ISS 2016 Statistics Paper-1 Solution: Question 18 (Sum of a Simple Random Sample)
Continuing our question-by-question walk through the ISS Statistics Paper-1 archive, we now arrive at Question 18 of the 2016 paper. This one moves away from the usual named distributions and drops us into the world of finite population sampling theory — a topic that trips up a lot of aspirants simply because the formulas look unfamiliar, even though the algebra behind them is quite elementary once you slow down and work it out by hand.
Quick Summary
Topic: Simple Random Sampling Without Replacement (SRSWOR) — mean and variance of a sample sum
Reference: ISS 2016, Statistics Paper-1, Question 18
Correct Answer: (c) Both 1 and 2
The Question
Suppose 5 tickets are drawn at a time from a box containing n tickets numbered 1, 2, 3, …, n. Let X be the sum of the numbers of the 5 tickets drawn.
Consider the following statements:
E(X) = 5(n + 1) / 2
V(X) = 5(n + 1)(n − 5) / 12
Which of the above is/are correct?
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Step 1: Understanding the Setup
We have a finite population: the numbers 1, 2, 3, …, n, sitting in a box as n distinct tickets. We draw 5 of them without replacement (once a ticket is drawn, it is not put back), and X is defined as the sum of the 5 numbers we end up with.
This is exactly the setting of classical sampling theory: a population of size N (here N = n), from which we draw a simple random sample of size k (here k = 5), without replacement. To keep the population size and the sample size from getting confused with each other, we will call the population size N in our working, and remember that N is what the question calls n. The sample size stays fixed at k = 5.
Two quantities describe the population completely:
The population mean: Ȳ = (1 + 2 + … + N) / N = N(N + 1) / 2N = (N + 1) / 2
The population variance (using the (N − 1) divisor that sampling theory always uses): S² = Σ(Yi − Ȳ)² / (N − 1)
Step 2: Computing the Population Variance S²
Before we can say anything about our sample of 5, we need S² for the full population {1, 2, …, N}. Start from the identity
Σ(Yi − Ȳ)² = ΣYi² − N Ȳ²
We need two standard sums, both of which are worth memorising cold because they appear constantly in this kind of question:
ΣYi = 1 + 2 + … + N = N(N + 1)/2
ΣYi² = 1² + 2² + … + N² = N(N + 1)(2N + 1)/6
Now substitute Ȳ = (N + 1)/2:
Σ(Yi − Ȳ)² = N(N + 1)(2N + 1)/6 − N · (N + 1)²/4
Factor out N(N + 1) from both terms:
= N(N + 1) [ (2N + 1)/6 − (N + 1)/4 ]
Put the bracket over a common denominator of 12:
(2N + 1)/6 = 2(2N + 1)/12 = (4N + 2)/12, and (N + 1)/4 = 3(N + 1)/12 = (3N + 3)/12
So the bracket becomes (4N + 2 − 3N − 3)/12 = (N − 1)/12.
Therefore:
Σ(Yi − Ȳ)² = N(N + 1)(N − 1)/12 = N(N² − 1)/12
Dividing by (N − 1) to get S²:
S² = [N(N² − 1)/12] / (N − 1) = N(N + 1)(N − 1) / [12(N − 1)] =N(N + 1)/12
Pro Tip: This cancellation of the (N − 1) factor is exactly the step where most students panic and start second-guessing their algebra, because it looks like it "shouldn't" simplify so cleanly. The way this is usually taught is to always factor out N(N + 1) as a pair the moment you see both a linear sum and a quadratic sum in the same expression — it is present in both ΣYi and ΣYi², so it will always factor out, and the (N−1)² cancels every single time for this population. Once you've drilled this particular derivation two or three times until it's automatic, you stop fearing this question type altogether.
Step 3: Checking Statement 1 — E(X)
In SRSWOR, a foundational result (usually proved early in any sampling theory course) is that the sample mean is an unbiased estimator of the population mean:
E(ȳ) = Ȳ, where ȳ is the mean of the k = 5 sampled values.
This holds regardless of sampling being without replacement, because each individual ticket, considered on its own, is equally likely to be any of the N tickets — symmetry guarantees this. Since X is just the sum of the 5 drawn values, X = 5ȳ, so:
E(X) = 5 · E(ȳ) = 5 · Ȳ = 5(N + 1)/2
Recalling that the question calls N by the letter n, this is exactly Statement 1: E(X) = 5(n + 1)/2.Statement 1 is correct.
Step 4: Checking Statement 2 — V(X)
The variance of the sample mean under SRSWOR (without replacement) is a standard formula that every sampling theory syllabus states explicitly:
V(ȳ) = (S²/k) · (N − k)/N
The factor (N − k)/N is called the finite population correction: it is the adjustment that accounts for the fact that we are not putting tickets back, so later draws are not independent of earlier ones. If sampling were done with replacement, this correction factor would simply be absent (i.e., it would effectively be 1).
