Statistical inference · Lesson 3 of 13

Standard error and the central limit theorem

Last lesson ended with a promise: the sample mean has a distribution, centred on the truth, whose spread shrinks as the sample grows. This lesson puts numbers on that spread, shows why the distribution turns into a bell whatever the population looks like, and turns the question around: how many households must you visit to be within a given distance of the truth?

← Statistical inference moduleLesson 3 of 13 · about 45 minutes

A question to hold on to

The standard error of the sample mean of a variable x is 20 when samples of size 25 are taken and 10 when samples of size 100 are taken. Quadrupling the sample size has only halved the error. What is the standard deviation of x?

DSE 2018Solved at the end ↓

New here?What is Ashanagar?The invented district behind this page's examples: 60 households, with incomes you can inspect.

Ashanagar is an invented district. The statistics lessons use it for every worked example, so you meet the same households again as the ideas build: first as a list of numbers, then as a spread, then as a population to sample from.

It has 60 households, each with a monthly income in rupees. Unless a page says otherwise, those 60 are the whole population, and a “sample” means a few of them drawn at random. Where a page needs two variables, it uses a smaller set of 12 households with the head’s years of schooling beside income.

The figures are illustrative, made up once and fixed. They are not official statistics.

Households: 60
Mean income: ₹28,783 a month
Median income: ₹25,000 a month
Standard deviation: ₹13,366
Lowest and highest: ₹10,500 and ₹72,000

Meet the full data set →

The idea

How far does a sample mean stray?

Last lesson, the six-household population produced fifteen possible sample means, and their spread told us how much a single sample could miss by. That spread has a name. The standard error of a statistic is the standard deviation of its sampling distribution: how much the statistic typically differs from the thing it estimates, from one sample to the next.

For the sample mean the answer is short, and it is the single most useful formula in this module.

The standard error of the sample mean

SE (X ˉ) = Var (X ˉ) = \frac{σ}{n}

Valid for independent draws, or a sample that is small against the population.

A random sample of 25 from a population with σ = 10 has a standard error of 10 ÷ √25 = 2 (CUET 2023). The standard error is in the same units as the data, and it is not the standard deviation of the population (which would be 10) or of the sample. It says how precisely the sample mean pins down μ.

The square root is the whole story. The error falls with √n, not with n, so a sample four times as large has only half the error. That is exactly the opening question: 20 at n = 25 and 10 at n = 100 is a halving for a quadrupling, and σ ÷ 5 = 20 gives σ = 100. We will solve it properly at the end.

Interactive

The standard error of a mean and a proportion

Drag the sample size in the lab. The solid curve is σ ÷ √n for the sixty Ashanagar incomes. It falls steeply at first and flattens: the tenth household buys a lot of precision and the fiftieth buys very little. The dashed curve is the same quantity when the units are not put back, which we met last lesson, and it falls all the way to zero when the sample is the whole population. A census has no sampling error.

Households in the sample, n9

σ ÷ √nwith the finite population correction (N = 60)

σ of the 60 incomes₹13,366

Standard error, σ ÷ √n₹4,455

With the correction₹4,142

Roughly ±2 SE (95%)₹8,119

To halve the error you would need 36 households, four times as many. Going from 9 to 18 cuts it only to 71% of what it was.

With a finite population

SE (X ˉ) = \frac{σ}{n} \frac{N - n}{N - 1}

Use it when the sample is a noticeable part of the population, as a rule of thumb more than about 5%.

Proportions

A proportion is a mean of zeros and ones. If a unit is a success with probability p, its standard deviation is √(p(1 − p)), so the sample proportion p̂ has

The standard error of a sample proportion

SE (p^) = \frac{p ( 1 - p )}{n}

Fair coin, 25 tosses: the standard deviation of the proportion of heads is √(0.5 × 0.5 ÷ 25) = √0.01 = 0.1 (JNU 2016). A survey of 400 likely voters in which 215 back the ruling party has p̂ = 0.5375, and, since p is unknown, the standard error is estimated with p̂ in its place: √(0.5375 × 0.4625 ÷ 400) = 0.025 (DSE 2007). The expression p(1 − p) is largest at p = 0.5, so a proportion near one half is the hardest to pin down, a fact we will use to plan sample sizes.

