ISS 2016 Statistics Paper-1 Solution: Question 8 (Poisson Distribution Skewness)

Continuing our question-by-question walk through the ISS Statistics Paper-1 previous year papers, today we pick up right where the last post left off and tackle Question 8 of the ISS 2016 Statistics Paper-1. This one looks intimidating at first glance because of the strange-looking probability equation, but once you break it into small, manageable steps, it turns out to be a fairly standard exercise in Poisson probabilities and moments.

Quick Summary

• Topic: Poisson distribution — solving for the parameter λ from a probability condition, then finding the coefficient of skewness

• Question Reference: ISS 2016, Statistics Paper-1, Question 8

• Correct Answer: Option (a), 1/√3

The Question, As It Appeared in the Paper

Let X be a Poisson variate with parameter λ such that P(X = 2) = 2P(X = 4) + 20P(X = 6). What is the coefficient of skewness?

• (a) 1/√3

• (b) 1

• (c) 1/2

• (d) -1/√3

Step 1: Write Down the Poisson Probability Formula

Before doing anything else, recall the basic formula for a Poisson distribution with parameter λ. If X follows a Poisson distribution, then for any whole number k = 0, 1, 2, 3, ...

P(X = k) = e-λ λk / k!

Here, e is the mathematical constant (approximately 2.71828), λ is the single parameter of the distribution (it also happens to be both the mean and the variance of a Poisson variable), and k! means "k factorial", i.e., k × (k-1) × (k-2) × ... × 1.

We are told that P(X = 2) = 2P(X = 4) + 20P(X = 6). Let us write out each term using the formula above.

P(X = 2) = e-λ λ2 / 2!

P(X = 4) = e-λ λ4 / 4!

P(X = 6) = e-λ λ6 / 6!

Step 2: Substitute Into the Given Equation

Plugging these into P(X = 2) = 2P(X = 4) + 20P(X = 6), we get:

e-λ λ2 / 2! = 2 × (e-λ λ4 / 4!) + 20 × (e-λ λ6 / 6!)

Notice that every single term on both sides has a factor of e-λ. Since e-λ is never zero (it's an exponential, so it's always strictly positive), we can safely cancel it out from every term. This is a very common simplification trick in Poisson-based questions — always look for the common e-λ factor first, because it disappears immediately and makes the rest of the algebra purely about powers of λ.

After cancelling e-λ, and writing out the factorials (2! = 2, 4! = 24, 6! = 720):

λ2 / 2 = 2λ4 / 24 + 20λ6 / 720

Simplify the fractions on the right-hand side. 2/24 reduces to 1/12, and 20/720 reduces to 1/36. So the equation becomes:

λ2 / 2 = λ4 / 12 + λ6 / 36

Step 3: Clear the Denominators

To get rid of the fractions, multiply every term by the least common multiple of 2, 12, and 36, which is 36.

36 × (λ2 / 2) = 36 × (λ4 / 12) + 36 × (λ6 / 36)

18λ2 = 3λ4 + λ6

Now, since λ is the parameter of a Poisson distribution, it must be strictly greater than 0 (a Poisson variable with λ = 0 is a degenerate case that always equals 0, which is not of interest here). Because λ is not zero, λ2 is also not zero, so we are allowed to divide every term of the equation by λ2 without losing any valid solutions.

18 = 3λ2 + λ4

Rearranging so everything is on one side:

λ4 + 3λ2 - 18 = 0

Step 4: Solve the Quadratic in λ²

This equation has λ raised only to even powers (4 and 2), which is the standard signal to substitute u = λ2. Doing that turns a fourth-degree equation into an ordinary quadratic equation in u:

u2 + 3u - 18 = 0

We can factor this directly. We need two numbers that multiply to -18 and add to 3 — those numbers are 6 and -3.

(u + 6)(u - 3) = 0

So u = -6 or u = 3.

Remember that u stands for λ2, and a square can never be negative for a real number λ. So u = -6 is rejected outright as impossible. That leaves only:

λ2 = 3

Taking the square root, and keeping only the positive root because λ must be a positive parameter:

λ = √3

Pro Tip: The single easiest place to lose marks on this question is at the very last algebra step — forgetting to reject u = -6 and instead trying to take the square root of a negative number, or accidentally keeping the negative square root of λ² = 3 and writing λ = -√3. Always pause after solving any equation for λ² or λ³ in a Poisson or Binomial question and ask "does this value actually make sense as a parameter?" before moving forward. This habit of sanity-checking each intermediate value against what the parameter is physically allowed to be is drilled repeatedly in a well-structured problem-solving routine — the kind of habit that coaching such as Sunrise Classes tries to build into students until it becomes automatic rather than an afterthought.

