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माना प्रेक्षण $x _{ i }(1 \leq i \leq 10)$ समीकरणों $\sum_{ i =1}^{10}\left( x _{ i }-5\right)=10$ तथा $\sum_{ i =1}^{10}\left( x _{ i }-5\right)^{2}=40$ को संतुष्ट करते है। यदि $\mu$ तथा $\lambda$ प्रेक्षणों $x _{1}-3, x _{2}-3, \ldots, x _{10}-3$ के क्रमशः माध्य तथा प्रसरण है, तो क्रमित युग्म $(\mu, \lambda)$ बराबर है
$(6, 6)$
$(3, 6)$
$(6, 3)$
$(3, 3)$
Solution
$\sum_{i=1}^{10}\left(x_{i}-5\right)=10$
$\Rightarrow$ Mean of observation $\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{3}\left(\mathrm{x}_{\mathrm{i}}-5\right)=1$
$\Rightarrow \mu=$ mean of observation $\left(\mathrm{x}_{\mathrm{i}}-3\right)$
$=\left(\text { mean of observation }\left(\mathrm{x}_{\mathrm{i}}-5\right)\right)+2$
$=1+2=3$
Variance of observation
$\mathrm{x}_{\mathrm{i}}-5=\frac{1}{10} \sum_{\mathrm{i}=1}^{10}\left(\mathrm{x}_{\mathrm{i}}-5\right)^{2}-\left(\mathrm{Mean} \text { of }\left(\mathrm{x}_{\mathrm{i}}-5\right)\right)^{2}=3$
$\Rightarrow \quad \lambda=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-3\right)$
$=$ variance of observation $\left(\mathrm{x}_{\mathrm{i}}-5\right)=3$ $\therefore \quad(\mu, \lambda)=(3,3)$
