Question

Let the observations $x_i\left(1\le i\le 10\right)$ satisfy the equations, $\sum _{i=1}^{10}\left(x_i-5\right)=10$ and $\sum _{i=1}^{10}{\left(x_i-5\right)}^2=40$. If $\mu$ and $\lambda$ are the mean and the variance of observations, $\left(x_1-3\right)\left(x_2-3\right)\dots \left(x_{10}-3\right)$, then the ordered pair $\left(\mu ,\lambda \right)$ is equal to:

• A. $\left(6,3\right)$
• B. $\left(3,6\right)$
• C. $\left(3,3\right)$
• D. $\left(6,6\right)$

Answer 1

The correct option is C

$\left(3,3\right)$

Explanation for the correct option:

Step 1: Expanding the given summation

Given that $\sum _{i=1}^{10}\left(x_i-5\right)=10$ and $\sum _{i=1}^{10}{\left(x_i-5\right)}^2=40$ are the two equations also $x_i\left(1\le i\le 10\right)$.

Consider, $\sum _{i=1}^{10}\left(x_i-5\right)=10$ and expand the summation.

$$\begin{gathered} \Rightarrow \sum _{i=1}^{10}\left(x_i-5\right)=10 \\ \underset{i=1}{\overset{10}{\Rightarrow \sum }}{x}_i-\sum _{i=1}^{10}5=10 \\ \underset{i=1}{\overset{10}{\Rightarrow \sum }}{x}_i-50=10 \\ \Rightarrow \sum _{i=1}^{10}{x}_i=60 \end{gathered}$$

Consider, $\sum _{i=1}^{10}{\left(x_i-5\right)}^2=40$ and expand the summation.

$$\begin{gathered} \Rightarrow \sum _{i=1}^{10}{\left(x_i-5\right)}^2=40 \\ \Rightarrow \sum _{i=1}^{10}{x_i}^2-\underset{i=1}{\overset{10}{10\sum }}{x}_i+\sum _{i=1}^{10}25=40 \\ \Rightarrow \sum _{i=1}^{10}{x_i}^2-10\left(60\right)+250=40 \\ \Rightarrow \sum _{i=1}^{10}{x_i}^2=390 \end{gathered}$$

Step 2: Finding the mean $\mu$

$\mu =\frac{1}{N}\left(\sum _{i=1}^{10}\left(x_i-3\right)\right)$

Substituting $N$ as $10$ and $\begin{array}{rcl}& & \sum _{i=1}^{10}{x}_i\end{array}$ as $60$ in the formula of mean.

$$\begin{gathered} \mu =\frac{1}{10}\left(\begin{array}{l}\sum _{i=1}^{10}\end{array}{x}_i-\sum _{i=1}^{10}3\right) \\ \Rightarrow \mu =\frac{1}{10}\left(60-30\right) \\ \Rightarrow \mu =\frac{1}{10}\left(30\right) \\ \Rightarrow \mu =3 \end{gathered}$$

Step 3: Finding the mean $\lambda$

$\lambda =\frac{1}{N}\begin{array}{l}\sum _{i=1}^{10}\end{array}{\left(\begin{array}{l}{x}_i-3\end{array}\right)}^2-{\left(\frac{1}{N}\begin{array}{l}\sum _{i=1}^{10}\end{array}\left(\begin{array}{l}{x}_i-3\end{array}\right)\right)}^2$

Substitute $N$ as $10$, $\frac{1}{N}\begin{array}{l}\sum _{i=1}^{10}\end{array}\left(\begin{array}{l}{x}_i-3\end{array}\right)$ as $3$, $\begin{array}{l}\sum _{i=1}^{10}{x}_i\end{array}$ as $60$ and $\begin{array}{l}\sum _{i=1}^{10}{x_i}^2\end{array}$as $390$ in the formula of variance.

$$\begin{gathered} \lambda =\frac{1}{N}\begin{array}{l}\sum _{i=1}^{10}\end{array}{\left(\begin{array}{l}{x}_i-3\end{array}\right)}^2-{\left(\frac{1}{N}\begin{array}{l}\sum _{i=1}^{10}\end{array}\left(\begin{array}{l}{x}_i-3\end{array}\right)\right)}^2 \\ \Rightarrow \lambda =\frac{1}{10}\left(\begin{array}{l}\sum _{i=1}^{10}\end{array}{x_i}^2-6\sum _{i=1}^{10}{x}_i+\sum _{i=1}^{10}9\right)-{\left(3\right)}^2 \\ \Rightarrow \lambda =\frac{1}{10}\left(390-6\times 60+90\right)-9 \\ \Rightarrow \lambda =\frac{1}{10}\left(120\right)-9 \\ \Rightarrow \lambda =3 \end{gathered}$$

Step 4: Finding the value of the ordered pair

Therefore, the ordered pair $\left(\mu ,\lambda \right)$ is equal to $\left(3,3\right)$

Hence, option (C) is the correct answer.