If sum(xi - a) = n and sum(xi - a)^2 = na, (n, a 1) then the standard deviation of n observations

Q: Wait, why does shifting the origin not change the standard deviation? Doesn't the spread change if I move the data?

That is a fantastic conceptual question! Think of standard deviation as the 'average distance' of data points from their mean. If you shift every single data point by the same constant $a$, the mean also shifts by exactly $a$. Consequently, the distance between any two points (or between a point and the mean) remains perfectly preserved. Imagine a group of people standing on a field; if everyone takes three steps forward, the distance between them doesn't change. That is why standard deviation is invariant under a change of origin.

Q: Why can't I just expand $\sum (x_i - a)^2$ and solve for $x_i$?

You absolutely could, but it would be a mathematical nightmare! Expanding that summation would lead to a complex expression involving $\sum x_i^2$, $\sum x_i$, and $n$. While you could technically solve it, you would be doing unnecessary heavy lifting. In JEE, the goal is to recognize the structure. By substituting $y_i = x_i - a$, you transform a messy problem into a clean, elegant one. Always look for the substitution that simplifies the algebra!

Q: I got $a-1$ as my answer, but the options say $\sqrt{a-1}$. Why is that?

Ah, the classic trap! Remember that the formula we used, $\sigma^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2$, gives you the *variance* ($\sigma^2$), not the standard deviation ($\sigma$). The standard deviation is defined as the positive square root of the variance. Always double-check what the question is asking for—it is a very common mistake to stop at the variance!

Question

If $\sum_{i=1}^{n} (x_i - a) = n$ and $\sum_{i=1}^{n} (x_i - a)^2 = na$, $(n, a > 1)$ then the standard deviation of $n$ observations $x_1, x_2, \dots, x_n$ is:

Options:

(A) $a-1$

(B) $n\sqrt{a-1}$

(C) $\sqrt{n(a-1)}$

(D) $\sqrt{a-1}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a vast field, and you have a collection of data points $x_1, x_2, \dots, x_n$ scattered across the ground. Your goal is to understand how spread out these points are; this is the essence of standard deviation.

In this problem, we are given two conditions:
$$\sum_{i=1}^{n} (x_i - a) = n$$
$$\sum_{i=1}^{n} (x_i - a)^2 = na$$

At first glance, these might look like abstract algebraic constraints, but they are actually a roadmap to the solution.

### The Power of Perspective: Change of Origin

The most beautiful part of statistics is realizing that the absolute position of your data does not matter as much as the relative distance between the points. This is the concept of **invariance under change of origin**.

If we shift all our data points by a constant $a$, the spread—the standard deviation—remains untouched. This is our secret weapon.

Let us define a new variable, $y_i = x_i - a$. By doing this, we are essentially shifting our coordinate system so that the data is centered around a new origin. Our new data set is $y_1, y_2, \dots, y_n$.

### Simplifying the Landscape

Now, let us look at our conditions through this new lens. The first condition, $\sum_{i=1}^{n} (x_i - a) = n$, becomes:
$$\sum_{i=1}^{n} y_i = n$$

The mean of our new data set, $\bar{y}$, is defined as $\bar{y} = \frac{\sum y_i}{n}$. Substituting our condition, we get:
$$\bar{y} = \frac{n}{n} = 1$$

Our data is now centered at $1$. Next, let us look at the second condition: $\sum_{i=1}^{n} (x_i - a)^2 = na$. With our substitution, this becomes:
$$\sum_{i=1}^{n} y_i^2 = na$$

We are looking for the mean of the squares of our new data, which is $\frac{\sum y_i^2}{n}$. Substituting our second condition, we get:
$$\frac{\sum y_i^2}{n} = \frac{na}{n} = a$$

### The Final Synthesis

We are now ready to calculate the variance, $\sigma^2$. The formula for variance is the mean of the squares minus the square of the mean:
$$\sigma^2 = \frac{\sum y_i^2}{n} - (\bar{y})^2$$

We have all the pieces of the puzzle: $\frac{\sum y_i^2}{n} = a$ and $\bar{y} = 1$. Plugging these in, we get:
$$\sigma^2 = a - (1)^2 = a - 1$$

Finally, the standard deviation $\sigma$ is the positive square root of the variance:
$$\sigma = \sqrt{\sigma^2} = \mathbf{\sqrt{a - 1}}$$

Through the simple act of shifting our perspective, we have turned a daunting algebraic problem into a clear, elegant result. This is the power of mathematical intuition.

Animated Solution for Mathematics - Statistics: If $\sum_{i = 1}^{n} (x_{i} - a) = n$ and $\sum_{i = 1}^{n} (x_{i} - a)^{2} = na$ , $(n, a > 1)$ then the standard deviation of $n$ observations $x_{1}, x_{2}, \dots, x_{n}$ is:

Select Answer:

Visualized Solution

Analyzing the Given Data

The Change of Origin Concept

Defining the New Variable $y_{i}$

Transforming the First Condition

Calculating the Mean of $y_{i}$

Transforming the Second Condition

Mean of Squares of $y_{i}$

The Variance Formula

Substituting Values into Variance

Final Standard Deviation

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Perspective

Simplifying the Landscape

The Final Synthesis

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