Central Limit Theorem & Confidence Intervals

Here is one of the strangest facts in all of probability. Take any population — skewed, discrete, lopsided, doesn’t matter — draw a sample, and compute the average. Repeat that a few thousand times. The histogram of those averages always pulls itself into a clean bell shape. The underlying data can look like anything; the act of averaging is what manufactures the normal curve. That fact is the Central Limit Theorem, and it is the engine behind every GATE question that hands you a sum or a sample mean and expects you to standardise and read Phi like the previous lesson.

The sample mean: mean μ, variance σ²/n

Take n independent draws X₁, …, Xₙ from a population with mean μ and variance σ². Their average is the sample mean x̄ = (X₁ + … + Xₙ) / n. Two facts hold for any population, even before the CLT:

• Mean of x̄ is μ — averaging does not shift the centre.
• Variance of x̄ is σ²/n — averaging shrinks the spread. So the standard deviation of the sample mean is σ/√n (the “standard error”).

More data gives a tighter estimate, and it tightens like √n, not like n. To halve the standard error you need four times the data.

The Central Limit Theorem

Central Limit Theorem. For independent, identically distributed X₁, …, Xₙ with mean μ and finite variance σ², as n grows the sample mean is approximately

x̄  ≈  Normal( μ ,  σ²/n )        equivalently       sum  ≈  Normal( nμ , nσ² )

regardless of the population’s original distribution. Once you accept that, every question becomes the same move you learned for the Normal: standardise, then read Φ.

                 value − mean              x̄ − μ                  sum − nμ
        z  =  ────────────────      =   ───────────       =     ────────────
                  std. dev                 σ/√n                    √(nσ²)

where Φ(z) = P(Z ≤ z) is the standard-normal table. A rule of thumb: n ≥ 30 is “large enough” for the approximation in most GATE problems.

Try it. Pick the most lopsided base population you can find — the exponential, the bimodal — and start drawing samples. At n = 1 the histogram of “sample means” still looks like the base population. Bump n up and watch the shape straighten into a bell, hugging N(μ, σ/√n) whatever the population was. That shrinking standard error σ/√n is exactly why bigger samples give tighter estimates — and why a 95% confidence interval gets narrower as n grows.

Confidence intervals (known σ)

The CLT also tells you how trustworthy an estimate is. If x̄ is approximately Normal(μ, σ²/n), then μ lies within a predictable band around x̄. A confidence interval for the mean, when the population σ is known, is

x̄  ±  z · (σ/√n)

The multiplier z comes from the standard normal and sets the confidence level:

• 90% confidence: z = 1.645
• 95% confidence: z = 1.96
• 99% confidence: z = 2.576

A higher confidence level needs a bigger z, which makes the interval wider — being more sure costs precision. The interval also narrows as n grows, again like √n.

How GATE asks this

A NAT or MCQ. The classic pattern hands you a sum or proportion built from many iid pieces, tells you to treat it as Normal, and either gives you a Φ value (often Φ(2) ≈ 0.9772 or Φ(1) ≈ 0.8413) to reach a decimal, or — as in GATE DA 2025 — asks you to pick the right Φ expression from four options. Either way you compute the mean and variance of the sum, standardise the endpoints, and combine the Φ readings; the 2025 sum-of-300-Bernoulli question is worked below. A close cousin asks for a 95% confidence interval given x̄, σ, and n, expecting x̄ ± 1.96·(σ/√n).

Worked example — a real GATE DA 2025 question

Let Y be the sum of 300 independent Bernoulli(0.25) random variables (each is 1 with probability 0.25, else 0). Using the normal approximation, P(60 ≤ Y ≤ 90) equals which of Φ(2) − Φ(−2), Φ(1) − Φ(−1), Φ(3) − Φ(−3), or Φ(90) − Φ(60)? (We also evaluate it numerically with Φ(2) ≈ 0.9772.)

Step 1 — mean and variance of the sum. A single Bernoulli(p) has mean p and variance p(1−p). For the sum of 300 of them:

mean      = 300 · 0.25            = 75
variance  = 300 · 0.25 · 0.75     = 56.25
std. dev  = √56.25                = 7.5

Step 2 — standardise both endpoints. Subtract the mean and divide by 7.5:

lower:  z = (60 − 75) / 7.5  =  −15 / 7.5  =  −2
upper:  z = (90 − 75) / 7.5  =   15 / 7.5  =  +2

Step 3 — read the Φ table. Using the symmetry Φ(−2) = 1 − Φ(2):

P(60 ≤ Y ≤ 90)  =  Φ(2) − Φ(−2)
                =  0.9772 − (1 − 0.9772)
                =  0.9772 − 0.0228
                =  0.9544

So the answer is the expression Φ(2) − Φ(−2), which evaluates to ≈ 0.9544. This is the real GATE DA 2025 question (an MCQ over the four Φ-expressions) — the whole solution is “mean, variance, standardise, subtract two Φ values.”

Quick check