Match List - I with List - II. List - I List - II (Population Mean (μ) and 1/N…
2024
Match List - I with List - II.
List - I | List - II |
|---|---|
(Population Mean (μ) and 1/N Σ Xi2) | (Population Standard Deviation (σ)) |
(A) μ=5, 1/N Σ Xi2 = 50 | (I) 8 |
(B) μ=4, 1/N Σ Xi2 = 52 | (II) 7 |
(C) μ=3, 1/N Σ Xi2 = 58 | (III) 6 |
(D) μ=6, 1/N Σ Xi2 = 100 | (IV) 5 |
Choose the correct answer from the options given below :
Answer: C. (A)-(IV), (B)-(III), (C)-(II), (D)-(I) — Concept — For a population, the variance is the mean of the squares minus the square of the mean: σ2 = (1/N)ΣXi2 − μ2. The standard deviation σ is the…
- A.
(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- B.
(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
- C.
(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- D.
(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
Show answer & explanation
Correct answer: C
Concept — For a population, the variance is the mean of the squares minus the square of the mean: σ2 = (1/N)ΣXi2 − μ2. The standard deviation σ is the non-negative square root of that variance, so σ = √( (1/N)ΣXi2 − μ2 ). Every row of List - I supplies exactly those two quantities, so each row is settled by one subtraction followed by one square root — no raw data values are needed.
Application — Apply the identity to each row of List - I in turn.
μ = 5 and (1/N)ΣXi2 = 50 → σ2 = 50 − 52 = 50 − 25 = 25, so σ = √25 = 5, which is the List - II entry (IV).
μ = 4 and (1/N)ΣXi2 = 52 → σ2 = 52 − 42 = 52 − 16 = 36, so σ = √36 = 6, which is the List - II entry (III).
μ = 3 and (1/N)ΣXi2 = 58 → σ2 = 58 − 32 = 58 − 9 = 49, so σ = √49 = 7, which is the List - II entry (II).
μ = 6 and (1/N)ΣXi2 = 100 → σ2 = 100 − 62 = 100 − 36 = 64, so σ = √64 = 8, which is the List - II entry (I).
Cross-check — Reverse the identity: adding μ2 back to σ2 must reproduce the mean of squares that was given.
List - I row | σ2 = mean of squares − μ2 | σ | Check: σ2 + μ2 |
|---|---|---|---|
(A) | 50 − 25 = 25 | 5 | 25 + 25 = 50 |
(B) | 52 − 16 = 36 | 6 | 36 + 16 = 52 |
(C) | 58 − 9 = 49 | 7 | 49 + 9 = 58 |
(D) | 100 − 36 = 64 | 8 | 64 + 36 = 100 |
Result — the standard deviations in List - I order are 5, 6, 7 and 8, so the matching is (A)-(IV), (B)-(III), (C)-(II), (D)-(I). A common slip is to forget the subtraction of μ2 and read the mean of squares itself as a variance; the identity above is what prevents it.