Substitute k = 5 and S² = N(N + 1)/12 from Step 2:
V(ȳ) = [N(N + 1)/12] / 5 · (N − 5)/N = (N + 1)(N − 5) / 60
Now since X = 5ȳ, and multiplying a random variable by a constant multiplies its variance by the square of that constant:
V(X) = 5² · V(ȳ) = 25 · (N + 1)(N − 5)/60 = 5(N + 1)(N − 5)/12
Replacing N with n as in the question, this is precisely Statement 2: V(X) = 5(n + 1)(n − 5)/12.Statement 2 is also correct.
Step 5: A Concrete Numerical Check
Algebra is convincing, but a small worked example makes it airtight. Take n = 6, so the box contains tickets {1, 2, 3, 4, 5, 6} and we draw 5 of them without replacement. There are exactly 6 possible samples of size 5 — one for each ticket that gets left out. The total of all six tickets is 1 + 2 + 3 + 4 + 5 + 6 = 21, so if the ticket numbered j is the one left out, the sample sum is X = 21 − j.
As j ranges over 1, 2, 3, 4, 5, 6 (each equally likely), X takes the values 20, 19, 18, 17, 16, 15 — that is, X is uniformly distributed on {15, 16, 17, 18, 19, 20}.
Direct calculation: E(X) = (15 + 16 + 17 + 18 + 19 + 20)/6 = 105/6 = 17.5
Formula from Statement 1: 5(n + 1)/2 = 5(7)/2 = 35/2 = 17.5 — matches exactly.
For the variance of a discrete uniform distribution on m consecutive integers, the standard result is (m² − 1)/12. Here m = 6 values, so V(X) = (36 − 1)/12 = 35/12.
Formula from Statement 2: 5(n + 1)(n − 5)/12 = 5(7)(1)/12 = 35/12 — matches exactly again.
Final Answer
Both Statement 1 and Statement 2 are correct, both algebraically and by direct numerical verification. This corresponds to option (c) Both 1 and 2, which is also exactly what the official ISS 2016 Series A answer key records for Question 18 (marked "C"). Our independently derived answer matches the official key.
Why This Question Matters
Finite population sampling theory questions like this one are a recurring fixture in ISS Paper-1, and they reward candidates who have internalised two formulas rather than memorised the final answer to any one problem: the unbiasedness of the sample mean, and the finite population correction in the variance of the sample mean. Once you know these two results cold, along with how to derive the population variance S² for the specific population {1, …, N}, this entire family of questions — regardless of how the numbers are dressed up — collapses into routine substitution. It is exactly the kind of question where a candidate who has practiced the derivation, rather than just the formula, finishes in under two minutes.
Frequently Asked Questions
Why does the population variance S² use (N − 1) in the denominator instead of N?
Sampling theory conventionally defines the population mean square as S² = Σ(Yi − Ȳ)²/(N − 1), not the (1/N) divisor you might expect from a basic statistics course. This convention exists because it makes the sample variance an unbiased estimator of S² under SRSWOR, which keeps all the downstream variance formulas clean and consistent.
What exactly is the finite population correction factor, and why does it matter here?
The factor (N − k)/N in the variance formula accounts for sampling without replacement: as tickets are removed from the box, the remaining draws are no longer independent of the ones already taken, which reduces the variability compared to sampling with replacement. Without this correction, you would be implicitly assuming an infinite population, which is not the case when only n tickets exist in total.
What happens to V(X) if n = 5 exactly?
If n = 5, all 5 tickets are drawn every single time, so X is not random at all — it always equals 1 + 2 + 3 + 4 + 5 = 15, a constant. Plugging n = 5 into the formula gives V(X) = 5(6)(0)/12 = 0, which correctly confirms zero variance for a constant.
Could this same logic be applied if the sample size were something other than 5?
Yes, entirely. If the sample size is k instead of 5, the same two-step derivation gives E(X) = k(N + 1)/2 and V(X) = k(N + 1)(N − k)/12, of which the question's formulas are simply the k = 5 special case. It is worth deriving the general-k version once so that any variant of this question, for any sample size, becomes immediate.
Is this topic important for the ISS Paper-1 syllabus overall?
Yes, sampling theory (SRSWOR, SRSWR, stratified and systematic sampling, and their associated mean/variance formulas) is a standing topic across ISS Paper-1 papers nearly every year, so getting comfortable with deriving these results from first principles, rather than trying to memorise every variant, pays off across multiple questions on the same paper.
What is the quickest way to avoid sign or arithmetic errors in this derivation?
Always keep N (population size) and k (sample size) as separate symbols during the derivation, even if the original question uses the same letter for something else, and substitute numbers back in only at the very last step; mixing them up midway is the single most common source of error in this type of question.
If any step above felt rushed, don't hesitate to work through the algebra on paper yourself, drop your doubts in the comments, and pass this along to a fellow ISS aspirant who might be stuck on the same sampling theory question — discussing it together is often the fastest way to make it stick.
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