Find the mean and variance for the data 6, 7, | vedclass

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

Question

(A) Given data: $6, 7, 10, 12, 13, 4, 8, 12$.Number of observations,$n = 8$.Mean,$\bar{x} = \frac{\sum x_i}{n} = \frac{6+7+10+12+13+4+8+12}{8} = \frac

Vedclass · Accepted Answer

Mean = $9$,Variance = $9.25$

Vedclass · Answer

(A) Given data: $6, 7, 10, 12, 13, 4, 8, 12$.Number of observations,$n = 8$.Mean,$\bar{x} = \frac{\sum x_i}{n} = \frac{6+7+10+12+13+4+8+12}{8} = \frac{72}{8} = 9$.Calculation of variance:						$x_i$			$(x_i - \bar{x})^2$							$6$			$(6-9)^2 = 9$							$7$			$(7-9)^2 = 4$							$10$			$(10-9)^2 = 1$							$12$			$(12-9)^2 = 9$							$13$			$(13-9)^2 = 16$							$4$			$(4-9)^2 = 25$							$8$			$(8-9)^2 = 1$							$12$			$(12-9)^2 = 9$							Sum			$74$			Variance,$\sigma^2 = \frac{1}{n} \sum (x_i - \bar{x})^2 = \frac{74}{8} = 9.25$.

Find the mean and variance for the data $6, 7, 10, 12, 13, 4, 8, 12$.

Similar Questions

If the variance of the first $n$ natural numbers is $10$ and the variance of the first $m$ even natural numbers is $16$,then $m + n$ is equal to

Let $X = \{x \in \mathbb{N} : 1 \le x \le 19\}$ and for some $a, b \in \mathbb{R}$,$Y = \{ax + b : x \in X\}$. If the mean and variance of the elements of $Y$ are $30$ and $750$,respectively,then the sum of all possible values of $b$ is

The standard deviation of the scores $505, 510, 515, 520, \ldots, 595$ is

Suppose values taken by a variable $x$ are such that $a \le x_i \le b$,where $x_i$ denotes the value of $x$ in the $i^{th}$ case for $i = 1, 2, ..., n$. Then:

Calculate the variance of the following continuous frequency distribution:
Class Interval $0$–$4$ $4$–$8$ $8$–$12$ $12$–$16$
Frequency $(f_i)$ $1$ $2$ $2$ $1$

Vedclass Test Series

Exam Paper Generator

Online Exam Module