For the frequency distribution :
Variate (x) : x₁ x₂ x₃ .... x₁₅
Frequency (f) : f₁ f₂ f₃ ...... f₁₅
where 0 < x₁ < x₂ < x₃ < ... < x₁₅ = 10 and
Σᵢ₌₁¹⁵ fᵢ > 0, the standard deviation cannot be :

Solution

Maths sir · 4 steps

Every value sits inside (0, 10], so the SD can never cross half of that range, which is 5.

1. Where the data lie. All the values satisfy 0 < xᵢ ≤ 10. So the largest value M and the smallest value m give a range M - m < 10.
2. Variance about any point. For any number c, Σ fᵢ (xᵢ - c)² = Σ fᵢ (xᵢ - x̄)² + N(x̄ - c)², where N = Σ fᵢ. So the spread measured about the mean is the smallest possible, and σ² ≤ 1/NΣ fᵢ (xᵢ - c)² for every c.
3. Pick the midpoint. Take c = (m + M)/2. Every value is within (M - m)/2 of this midpoint, so each (xᵢ - c)² ≤ ((M - m)/2)².

   σ ≤ (M - m)/2 < 10/2 = 5

4. So. The standard deviation is always less than 5. The values 1, 2 and 4 can all occur, but 6 never can, so the answer is (a).

The idea to keep. Standard deviation can never exceed half the range. Once you know the interval the data lives in, you know the largest SD the data can have.

Answer from the NTA final answer key. Solution by PrepWiser.