ISS 2016 Statistics Paper-1 Solution: Question 61 (Simpson's 1/3 Rule Integration)

Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, today we pick up right where we left off in the 2016 paper and move into the Numerical Analysis section with a classic on numerical integration. If you have been following this series, you already know the drill: restate the question exactly as it appeared, solve it from first principles, and check it against the official key.

Quick Summary

• Topic: Numerical Integration – Simpson's 1/3 Rule

• Question Reference: ISS 2016, Statistics Paper-1, Question 61

• Correct Answer: (a) 10

The Question

Consider the following data:

• x = 3, y(x) = −1

• x = 6, y(x) = 1

• x = 9, y(x) = 2

• x = 12, y(x) = 3

What is the value of ∫312 y(x) dx, when computed by Simpson's 1/3 rule?

• (a) 10

• (b) 9

• (c) 6

• (d) 3

Step 1: Write Down What Simpson's 1/3 Rule Actually Says

Simpson's 1/3 rule approximates a definite integral by fitting parabolas through consecutive triples of points. For data points x0, x1, x2, …, xn that are equally spaced with common interval width h, the rule is written as:

∫ y dx ≈ (h/3) × [ (y0 + yn) + 4 × (sum of y at odd-numbered points) + 2 × (sum of y at even-numbered points, excluding the first and last) ]

The first thing a novice should do with any numerical integration question is simply list out x0, x1, x2, x3, … and label each corresponding y value. Here we have exactly four points, so let us index them from 0 to 3:

• x0 = 3, y0 = −1

• x1 = 6, y1 = 1

• x2 = 9, y2 = 2

• x3 = 12, y3 = 3

Step 2: Find the Common Interval Width h

Since the x values are equally spaced, h is simply the gap between any two consecutive x values:

h = x1 − x0 = 6 − 3 = 3

Check this against the other gaps: 9 − 6 = 3, and 12 − 9 = 3. All equal, so h = 3 is confirmed and the data is suitable for Simpson's rule.

Step 3: Apply the Weight Pattern (1, 4, 2, 1)

In Simpson's 1/3 rule, every point gets a weight before being added up: the first and last points get weight 1, points at odd positions (y1, y3, y5, …) get weight 4, and interior points at even positions (y2, y4, …) get weight 2. With four points indexed 0 to 3, that gives us weights 1, 4, 2, 1 respectively:

• y0 = −1, weight 1 → contributes −1

• y1 = 1, weight 4 → contributes 4

• y2 = 2, weight 2 → contributes 4

• y3 = 3, weight 1 → contributes 3

Adding these weighted contributions together:

Sum = (y0 + y3) + 4y1 + 2y2 = (−1 + 3) + 4(1) + 2(2) = 2 + 4 + 4 = 10

Step 4: Multiply by h/3

Now plug this sum into the full formula:

∫312 y(x) dx ≈ (h/3) × Sum = (3/3) × 10 = 1 × 10 = 10

So the value of the integral, computed using Simpson's 1/3 rule, is 10.

Pro Tip: Textbooks usually state Simpson's 1/3 rule with a quiet footnote that the number of sub-intervals, n, must be

Step 5: Confirm Against the Official Answer Key

The official ISS 2016 Statistics Paper-1 answer key (Series A) lists option (a) as the correct response for Question 61. Our independent step-by-step computation also arrives at 10, which is option (a).The two match — the final answer is confirmed as (a) 10.

Why This Question Matters

Numerical integration questions like this one test whether you can mechanically and accurately apply a formula under exam time pressure, without a calculator doing the heavy lifting. Simpson's 1/3 rule appears repeatedly across ISS, IIT JAM(MS), and GATE(ST) numerical analysis sections, often disguised with different data sets but testing the exact same weight-pattern logic. Being fluent with indexing points correctly, identifying h, and applying weights without hesitation saves precious minutes that matter a great deal in a 200-mark, two-hour paper.

Frequently Asked Questions

Why does Simpson's 1/3 rule use the weights 1, 4, 2, 4, 2, …, 4, 1?

These weights come directly from integrating the parabola that is fitted through each consecutive pair of sub-intervals. When you work out that integration algebraically, the middle point of each pair always ends up contributing four times as much as the shared endpoints, which is where the characteristic 4 and 2 pattern originates.

What happens if the x values are not equally spaced?

Simpson's 1/3 rule assumes a constant interval width h between consecutive x values. If the spacing is unequal, this rule cannot be applied directly, and you would instead need methods designed for unequal intervals, or you would first interpolate the data onto an equally spaced grid.

Is Simpson's 1/3 rule more accurate than the trapezoidal rule?

Yes, generally. Simpson's 1/3 rule fits parabolas rather than straight lines through the data, so it captures curvature in the underlying function more accurately, which usually results in a smaller error for the same number of data points.

Why is an even number of sub-intervals required in theory?

Because Simpson's 1/3 rule processes the data two sub-intervals at a time, fitting one parabola across each pair. If the total number of sub-intervals is odd, one sub-interval is left over and cannot be paired, which breaks the clean derivation of the rule.

How do I quickly check if my spacing is correct before applying the formula?

Subtract each consecutive pair of x values and confirm they are all identical. If even one gap differs, double check the data before proceeding, since an incorrect h value will scale your entire final answer incorrectly.

Could a negative y value like y0 = −1 cause an error?

Not at all. Simpson's 1/3 rule works exactly the same way regardless of the sign of the function values; you simply carry the negative sign through the weighted sum carefully, as we did in Step 3 above.

If anything in this solution feels unclear, or if you spotted a step you would have approached differently, drop a comment below and share this post with a fellow ISS aspirant who might be revising the same topic this week.