Step 5: Recall the Moments of a Poisson Distribution

Now that we know λ = √3, we need the coefficient of skewness. To get there, we first need two central moments of the Poisson distribution: the second central moment (which is just the variance, usually written μ2) and the third central moment (written μ3).

A very useful fact about the Poisson distribution, which is worth memorising completely, is that all of its cumulants are equal to λ. In particular:

• Mean = λ

• Variance, μ2 = λ

• Third central moment, μ3 = λ

This is a special property of the Poisson distribution — it does not hold for most other distributions, so it is worth remembering specifically for Poisson-based questions like this one.

Step 6: Compute the Coefficient of Skewness

The moment-based coefficient of skewness (often denoted β1, and read as "beta-one") is defined as:

β1 = μ32 / μ23

Substituting μ2 = λ and μ3 = λ for our Poisson variable:

β1 = λ2 / λ3 = 1 / λ

We already found λ = √3, so:

β1 = 1 / √3

This matches option (a) exactly.

Final Answer: (a) 1/√3 — and this agrees fully with the official ISS 2016 answer key, which also lists option (a) as correct for Question 8.

Why This Question Matters

This question is a good example of how ISS Paper-1 likes to combine two separate skills into one item: first, manipulating the Poisson probability mass function algebraically to pin down an unknown parameter, and second, recalling a specific structural property of a named distribution (here, that all Poisson cumulants equal λ) to finish the question quickly. If you only know the pmf formula but not the moment properties, you would be stuck computing μ3 from scratch using summations, which is far slower under exam time pressure. Building a small mental table of mean, variance, and skewness/kurtosis formulas for each standard distribution — Binomial, Poisson, Exponential, Normal, Gamma — before the exam pays off directly in questions exactly like this one.

Frequently Asked Questions

Why do we cancel e-λ from both sides so early in the solution?

Every term of the Poisson pmf formula P(X = k) = e-λλk/k! contains the same factor e-λ, regardless of the value of k. Since e-λ is a positive number for any real λ, it can never be zero, so dividing it out of every term in an equation is always a valid, safe step that immediately simplifies a probability equation into a pure polynomial equation in λ.

Why did we reject u = -6 when solving the quadratic?

We substituted u = λ2 specifically to simplify the algebra, but that substitution only makes sense if u itself is non-negative, since a real number squared can never be negative. Since -6 is negative, it cannot correspond to any real value of λ, so it is discarded as an extraneous root introduced by the substitution.

Is the Poisson distribution always positively skewed?

Yes. Since both μ2 and μ3 equal λ, which is always a positive quantity for a valid Poisson parameter, the coefficient of skewness 1/λ (or its square-root form) is always positive. This means every Poisson distribution has a longer tail on the right-hand side, and it only becomes closer to symmetric as λ grows very large.

What is the difference between β1 and γ1 as measures of skewness?

β1 is defined as μ32/μ23 and is always non-negative because it involves a squared term in the numerator. γ1, sometimes written as the "moment coefficient of skewness", is the signed square root of β1 (with the sign taken from μ3), so it can be positive or negative depending on the direction of the skew. This question asks specifically for the coefficient of skewness in the β1 form, which is what produces the clean value 1/√3 among the given options.

Do I need to memorise all four central moments of the Poisson distribution?

For MCQ-style exams like ISS Paper-1, it is enough to remember that μ2 = λ and μ3 = λ, since these two are what feed directly into the skewness formula. The fourth central moment (used for kurtosis questions) is μ4 = λ + 3λ2, which is worth memorising separately if kurtosis-based questions come up in the same paper.

What if I get a different value of λ by making an arithmetic slip — how would I know?

A good check is to verify that your value of λ actually satisfies the original probability condition P(X = 2) = 2P(X = 4) + 20P(X = 6) before moving on to compute the skewness. Plugging λ2 = 3 back into the simplified equation 18 = 3λ2 + λ4 gives 18 = 3(3) + 9 = 9 + 9 = 18, which checks out, confirming the algebra is correct before building further calculations on top of it.

If anything in this derivation felt unclear, or if you solved it using a slightly different route, drop a comment below — working through these small disagreements is often where the real learning happens. And if you found this useful, do share it with a fellow ISS aspirant who might be stuck on the same question.