Match LIST-I with LIST-II. LIST-I: Population mean (μ) and (1/N)Σxi2 LIST-II:…
2024
Match LIST-I with LIST-II.
LIST-I: Population mean (μ) and (1/N)Σxi2 | LIST-II: Population standard deviation (σ) |
|---|---|
A. μ = 7; (1/N)Σxi2 = 85 | I. 9 |
B. μ = 3; (1/N)Σxi2 = 58 | II. 8 |
C. μ = 7; (1/N)Σxi2 = 113 | III. 7 |
D. μ = 6; (1/N)Σxi2 = 117 | IV. 6 |
Choose the correct answer from the options given below:
Answer: D. A–IV, B–III, C–II, D–I — ConceptFor a population, variance equals the second raw moment minus the square of the population mean: σ2 = (1/N)Σxi2 − μ2. Population standard deviation is…
- A.
A–I, B–II, C–III, D–IV
- B.
A–II, B–III, C–IV, D–I
- C.
A–III, B–IV, C–I, D–II
- D.
A–IV, B–III, C–II, D–I
Show answer & explanation
Correct answer: D
Concept
For a population, variance equals the second raw moment minus the square of the population mean: σ2 = (1/N)Σxi2 − μ2.
Population standard deviation is the non-negative square root of the variance: σ = √[(1/N)Σxi2 − μ2].
Application
For A, variance = 85 − 72 = 36, so σ = √36 = 6. Therefore A matches IV.
For B, variance = 58 − 32 = 49, so σ = √49 = 7. Therefore B matches III.
For C, variance = 113 − 72 = 64, so σ = √64 = 8. Therefore C matches II.
For D, variance = 117 − 62 = 81, so σ = √81 = 9. Therefore D matches I.
Cross-check
Squaring the matched standard deviations and adding μ2 reproduces the given second moments: 62 + 72 = 85, 72 + 32 = 58, 82 + 72 = 113, and 92 + 62 = 117.
Hence the required matching is A–IV, B–III, C–II, D–I.