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)

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

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

${x_i}$	$3$	$8$	$13$	$18$	$23$
${f_i}$	$7$	$10$	$15$	$10$	$6$

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 variance of the first $n$ natural numbers.

Find the standard deviation for the following data:

${x_i}$ $3$ $8$ $13$ $18$ $23$

${f_i}$ $7$ $10$ $15$ $10$ $6$

The mean and variance of a set of $15$ numbers are $12$ and $14$ respectively. The mean and variance of another set of $15$ numbers are $14$ and $\sigma^2$ respectively. If the variance of all the $30$ numbers in the two sets is $13$,then $\sigma^2$ is equal to $.........$.

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:

In an experiment with $15$ observations on $x$,we have $\sum x^2 = 2830$ and $\sum x = 170$. One observation that was $20$ was found to be wrong and was replaced by the correct value $30$. The corrected variance 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 $\frac{n(4n^2 - 1)}{3}$.