The outcome of each of 30 items was observed; 10 items gave an outcome 1/2 – d each, 10 items gave o

Q: Why can't I just use the formula $\sigma^2 = E[X^2] - (E[X])^2$?

You absolutely can! It is a perfectly valid method. However, in problems with high symmetry, calculating the deviations directly from the mean is often faster and less prone to arithmetic errors. It helps you see the 'spread' of the data more intuitively.

Q: Wait, why is the mean $\frac{1}{2}$? Do I have to calculate it every time?

In this specific case, yes, calculating it confirms our intuition. But notice that for any symmetric distribution, the mean is the center of symmetry. Recognizing this is a superpower in JEE exams—it saves precious seconds!

Q: Does the frequency of the middle group matter?

It matters for the total count $N$, which is $30$. But because the middle group's value is exactly the mean, its deviation is zero. It contributes nothing to the variance, which is a beautiful way of saying that the middle group is 'stable' and doesn't add to the spread.

Question

The outcome of each of 30 items was observed; 10 items gave an outcome $\frac{1}{2} - d$ each, 10 items gave outcome $\frac{1}{2}$ each and the remaining 10 items gave outcome $\frac{1}{2} + d$ each. If the variance of this outcome data is $\frac{4}{3}$ then $|d|$ equals :

Options:

(A) 2

(B) $\frac{\sqrt{5}}{2}$

(C) $\frac{2}{3}$

(D) $\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Beauty of Symmetry: A Statistical Journey

Welcome, future engineer. Today, we are going to dive into the elegant world of statistics.

Often, students see a problem like this and immediately reach for the most complex formulas, but I want you to pause. Look at the data points: $\frac{1}{2}-d$, $\frac{1}{2}$, and $\frac{1}{2}+d$.

Do you see the rhythm? They are perfectly balanced around the center. This is not a coincidence; it is a gift.

In physics and mathematics, symmetry is your best friend. It is the secret shortcut that separates the frantic calculator from the master problem-solver.

### The Mean: The Center of Gravity

Before we can understand the spread, we must find the center. The arithmetic mean, $\bar{x}$, is the center of gravity of our data.

We have three groups of ten items each. The total number of items is $N = 30$.

To find the mean, we sum all the values and divide by $N$. Instead of brute-forcing the multiplication, let us use the symmetry we observed:

$$10\left(\frac{1}{2} - d\right) + 10\left(\frac{1}{2}\right) + 10\left(\frac{1}{2} + d\right)$$

When we factor out the $10$, we get:

$$10 \left[ \left(\frac{1}{2} - d\right) + \frac{1}{2} + \left(\frac{1}{2} + d\right) \right]$$

Look at that! The $-d$ and $+d$ cancel out perfectly, leaving us with $10 \left[ \frac{3}{2} \right]$.

Dividing by $30$, we get $\bar{x} = \frac{1}{2}$. Our mean is exactly the middle value. It feels right, doesn't it?

### The Variance: Measuring the Spread

Now, we move to the variance, $\sigma^2$. Variance is simply a measure of how far our data points are from the mean.

It is the average of the squared deviations. The formula is:

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

Let us calculate the deviations for each group. For the first group, the deviation is $(\frac{1}{2} - d) - \frac{1}{2} = -d$.

For the third group, it is $(\frac{1}{2} + d) - \frac{1}{2} = d$. The middle group sits right on the mean, so its deviation is $0$.

When we square these, the negative sign disappears: $(-d)^2 = d^2$ and $d^2 = d^2$. The middle group contributes nothing to the variance.

### The Final Calculation

We are almost there. The sum of squared deviations is $10(-d)^2 + 10(0)^2 + 10(d)^2$, which simplifies to $20d^2$.

Dividing by the total number of items, $30$, we get:

$$\sigma^2 = \frac{20d^2}{30} = \frac{2d^2}{3}$$

The problem tells us that the variance is $\frac{4}{3}$. So, we set:

$$\frac{2d^2}{3} = \frac{4}{3}$$

The denominators cancel out, leaving $2d^2 = 4$, which means $d^2 = 2$. Finally, taking the square root, we find **$|d| = \sqrt{2}$**.

It is elegant, it is clean, and it is finished. Remember, in JEE Advanced, the math is rarely about brute force; it is about finding the path of least resistance through the beauty of the structure.

Select Answer:

Visualized Solution

Visualizing the Data Groups

The Concept of Arithmetic Mean

Setting up the Mean Equation

Simplifying the Summation

Final Value of Mean $\overset{x}{ˉ}$

The Concept of Variance $σ^{2}$

Calculating Deviations

Setting up the Variance Equation

Simplifying the Variance

Equating with Given Variance

Isolating the Variable $d^{2}$

Finding the Absolute Value $∣ d ∣$

The Sigma Insight: Variance and Standard Deviation

The Beauty of Symmetry

The Mean

The Variance

The Final Calculation

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