ધારોકે ગણ A અને B બન્ને માં 5 ઘટકો છે.ધારોકે | vedclass

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Question

$\omega A=\left\{a_1, a_2, a_3, a_4, a_5ight\}$

$B=\left\{b_1, b_2, b_3, b_4, b_5ight\}$

$	ext { Given, } \sum_{ i =1}^3 ai =25, \sum_{ i =

Vedclass · Accepted Answer

$38$

Vedclass · Answer

$\omega A=\left\{a_1, a_2, a_3, a_4, a_5ight\}$

$B=\left\{b_1, b_2, b_3, b_4, b_5ight\}$

$	ext { Given, } \sum_{ i =1}^3 ai =25, \sum_{ i =1}^3 bi =40$

$\frac{\sum_{ i =1}^5 a _{ i }^2}{5}-\left(\frac{\sum_{ i =1}^5 a _{ i }}{5}ight)^2=12, \frac{\sum_{ i =1}^5 b _{ i }^2}{5}-\left(\frac{\sum_{ i =1}^5 b _{ i }}{5}ight)^2=20$

$\sum_{ i =1}^5 a _{ i }^2=185 \quad, \quad \sum_{ i =1}^5 b _{ i }^2=420$

$	ext { Now, } C =\left\{ C _1, C _2, \ldots . C _{10}ight\}$

$	ext { s.f. } C_i=a_i=3 	ext { or } b_i+2$

$	herefore 	ext { Mean of } C , \overline{ C }=\frac{\left(\sum a _{ i }-15ight)+\left(\sum b _{ i }+10ight)}{10}$

$\overline{ C }=\frac{10+50}{10}=6$

$	herefore \quad \sigma^2=\frac{\sum \limits_{ i =1}^{10} C _{ i }^2}{10}=(\overline{ C })^2$

$=\frac{\sum\left( a _{ i }-3ight)^2+\sum\left( b _{ i }+2ight)^2}{10}-(6)^2$

$=\frac{\sum a _{ i }{ }^2+\sum b _{ i }{ }^2-6 \sum a _{ i }+4 \sum b _{ i }+65}{10}-36$

$=\frac{185+420-150+160+65}{10}-36$

$=32$

$	herefore \quad$ Mean + Variance $=\overline{ C }+\sigma^2=6+32=38$

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