Let 9

Question

$9=x_1 < x_2 < \ldots \ldots < x_7$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{	ext {new }}=

Vedclass · Accepted Answer

$34$

Vedclass · Answer

$9=x_1 < x_2 < \ldots \ldots < x_7$

$9,9+d, 9+2 d, \ldots \ldots .9+6 d$

$0, d, 2 d, \ldots \ldots \cdot 6$

$\bar{x}_{	ext {new }}=\frac{21 d }{7}=3 d$

$16=\frac{1}{7}\left(0^2+1^2+\ldots \ldots+6^2ight) d^2-9 d^2$

$=\frac{1}{7}\left(\frac{6 	imes 7 	imes 13}{6}ight) d ^2-9 d ^2$

$16=4 d^2$

$d^2=4$

$d=2$

$\bar{x}+x_6=6+9+10+9$

${x_i}$	$4$	$8$	$11$	$17$	$20$	$24$	$32$
${f_i}$	$3$	$5$	$9$	$5$	$4$	$3$	$1$

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

Let $9 < x_1 < x_2 < \ldots < x_7$ be in an $A.P.$ with common difference $d$. If the standard deviation of $x_1, x_2 \ldots$, $x _7$ is $4$ and the mean is $\overline{ x }$, then $\overline{ x }+ x _6$ is equal to:

Similar Questions

The mean and variance of the marks obtained by the students in a test are $10$ and $4$ respectively. Later, the marks of one of the students is increased from $8$ to $12$ . If the new mean of the marks is $10.2.$ then their new variance is equal to :

Find the variance and standard deviation for the following data:

${x_i}$ $4$ $8$ $11$ $17$ $20$ $24$ $32$

${f_i}$ $3$ $5$ $9$ $5$ $4$ $3$ $1$

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