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Question
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.
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Solution
Since the required mark is at the end of the table
We apply backward interpolation formula.
Let the marks be x and No. of candidates be y.
| x | y | `Deltay` | `Delta^2y` | `Delta^3y` | `Delta^4y` |
| Below 20 | 41 | ||||
| 62 | |||||
| Below 40 | 103 | 3 | |||
| 65 | – 18 | ||||
| Below 60 | 168 | – 15 | 0 | ||
| 50 | – 18 | ||||
| Below 80 | 218 | – 33 | |||
| 17 | |||||
| Below 100 | 235 |
`y_((x = x_0 + "nh")) = y_"n" + "n"/(1!) ∇y_"n" + ("n"("n" + 1))/(2!) ∇^2y_"n" + ("n"("n" + 1)("n" + 2))/(3!) Delta^3y_"n" + ..........`
To find y at x = 70
x = x0 + nh
⇒ 70 = 100 + n(20)
70 – 100 = 20n
20n = – 30
⇒ n = `(- 30)/20`
n = – 1.5
`y_((x = 70)) = 235 + ((- 1.5))/(1!) (17) + ((-1.5)(- 1.5 + 1))/(2!) (- 33) + ((- 1.5)(-1.5 + 1)(-1.5 + 2))/(3!) (- 18) 6 ((1.5)(-1.5 + 1)(-1.5 + 2)(-1.5 + 3))/(4!) (0) +`
= `235 - 25.5 + ((-1.5)(-0.5)(-33))/2 + ((-1.5)(-0.5)(0.5))/6 (-18)`
= 235 – 25.5 – 12.375 – 1.125
= 235 – 39
= 196
∴ 196 candidates secured less than 70 marks
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