Match List-I with List-II
| List-I (Yearly data and moving average) | List-II (Number of moving averages obtained) | ||
| (A) | 3-year moving averages from the data of 10 years | (I) | 7 |
| (B) | 5-year moving averages from the data of 13 years | (II) | 8 |
| (C) | 4-year moving averages (before centring) from the data of 10 years | (III) | 6 |
| (D) | 4-year centred moving averages from the data of 10 years | (IV) | 9 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
From n yearly values, the first k-year moving average starts at year 1 and the last one starts at year n − k + 1, so there are n − k + 1 moving averages. Centring then joins each pair of neighbouring averages, which gives one value fewer.
(A) 3-year moving averages from the data of 10 years
This matches (II) 8.
\(10 - 3 + 1 = 8\)
(B) 5-year moving averages from the data of 13 years
This matches (IV) 9.
\(13 - 5 + 1 = 9\)
(C) 4-year moving averages (before centring) from the data of 10 years
This matches (I) 7.
\(10 - 4 + 1 = 7\)
(D) 4-year centred moving averages from the data of 10 years
This matches (III) 6. The 7 averages of (C) form 6 neighbouring pairs.
\(7 - 1 = 6\)