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The population of a city in a censes taken once in 10 years is given below. Estimate the population in the year 1955. Year 1951 1961 1971 1981 Population inlakhs 35 42 58 84

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प्रश्न

The population of a city in a censes taken once in 10 years is given below. Estimate the population in the year 1955.

Year 1951 1961 1971 1981
Population in
lakhs
35 42 58 84
सारिणी
योग
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उत्तर

Let the year be x and population be y.

To find the population for the year 1955.

i.e. The value of y at x = 1955

Since the value of y is required near the beginning of the table

We use the Newton’s forward interpolation formula.

`y_((x  =  x_+  "nh")) = y_0 + "n"/(1!) Deltay_0 + ("n"("n" - 1))/(2!) Delta^2y_0 + ("n"("n" - 1)("n" - 2))/(3!) Delta^3y_0 + .........`

To find y at x = 1955

∴ x0 + nh = 1955, x0 = 1951, h = 10

⇒ 1951 + n(10) = 1955

10n = 1955 – 1951

⇒ 10n = 4

n = `4/10` = 0.4

x y `Deltay` `Delta^2y` `Delta^3y`
1951 35      
    7    
1961 42   9  
    16   1
1971 58   10  
    26    
1981 84      

y = `35 + 0.4/(1!) (7) + ((0.4)(0.4 - 1))/(2!) (9) + (0.4(0.4 - 1)(0.4 - 2))/(3!)`

= `35 + 2.8/1 + ((0.4)(- 0.6)(9))/2 + ((0.4)(- 0.6)(- 1.6))/6`

y = 35 + 2.8 – 1.08 + 0.064

= 37.864 – 1.08

y = 36.784

∴ Population in the year 1955 is 36.784 (lakhs)

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Interpolation
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 5: Numerical Methods - Exercise 5.2 [पृष्ठ ११९]

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सामाचीर कलवी Business Mathematics and Statistics [English] Class 12 TN Board
अध्याय 5 Numerical Methods
Exercise 5.2 | Q 4 | पृष्ठ ११९

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