Statistics: PYQ Walkthrough (2)

Mean $= \dfrac{x_1 + x_2 + x_3 + x_4 + x_5}{5} = 4$, so the sum is 20.

Three given: $1 + 3 + 8 = 12$. Let the other two be $a$ and $b$. Then $a + b = 20 - 12 = 8$.

Variance $= \dfrac{1}{n}\sum(x_i - \bar{x})^2 = 5$, so $\sum(x_i - 4)^2 = 25$.

For the three given values: $(1-4)^2 + (3-4)^2 + (8-4)^2 = 9 + 1 + 16 = 26$.

Wait — this already exceeds 25. So the unknown two contribute negatively? No — squares can’t be negative. Let me re-read the problem.

The problem must have variance $\geq$ 26/5 just from the three known values. So variance 5 might be impossible. Let me re-examine: perhaps the data given is in a different setup.

Re-checking: $\sum(x_i - 4)^2 = 5 \times 5 = 25$. From the three given: $26$. So the constraint is $26 + (a-4)^2 + (b-4)^2 = 25$, which is impossible since squares are non-negative.

This means the problem as stated has no solution — which suggests we should use the formula $\text{Var} = \dfrac{1}{n}\sum x_i^2 - \bar{x}^2$ instead, and check using $\sum x_i^2$.

$\text{Var} = \dfrac{1}{n}\sum x_i^2 - \bar{x}^2 \Rightarrow 5 = \dfrac{1}{5}\sum x_i^2 - 16 \Rightarrow \sum x_i^2 = 105$.

From given: $1^2 + 3^2 + 8^2 = 1 + 9 + 64 = 74$.

So $a^2 + b^2 = 105 - 74 = 31$. Combined with $a + b = 8$: $(a+b)^2 = 64 = a^2 + 2ab + b^2 = 31 + 2ab$, so $ab = 16.5$.

The pair $(a, b)$ satisfies $a + b = 8$ and $ab = 16.5$. The quadratic $t^2 - 8t + 16.5 = 0$ has discriminant $64 - 66 = -2 < 0$ — no real solutions.

This particular numerical setup yields no real $a, b$ — illustrating that not every combination of mean, variance, and partial data is feasible. In standard CBSE problems with valid numbers, you’d get real answers. For example, if the variance were 9.2 instead of 5, $\sum x_i^2 = 5(9.2 + 16) = 126$, giving $a^2 + b^2 = 52$, $ab = 6$, $a+b=8$, so quadratic $t^2 - 8t + 6 = 0$ giving $t = 4 \pm \sqrt{10}$.

Standard deviation = $\sqrt{\text{variance}} = \sqrt{5} \approx 2.24$.