Central Limit Theorem & Confidence Intervals

Here is one of the strangest facts in all of probability, and the answer to the question the last few lessons kept circling. Take any population you like — wildly skewed, lumpy, lopsided, it does not matter — draw a sample and compute its average. Do that a few thousand times and collect the averages. Their histogram always pulls itself into a clean bell. The underlying data can look like anything at all; the very act of averaging manufactures the normal curve.

The sample mean: centre μ, spread 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 bell appears:

• The mean of x̄ is μ — averaging does not move the centre.
• The variance of x̄ is σ²/n — averaging shrinks the spread, so the standard deviation of the sample mean is σ/√n, called the standard error.

More data buys 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

The Central Limit Theorem says it precisely. For independent draws 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 own shape. Once you accept that, every such question becomes the move from the normal lesson — standardise, then read Φ:

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

A rule of thumb: n ≥ 30 is “large enough” for most GATE problems. Pick a lopsided base population in the simulator below — the exponential, the bimodal — and raise n: watch the histogram of sample means straighten into a bell that hugs σ/√n, narrowing as n climbs.

Confidence intervals (known σ)

The CLT also says how much to trust an estimate. Since x̄ is approximately Normal(μ, σ²/n), the true mean μ lies within a predictable band around x̄. With the population σ known, a confidence interval for the mean is

x̄  ±  z · (σ/√n)

The multiplier z comes from the standard normal and fixes the confidence level: z = 1.645 for 90%, 1.96 for 95%, 2.576 for 99%. A higher confidence needs a bigger z, so the interval gets wider — being more sure costs precision — and it narrows as n grows, again like √n.

A worked example — a real GATE DA 2025 question

Let Y be the sum of 300 independent Bernoulli(0.25) variables. Using the normal approximation, find P(60 ≤ Y ≤ 90). Use Φ(2) ≈ 0.9772.

First the mean and variance of the sum — n copies of a Bernoulli(p), whose mean is p and variance p(1−p):

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

Now standardise both ends and read Φ, using the symmetry Φ(−2) = 1 − Φ(2):

z_low  = (60 − 75)/7.5 = −2          z_high = (90 − 75)/7.5 = +2

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

So ≈ 0.9544. A single Bernoulli is nothing like a Normal — it is only 0 or 1 — yet the CLT makes their sum approximately Normal, which is the whole reason standardise-and-read-Φ works here.

A question to carry forward

The confidence interval used the sample to estimate the mean. But often you do not want an estimate — you want a decision: has the new process really changed the average, or is the wobble just sampling noise? Here is the thread onward: how do you turn a sample into a verdict, and bound the chance of getting that verdict wrong?

In one breath

• Sample mean x̄ of n iid draws: mean μ, variance σ²/n, so standard error = σ/√n (variance ÷ n, SD ÷ √n).
• CLT: for large n (rule of thumb n ≥ 30), x̄ ≈ Normal(μ, σ²/n) and the sum ≈ Normal(nμ, nσ²) — whatever the population shape.
• Then it is the normal lesson: standardise z = (x̄ − μ)/(σ/√n) and read Φ (2025: sum of 300 Bernoulli → Φ(2) − Φ(−2) ≈ 0.9544).
• Confidence interval (known σ): x̄ ± z·(σ/√n) — z = 1.645 (90%), 1.96 (95%), 2.576 (99%); higher confidence → wider.
• Precision tightens like √n: 4× the data halves the standard error.

Practice