Q.The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12, 14. Find the remaining two observations.

Concept: Mean-Variance of Natural Numbers – Use the mean to get the sum of all observations, then use the variance to get the sum of squares. Solve the resulting system for the two unknowns.

Let the missing observations be $a$ and $b$.

Step 1: Use the mean.

Mean = 8, so total sum = $7 \times 8 = 56$.

Given five numbers: $2 + 4 + 10 + 12 + 14 = 42$.

Thus $a + b = 56 - 42 = 14$.

Step 2: Use the variance.

Variance = 16, so $\frac{\sum x_i^2}{7} - (\text{mean})^2 = 16$.

Hence $\sum x_i^2 = 7 \times (16 + 64) = 7 \times 80 = 560$.

Sum of squares of given five: $2^2 + 4^2 + 10^2 + 12^2 + 14^2 = 4 + 16 + 100 + 144 + 196 = 460$.

Thus $a^2 + b^2 = 560 - 460 = 100$.

Step 3: Solve the system.

We have $a + b = 14$ and $a^2 + b^2 = 100$.

Using $(a+b)^2 = a^2 + b^2 + 2ab$, we get $196 = 100 + 2ab \Rightarrow ab = 48$.

So $a$ and $b$ are roots of $t^2 - 14t + 48 = 0$, i.e. $(t-6)(t-8)=0$.

We have seven observations in total. Five are given: 2, 4, 10, 12, 14. Let the two unknown observations be $a$ and $b$. The mean of all seven is 8, and the variance is 16.

The core idea is simple: the mean gives us one linear relation between $a$ and $b$, and the variance gives us another (quadratic) relation. Solving these simultaneously will pin down the pair.

1. Use the mean to get the sum of the unknowns.

   The mean of 7 observations is 8, so the total sum is:

The sum of the five known observations is:

Therefore:

So $a + b = 14$. That’s our first equation.

2. Use the variance to get the sum of squares of the unknowns.

   Variance formula for ungrouped data (population variance, as used in most class‑11/12 contexts) is:

Here $n=7$, $\bar{x}=8$, $\sigma^2=16$. So:

Multiply through:

Compute the squared deviations for the five known values:

• $2$: $(2-8)^2 = (-6)^2 = 36$
• $4$: $(4-8)^2 = (-4)^2 = 16$
• $10$: $(10-8)^2 = 2^2 = 4$
• $12$: $(12-8)^2 = 4^2 = 16$
• $14$: $(14-8)^2 = 6^2 = 36$

Sum of these five squared deviations:

Let the squared deviations for $a$ and $b$ be $(a-8)^2$ and $(b-8)^2$. Then:

So:

3. Solve the system.

   We have:

A neat trick: let $u = a-8$ and $v = b-8$. Then $a = u+8$, $b = v+8$, and $a+b = u+v+16 = 14$ gives $u+v = -2$. The second equation becomes $u^2 + v^2 = 4$.

Now we have $u+v = -2$ and $u^2+v^2 = 4$. Recall the identity:

Substitute:

So $u$ and $v$ are two numbers whose sum is $-2$ and product is $0$. That means one of them is $0$ and the other is $-2$.

• If $u = 0$, then $v = -2$: $a = 8$, $b = 6$.
• If $u = -2$, then $v = 0$: $a = 6$, $b = 8$.

Either way, the two missing observations are 6 and 8.