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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 - Business Mathematics and Statistics

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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

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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