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

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: \(c = 7\)