Question bank — Central Limit Theorem — statement, proof sketch, significance

False. The data keeps its original shape forever; only the sampling distribution of the mean or sum (a derived quantity) becomes normal.

False. The entire power of the theorem is that the inputs can be anything — dice, coins, exponentials — as long as their variance is finite; normality is an output, not a requirement.

True. A sum of independent normals is exactly normal, so no limit is needed — the CLT's approximation becomes an equality for every finite $n$.

False — that is the Law of Large Numbers. The CLT is a finer statement about the shape of the fluctuations around $\mu$ once you magnify them by $\sqrt{n}$.

False. The standard error is $\sigma /\sqrt{n}$, so to halve it you need to quadruple $n$ — error shrinks like $1/\sqrt{n}$, not $1/n$.

False. The Cauchy has infinite variance (and no defined mean), so a core hypothesis fails; the average of $n$ Cauchy variables is still Cauchy with the same spread, no matter how large $n$ is.

False. The classical CLT needs independence; strongly dependent variables can converge to a non-normal limit or fail to stabilize. Weak-dependence versions exist but require extra conditions.

True. $\mathrm{Var}(\sum X_i)=n{\sigma }^2$ so $\mathrm{SD}=\sigma \sqrt{n}$ — this $\sqrt{n}$ growth is exactly why we divide by $\sqrt{n}$ to stabilize the limit.

False. That histogram estimates the original distribution of $X$. The CLT is about the histogram of many averages, each computed from its own fresh sample.

False. The limiting shape uses only the first two moments; higher moments (via skewness) affect only the speed of convergence, as quantified by Berry–Esseen below.

The skew lives in the individual $X_i$; averaging 50 of them cancels most of the skew, and ${\bar{X}}_{50}$ is well-approximated by a bell. Skewness slows convergence but does not forbid it for finite variance.

You forgot the continuity correction. A discrete count is being covered by a continuous curve, so use $50.5$ (the right edge of the bar) to capture its width; ignoring it biases the probability.

You mixed two different limits. As a distribution ${\bar{X}}_n$ does collapse to $\mu$ (that's LLN); the CLT rescales by $\sqrt{n}$ first so the spread stays fixed at $1$, giving the non-degenerate $N(0,1)$.

The "$n\ge 30$" rule of thumb assumes mild skew. Heavily skewed or heavy-tailed distributions may need hundreds or thousands of terms before the approximation is good.

Convergence in distribution means $P(Z_n\le z)\to \Phi (z)$ — it approaches but never equals $\Phi (z)$ for any finite $n$. The error is small, not zero.

Independence makes characteristic functions multiply, giving a power $[{\phi }_Y(t/\sqrt{n}){]}^n$, not a product with $n$. The $n$ is an exponent, not a coefficient.

We drop it because we standardized $Y=(X-\mu )/\sigma$, making $E[Y]=0$ exactly. It vanishes by construction, not by being negligible.

Dividing by $n$ over-squashes the deviation to $0$ (the LLN regime); dividing by $1$ lets it blow up to infinity. Only $\sqrt{n}$, the actual standard deviation of the sum, holds the spread fixed for a stable limit.

For independent sums, densities combine by messy convolution, but characteristic functions simply multiply. The CF is also a unique fingerprint, and pointwise CF convergence implies distributional convergence via Lévy's continuity theorem.

The Taylor expansion of ${\phi }_Y$ near $0$ is $1-\frac{s^2}{2}+o(s^2)$ (here $s$ is the expansion's dummy argument); the constant term and the $s^2$ term (fixed by mean $0$, variance $1$) survive the $n$-th power limit after $s=t/\sqrt{n}$, while everything higher is swept into $o(1/n)$.

It's the classic exponential limit $(1+a_n/n{)}^n\to e^a$ with $a=-t^2/2$. This is why the bell curve, whose CF is $e^{-t^2/2}$, emerges no matter what the inputs were.

Rare extreme values dominate the sum and refuse to average out. If the tail is so heavy that ${\sigma }^2=\infty$, there is no stabilizing scale and the CLT genuinely fails, as with the Cauchy Distribution.

A $\text{Bin}(n,p)$ variable is a sum of $n$ i.i.d. Bernoulli$(p)$ trials — exactly the setup the CLT needs — so its distribution approaches a bell (the de Moivre–Laplace theorem).

They assume the standardized sample mean is approximately $N(0,1)$, which is precisely what the CLT delivers even when the underlying data are non-normal.

Measurement error is typically a sum of many tiny independent effects; the CLT says such sums are approximately Gaussian regardless of each effect's individual distribution.

Yes: the Berry–Esseen theorem says ${\sup }_z|P(Z_n\le z)-\Phi (z)|\le \frac{C\rho }{{\sigma }^3\sqrt{n}}$, where $\rho =E|X-\mu {|}^3$ is the third absolute moment and $C\approx 0.47$. The error shrinks like $1/\sqrt{n}$ and grows with skewness — the quantitative reason skewed data need larger $n$.

Then every $X_i$ equals $\mu$ with probability one; there is no randomness to average, $Z_n$ is undefined (division by zero), and the theorem's hypothesis ${\sigma }^2>0$ is violated.

The classical CLT does not apply. The properly scaled sum may converge to a non-Gaussian stable law instead of a normal, and the $\sqrt{n}$ scaling is generally wrong.

Yes, under an extra uniformity condition — the Lindeberg–Feller Condition — which ensures no single term dominates the sum. Without it, one large-variance term can prevent normality.

It stays exactly Cauchy with the same scale — it does not concentrate and never becomes normal, a dramatic failure of both the Law of Large Numbers and the CLT.

Generally no. With so few terms the shape of the original distribution still dominates; the bell only emerges as $n$ grows, faster for symmetric light-tailed inputs and slower for skewed ones.

Any $n\ge 1$ — a sum or mean of independent normals is exactly normal, so no minimum sample size is needed in this special case.

A lattice distribution puts all its mass on evenly spaced points (e.g. integers), so its CDF is a staircase. Fitting a smooth normal curve to a staircase means each jump should be split at the midpoint, i.e. shift the cutoff by $\pm 0.5$. A non-lattice continuous variable already has a smooth CDF, so no such correction is needed.