← 2013 Paper 2

UPSC 2013 Maths Optional Paper 2 Q5c — Step-by-Step Solution

10 marks · Section B

Newton's forward difference interpolation · Numerical Analysis · asked 3× in 13 yrs · Read the full method →

Question

In an examination, the number of students who obtained marks between certain limits were given in the following table:

Marks30–4040–5050–6060–7070–80
No. of Students3142513531

Using Newton forward interpolation formula, find the number of students whose marks lie between 45 and 50.

Technique

Newton forward interpolation on the cumulative frequency, then differencing.

Solution

Strategy. Convert the frequency distribution to a cumulative count F(M)=F(M)= (number of students with marks <M<M). Use Newton forward interpolation to estimate F(45)F(45). Then the students in [45,50][45,50] count is F(50)−F(45)F(50)-F(45).

Step 1 — Cumulative table

MM303040405050606070708080
F(M)F(M)0031317373124124159159190190

(Each F(Mi+1)=F(Mi)+(frequency in [Mi,Mi+1))F(M_{i+1})=F(M_i)+\text{(frequency in }[M_i,M_{i+1})).)

Step 2 — Forward-difference table

Step size h=10h=10, base point x0=30x_0=30.

xxFFΔF\Delta FΔ2F\Delta^{2}FΔ3F\Delta^{3}FΔ4F\Delta^{4}FΔ5F\Delta^{5}F
3003111−2-2−23-236060
4031429−25-253737
507351−16-161212
6012435−4-4
7015931
80190

Step 3 — Newton forward formula at x=45x=45

With u=(45−30)/10=1.5u=(45-30)/10=1.5:

F(x0+uh)=F0+uΔF0+(u2)Δ2F0+(u3)Δ3F0+(u4)Δ4F0+(u5)Δ5F0.F(x_0+uh)=F_0+u\Delta F_0+\binom{u}{2}\Delta^{2}F_0+\binom{u}{3}\Delta^{3}F_0+\binom{u}{4}\Delta^{4}F_0+\binom{u}{5}\Delta^{5}F_0.

Binomial coefficients (continuous form):

(1.51)=1.5,    (1.52)=1.5⋅0.52=0.375,\binom{1.5}{1}=1.5,\;\;\binom{1.5}{2}=\frac{1.5\cdot 0.5}{2}=0.375, (1.53)=1.5⋅0.5⋅(−0.5)6=−0.0625,    (1.54)=1.5⋅0.5⋅(−0.5)⋅(−1.5)24=0.02344,\binom{1.5}{3}=\frac{1.5\cdot 0.5\cdot(-0.5)}{6}=-0.0625,\;\;\binom{1.5}{4}=\frac{1.5\cdot 0.5\cdot(-0.5)\cdot(-1.5)}{24}=0.02344, (1.55)=1.5⋅0.5⋅(−0.5)⋅(−1.5)⋅(−2.5)120=−0.01172.\binom{1.5}{5}=\frac{1.5\cdot 0.5\cdot(-0.5)\cdot(-1.5)\cdot(-2.5)}{120}=-0.01172.

Term-by-term contributions:

TermValueContribution
000000
uΔF0u\Delta F_01.5×311.5\times 3146.546.5
(u2)Δ2F0\binom{u}{2}\Delta^{2}F_00.375×110.375\times 114.1254.125
(u3)Δ3F0\binom{u}{3}\Delta^{3}F_0−0.0625×(−2)-0.0625\times(-2)0.1250.125
(u4)Δ4F0\binom{u}{4}\Delta^{4}F_00.02344×(−23)0.02344\times(-23)−0.539-0.539
(u5)Δ5F0\binom{u}{5}\Delta^{5}F_0−0.01172×60-0.01172\times 60−0.703-0.703

Sum: 0+46.5+4.125+0.125−0.539−0.703≈49.510+46.5+4.125+0.125-0.539-0.703\approx 49.51.

Step 4 — Number of students in [45,50][45,50]

F(50)−F(45)=73−49.51=23.49≈23.F(50)-F(45)=73-49.51=23.49\approx 23.

Answer

  ≈23 students have marks between 45 and 50.  \boxed{\;\approx 23\text{ students have marks between 45 and 50.}\;}
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