Find the mean and variance for the first n na | vedclass

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

Question

(A) The mean of the first $n$ natural numbers is calculated as follows:Mean $= \frac{	ext{Sum of all observations}}{	ext{Number of observations}} =

Vedclass · Accepted Answer

Mean $= \frac{n+1}{2}$,Variance $= \frac{n^2-1}{12}$

Vedclass · Answer

(A) The mean of the first $n$ natural numbers is calculated as follows:Mean $= \frac{	ext{Sum of all observations}}{	ext{Number of observations}} = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}$Variance $(\sigma^2)$ is given by $\frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2$.Using the formula $\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2$:$\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$$\sigma^2 = \frac{1}{n} \left[ \frac{n(n+1)(2n+1)}{6} ight] - \left( \frac{n+1}{2} ight)^2$$\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$$\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$$\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12}$$\sigma^2 = \frac{(n+1) [4n+2 - 3n-3]}{12} = \frac{(n+1)(n-1)}{12} = \frac{n^2-1}{12}$Thus,the mean is $\frac{n+1}{2}$ and the variance is $\frac{n^2-1}{12}$.

Find the mean and variance for the first $n$ natural numbers.

Similar Questions

Variance of first $2n$ natural numbers is

If $\sum_{i=1}^{n}(x_{i}-a)=n$ and $\sum_{i=1}^{n}(x_{i}-a)^{2}=na$,where $n, a > 1$,then the standard deviation of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ is

The raw data $x_1, x_2, \ldots, x_{n}$ is an $A$.$P$. with common difference $d$ and first term $0$. If $\bar{x}$ and $\sigma^2$ are the mean and variance of $x_{i}, i=1, 2, \ldots, n$,then $\sigma^2$ is:

The mean and standard deviation of six observations are $8$ and $4,$ respectively. If each observation is multiplied by $3,$ find the new mean and new standard deviation of the resulting observations.

The variance of the data $2, 3, 5, 11, 13, 17, 19$ is nearly

Vedclass Test Series

Exam Paper Generator

Online Exam Module