If the variance of the frequency distribution is 160, then the value of \( c \in \mathbb{N} \) is: \[ \begin{array}{|c|c|c|c|c|c|c|} \hline x & c & 2c & 3c & 4c & 5c & 6c \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} \]
- 5
- 8
- 7
- 6
The Correct Option is C
Approach Solution - 1
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.
- Calculate the total frequency: \(\sum f = 2 + 1 + 1 + 1 + 1 + 1 = 7\).
- Calculate the sum of values: \(\sum fx = 2c + 2c + 3c + 4c + 5c + 6c = 22c\).
- 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 \(\text{Variance} = \frac{\sum f(x_i - \bar{x})^2}{\sum f}\).
- 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, \(\sum fx^2 = 2c^2 + 4c^2 + 9c^2 + 16c^2 + 25c^2 + 36c^2 = 92c^2\).
- Substitute into the variance formula and set it equal to 160:
- Reorganize and simplify the equation:
- Calculate \(c\)\):
Hence, the correct value of \(c\)\) in natural numbers is 7. Therefore, the correct answer is 7.
Approach Solution -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: \(c = 7\)
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