Q.The mean and variance of eight observations are 9 and 9.25, respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Concept: Mean & Variance of Natural Numbers — use the given mean to get a sum constraint, then the variance to get a sum-of-squares constraint, and solve the resulting system.

Let the two unknown observations be $x$ and $y$.

Step 1: Use the mean.

Mean = 9, so total sum = $8 \times 9 = 72$.

Sum of known six = $6 + 7 + 10 + 12 + 12 + 13 = 60$.

Hence $x + y = 72 - 60 = 12$.

Step 2: Use the variance.

Variance $\sigma^2 = 9.25$, so

$\frac{\sum x_i^2}{8} - (\text{mean})^2 = 9.25$

$\Rightarrow \frac{\sum x_i^2}{8} - 81 = 9.25$

$\Rightarrow \sum x_i^2 = 8 \times 90.25 = 722$.

Sum of squares of known six:

$6^2 + 7^2 + 10^2 + 12^2 + 12^2 + 13^2 = 36 + 49 + 100 + 144 + 144 + 169 = 642$.

Thus $x^2 + y^2 = 722 - 642 = 80$.

Step 3: Solve the system.

We have $x + y = 12$ and $x^2 + y^2 = 80$.

Using $(x+y)^2 = x^2 + y^2 + 2xy$,

$144 = 80 + 2xy \Rightarrow 2xy = 64 \Rightarrow xy = 32$.

So $x$ and $y$ are roots of $t^2 - 12t + 32 = 0$, giving $t = 4$ or $t = 8$.

Concept & Intuition

When you know the mean and variance of a dataset, you effectively know two things: the total sum of all observations, and the total sum of their squares. The mean gives you the sum directly. The variance, being the average of squared deviations from the mean, can be rewritten in terms of the sum of squares and the mean. This gives you a second equation. With two unknowns, you can solve.

Here, we have eight observations, six known and two unknown. Let the missing observations be $x$ and $y$. We'll use the given mean and variance to find $x+y$ and $x^2+y^2$, then solve for $x$ and $y$ individually.

Step-by-step solution

1. Use the mean to find the sum of all eight observations.

   Mean $= 9$, so total sum $= 8 \times 9 = 72$.

   The sum of the six known observations is:

Therefore:

2. Use the variance to find the sum of squares of all eight observations.

   Variance $\sigma^2 = 9.25$. For ungrouped data:

Substituting:

3. Find the sum of squares of the known observations.

Hence:

4. Solve for $x$ and $y$.

   We have:

Recall the identity:

Now $x$ and $y$ are the roots of the quadratic:

Factorising:

So $t = 4$ or $t = 8$.