Question:medium

The variance \(\sigma^2\) of the data

Is _______.

 

Updated On: Jan 13, 2026
Show Solution

Correct Answer: 29

Solution and Explanation

Summations:

\[ \sum f_i = 22, \quad \sum f_i x_i = 176, \quad \sum f_i x_i^2 = 2048. \]

Mean calculation \( \bar{x} \):

\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{176}{22} = 8. \]

Variance calculation \( \sigma^2 \):

\[ \sigma^2 = \frac{1}{N} \sum f_i x_i^2 - \bar{x}^2, \]

where \( N = \sum f_i \).

Value substitution:

\[ \sigma^2 = \frac{1}{22} \times 2048 - (8)^2 = \frac{2048}{22} - 64. \]

Computation:

\[ \frac{2048}{22} = 93.09090909 \quad \text{and} \quad \sigma^2 = 93.09090909 - 64 = 29.09090909. \]

The variance is: 29.09090909

Was this answer helpful?
1

Top Questions on Mean and Variance of Random variables


Questions Asked in JEE Main exam