When two independent samples are compared, their variances add. For the difference of two sample proportions drawn without replacement from populations of sizes N₁ and N₂, the standard deviation is the square root of the sum of two corrected variances (IIT JAM 2022):

Difference of two independent proportions

\frac{p _{1} q _{1}}{n _{1}} \cdot \frac{N _{1} - n _{1}}{N _{1} - 1} + \frac{p _{2} q _{2}}{n _{2}} \cdot \frac{N _{2} - n _{2}}{N _{2} - 1}

Two details are exam favourites. The variances are added, never subtracted, even for a difference, because uncertainty from two independent samples piles up. And each sample gets its own correction factor. When the populations are large, the factors are 1 and the formula reduces to the familiar √(p₁q₁/n₁ + p₂q₂/n₂).

Derive it

Where the standard error σ ÷ √n comes from

We want to show

For n independent draws with variance σ², the sample mean has

Var (X ˉ) = \frac{σ ^{2}}{n}

, so its standard error is σ ÷ √n.

Write the mean as a sum.
$X ˉ = \frac{1}{n} (X_{1} + X_{2} + \dots + X_{n})$
Everything about the mean now reduces to two questions: what is the variance of a sum, and what does dividing by n do to a variance?
The variance of a sum of independent draws is the sum of the variances.
From the rule for combinations in the correlation lesson, Var(aX + bY) = a²σ_X² + b²σ_Y² + 2ab Cov(X, Y). Independent draws have Cov = 0, so the cross terms vanish, and with a = b = 1 and n terms
$Var (X_{1} + \dots + X_{n}) = σ^{2} + \dots + σ^{2} = n σ^{2}$
The sum gets more spread out as draws are added, but only like √n, because the variance grows like n. The left picture shows it. The Ashanagar incomes have σ² = 178.6, so four draws added together have a variance near 4 × 178.6 = 714.6.
Dividing by n divides the variance by n².
Multiply every value by a constant c and every deviation from the mean is multiplied by c, so every squared deviation by c². Hence Var(cY) = c²Var(Y). With c = 1 ÷ n:
$Var (X ˉ) = \frac{1}{n ^{2}} \cdot n σ^{2} = \frac{σ ^{2}}{n}, SE (X ˉ) = \frac{σ}{n}$
For four draws, 178.6 ÷ 4 = 44.7, a standard error of 6.68 (₹ thousand) against a population standard deviation of 13.37.
Why the square root matters.
The variance falls like 1 ÷ n, so the standard error falls like 1 ÷ √n. To halve the error you need four times the sample, to cut it to a tenth you need a hundred times. This is the fact behind DSE 2018: quadrupling the sample size only halved the standard error. It is also why precision gets expensive: the tenth household buys a lot, the fiftieth buys little.
When the draws are not independent: sampling without replacement.
Draw n households from a population of N without putting any back, and the draws are linked: if the first household is rich, the others are a little less likely to be. Their covariance is the same for every pair, call it c. Take all N households. Their total is fixed, so its variance is zero, and the variance formula gives
$0 = N σ^{2} + N (N - 1) c ⟹ c = - \frac{σ ^{2}}{N - 1}$
Now add n draws, which have n variances and n(n − 1) ordered pairs of covariances:
$Var (\sum X_{i}) ⟹ Var (X ˉ) = n σ^{2} - n (n - 1) \frac{σ ^{2}}{N - 1} = n σ^{2} \frac{N - n}{N - 1} = \frac{σ ^{2}}{n} \cdot \frac{N - n}{N - 1}$
That is the finite population correction of the sampling lesson, derived. With N = 6 and n = 2 the covariance is −σ² ÷ 5 and the factor is (6 − 2) ÷ (6 − 1) = 0.8. Take n = N and it is 0: a census has no sampling error.

Step 1 of 5

Interactive

The law of large numbers and the central limit theorem

Two theorems describe what happens to the sample mean as n grows. They are often confused, and examiners put them side by side.

The law of large numbers says the sample mean gets closer to μ. Since its standard error σ/√n shrinks to zero, X̄ converges to the population mean: it is a consistent estimator of μ. That is a statement about where X̄ ends up, and says nothing about its shape.

