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

Answer: (C)

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: $c = 7$