The variance of the first n natural numbers i | vedclass

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

Question

(A) The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2$. F

Vedclass · Accepted Answer

$\frac{n^2 - 1}{12}$

Vedclass · Answer

(A) The variance $\sigma^2$ is given by the formula $\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} x_i^2 - \left( \frac{1}{n} \sum_{i=1}^{n} x_i \right)^2$. For the first $n$ natural numbers,$\sum x_i = \frac{n(n+1)}{2}$ and $\sum x_i^2 = \frac{n(n+1)(2n+1)}{6}$. Substituting these values: $\sigma^2 = \frac{n(n+1)(2n+1)}{6n} - \left( \frac{n(n+1)}{2n} \right)^2$ $\sigma^2 = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4}$ $\sigma^2 = \frac{2(n+1)(2n+1) - 3(n+1)^2}{12}$ $\sigma^2 = \frac{(n+1) [2(2n+1) - 3(n+1)]}{12}$ $\sigma^2 = \frac{(n+1) [4n + 2 - 3n - 3]}{12}$ $\sigma^2 = \frac{(n+1)(n-1)}{12} = \frac{n^2 - 1}{12}$.

The variance of the first $n$ natural numbers is

Similar Questions

If the variance of $10$ natural numbers $1, 1, 1, \dots, 1, k$ is less than $10$,then the maximum possible value of $k$ is ...... .

Let the observations $x_{i} (1 \leq i \leq 10)$ satisfy the equations $\sum_{i=1}^{10}(x_{i}-5)=10$ and $\sum_{i=1}^{10}(x_{i}-5)^{2}=40$. If $\mu$ and $\lambda$ are the mean and the variance of the observations $x_{1}-3, x_{2}-3, \dots, x_{10}-3$,then the ordered pair $(\mu, \lambda)$ is equal to:

The variance of the data $1001, 1003, 1006, 1007, 1009, 1010$ is:

Which of the following sets of data has the least standard deviation?

Let ${x_1}, {x_2}, \ldots, {x_n}$ be $n$ observations,and let $\bar x$ be their arithmetic mean and ${\sigma ^2}$ be the variance.
Statement-$1$: The variance of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $4{\sigma ^2}$.
Statement-$2$: The arithmetic mean of $2{x_1}, 2{x_2}, \ldots, 2{x_n}$ is $4\bar x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module