The variance of first 50 even natural numbers | vedclass

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

Question

$2,4,6,8,......,98,100$

${\sigma ^2} = \frac{{\sum x_1^2}}{n} - {\left( {\overline {.x} } \right)^2}$

$\frac{{{2^2} + {4^2} + {6^2} + .... + {{1

Vedclass · Accepted Answer

$833$

Vedclass · Answer

$2,4,6,8,......,98,100$

${\sigma ^2} = \frac{{\sum x_1^2}}{n} - {\left( {\overline {.x} } ight)^2}$

$\frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$$ - {\left( {\frac{{2 + 4 + 6 + .... + 100}}{{50}}} ight)^2}$

${i_1} = \frac{{{2^2} + {4^2} + {6^2} + .... + {{100}^2}}}{{50}}$

$ = {2^2}\frac{{{1^2} + {2^2} + {3^2} + ... + {{50}^2}}}{{50}}$

$ = \frac{{{2^2}}}{{50}} 	imes 50\left( {50 + 1} ight)\left( {100 + 1} ight)$

$ = 3434$

${i_2} = {\left( {\frac{{2 + 4 + 6 + ..... + 100}}{{50}}} ight)^2}$

$ = {\left( {\frac{{50 	imes \frac{{2 + 100}}{2}}}{{50}}} ight)^2}$

$ = {\left( {51} ight)^2}$

${\sigma ^2} = 3434 - 2661 = 833$

The variance of first $50$ even natural numbers is

Similar Questions

The variance of $20$ observation is $5$ . If each observation is multiplied by $2$ , then the new variance of the resulting observations, is

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50$. The correct mean and $S.D.$ respectively are...

Mean and standard deviation of 100 observations were found to be 40 and 10 , respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Given that $\bar{x}$ is the mean and $\sigma^{2}$ is the variance of $n$ observations $x_{1}, x_{2}, \ldots, x_{n}$ Prove that the mean and variance of the observations $a x_{1}, a x_{2}, a x_{3}, \ldots ., a x_{n}$ are $a \bar{x}$ and $a^{2} \sigma^{2},$ respectively, $(a \neq 0)$

The variance of the first $n$ natural numbers is