(D) The terms are $x_i = (i-1)d$ for $i=1, 2, \ldots, n$.$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (i-1)d = \frac{d}{n} \frac{(n-1)n}{2} = \frac{(n-1)d}{2

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(D) The terms are $x_i = (i-1)d$ for $i=1, 2, \ldots, n$.$\bar{x} = \frac{1}{n} \sum_{i=1}^{n} (i-1)d = \frac{d}{n} \frac{(n-1)n}{2} = \frac{(n-1)d}{2}$.$\sum x_i^2 = \sum_{i=1}^{n} (i-1)^2 d^2 = d^2 \sum_{k=0}^{n-1} k^2 = d^2 \frac{(n-1)n(2n-1)}{6}$.$\sigma^2 = \frac{1}{n} \sum x_i^2 - (\bar{x})^2 = \frac{d^2(n-1)(2n-1)}{6} - \frac{(n-1)^2 d^2}{4}$.$\sigma^2 = \frac{d^2(n-1)}{2} \left[ \frac{2n-1}{3} - \frac{n-1}{2} \right] = \frac{d^2(n-1)}{2} \left[ \frac{4n-2-3n+3}{6} \right] = \frac{d^2(n-1)(n+1)}{12} = \frac{(n^2-1)d^2}{12}$.

Class 11 Mathematics Statistics Variance and Standard Deviation

Medium

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:

MHT CET 2023 All MHT CET

A
$\frac{(n^2+1) d^2}{24}$
B
$\frac{(n^2-1) d^2}{24}$
C
$\frac{(n^2+1) d^2}{12}$
D
$\frac{(n^2-1) d^2}{12}$

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Variance and Standard Deviation

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If the variance of the numbers $2, 3, 11$ and $x$ is $\frac{49}{4}$,then the values of $x$ are

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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 $\frac{n(4n^2 - 1)}{3}$.

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Let $X_{1}, X_{2}, \ldots, X_{18}$ be eighteen observations such that $\sum_{i=1}^{18}(X_{i}-\alpha)=36$ and $\sum_{i=1}^{18}(X_{i}-\beta)^{2}=90$,where $\alpha$ and $\beta$ are distinct real numbers. If the standard deviation of these observations is $1$,then the value of $|\alpha-\beta|$ is ...... .

DifficultJEE MAIN 2021

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Find the mean,variance,and standard deviation using the short-cut method for the following data:

Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$

No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

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The mean and standard deviation of $100$ items are $50$ and $4$ respectively. The sum of the squares of all the items is:

MediumKCET 2019

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Height in cms	$70-75$	$75-80$	$80-85$	$85-90$	$90-95$	$95-100$	$100-105$	$105-110$	$110-115$
No. of children	$3$	$4$	$7$	$7$	$15$	$9$	$6$	$6$	$3$

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:

Similar Questions

If the variance of the numbers $2, 3, 11$ and $x$ is $\frac{49}{4}$,then the values of $x$ are

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 $\frac{n(4n^2 - 1)}{3}$.

Find the mean,variance,and standard deviation using the short-cut method for the following data: Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$ No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$

The mean and standard deviation of $100$ items are $50$ and $4$ respectively. The sum of the squares of all the items 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 $\frac{n(4n^2 - 1)}{3}$.

Find the mean,variance,and standard deviation using the short-cut method for the following data:

Height in cms $70-75$ $75-80$ $80-85$ $85-90$ $90-95$ $95-100$ $100-105$ $105-110$ $110-115$

No. of children $3$ $4$ $7$ $7$ $15$ $9$ $6$ $6$ $3$