Standardizing Sample Means With the Central Limit Theorem

Why Standardize a Sample Mean?

A sample mean describes the average of several observations, but its variability is not the same as the variability of one observation. As the sample size increases, the sample mean typically concentrates more closely around the population mean. The central limit theorem (CLT) turns that fact into a practical calculation: after centering and scaling a sample mean, its distribution can often be approximated by the standard normal distribution.

For actuarial problems, the essential steps are to identify the population mean and variance, calculate the sample mean's standard error, and convert the event's bounds to standard normal scores. A frequent error is to divide the population standard deviation by the sample size rather than by its square root.

Definition: Let $X_1,\ldots,X_n$ be independent and identically distributed random variables with finite population mean $E[X_i]=\mu$ and finite, positive population variance $\operatorname{Var}(X_i)=\sigma^2$. Their sample mean is $$ \bar{X}=\frac{1}{n}\sum_{i=1}^n X_i. $$ Its mean is $\mu$, its variance is $\sigma^2/n$, and its standard error is $\sigma/\sqrt{n}$.

The CLT for a Sample Mean

Write $\Phi(z)=P(Z\le z)$, where $Z\sim N(0,1)$. Under the independent, identically distributed assumptions above, the central limit theorem states that

\frac{\bar{X}-\mu}{\sigma/\sqrt{n}}\ \xrightarrow{d}\ N(0,1) \quad\text{as }n\to\infty.

The notation $\xrightarrow{d}$ means convergence in distribution. For a sufficiently large sample, this result supports approximating the standardized sample mean by a standard normal random variable. It does not say that every finite-sample distribution of $\bar{X}$ is exactly normal. If the population itself is normal, however, the sample mean is exactly normal for every positive integer $n$.

The sample mean's standard error, $\sigma/\sqrt n$, measures the spread of sample means around $\mu$. It decreases as $n$ increases. It is not the population standard deviation $\sigma$, which describes the spread of individual observations.

Why the Standardization Works

The CLT is stated for standardized sums. The link to a sample mean follows from the definitions and the rules for means and variances of independent random variables.

Center the observations.
Each $X_i-\mu$ has mean $0$ and variance $\sigma^2$. Independence gives the sum $\sum_{i=1}^n(X_i-\mu)$ mean $0$ and variance $n\sigma^2$.

Scale the sum.
Dividing the centered sum by its standard deviation $\sigma\sqrt n$ gives $\frac{\sum_{i=1}^n(X_i-\mu)}{\sigma\sqrt n}$, which the CLT approximates by $N(0,1)$ for large $n$.

Rewrite it as a sample mean.
Since $\sum_{i=1}^n(X_i-\mu)=n(\bar X-\mu)$, the standardized sum equals $\frac{\bar X-\mu}{\sigma/\sqrt n}$. Thus the sample mean uses standard error $\sigma/\sqrt n$.

This derivation also explains why the square root appears: the variance of a sum of $n$ independent observations is $n\sigma^2$, so its standard deviation is $\sigma\sqrt n$. Dividing the sum by $n$ to form the mean then gives standard deviation $\sigma/\sqrt n$.

Converting Events to Standard Normal Probabilities

For an event about $\bar X$, standardize each boundary by subtracting $\mu$ and dividing by $\sigma/\sqrt n$. Because the standard error is positive, the direction of an inequality does not change. For example, an interval event becomes

P(a\le \bar X\le b) \ \approx\ \Phi\!\left(\frac{b-\mu}{\sigma/\sqrt n}\right) - \Phi\!\left(\frac{a-\mu}{\sigma/\sqrt n}\right).

For a lower-tail event $P(\bar X\le b)$, use $\Phi((b-\mu)/(\sigma/\sqrt n))$. For an upper-tail event $P(\bar X>b)$, use $1-\Phi((b-\mu)/(\sigma/\sqrt n))$. Keep the original event visible until both standardized bounds have been calculated; this helps prevent reversing a tail or subtracting the CDF values in the wrong order.

Worked Example: An Interval for an Exponential Sample Mean

Suppose hypothetical repair times $X_1,\ldots,X_{64}$, in hours, are independent and identically distributed exponential random variables with rate $\lambda=0.5$ per hour. Find the CLT approximation to $P(1.75<\bar X<2.50)$. This is a probability model only; no interest rate or mortality assumption applies.

Method and moments. For an exponential distribution with rate $\lambda$, the mean is $1/\lambda$ and the variance is $1/\lambda^2$. Thus $\mu=1/0.5=2$ hours and $\sigma^2=1/(0.5)^2=4$ hours squared, so $\sigma=2$ hours.

Calculate the standard error.

\operatorname{SE}(\bar X)=\frac{\sigma}{\sqrt n} =\frac{2}{\sqrt{64}}=\frac{2}{8}=0.25\text{ hour}.

Standardize both bounds.

z_{\text{lower}}=\frac{1.75-2}{0.25}=-1, \qquad z_{\text{upper}}=\frac{2.50-2}{0.25}=2.

Find the probability. Using $\Phi(2)=0.9772$ and $\Phi(-1)=0.1587$,

P(1.75<\bar X<2.50) \approx \Phi(2)-\Phi(-1) =0.9772-0.1587 =0.8185\approx 0.8186.

The answer is approximately 0.8186, or 81.86%. The bounds lie one standard error below and two standard errors above the mean, which agrees with the standardized interval $(-1,2)$. The exponential population is not normal, so this is a CLT approximation, not an exact probability.

