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यदि बारंबारता बंटन
| $x_i$ | $2$ | $4$ | $6$ | $8$ | $10$ | $12$ | $14$ | $16$ |
| $f_i$ | $4$ | $4$ | $\alpha$ | $15$ | $8$ | $\beta$ | $4$ | $5$ |
के माध्य तथा प्रसरण क्रमशः $9$ तथा $15.08$ हैं, तो $\alpha^2+\beta^2-\alpha \beta$ का मान है________________
$24$
$23$
$25$
$22$
Solution
$N=\sum f_i=40+\alpha+\beta$
$\sum f_i x_i=360+6 \alpha+12 \beta$
$\sum f _{ i } x _{ i }^2=3904+36 \alpha+144 \beta$
$\operatorname{Mean}(\overline{ x })=\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}=9$
$\Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta)$
$3 \alpha=3 \beta \Rightarrow \alpha=\beta$
$\sigma^2=\frac{\sum f _{ i } x _1^2}{\sum f _{ i }}-\left(\frac{\sum f _{ i } x _{ i }}{\sum f _{ i }}\right)^2$
$\Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(\overline{ x })^2=15.08$
$\Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}-(9)^2=15.08$
$\Rightarrow \alpha=5$
Now, $\alpha^2+\beta^2-\alpha \beta=\alpha^2=25$
Similar Questions
निम्नलिखित आँकड़ों के लिए माध्य व प्रसरण ज्ञात कीजिए।
| ${x_i}$ | $6$ | $10$ | $14$ | $18$ | $24$ | $28$ | $30$ |
| ${f_i}$ | $2$ | $4$ | $7$ | $12$ | $8$ | $4$ | $3$ |
