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Question
Following are the population of a district
| Year (x) | 1881 | 1891 | 1901 | 1911 | 1921 | 1931 |
| Population (y) Thousands |
363 | 391 | 421 | - | 467 | 501 |
Find the population of the year 1911
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Solution
Since only five values of fix) are given, the polynomial which fits the data is of degree four.
Hence fifth differences are zeros.
i.e. Δ5y0 = 0
(E – 1)5y0 = 0
(E5 – 5E4 + 10E3 – 10E2 + 5E – 1)y0 = 0
E5y0 – 5E4y0 + 10E3y0 – 10E2y0 + 5Ey0 – y0 = 0
y5 – 5y4 + 10y3 – 10y2 + 5y1 – y0 = 0
501 – 5(467) + 10(y3) – 10(421) + 5(391) – 363 = 0
2456 – 6908 + 10y3 = 0
– 4452 + 10y3 = 0
⇒ 10y3 = 4452
y = `4452/10` = 445.2
The population of the year 1911 is 445.2 thousands
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