Worked Example: An Upper Tail for a Uniform Sample Mean

A hypothetical sensor's independent measurement errors $X_1,\ldots,X_{36}$ each have a uniform distribution on $(0,12)$ units. Find the CLT approximation to the probability that the sample mean exceeds 6.8 units. The distribution is $X_i\sim\operatorname{Uniform}(0,12)$; no interest rate or mortality assumption applies.

Find the population moments. For a uniform distribution on $(a,b)$, the mean is $(a+b)/2$ and the variance is $(b-a)^2/12$. Therefore,

\mu=\frac{0+12}{2}=6, \qquad \sigma^2=\frac{(12-0)^2}{12}=\frac{144}{12}=12, \qquad \sigma=\sqrt{12}.

Calculate the standard error.

\operatorname{SE}(\bar X)=\frac{\sqrt{12}}{\sqrt{36}} =\frac{\sqrt{12}}{6} =\frac{1}{\sqrt 3} \approx 0.5774.

Standardize the cutoff.

z=\frac{6.8-6}{1/\sqrt 3}=0.8\sqrt 3\approx 1.3856.

This is an upper-tail event, so use the complement of the standard normal CDF. With $\Phi(1.3856)\approx 0.9170$,

P(\bar X>6.8)\approx 1-\Phi(1.3856) =1-0.9170 =0.0830.

The probability is approximately 0.0830. Since 6.8 is above the population mean of 6, a probability below one-half is reasonable. The uniform population is not itself normal, so the CLT supplies an approximation.

Worked Example: An Interval for a Gamma-Distributed Claim Mean

Suppose hypothetical claim amounts $X_1,\ldots,X_{36}$, measured in thousands of dollars, are independent and identically distributed with a gamma distribution having shape $\alpha=100/9$ and rate $\lambda=5/18$. Find the CLT approximation to $P(38\le\bar X\le43)$. No discounting or mortality model is involved.

Identify the moments. For a gamma distribution parameterized by shape $\alpha$ and rate $\lambda$, $E[X]=\alpha/\lambda$ and $\operatorname{Var}(X)=\alpha/\lambda^2$. Therefore,

\mu=\frac{100/9}{5/18}=40, \qquad \sigma^2=\frac{100/9}{(5/18)^2}=144, \qquad \sigma=12.

The units are thousands of dollars for $\mu$ and $\sigma$, and thousands of dollars squared for the variance.

Calculate the standard error and standardized bounds.

\operatorname{SE}(\bar X)=\frac{12}{\sqrt{36}}=2, \qquad z_{\text{lower}}=\frac{38-40}{2}=-1, \qquad z_{\text{upper}}=\frac{43-40}{2}=1.5.

Subtract the CDF values. Using $\Phi(1.5)=0.9332$ and $\Phi(-1)=0.1587$,

P(38\le\bar X\le43) \approx \Phi(1.5)-\Phi(-1) =0.9332-0.1587 =0.7745.

The estimated probability is 0.7745. The lower and upper bounds are one and one-and-a-half standard errors from the mean, respectively, so their standardized positions and the interval probability are consistent. The gamma distribution is generally not normal, making this a CLT approximation.

Common Mistake: Dividing by the Sample Size

The standard error of a sample mean is $\sigma/\sqrt n$, not $\sigma/n$. For the gamma example, using the incorrect value $12/36=1/3$ would produce standardized bounds

\frac{38-40}{1/3}=-6, \qquad \frac{43-40}{1/3}=9.

That incorrect calculation would suggest an interval probability extremely close to 1. The correct standard error is $12/\sqrt{36}=2$, giving bounds $-1$ and $1.5$, and an estimated probability of 0.7745. The incorrect standard error is too small because it divides by $n$ rather than by $\sqrt n$.

Another trap is to treat the CLT approximation as exact for every population and sample size. The sample mean is exactly normal for every $n$ only when the observations are normally distributed. For a nonnormal population, the CLT gives an approximation whose quality depends on the sample size and the population's shape; there is no universal sample-size cutoff that guarantees a particular accuracy.

Key takeaway: For independent, identically distributed observations with mean $\mu$ and finite positive variance $\sigma^2$, standardize a sample mean using $(\bar X-\mu)/(\sigma/\sqrt n)$. Apply the standard normal CDF to the resulting bounds, while distinguishing an exact normal result from a CLT approximation.

Check Your Understanding

For each problem, use the CLT when the population is not normal and report probabilities to four decimal places.

Independent, identically distributed observations have mean 10 and standard deviation 4. For $n=16$, approximate $P(9<\bar X<11)$. (Answer: 0.6827)
Independent observations have an exponential distribution with mean 5 and variance 25. For $n=100$, approximate $P(\bar X>5.8)$. (Answer: 0.0548)
A nonnormal population has mean 80 and variance 225. For an independent sample of size 25, approximate $P(\bar X<74\text{ or }\bar X>86)$. (Answer: 0.0455)
Independent, identically distributed measurements have mean 12 and standard deviation 6. For $n=64$, approximate $P(11.25\le\bar X\le13.5)$. (Answer: 0.8186)
A population has mean $\mu$, standard deviation 8, and a nonnormal distribution. Using the CLT, find the smallest integer sample size $n$ for which the approximate probability that $\bar X$ lies within 1 unit of $\mu$ is at least 0.95. Use the central 95% standard normal cutoff $1.96$. (Answer: 246)

What You'll Learn