The variance of the following distribution is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:Marks$x$$f$$f \cdot x$$f \cdot x^2$$1-3$$2$$40$$80$$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) To find the variance,we first calculate the mid-interval values $(x)$ and the frequency $(f)$:Marks$x$$f$$f \cdot x$$f \cdot x^2$$1-3$$2$$40$$80$$160$$3-5$$4$$30$$120$$480$$5-7$$6$$20$$120$$720$$7-9$$8$$10$$80$$640$Total$100$$400$$2000$The mean $\bar{x} = \frac{\sum f \cdot x}{\sum f} = \frac{400}{100} = 4$.The variance $\sigma^2 = \frac{\sum f \cdot x^2}{\sum f} - (\bar{x})^2$.$\sigma^2 = \frac{2000}{100} - (4)^2 = 20 - 16 = 4$.

The variance of the following distribution is:
Marks $1-3$ $3-5$ $5-7$ $7-9$
Number of students $40$ $30$ $20$ $10$

Similar Questions

The variance of the discrete data $3, 4, 5, 6, 7, 8, 10, 13$ is

The mean and $S.D.$ of $1, 2, 3, 4, 5, 6$ are

The variance of the variates $112, 116, 120, 125, 132$ about their $A.M.$ is

If $\sum\limits_{i = 1}^{18} {(x_i - 8) = 9}$ and $\sum\limits_{i = 1}^{18} {(x_i - 8)^2 = 45}$,then the standard deviation of $x_1, x_2, \dots, x_{18}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The variance of the following distribution is: Marks$1-3$$3-5$$5-7$$7-9$Number of students$40$$30$$20$$10$