Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$
Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$
- [AIEEE 2012]
- A
Statement $1$ is true, Statement $2$ is false.
- B
Statement $1$ is true, Statement $2$ is true;
Statement $2$ is not a correct explanation for Statement $1$. - C
Statement $1$ is false, Statement $2$ is true.
- D
Statement $1$ is true, Statement $2$ is true,
Statement $2$ is a correct explanation for Statement $1$.
Similar Questions
Calculate the mean, variance and standard deviation for the following distribution:
Class
$30-40$
$40-50$
$50-60$
$60-70$
$70-80$
$80-90$
$90-100$
$f_i$
$3$
$7$
$12$
$15$
$8$
$3$
$2$
Calculate the mean, variance and standard deviation for the following distribution:
| Class | $30-40$ | $40-50$ | $50-60$ | $60-70$ | $70-80$ | $80-90$ | $90-100$ |
| $f_i$ | $3$ | $7$ | $12$ | $15$ | $8$ | $3$ | $2$ |
Determine mean and standard deviation of first n terms of an $A.P.$ whose first term is a and common difference is d.
Determine mean and standard deviation of first n terms of an $A.P.$ whose first term is a and common difference is d.
Let $\mathrm{X}$ be a random variable with distribution.
$\mathrm{x}$
$-2$
$-1$
$3$
$4$
$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$
$\frac{1}{5}$
$\mathrm{a}$
$\frac{1}{3}$
$\frac{1}{5}$
$\mathrm{~b}$
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
Let $\mathrm{X}$ be a random variable with distribution.
| $\mathrm{x}$ | $-2$ | $-1$ | $3$ | $4$ | $6$ |
| $\mathrm{P}(\mathrm{X}=\mathrm{x})$ | $\frac{1}{5}$ | $\mathrm{a}$ | $\frac{1}{3}$ | $\frac{1}{5}$ | $\mathrm{~b}$ |
If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :
- [JEE MAIN 2021]
The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is
The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is
The diameters of circles (in mm) drawn in a design are given below:
Diameters
$33-36$
$37-40$
$41-44$
$45-48$
$49-52$
No. of circles
$15$
$17$
$21$
$22$
$25$
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]
The diameters of circles (in mm) drawn in a design are given below:
| Diameters | $33-36$ | $37-40$ | $41-44$ | $45-48$ | $49-52$ |
| No. of circles | $15$ | $17$ | $21$ | $22$ | $25$ |
Calculate the standard deviation and mean diameter of the circles.
[ Hint : First make the data continuous by making the classes as $32.5-36.5,36.5-40.5,$ $40.5-44.5,44.5-48.5,48.5-52.5 $ and then proceed.]