ISS 2016 Statistics Paper-1 Solution: Question 58 (Weddle's Rule vs Simpson's Rule)
Continuing our question-by-question march through the ISS Statistics Paper-1 archive, today's stop is Question 58 from the 2016 paper. It's not a calculation-heavy item at all — it's a conceptual "true or false" question on Numerical Analysis that quietly separates candidates who have memorized a formula from those who actually understand what it does and why it works.
Quick Summary
Topic:Numerical Integration — Weddle's Rule and its properties, compared with Simpson's Rule
Question Reference:ISS 2016, Statistics Paper-1, Question 58
Correct Answer:Option (a) — Statements 2 and 3 only
The Question As It Appeared In The Paper
Q58.Which of the following statements are correct?
Weddle's rule requires at least six consecutive values of the function.
If f(x) is a polynomial of 4th degree, then Weddle's rule gives exact result.
In general, Weddle's rule is more accurate than Simpson's rule.
Select the correct answer using the code given below:
(a) 2 and 3 only
(b) 1 and 3 only
(c) 1 and 2 only
(d) 1, 2 and 3
Step 1: Write Down What Weddle's Rule Actually Is
Before judging any of the three statements, we need the rule itself in front of us. Weddle's rule is a method for approximating a definite integral ∫f(x)dx by splitting the interval [x0, xn] into n equal strips of width h each, and then grouping those strips six at a time.
For one such group — the ordinates y0, y1, y2, y3, y4, y5, y6 — Weddle's rule states:
∫x0x6 f(x) dx ≈ (3h/10) × [y0 + 5y1 + y2 + 6y3 + y4 + 5y5 + y6]
Notice immediately how many separate pieces of data are sitting inside that bracket: y0 through y6. That is seven numbers, not six. Keep this count in your head — it is exactly what decides Statement 1.
Step 2: Checking Statement 1 — How Many Function Values Are Really Needed?
Statement 1 claims: "Weddle's rule requires at least six consecutive values of the function."
From the formula above, one full application of Weddle's rule spans six equal subintervals (x0 to x6), but to form those six subintervals you need seven boundary points: x0, x1, x2, x3, x4, x5, x6. Correspondingly, you need the function evaluated at all seven of those points: y0, y1, ..., y6.
So the precise, textbook-correct condition is:the total number of subintervals must be a multiple of six, which in its smallest case (one block) means seven consecutive function values are required, not six. Six values would only give you five subintervals, and Weddle's formula simply has no slot for a five-strip block — the 6, in "y0 + 5y1 + y2 +6y3 + y4 + 5y5 + y6", is the weight on the exact middle ordinate of a seven-point block, and there is no version of the rule that works on only six points.
Because the statement understates the true requirement by one,Statement 1 is incorrect.
Pro Tip: The single most common slip students make on this topic is confusing "number of intervals" with "number of points." Whenever a numerical-methods statement talks about how many values or ordinates a rule "requires," redraw the x-axis, mark the points physically, and count the dots — not the gaps between them. This is exactly the kind of habit that gets drilled into students until it becomes automatic reflex rather than something you pause to think about mid-exam.
Step 3: Checking Statement 2 — Is Weddle's Rule Exact For A 4th-Degree Polynomial?
Statement 2 claims Weddle's rule gives an exact result whenever f(x) is a polynomial of degree 4. Rather than just quoting this as a textbook fact, let's actually test it with real numbers, step by step, so you can see for yourself that it holds.
Take the simplest 4th-degree polynomial: f(x) = x4. Choose h = 1 and the seven equally spaced points x0 = 0, x1 = 1, x2 = 2, x3 = 3, x4 = 4, x5 = 5, x6 = 6. Compute each ordinate:
y0 = 04 = 0
y1 = 14 = 1
y2 = 24 = 16
y3 = 34 = 81
y4 = 44 = 256
y5 = 54 = 625
y6 = 64 = 1296
Now plug these into Weddle's bracket, term by term. We want y0 + 5y1 + y2 + 6y3 + y4 + 5y5 + y6, which works out as follows:
0 + 5(1) + 16 + 6(81) + 256 + 5(625) + 1296 = 0 + 5 + 16 + 486 + 256 + 3125 + 1296 =5184
Multiply by (3h/10) with h = 1:
Weddle's estimate = (3 × 1 / 10) × 5184 = 0.3 × 5184 =1555.2
Now compute the true value of the integral directly using ordinary calculus, so we have something honest to compare against:
∫06 x4 dx = [x5/5]06 = 65/5 − 0 = 7776/5 =1555.2
The two numbers match exactly, with no approximation error at all. This isn't a coincidence of this particular polynomial — it's a structural feature of Weddle's rule: although it is built from a six-interval interpolation argument, it turns out to be exact for every polynomial up to degree five, one degree higher than you'd naively expect. A 4th-degree polynomial comfortably sits inside that range, so it is reproduced with zero error every time.
