Statement-1: The variance of the first n even | vedclass

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(B) For Statement-$1$: The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \cdot n(n+1)

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true. Statement-$2$ is not a correct explanation for Statement-$1$.

Vedclass · Answer

(B) For Statement-$1$: The first $n$ even natural numbers are $2, 4, 6, \dots, 2n$.The mean $\bar{x} = \frac{2(1+2+\dots+n)}{n} = \frac{2 \cdot n(n+1)}{2n} = n+1$.The variance $\sigma^2 = \frac{1}{n} \sum_{i=1}^n (2i)^2 - (n+1)^2 = \frac{4}{n} \cdot \frac{n(n+1)(2n+1)}{6} - (n+1)^2$.$\sigma^2 = \frac{2(n+1)(2n+1)}{3} - (n+1)^2 = (n+1) [\frac{4n+2-3n-3}{3}] = \frac{(n+1)(n-1)}{3} = \frac{n^2-1}{3}$.Thus,Statement-$1$ is true.For Statement-$2$: The sum of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1) = 2 \cdot \frac{n(n+1)}{2} - n = n^2+n-n = n^2$.The sum of the squares of the first $n$ odd natural numbers is $\sum_{k=1}^n (2k-1)^2 = \sum_{k=1}^n (4k^2 - 4k + 1) = 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n$.$= \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} = \frac{n[2(2n^2+3n+1) - 6n - 6 + 3]}{3} = \frac{n(4n^2+6n+2-6n-3)}{3} = \frac{n(4n^2-1)}{3}$.Statement-$2$ is true,but it does not explain the variance calculation in Statement-$1$ directly. Therefore,Statement-$2$ is not the correct explanation for Statement-$1$.

$x_i$	$92$	$93$	$97$	$98$	$102$	$104$	$109$
$f_i$	$3$	$2$	$3$	$2$	$6$	$3$	$3$

Class	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$2$	$3$	$x$	$5$	$4$

Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.
Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.

Similar Questions

Find the mean and variance for the following data:

$x_i$ $92$ $93$ $97$ $98$ $102$ $104$ $109$

$f_i$ $3$ $2$ $3$ $2$ $6$ $3$ $3$

The mean and variance of six observations are $6$ and $12$ respectively. If each observation is multiplied by $3$,then the new variance of the resulting observations is:

If the mean of the frequency distribution is $28$,then its variance is $........$.

Class $0-10$ $10-20$ $20-30$ $30-40$ $40-50$

Frequency $2$ $3$ $x$ $5$ $4$

Assertion $(A)$: The variance of the first $n$ odd natural numbers is $\frac{n^2-1}{3}$.
Reason $(R)$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2-1)}{3}$.
Which of the following alternatives is correct?

$A$ data consists of $n$ observations $x_1, x_2, ......, x_n$. If $\sum_{i=1}^n (x_i + 1)^2 = 9n$ and $\sum_{i=1}^n (x_i - 1)^2 = 5n$,then the standard deviation of this data is

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Statement-$1$: The variance of the first $n$ even natural numbers is $\frac{n^2 - 1}{3}$.Statement-$2$: The sum of the first $n$ odd natural numbers is $n^2$ and the sum of the squares of the first $n$ odd natural numbers is $\frac{n(4n^2 - 1)}{3}$.