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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.
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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