Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :
Let the observations $\mathrm{x}_{\mathrm{i}}(1 \leq \mathrm{i} \leq 10)$ satisfy the equations, $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)=10$ and $\sum\limits_{i=1}^{10}\left(x_{i}-5\right)^{2}=40$ If $\mu$ and $\lambda$ are the mean and the variance of the observations, $\mathrm{x}_{1}-3, \mathrm{x}_{2}-3, \ldots ., \mathrm{x}_{10}-3,$ then the ordered pair $(\mu, \lambda)$ is equal to :
- [JEE MAIN 2020]
- A
$(6, 6)$
- B
$(3, 6)$
- C
$(6, 3)$
- D
$(3, 3)$
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- [JEE MAIN 2020]
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- [JEE MAIN 2023]
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$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Marks } & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text { Frequency } & x-2 & x & x^{2} & (x+1)^{2} & 2 x & x+1 \\ \hline \end{array}$
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