The central limit theorem says something much stronger about the shape. Take independent draws from any population with a finite mean μ and variance σ². Standardise the sample mean by subtracting μ and dividing by its standard error. As n grows, the distribution of the result approaches the standard normal.

The central limit theorem

\frac{X ˉ - μ}{σ / n} d N (0, 1) as n \to \infty

Convergence is in distribution. The population itself need not be normal.

The lab draws two thousand samples from a population and plots their means. None of the four populations is bell-shaped: Ashanagar’s incomes lean right, the lottery ticket is almost all zeros with a rare ₹100, the two camps have two peaks, and a die is flat. Slide n upwards and watch the bars settle onto the dashed normal curve with mean μ and standard deviation σ/√n.

Sample size, n1Each of the 2,000 samples takes 1 draw with replacement

The population (₹ thousand)

2,000 sample means, n = 1; dashed: normal(μ, σ/√n)

Population mean μ · SD σ28.78 · 13.37

σ ÷ √n (theory)13.366

SD of the 2,000 means13.127

Skewness: population → means0.88 → 0.88

With n = 1 each “mean” is a single draw, so the bars simply copy the population. Increase n.

Four things to look for as you move the slider:

The centre never moves. The bars stay centred on μ at every n, which is unbiasedness.
The spread shrinks as 1/√n. The SD of the means tracks the theory readout at every n, and going from n = 10 to n = 100 narrows it by a factor near 3.2.
The shape changes. The skewed populations lose their lean as n grows, and the skewness of the means falls towards zero.
The speed depends on the population. The die and the two-camp population are already bell-shaped by n = 5. The incomes need about 30. The lottery ticket is still visibly lopsided at n = 30.

Pause and predict

The population is not normal. For a large sample, the distribution of the sample mean is

Several past questions test exactly this wording. A sample mean of independent trials with success probability p approximates a normal distribution with mean p; it does not have one (DSE 2019), because a mean of zeros and ones takes finitely many values. The assertion that the distribution of the sample mean tends to normal for very large n has the central limit theorem as its correct reason (UGC NET 2023). IIT JAM 2024 asks which of four statements about a sample from a distribution with mean μ and non-zero variance σ² are correct, and three are: the standard deviation of the sample mean is σ/√n, the standardised sum tends to a standard normal, and the sample mean is a consistent estimator of μ. The fourth, that (n − 1)S²/σ² follows a chi-square distribution, is true only when the data are normal.

What the theorem needs

Independent draws from a population with a finite variance, and a large n. It does not need the population to be normal. If the population is normal, the sample mean is exactly normal for every n, however small, and the theorem is not needed.

Derive it

Why sums turn into a bell: the central limit theorem, built by adding

We want to show

Whatever the shape of one draw, once its mean and variance are standardised the distribution of the sum of k independent draws approaches the normal curve as k grows.

Know what the sum must look like in centre and spread.
From the standard error derivation, k independent draws with mean μ and variance σ² have a sum with mean kμ and variance kσ². Standardise: Z_k = (S_k − kμ) ÷ (σ√k). Whatever the population, Z_k has mean 0 and variance 1. Centre and spread are therefore fixed, and the only open question is the shape.
Build the sum's distribution by adding one draw at a time.
If S has probabilities P(S = s) and one more independent draw X is added, the new sum equals t in several ways: S = s and X = t − s, for every s. So
$P (S + X = t) = s \sum P (S = s) P (X = t - s)$
This operation is called convolution. For one fair die, each total from 1 to 6 has probability 1 ÷ 6. For two dice, total 7 arises in 6 ways and total 2 in one way, giving the triangle. Each pass spreads the mass out and rounds off whatever corner was there.
Read the picture.
Choose the fair die and press the slider up. One die has a flat top, and the largest gap between its cumulative probabilities and the normal curve is 0.054. Two dice: 0.016. Ten dice: 0.003. Thirty: 0.001. Pick the very lopsided die and the same thing happens more slowly, from 0.254 at one draw to 0.036 at thirty. The shape forgets where it started, and the speed of forgetting depends on how skewed the start was.
Why the bell and not something else?
Suppose the standardised sums settle on some limiting shape F. Then the sum of 2k draws is two independent sums of k draws added, and its standardised form is Z_2k = (Z_k + Z_k′) ÷ √2. In the limit F must therefore be a shape that reproduces itself when two independent copies are added and divided by √2. The normal curve does exactly that: if Z and Z′ are independent standard normals, then (Z + Z′) ÷ √2 has variance (1 + 1) ÷ 2 = 1 and is again normal, because a sum of independent normals is normal. A classical theorem says it is the only shape with finite variance that survives this operation.
What the theorem does and does not say.
It says the shape of the sum or the mean approaches the bell as k grows, for any population with a finite variance. It does not say how fast, and the picture shows that the speed depends on the starting shape. It also does not say the data themselves become normal: a single die stays a die, and only sums and means smooth out. A proof needs more machinery (moment generating functions or characteristic functions), and what this derivation gives is the reason the limit is the normal curve and not another one.

