The mean and variance of 5 observations are respectively 4.4 and 8.24. If three observation are 1,2 and 4 then find the remaining two observations.

Explanation

To solve the problem step by step, we need to find the remaining two observations, given that the mean and variance of five observations are 4.4 and 8.24, respectively, and three of the observations are 1, 2, and 4.

### Step 1: Set up the equations for the mean
Let the two unknown observations be \( x \) and \( y \). The mean of the five observations can be expressed as:

\[ \text{Mean} = \frac{\text{Sum of observations}}{\text{Number of observations}} = 4.4 \]

The sum of the observations is:

\[ 1 + 2 + 4 + x + y = 7 + x + y \]

Since there are 5 observations, we can set up the equation:

\[ \frac{7 + x + y}{5} = 4.4 \]

### Step 2: Solve for \( x + y \)
Multiplying both sides by 5 gives:

\[ 7 + x + y = 22 \]

Subtracting 7 from both sides results in:

\[ x + y = 15 \quad \text{(Equation 1)} \]

### Step 3: Set up the equation for the variance
The variance is given by the formula:

\[ \text{Variance} = \frac{\text{Sum of squares of observations}}{\text{Number of observations}} - \text{Mean}^2 \]

Calculating the sum of squares of the known observations:

\[ 1^2 + 2^2 + 4^2 = 1 + 4 + 16 = 21 \]

Thus, the equation for variance can be set up as:

\[ \frac{21 + x^2 + y^2}{5} - (4.4)^2 = 8.24 \]

Calculating \( (4.4)^2 \):

\[ (4.4)^2 = 19.36 \]

Now substituting this back into the variance equation:

\[ \frac{21 + x^2 + y^2}{5} - 19.36 = 8.24 \]

### Step 4: Solve for \( x^2 + y^2 \)
Adding \( 19.36 \) to both sides gives:

\[ \frac{21 + x^2 + y^2}{5} = 27.6 \]

Multiplying both sides by 5 results in:

\[ 21 + x^2 + y^2 = 138 \]

Subtracting 21 from both sides gives:

\[ x^2 + y^2 = 117 \quad \text{(Equation 2)} \]

### Step 5: Use the equations to find \( x \) and \( y \)
Now we have two equations:

1. \( x + y = 15 \)
2. \( x^2 + y^2 = 117 \)

We can use the identity:

\[ (x + y)^2 = x^2 + y^2 + 2xy \]

Substituting the values we have:

\[ 15^2 = 117 + 2xy \]

Calculating \( 15^2 \):

\[ 225 = 117 + 2xy \]

Now, solving for \( 2xy \):

\[ 2xy = 225 - 117 = 108 \]

Thus:

\[ xy = 54 \quad \text{(Equation 3)} \]

### Step 6: Solve the system of equations
Now we have a system of equations:

1. \( x + y = 15 \)
2. \( xy = 54 \)

We can express \( y \) in terms of \( x \):

\[ y = 15 - x \]

Substituting into Equation 3:

\[ x(15 - x) = 54 \]

Expanding this gives:

\[ 15x - x^2 = 54 \]

Rearranging leads to:

\[ x^2 - 15x + 54 = 0 \]

### Step 7: Solve the quadratic equation
Now we can solve the quadratic equation using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -15, c = 54 \):

\[ x = \frac{15 \pm \sqrt{(-15)^2 - 4 \cdot 1 \cdot 54}}{2 \cdot 1} \]

Calculating the discriminant:

\[ x = \frac{15 \pm \sqrt{225 - 216}}{2} \]

\[ x = \frac{15 \pm \sqrt{9}}{2} \]

\[ x = \frac{15 \pm 3}{2} \]

Calculating the two possible values for \( x \):

1. \( x = \frac{18}{2} = 9 \)
2. \( x = \frac{12}{2} = 6 \)

### Step 8: Find corresponding \( y \) values
Using \( x + y = 15 \):

If \( x = 9 \):

\[ y = 15 - 9 = 6 \]

If \( x = 6 \):

\[ y = 15 - 6 = 9 \]

Thus, the remaining two observations are \( 9 \) and \( 6 \).

### Final Answer
The remaining two observations are **6 and 9**.