13.StatisticsMedium

જો એક વિતરણ માટે $\Sigma(x-5)=3, \Sigma(x-5)^{2}=43$ અને વસ્તુઓની સંખ્યા $18$ હોય તો તેનો મધ્યક અને પ્રમાણિત વિચલન મેળવો

Solution

Given, $n=18, \Sigma(x-5)=3$ and $\Sigma(x-5)^{2}=43$

$\therefore \quad \operatorname{Mean}=A+\frac{\Sigma(x-5)}{18}=5+\frac{3}{18}=5+0.1666=5.1666=5.17$

and $\quad SD =\sqrt{\frac{\Sigma(x-5)^{2}}{n}-\left(\frac{\Sigma(x-5)}{n}\right)^{2}}=\sqrt{\frac{43}{18}-\left(\frac{3}{18}\right)^{2}}$

$=\sqrt{2.3889-(0.166)^{2}}=\sqrt{2.3889-0.0277}=1.53$

જો આવૃત્તિ વિતરણ

$X_i$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

Frequency $f_i$ $3$ $6$ $16$ $\alpha$ $9$ $5$ $6$

નું વિચરણ $3$ હોય, તો $\alpha=..............$

Medium

Medium

Difficult

Easy

$X_i$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
Frequency $f_i$	$3$	$6$	$16$	$\alpha$	$9$	$5$	$6$

વર્ગ	$0-30$	$30-60$	$60-90$	$90-120$	$120-150$	$50-180$	$180-210$
આવૃત્તિ	$2$	$3$	$5$	$10$	$3$	$5$	$2$

$\mathrm{x}$	$\mathrm{x}_{1}=2$	$\mathrm{x}_{2}=6$	$\mathrm{x}_{3}=8$	$\mathrm{x}_{4}=9$
$\mathrm{f}$	$4$	$4$	$\alpha$	$\beta$