Let xi (1

Q: Wait, why does $p$ not affect the standard deviation? Shouldn't shifting the data change the values?

Think of standard deviation as the 'average distance' of points from their mean. If you shift every single point by $p$, the mean also shifts by $p$. The distance between any two points (or between a point and the mean) remains exactly the same. Thus, the spread is invariant. It is a beautiful property of statistics!

Q: Can I solve this by finding the mean $\bar{x}$ first?

You could, but it is a much longer path. You would need to find $\bar{x}$ in terms of $p$, then calculate $\sum (x_i - \bar{x})^2$. By using the substitution $y_i = x_i - p$, you bypass the need to know $p$ entirely. Efficiency is the hallmark of a JEE topper!

Q: Is it possible for the standard deviation to be negative?

Never! Standard deviation is the square root of variance, which is a sum of squares. It is always non-negative. If you ever get a negative number under your square root, stop and recheck your arithmetic—you have likely swapped the terms in the formula.

Question

Let $x_i (1 \leq i \leq 10)$ be ten observations of a random variable $x$. If $\sum_{i=1}^{10} (x_i - p) = 3$ and $\sum_{i=1}^{10} (x_i - p)^2 = 9$ where $0 
eq p \in R$, then the standard deviation of these observations is

Options:

(A) $7/10$

(B) $9/10$

(C) $\sqrt{3/5}$

(D) $4/5$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer! Today, we are not just crunching numbers; we are peeling back the layers of a statistical problem to reveal the elegant simplicity hidden underneath.

When you first look at the problem, you see $x_i$ and a mysterious constant $p$. It is natural to feel a moment of hesitation. You might ask, "How can I find the standard deviation if I do not know the value of $p$?"

This is the first trap of the JEE Advanced—the illusion of missing information. But here is the secret: in statistics, the absolute values of your data points matter far less than their relative positions.

### The Power of Substitution

Let us transform our perspective. We are given ten observations, $n = 10$, and two powerful summations:

$$\sum_{i=1}^{10} (x_i - p) = 3 \quad 	ext{and} \quad \sum_{i=1}^{10} (x_i - p)^2 = 9$$

The term $(x_i - p)$ is screaming for a substitution. Let us define a new variable, $y_i = x_i - p$. By doing this, we are essentially shifting the origin of our coordinate system.

Imagine you are standing on a field, and you move your entire group of data points by a distance $p$. Does the distance between the people in your group change? Of course not!

This brings us to the golden rule of statistics: **Standard deviation is independent of the change of origin.** Therefore, the standard deviation of $x$, denoted as $\sigma_x$, is identical to the standard deviation of $y$, denoted as $\sigma_y$. We have effectively eliminated the unknown $p$ from our problem entirely.

### The Computational Engine

Now that we have simplified our variables, the path forward is clear. We are working with the variable $y_i$, where $\sum y_i = 3$ and $\sum y_i^2 = 9$.

To find the standard deviation, we reach for our most reliable tool, the computational formula:

$$\sigma_y = \sqrt{\frac{\sum y_i^2}{n} - \left(\frac{\sum y_i}{n}ight)^2}$$

This formula is the heartbeat of variance calculations. It elegantly relates the mean of the squares to the square of the mean. Let us plug in our values: $n = 10$, $\sum y_i = 3$, and $\sum y_i^2 = 9$.

Substituting these into our equation, we get:

$$\sigma_y = \sqrt{\frac{9}{10} - \left(\frac{3}{10}ight)^2}$$

### Final Calculation

Now, take a breath. Do not rush the arithmetic. We have $\frac{9}{10} = 0.9$ and $\left(\frac{3}{10}ight)^2 = \frac{9}{100} = 0.09$.

The expression inside the square root becomes $0.9 - 0.09$. Think of this as $0.90 - 0.09$, which equals $0.81$.

Finally, we calculate the square root:

$$\sigma_y = \sqrt{0.81} = 0.9$$

Converting this back to a fraction, we get **$9/10$**. You have successfully navigated the trap, applied the invariance principle, and executed the calculation with precision. This is how you conquer the JEE!

Animated Solution for Mathematics - Statistics: Let $x_{i} (1 \leq i \leq 10)$ be ten observations of a random variable $x$ . If $\sum_{i = 1}^{10} (x_{i} - p) = 3$ and $\sum_{i = 1}^{10} (x_{i} - p)^{2} = 9$ where $0 \neq = p \in R$ , then the standard deviation of these observations is

Select Answer:

Visualized Solution

Given Observations

Change of Origin

Property of Standard Deviation

Sums in terms of $y_{i}$

Formula for $σ_{y}$

Substitution

Raw Equation

First Term Calculation

Second Term Calculation

Subtraction Setup

Difference

Square Root

Final Conclusion

The Sigma Insight: Variance and Standard Deviation

Analyzing the Setup

The Power of Substitution

The Computational Engine

Final Calculation

Similar Questions

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