Step 1 of 5

A caution

How large is large enough?

Many textbooks say “n ≥ 30 is large”. As a rule of thumb for moderately skewed data it is fine. As a statement about the theorem it is wrong: the central limit theorem is a statement about the limit, and prescribes no minimum sample size (DSE 2011). The lab shows why. The right n depends on how far the population is from symmetric.

Symmetric, light-tailed populations (a die, two equal camps) need very few draws.
Moderately skewed populations such as the Ashanagar incomes, whose skewness is 0.88, need a few dozen. The skewness of the mean is the population skewness divided by √n, so at n = 30 it is still about 0.16.
Heavily skewed or rare-event populations need hundreds. A proportion with p near 0 or 1 needs both np and n(1 − p) to be reasonably large, commonly at least 10 or so, before the normal curve fits.

For the same reason, a bigger sample is better for a reason that is not the central limit theorem. A sample of 30 gives a more precise estimate of average height than a sample of 20 because its standard error is smaller, not because 20 is too few for a theorem to work and not because the smaller sample is more “biased” (DSE 2011). Both samples are unbiased if they are drawn at random. They differ in precision.

Turning it round

How big a sample do you need?

So far n was given and the error was the answer. Planning a survey reverses it: you decide how close you want to be, and the formula tells you the cost. If you want the sample mean to be within a margin E of μ with probability 95%, then 1.96 standard errors must not exceed E. Solving for n:

Sample size for a mean

n = (\frac{z σ}{E})^{2}

Round up, never down. z is 1.96 for 95% confidence, 1.645 for 90% and 2.576 for 99%.

UGC NET June 2025 · hard

A firm wants to estimate the mean time for a manual operation. There should be 0.95 confidence that the error will not exceed 2 minutes. A time-and-motion study gives σ = 16 minutes. What sample size is needed?

Pick z.
95% confidence gives z = 1.96.
Substitute.
n = (1.96 × 16 ÷ 2)² = 15.68² = 245.86.
Round up.
A sample of 245 would fall just short of the precision wanted, so n = 246.

Answer

n = 246, option D. The other options are not what the formula gives for these figures. Always round up: 245.86 needs 246 respondents, not 245.

For a proportion the standard deviation is not known in advance, but it can never exceed 0.5, since p(1 − p) peaks at 0.25. Using p = 0.5 gives a worst-case sample size that is safe whatever the true proportion turns out to be.

Sample size for a proportion

n = \frac{z ^{2} p ( 1 - p )}{E ^{2}} \leq \frac{z ^{2}}{4 E ^{2}}

For a margin of error of 3 percentage points at 95% confidence, the worst case is 1.96² × 0.25 ÷ 0.03² = 1,067.1, so 1,068 respondents. This is why opinion polls so often sample about a thousand people. Notice what is missing: the size of the population. A sample of 1,068 gives the same margin whether the electorate is a lakh or a billion, until the sample becomes a sizeable part of the population and the correction applies. Halving the margin to 1.5 points costs four times as many people, 4,269, which is the square root law again.

Watch out for

The mistakes examiners are fishing for

Standard error against standard deviation. σ describes the spread of individual values. σ/√n describes the spread of the mean. Using σ when the question asks for the standard error of a mean is the commonest slip.
Four times the sample, half the error. Not a quarter.
Rounding the sample size down. Always round up, even from 245.1.
Believing the CLT sets a minimum n of 30. It describes a limit. The sample size needed depends on the population’s shape.
Subtracting variances for a difference. For independent samples, Var(X̄₁ − X̄₂) = Var(X̄₁) + Var(X̄₂).

