In a series of 2n observations, half of them equal a and remaining half equal -a. If the standard de

Q: Wait, why is the mean zero? I thought we needed to calculate it using the sum.

Great question! Because the dataset is perfectly symmetric—half the values are $a$ and half are $-a$—the positive and negative contributions cancel out exactly. The mean is $\bar{x} = \frac{n(a) + n(-a)}{2n} = 0$. Symmetry is your best friend in statistics, as it simplifies the center of gravity calculation instantly!

Q: Why does the $n$ disappear in the variance calculation?

It is a beautiful cancellation! When you calculate the variance, the $n$ in the numerator (from the sum of squares) and the $n$ in the denominator (the total count $2n$) cancel out. It shows that for this specific distribution, the spread depends only on the magnitude of the values, not the number of observations.

Q: Why is the answer $|a|$ and not just $a$?

Standard deviation is a measure of dispersion, which must always be non-negative. Even if $a$ is negative, the distance from the mean is positive. Squaring $a$ makes it positive, and taking the square root gives us the magnitude, $|a|$. Always watch out for that absolute value trap in competitive exams!

Question

In a series of 2n observations, half of them equal a and remaining half equal -a. If the standard deviation of the observations is 2, then $|a|$ equals.

Options:

(A) $\frac{\sqrt{2}}{n}$

(B) $\sqrt{2}$

(C) 2

(D) $\frac{1}{n}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a number line. To your left, at position $-a$, there is a cluster of $n$ points. To your right, at position $a$, there is an identical cluster of $n$ points.

This is the heart of our problem: a perfectly balanced, symmetric distribution. When you see such symmetry in a JEE problem, your first instinct should be to look for the center of gravity.

The arithmetic mean, $\bar{x}$, is defined as the sum of all observations divided by the total number of observations, $N$. Here, $N = 2n$.

Calculating the mean is straightforward:

$$\bar{x} = \frac{n(a) + n(-a)}{2n}$$

Since $n(a) + n(-a) = na - na = 0$, the mean is exactly zero. This is the anchor of our entire calculation.

### The Variance: Measuring the Spread

Now that we know the mean is zero, we move to the variance, $\sigma^2$. The variance measures the average squared distance of each data point from the mean.

The formula is:

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

Substituting our values, we get:

$$\sigma^2 = \frac{n(a - 0)^2 + n(-a - 0)^2}{2n}$$

Notice how the mean being zero simplifies the expression inside the parentheses to just $a$ and $-a$. Squaring these terms, we get:

$$\sigma^2 = \frac{n(a)^2 + n(-a)^2}{2n}$$

Since any real number squared is positive, $(-a)^2$ becomes $a^2$. Thus, the numerator simplifies to $na^2 + na^2 = 2na^2$.

Dividing by the total number of observations, $2n$, we find:

$$\sigma^2 = \frac{2na^2}{2n} = a^2$$

The $n$ cancels out, leaving us with a clean, elegant result: the variance is simply $a^2$.

### The Final Revelation

We are almost there. The problem provides the standard deviation, $\sigma$, which is the square root of the variance.

Therefore:

$$\sigma = \sqrt{\sigma^2} = \sqrt{a^2}$$

Here is where the trap lies: $\sqrt{a^2}$ is not just $a$; it is the absolute value, $|a|$. Standard deviation must always be a non-negative quantity.

Since the problem states the standard deviation is 2, we have $|a| = 2$.

This result is powerful because it tells us that for any dataset balanced at $\pm a$, the standard deviation is simply the distance of the points from the mean. You have successfully navigated the symmetry, the variance, and the final absolute value constraint.

**The final result is $|a| = 2$.**

Animated Solution for Mathematics - Statistics: In a series of 2n observations, half of them equal a and remaining half equal -a. If the standard deviation of the observations is 2, then $∣ a ∣$ equals.

Select Answer:

Visualized Solution

Visualizing the Data Distribution

Grouping the Observations

Formula for Arithmetic Mean

Substituting Values for Mean

Calculating the Mean

Formula for Variance

Substituting Values for Variance

Simplifying the Numerator

Expanding the Squares

Final Variance Calculation

Standard Deviation from Variance

Expressing Standard Deviation

Equating to Given Value

Key Takeaway

The Sigma Insight: Measures of Dispersion

Analyzing the Setup

The Variance

The Final Revelation

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