us call this target mean $M$. Mathematically, we define it as:

$$ M = \frac{\sum_{i=1}^{50} (x_i - 4)^2}{50} $$

Using the algebraic identity $(a-b)^2 = a^2 - 2ab + b^2$, we expand the term:

$$ (x_i - 4)^2 = x_i^2 - 8x_i + 16 $$

Substituting this back into our expression for $M$, we get:

$$ M = \frac{\sum (x_i^2 - 8x_i + 16)}{50} $$

### The Elegance of Linearity

We use the linearity property of summation to break this fraction into three manageable parts:

$$ M = \frac{\sum x_i^2}{50} - 8\left(\frac{\sum x_i}{50}ight) + \frac{\sum_{i=1}^{50} 16}{50} $$

Look at each term individually:
1. The first term is the mean of the squares, which we already found to be $512$.
2. The second term contains $\frac{\sum x_i}{50}$, which is simply the original mean, $16$.
3. The third term is the sum of the constant $16$ added $50$ times, divided by $50$, which leaves us with $16$.

### The Final Assembly

Now, it is just simple arithmetic. We substitute our values into the expanded expression:

$$ M = 512 - 8(16) + 16 $$

First, calculate $8 	imes 16 = 128$. Then, perform the subtraction and addition:

$$ M = 512 - 128 + 16 = 384 + 16 = 400 $$

And there we have it! The mean of the new squared observations is **400**. By carefully breaking down the variance formula and expanding the target expression, we transformed a complex-looking problem into a straightforward calculation.

Animated Solution for Mathematics - Statistics: If both the mean and the standard deviation of 50 observations $x_{1}, x_{2}, ..., x_{50}$ are equal to 16, then the mean of $(x_{1} - 4)^{2}, (x_{2} - 4)^{2}, ..., (x_{50} - 4)^{2}$ is :

Select Answer:

Visualized Solution

Given Parameters

Variance Formula

Substitute Values

Calculate Mean of Squares

Target Expression

Expand the Quadratic Term

Distribute the Summation

Substitute Known Values

Final Arithmetic Calculation

The Final Answer

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Variance Key

The Transformation

The Elegance of Linearity

The Final Assembly

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