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 objective is to determine the natural number \( c \) for which the variance of the provided frequency distribution equals 160. The distribution is presented in the table below:

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

First, we compute the mean, denoted as \( \bar{x} \), of the distribution.

1. The total frequency is calculated as: \( \sum f = 2 + 1 + 1 + 1 + 1 + 1 = 7 \).
2. The sum of the products of \( f \) and \( x \) is: \( \sum fx = (2 \times c) + (1 \times 2c) + (1 \times 3c) + (1 \times 4c) + (1 \times 5c) + (1 \times 6c) = 2c + 2c + 3c + 4c + 5c + 6c = 22c \).
3. The mean is then found by: \( \bar{x} = \frac{\sum fx}{\sum f} = \frac{22c}{7} \).

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

1. We calculate \( \sum fx^2 \):
   • The terms \( fx^2 \) are: \( (2 \times c^2) = 2c^2 \), \( (1 \times (2c)^2) = 4c^2 \), \( (1 \times (3c)^2) = 9c^2 \), \( (1 \times (4c)^2) = 16c^2 \), \( (1 \times (5c)^2) = 25c^2 \), \( (1 \times (6c)^2) = 36c^2 \).
   • Thus, \( \sum fx^2 = 2c^2 + 4c^2 + 9c^2 + 16c^2 + 25c^2 + 36c^2 = 92c^2 \).
2. We substitute these values into the variance formula and equate it to 160:

\[\frac{\sum fx^2}{\sum f} - (\bar{x})^2 = 160\]\[\frac{92c^2}{7} - \left(\frac{22c}{7}\right)^2 = 160\]

1. The equation is reorganized and simplified:

\[ \frac{92c^2}{7} - \frac{484c^2}{49} = 160 \] \[ \frac{644c^2 - 484c^2}{49} = 160 \] \[ \frac{160c^2}{49} = 160 \] \[ 160c^2 = 49 \times 160 \]

1. Solving for \( c \):

\[ c^2 = \frac{49 \times 160}{160} \] \[ c^2 = 49 \] \[ c = \pm 7 \]

Since \( c \) must be a natural number, we select the positive value. Therefore, the value of \( c \) is 7. The correct answer is 7.