The mean and variance of 8 observations are 1 | vedclass

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

Question

$\bar{x}=10$

$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$

since, variance is independent of origin. So, we subtract 10 from each

Vedclass · Accepted Answer

$7$

Vedclass · Answer

$\bar{x}=10$

$\Rightarrow \bar{x}=\frac{63+a+b}{8}=10 \Rightarrow a+b=17$

since, variance is independent of origin. So, we subtract 10 from each observation.

$So , \sigma^{2}=13.5=\frac{79+( a -10)^{2}+( b -10)^{2}}{8}-(10-10)^{2}$

$\Rightarrow a ^{2}+ b ^{2}-20( a + b )=-171$

$\Rightarrow a ^{2}+ b ^{2}=169 \quad \ldots(2)$

From

$(i) and (ii)$ $; a=12 \& b=5$

Classes	$0-30$	$30-60$	$60-90$	$90-120$	$120-150$	$50-180$	$180-210$
$f_i$	$2$	$3$	$5$	$10$	$3$	$5$	$2$

class	$0-10$	$10-20$	$20-30$	$30-40$
Freq	$1$	$3$	$4$	$2$

The mean and variance of $8$ observations are $10$ and $13.5,$ respectively. If $6$ of these observations are $5,7,10,12,14,15,$ then the absolute difference of the remaining two observations is

Similar Questions

The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation is$.....$

The $S.D$. of the first $n$ natural numbers is

Find the mean and variance for the following frequency distribution.

Classes $0-30$ $30-60$ $60-90$ $90-120$ $120-150$ $50-180$ $180-210$

$f_i$ $2$ $3$ $5$ $10$ $3$ $5$ $2$

What is the standard deviation of the following series

class $0-10$ $10-20$ $20-30$ $30-40$

Freq $1$ $3$ $4$ $2$