Question

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

• A. 5
• B. 8
• C. 7
• D. 6

Answer 1

The Correct Option is C

To solve the given problem, we need to find the natural number \( c \) such that the variance of the frequency distribution is 160. The distribution table is as follows:

x	c	2c	3c	4c	5c	6c
f	2	1	1	1	1	1

Let's first find the mean, \(\bar{x}\), of the distribution.

1. Calculate the total frequency: \(\sum f = 2 + 1 + 1 + 1 + 1 + 1 = 7\).
2. Calculate the sum of values: \(\sum fx = 2c + 2c + 3c + 4c + 5c + 6c = 22c\).
3. Calculate the mean: \(\bar{x} = \frac{\sum fx}{\sum f} = \frac{22c}{7}\).

Given that the variance is 160, we will calculate it using the formula \(\text{Variance} = \frac{\sum f(x_i - \bar{x})^2}{\sum f}\).

1. Calculate \(\sum f x^2\):
   • \((fx^2)_1 = 2c^2\), \((fx^2)_2 = 4c^2\), \((fx^2)_3 = 9c^2\), \((fx^2)_4 = 16c^2\), \((fx^2)_5 = 25c^2\), \((fx^2)_6 = 36c^2\).
   • So, \(\sum fx^2 = 2c^2 + 4c^2 + 9c^2 + 16c^2 + 25c^2 + 36c^2 = 92c^2\).
2. Substitute into the variance formula and set it equal to 160: 

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

1. Reorganize and simplify the equation: 

\[\frac{92c^2 - \frac{484c^2}{7}}{7} = 160 \] \[ \frac{164c^2}{49} = 160 \] \[ 164c^2 = 49 \times 160\]

1. Calculate \(c\)\): 

\[c^2 = \frac{7840}{164} \] \[ c^2 ≈ 47.8 \] \[ c ≈ 7\]

Hence, the correct value of \(c\)\) in natural numbers is 7. Therefore, the correct answer is 7.

Answer 2

The variance formula for a frequency distribution is:

\[ \text{Variance} = \frac{\sum fx^2}{\sum f} - \left( \frac{\sum fx}{\sum f} \right)^2. \]

From the table, we calculate \(\sum f\), \(\sum fx\), and \(\sum fx^2\):

\(x\)	$f$	\(f \times x\)	\(f \times x^2\)
c	$2$	2c	$2c^2$
2c	$1$	2c	$4c^2$
3c	$1$	3c	$9c^2$
4c	$1$	4c	$16c^2$
5c	$1$	5c	$25c^2$
6c	$1$	6c	$36c^2$
Total	7	22c	$92c^2$

Step 1: Variance formula. Substitute into the formula:

\[ \text{Variance} = \frac{\sum fx^2}{\sum f} - \left( \frac{\sum fx}{\sum f} \right)^2. \]

Substitute \(\sum f = 7\), \(\sum fx = 22c\), and \(\sum fx^2 = 92c^2\):

\[ \text{Variance} = \frac{92c^2}{7} - \left( \frac{22c}{7} \right)^2. \]

Simplify:

\[ \text{Variance} = \frac{92c^2}{7} - \frac{(22c)^2}{7^2}. \] \[ \text{Variance} = \frac{92c^2}{7} - \frac{484c^2}{49}. \]

Take the LCM of 7 and 49:

\[ \text{Variance} = \frac{(92 \cdot 7)c^2 - 484c^2}{49}. \] \[ \text{Variance} = \frac{644c^2 - 484c^2}{49}. \] \[ \text{Variance} = \frac{160c^2}{49}. \]

Step 2: Set variance to 160. The problem states that the variance is 160. Therefore:

\[ \frac{160c^2}{49} = 160. \]

Simplify:

\[ 160c^2 = 160 \cdot 49. \] \[ c^2 = 49 \implies c = 7. \]

Final Answer: \(c = 7\)