Advertisements
Advertisements
Question
Using interpolation estimate the business done in 1985 from the following data
| Year | 1982 | 1983 | 1984 | 1986 |
| Business done (in lakhs) |
150 | 235 | 365 | 525 |
Advertisements
Solution
Here the intervals are unequal.
By Lagrange’s formula we have,
x0 = 1982
x1 = 1983
x2 = 1984
x3 = 1986
y0 = 150
y1 = 235
y2 = 365
y3 = 525 and x = 1985.
y = `"f"(x) = ((x - x_1)(x - x_2)(x - x_3))/((x_0 - x_1)(x_0 - x_2)(x_0 - x_3)) xx y_0 + ((x - x_0)(x - x_2)(x - x_3))/((x_1 - x_0)(x_1 - x_2)(x_1 - x_3)) xx y_1 + ((x - x_0)(x - x_1)(x - x_3))/((x _2 - x_0)(x_2 - x_1)(x_2 - x_3)) xx y_2 + ((x - x_0)(x - x_1)(x - x_2))/((x_3 - x_0)(x_3 - x_1)(x_3 - x_2)) xx y_3`
y = `((1985 - 1983)(1985 - 1984)(1985 - 1986))/((1982 - 1983)(1982 - 1984)(1982 - 1986)) xx 150 + ((1985 - 1982)(1985 - 1984)(1985 - 1986))/((1983 - 1982)(1983 - 1984)(1983 - 1986)) xx 235 ((198 - 982)(1985 - 1983)(985 - 1986))/((984 - 1982)(1984 - 1983)(1984 - 1986)) xx 365 + ((1985 - 1982)(1985 - 1983)(1985 - 1984))/((1986 - 1982)(1986 - 1983)(1986 - 1984)) xx 525`
= `((2)(1)(-1))/((-1)(-2)(-4)) xx 150 + ((3)(1)(-1))/((1)(-1)(-3)) xx 235 + ((3)(2)(-1))/((2)(1)(-)) xx 365 + ((3)(2)(1))/((4)(3)(2)) xx 525`
= `(-2 xx 150)/(-8) + (-3 xx 235)/3 + (-6 xx 365)/(-4) xx (6 xx 525)/24`
= 37.5 – 235 + 547.5 + 131.25
= 716.125 – 235
= 481.25
∴ Business done in the year 1985 is 481.25 lakhs.
APPEARS IN
RELATED QUESTIONS
Using Newton’s forward interpolation formula find the cubic polynomial.
| x | 0 | 1 | 2 | 3 |
| f(x) | 1 | 2 | 1 | 10 |
In an examination the number of candidates who secured marks between certain intervals was as follows:
| Marks | 0 - 19 | 20 - 39 | 40 - 59 | 60 - 79 | 80 - 99 |
| No. of candidates |
41 | 62 | 65 | 50 | 17 |
Estimate the number of candidates whose marks are less than 70.
The following data gives the melting point of a alloy of lead and zinc where ‘t’ is the temperature in degree c and P is the percentage of lead in the alloy.
| P | 40 | 50 | 60 | 70 | 80 | 90 |
| T | 180 | 204 | 226 | 250 | 276 | 304 |
Find the melting point of the alloy containing 84 percent lead.
Using interpolation estimate the output of a factory in 1986 from the following data.
| Year | 1974 | 1978 | 1982 | 1990 |
| Output in 1000 tones |
25 | 60 | 80 | 170 |
Choose the correct alternative:
For the given points (x0, y0) and (x1, y1) the Lagrange’s formula is
Choose the correct alternative:
Lagrange’s interpolation formula can be used for
Choose the correct alternative:
If f(x) = x2 + 2x + 2 and the interval of differencing is unity then Δf(x)
Choose the correct alternative:
For the given data find the value of Δ3y0 is
| x | 5 | 6 | 9 | 11 |
| y | 12 | 13 | 15 | 18 |
Find f(0.5) if f(– 1) = 202, f(0) = 175, f(1) = 82 and f(2) = 55
The area A of circle of diameter ‘d’ is given for the following values
| D | 80 | 85 | 90 | 95 | 100 |
| A | 5026 | 5674 | 6362 | 7088 | 7854 |
Find the approximate values for the areas of circles of diameter 82 and 91 respectively