Solving the opening question

The opening question, solved

DSE 2018 · easy

The standard error of the sample mean is 20 at n = 25 and 10 at n = 100. Find the standard deviation of x.

Write the formula for each.
σ ÷ √25 = 20 and σ ÷ √100 = 10.
Solve the first.
σ = 20 × 5 = 100.
Check the second.
100 ÷ 10 = 10. ✓

Answer

σ = 100, option C. Either equation alone is enough. The second line of the question, “quadrupling the sample size has only halved the error”, is the square root law stated in words.

Revise

The lesson on one screen

Standard error: The standard deviation of a statistic over repeated samples. For the mean, σ/√n. Four times the sample, half the error.
Finite population: Multiply the variance by (N − n)/(N − 1), or the SE by its square root. Worth using when n is more than about 5% of N.
Proportions: SE(p̂) = √(p(1 − p)/n), largest at p = 0.5. For a difference of independent samples, add the variances.
Law of large numbers: X̄ → μ: the sample mean is consistent. Says nothing about shape.
Central limit theorem: (X̄ − μ)/(σ/√n) → N(0, 1) for any population with finite variance. The population need not be normal. It sets no minimum n.
Sample size: Mean: n = (zσ/E)². Proportion: n = z²p(1 − p)/E², worst case p = 0.5. Round up. The population size does not enter.

Practice

Check yourself

0 of 13 correct so far

Compute

Standard error of a mean

A population has σ = 15. Find the standard error of the mean of a random sample of 36.

Halving the error

A survey of 50 households has a standard error of ₹800. How many households would give a standard error of ₹400?

Households

Standard error of a proportion

In a sample of 100 households, 30% own a refrigerator. Estimate the standard error of the sample proportion, using p̂ = 0.3.

With the correction

A sample of 12 is drawn without replacement from Ashanagar’s 60 households, whose income standard deviation is ₹13,366. Find the standard error of the mean, in rupees, with the finite population correction.

SE (₹)

Sample size for a mean

You want the sample mean to be within 3 of μ with 90% confidence, and σ = 15. Take z = 1.645. What is the smallest sample size?

Sample size for a proportion

How many respondents guarantee a margin of error of ±3 percentage points at 95% confidence, whatever the true proportion? Use z = 1.96 and the worst case p = 0.5.

Two independent means

Sample 1 has n = 25 and σ = 10. Sample 2 has n = 36 and σ = 12. Find the standard error of the difference of the two sample means.

The CLT at work

Weekly spending has mean 28 and standard deviation 10. For a sample of 25, find P(sample mean > 30). Use the normal approximation.

Probability

Quick questions

1A sample of 100 has a standard error of 4. A sample of what size has a standard error of 2?

2According to the central limit theorem, as n grows the distribution of the sample mean approaches

3Which of these does the central limit theorem not require?

4For a fixed sample size, the standard error of a sample proportion is largest when p equals

5Which change reduces the standard error of a sample mean?

Real questions

Past papers, by level

Foundation

Definitions and one-step calculations. Most CUET and Hyderabad questions sit here.

5 questionsPractise →

Standard

Joint tables, standard errors and statement questions, up to JNU and IIT JAM level.

4 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 13: Sampling and the standard error.

2 questionsLesson 13 →

All 9 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 2Samples, sampling methods and the sampling distribution Lesson 4 →Estimators: bias, variance, consistency, sufficiency

Standard error and the central limit theorem

How far does a sample mean stray?

The standard error of a mean and a proportion

Proportions

Where the standard error σ ÷ √n comes from

The law of large numbers and the central limit theorem

Why sums turn into a bell: the central limit theorem, built by adding

How large is large enough?

How big a sample do you need?

The mistakes examiners are fishing for

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Standard error and the central limit theorem

How far does a sample mean stray?

The standard error of a mean and a proportion

Proportions

Where the standard error σ ÷ √n comes from

The law of large numbers and the central limit theorem

Why sums turn into a bell: the central limit theorem, built by adding

How large is large enough?

How big a sample do you need?

The mistakes examiners are fishing for

The opening question, solved

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next