9. The mean and variance of 5 observations are 7 and 11.6 . If three the observations are 7,9 and 12 , find the remaining two observatio [Ans. : 5 ,

Given: - Mean of 5 observations = 7 - Variance of 5 observations = 11.6 - Three observations are 7, 9, and 12 Let's denote the remaining two observations as \( x \) and \( y \). The formula for the mean of a set of observations is: \[ \text{Mean} = \frac{x_1 + x_2 + x_3 + x_4 + x_5}{5} \] The formula for the variance of a set of observations is: \[ \text{Variance} = \frac{(x_1 - \text{Mean})^2 + (x_2 - \text{Mean})^2 + (x_3 - \text{Mean})^2 + (x_4 - \text{Mean})^2 + (x_5 - \text{Mean})^2}{5} \] We are given that the mean is 7 and the variance is 11.6. We can use these values to find the remaining two observations. First, let's find the sum of the five observations: \[ 7 = \frac{7 + 9 + 12 + x + y}{5} \] \[ 35 = 28 + x + y \] \[ x + y = 7 \] Next, let's find the variance using the given variance value: \[ 11.6 = \frac{(7-7)^2 + (9-7)^2 + (12-7)^2 + (x-7)^2 + (y-7)^2}{5} \] \[ 11.6 = \frac{0 + 4 + 25 + (x-7)^2 + (y-7)^2}{5} \] \[ 11.6 = \frac{29 + (x-7)^2 + (y-7)^2}{5} \] \[ 58 = 29 + (x-7)^2 + (y-7)^2 \] \[ 29 = (x-7)^2 + (y-7)^2 \] Now, we have two equations: 1. \( x + y = 7 \) 2. \( (x-7)^2 + (y-7)^2 = 29 \) We can solve these two equations to find the values of \( x \) and \( y \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}x+y=7\\\left(x-7\right)^{2}+\left(y-7\right)^{2}=29\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=7-y\\\left(x-7\right)^{2}+\left(y-7\right)^{2}=29\end{array}\right.\) - step2: Substitute the value of \(x:\) \(\left(7-y-7\right)^{2}+\left(y-7\right)^{2}=29\) - step3: Simplify: \(2y^{2}-14y+49=29\) - step4: Move the expression to the left side: \(2y^{2}-14y+49-29=0\) - step5: Subtract the numbers: \(2y^{2}-14y+20=0\) - step6: Factor the expression: \(2\left(y-5\right)\left(y-2\right)=0\) - step7: Divide the terms: \(\left(y-5\right)\left(y-2\right)=0\) - step8: Separate into possible cases: \(\begin{align}&y-5=0\\&y-2=0\end{align}\) - step9: Solve the equation: \(\begin{align}&y=5\\&y=2\end{align}\) - step10: Calculate: \(y=5\cup y=2\) - step11: Rearrange the terms: \(\left\{ \begin{array}{l}x=7-y\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=7-y\\y=2\end{array}\right.\) - step12: Calculate: \(\left\{ \begin{array}{l}x=2\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=2\end{array}\right.\) - step13: Check the solution: \(\left\{ \begin{array}{l}x=2\\y=5\end{array}\right.\cup \left\{ \begin{array}{l}x=5\\y=2\end{array}\right.\) - step14: Rewrite: \(\left(x,y\right) = \left(2,5\right)\cup \left(x,y\right) = \left(5,2\right)\) The solutions to the system of equations are: 1. \( x = 2 \) and \( y = 5 \) 2. \( x = 5 \) and \( y = 2 \) Therefore, the remaining two observations are 2 and 5.