Advertisements
Advertisements
प्रश्न
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
Advertisements
उत्तर
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
APPEARS IN
संबंधित प्रश्न
Evaluate Δ(log ax)
If f(x) = x2 + 3x than show that Δf(x) = 2x + 4
Choose the correct alternative:
If c is a constant then Δc =
Choose the correct alternative:
If m and n are positive integers then Δm Δn f(x)=
Choose the correct alternative:
If ‘n’ is a positive integer Δn[Δ-n f(x)]
Choose the correct alternative:
E f(x) =
Choose the correct alternative:
∇f(a) =
If f(x) = eax then show that f(0), Δf(0), Δ2f(0) are in G.P
Prove that (1 + Δ)(1 – ∇) = 1
Prove that EV = Δ = ∇E
