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ધારો કે અવલોકનો $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ એ સમીકરણો $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ અને $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ નું સમાધાન કરે છે. જો $\mu$ અને $\lambda$ એ અનુક્રમે અવલોકનો $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ નો મધ્યક અને વિચરણ હોય તો ક્રમયુક્ત જોડ $(\mu, \lambda)$ મેળવો.
$(6, 6)$
$(3, 6)$
$(6, 3)$
$(3, 3)$
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 $\left(\mathrm{x}_{\mathrm{i}}-5\right)=3$ $\therefore \quad(\mu, \lambda)=(3,3)$
Similar Questions
ટૂંકી રીતનો ઉપયોગ કરીને મધ્યક અને પ્રમાણિત વિચલન શોધો.
| ${x_i}$ | $60$ | $61$ | $62$ | $63$ | $64$ | $65$ | $66$ | $67$ | $68$ |
| ${f_i}$ | $2$ | $1$ | $12$ | $29$ | $25$ | $12$ | $10$ | $4$ | $5$ |