Statement 2 is correct.
Step 4: Checking Statement 3 — Is Weddle's Rule Generally More Accurate Than Simpson's Rule?
To compare accuracy fairly, look at what degree of polynomial each rule can integrate exactly, because that is what controls how fast the leftover error shrinks as h gets smaller:
Simpson's 1/3 rule (applied over two strips, x0 to x2) is exact only up to degree 3 polynomials. Its leftover error term behaves like h5 times the function's 4th derivative.
Weddle's rule (applied over six strips, x0 to x6), as we just confirmed numerically above, is exact up to degree 5. Its leftover error term behaves like h7 times the function's 6th derivative.
Because the error order is higher (h7 instead of h5), for a reasonably smooth function and a small enough strip width h, Weddle's rule's error shrinks to zero much faster than Simpson's rule's error does. That is precisely what "more accurate in general" means in numerical analysis — not that it is better in literally every possible edge case, but that as a rule of thumb, for well-behaved functions, it converges faster.
Statement 3 is correct.
Step 5: Putting It Together — The Final Answer
Statement 1 is false (it needs seven values, not six). Statement 2 is true (confirmed both by the general theory and by our own direct numeric check above). Statement 3 is true (higher error order means faster convergence in general).
So the statements that are correct are 2 and 3 only, which is option (a).
This matches the official answer key for the 2016 ISS Statistics Paper-1 (Series A), which also records the answer to Question 58 as (a). Our independent derivation and the official key are in full agreement.
Why This Question Matters
Numerical analysis questions in the ISS exam rarely ask you to just crunch a Weddle's-rule calculation; far more often, like here, they test whether you actually understand the small print behind a named rule — how many points it needs, what degree of polynomial it reproduces exactly, and how it stacks up against a rival method. A candidate who has only memorized "Weddle's rule: (3h/10)[y0+5y1+y2+6y3+y4+5y5+y6]" without ever asking why those particular coefficients exist will get statements like these wrong roughly half the time, since three of the four options differ only in whether Statement 1 is included.
Frequently Asked Questions
What exactly is Weddle's Rule used for?
Weddle's rule is a numerical method for estimating the value of a definite integral when you only have the function's values at a set of equally spaced points, rather than a closed-form antiderivative. It is one of several such "quadrature" rules, alongside the Trapezoidal rule and Simpson's rules.
Why must the number of intervals be a multiple of six?
Weddle's formula is built around one specific repeating pattern of coefficients — 1, 5, 1, 6, 1, 5, 1 — that spans exactly seven consecutive ordinates (six subintervals). To cover a longer interval, you simply repeat this seven-point block end to end, which only works cleanly if the total number of subintervals divides evenly into groups of six.
How is Weddle's Rule different from Simpson's Rule in practice?
Simpson's 1/3 rule groups ordinates two subintervals at a time and is exact up to cubic polynomials, while Weddle's rule groups them six subintervals at a time and is exact up to quintic (degree 5) polynomials. The larger grouping and higher exactness generally make Weddle's rule converge faster for smooth functions, at the cost of needing more data points per block.
Is Weddle's Rule always more accurate than Simpson's Rule, without exception?
No, "in general" is doing real work in that statement. For specific functions, interval widths, or pathological cases, Simpson's rule can occasionally outperform Weddle's rule. The exam statement is correct as a general tendency based on error order, not as an absolute universal law.
What happens if the data I have doesn't come in a multiple of six intervals?
In that situation, Weddle's rule cannot be applied directly to the whole dataset. The usual practical fix is to apply Weddle's rule to as many complete six-interval blocks as possible and handle the small leftover portion with a different rule, such as the Trapezoidal or Simpson's rule, sized to fit the remaining intervals.
Why is Weddle's Rule exact for degree 5 and not just degree 4 like the question's statement suggests?
The question's Statement 2 only asks about degree 4, which is true, but it understates how good the rule actually is. Weddle's rule's coefficients happen to cancel out the error terms one degree further than a naive six-point interpolation would suggest, a well-known quirk of its particular coefficient pattern, which is why it reproduces not just degree 4 but also degree 5 polynomials exactly.
If any step above felt rushed, or you solved this one differently, drop a comment below — and if you know a fellow ISS aspirant who is working through the same 2016 Paper-1 set, send this one their way, since numerical analysis is exactly the kind of topic that rewards working through a few explained examples like this rather than memorizing the formula in isolation.
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