CUET PG 2024 — Computer PYQ

Given:
Mean of 5 data = 5.2
Variance = 27.296
Three of the data points are: 1, 3, and 6
Formula used:
Mean = $\frac{\text{Sum of all data}}{n}$
Variance = $\frac{\sum (x_i-\bar{x}{)}^2}{n}$

Calculation:

Mean = 5.2 = $\frac{1+3+6+x_1+x_2}{5}$

$\Rightarrow$ 5.2 × 5 = 1 + 3 + 6 + x₁ + x₂

$\Rightarrow$ 26 = 10 + x₁ + x₂

$\Rightarrow$ x₁ + x₂ = 16

Variance = 27.296 = $\frac{(1-5.2{)}^2+(3-5.2{)}^2+(6-5.2{)}^2+(x_1-5.2{)}^2+(x_2-5.2{)}^2}{5}$

27.296 × 5 = (1 - 5.2)² + (3 - 5.2)² + (6 - 5.2)² + (x₁ - 5.2)² + (x₂ - 5.2)²

$\Rightarrow$ 136.48 = (-4.2)² + (-2.2)² + (0.8)² + (x₁ - 5.2)² + (x₂ - 5.2)²

$\Rightarrow$ 136.48 = 17.64 + 4.84 + 0.64 + (x₁ - 5.2)² + (x₂ - 5.2)²

$\Rightarrow$ 136.48 = 23.12 + (x₁ - 5.2)² + (x₂ - 5.2)²

$$=(x_1-5.2{)}^2+(x_2-5.2{)}^2=136.48-23.12=113.36$$
Let $x_1=16-x_2$ (from Step 1)
Substitute this in the equation:
$$<br>=(x_1-5.2{)}^2+(x_2-5.2{)}^2=113.36$$
By substituting the values and solving, we get the values of $x_1$ and $x_2$ as:
$$x_1=9,x_2=7$$

∴ The other two data points are 9 and 7