If the variance of the frequency distribution x: c, 2c, 3c, 4c, 5c, 6c; f: 2, 1, 1, 1, 1, 1 is 160,

Q: Wait, why do we reject the negative root here? $c^2 = 49$ has two solutions, $7$ and $-7$.

That is a fantastic observation! In pure algebra, $c^2 = 49$ indeed gives $c = \pm 7$. However, the problem explicitly states that $c \in \mathbb{N}$, meaning $c$ must be a natural number (a positive integer). Therefore, we discard the negative root. Always keep an eye on the domain constraints in JEE problems!

Q: Why can't we just use the definition $\frac{\sum f_i(x_i - \bar{x})^2}{N}$?

You absolutely can! The definition is conceptually beautiful. However, calculating the mean $\bar{x}$ first and then subtracting it from every $x_i$ often introduces rounding errors or messy fractions. The computational formula $\frac{\sum f_i x_i^2}{N} - \bar{x}^2$ is the 'pro' move because it keeps the algebra cleaner and faster.

Q: Is there a property of variance I could use to solve this faster?

Yes! If you notice that $x_i = c \cdot y_i$ where $y_i \in \{1, 2, 3, 4, 5, 6\}$, you can use the property $Var(cX) = c^2 Var(X)$. Calculate the variance of the distribution with $x_i \in \{1, 2, 3, 4, 5, 6\}$ and frequencies $\{2, 1, 1, 1, 1, 1\}$, let's call this $\sigma_y^2$. Then, simply solve $c^2 \sigma_y^2 = 160$. It is a great way to verify your answer!

Question

If the variance of the frequency distribution $$\begin{array}{|c|c|c|c|c|c|c|} \hline x & c & 2c & 3c & 4c & 5c & 6c \ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \ \hline \end{array}$$ is 160, then the value of $c \in \mathbb{N}$ is

Options:

(A) 7

(B) 8

(C) 5

(D) 6

Sigma Percentile · Accepted Answer

### The Elegance of Statistical Scaling

Welcome, fellow traveler on the JEE journey. Today, we are not just solving a statistics problem; we are uncovering the hidden symmetry within a frequency distribution.

When you look at the table provided, do not just see numbers. See a pattern. We have observations $x_i$ that are all multiples of a mysterious natural number $c$, and frequencies $f_i$ that dictate how often these values appear.

Our mission is to unmask $c$, given that the variance of this distribution is $160$.

### Phase 1: The Foundation

Before we dive into the variance, we must understand the 'size' of our dataset. We calculate the total frequency $N$ by summing the second row:

$$N = 2 + 1 + 1 + 1 + 1 + 1 = 7$$

This $N$ is the bedrock of our calculations. It tells us how many data points we are dealing with, and it will serve as our denominator for both the mean and the variance.

### Phase 2: The Computational Arsenal

To find the variance, we rely on the computational formula:

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

This formula is a lifesaver. It allows us to calculate the variance without needing to find the mean first. Let us find the two essential sums.

First, the sum of products $\sum f_i x_i$. We multiply each observation by its frequency:

$$\sum f_i x_i = 2(c) + 1(2c) + 1(3c) + 1(4c) + 1(5c) + 1(6c) = 22c$$

This represents the total sum of all observations.

Next, the sum of squares $\sum f_i x_i^2$. We square each observation and multiply by its frequency:

$$\sum f_i x_i^2 = 2(c)^2 + 1(2c)^2 + 1(3c)^2 + 1(4c)^2 + 1(5c)^2 + 1(6c)^2 = 92c^2$$

### Phase 3: The Algebraic Dance

Now, we substitute these into our variance formula. We are given $\sigma^2 = 160$.

$$160 = \frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2$$

Expanding the squared term, we get:

$$160 = \frac{92c^2}{7} - \frac{484c^2}{49}$$

To subtract these, we find a common denominator of $49$. Multiplying the first term by $\frac{7}{7}$, we get:

$$\frac{644c^2}{49} - \frac{484c^2}{49} = \frac{160c^2}{49}$$

Look at the beauty of the result:

$$160 = \frac{160c^2}{49}$$

The $160$ on both sides cancels out perfectly, leaving us with $c^2 = 49$. Since $c$ must be a natural number, we conclude **$c = 7$**.

This problem is a reminder that in the heat of an exam, if you stay calm and systematic, the algebra will often simplify in ways that feel almost magical. Keep practicing, keep visualizing, and trust the math!

Animated Solution for Mathematics - Statistics: If the variance of the frequency distribution $x f c 2 2 c 1 3 c 1 4 c 1 5 c 1 6 c 1$ is 160, then the value of $c \in N$ is

Select Answer:

Visualized Solution

Analyze the Frequency Distribution

Calculate Total Frequency $N$

Calculate $\sum f_{i} x_{i}$

Calculate $\sum f_{i} x_{i}^{2}$

The Variance Formula

Substitute the Values

Simplify the Expression

Solve for $c$

The Sigma Insight: Variance and Standard Deviation

The Elegance of Statistical Scaling

Phase 1

Phase 2

Phase 3

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Animated Solution for Mathematics - Statistics: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the variance of the frequency distribution xf​c2​2c1​3c1​4c1​5c1​6c1​​ is 160, then the value of c∈N is