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Worked CLT Problems With Counts and Sample Means

Learn to choose the right normal approximation, standardize a count or sample mean, and apply continuity corrections to obtain accurate probabilities.

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What You'll Learn

  • Standardize a sample mean using its mean and standard error.
  • Translate a sum threshold into an equivalent sample-mean event.
  • Approximate binomial probabilities with a continuity-corrected normal distribution.
  • Approximate Poisson count probabilities using the rate and exposure.
  • Select the correct half-unit boundary for interval and tail events.
  • Check model assumptions and identify when a normal approximation may be unreliable.

Using the CLT to Solve Probability Problems

A central limit theorem problem is not solved by writing “approximately normal” and stopping there. You must identify the random quantity, determine its mean and variance, translate the event into the correct scale, and standardize the relevant boundaries. For counts, integer boundaries usually need a continuity correction; for sample means, the standard error—not the population standard deviation—is used.

The central limit theorem (CLT) makes it possible to approximate probabilities for sums or averages of many independent observations even when the individual observations are not normally distributed. Binomial and Poisson counts also have useful normal approximations when their parameters are sufficiently large. This tutorial develops a consistent method for these exam-style calculations, aligned with the SOA Exam P (Probability) syllabus.

Definition: If \(X_1,\ldots,X_n\) are independent and identically distributed random variables with finite mean \(\mu\) and finite, positive variance \(\sigma^2\), their sample mean is \(\bar X=(X_1+\cdots+X_n)/n\). For large \(n\), the CLT approximates \(\bar X\) by a normal random variable with mean \(\mu\) and variance \(\sigma^2/n\). Equivalently, the standardized quantity \((\bar X-\mu)/(\sigma/\sqrt n)\) is approximately standard normal.

The standard deviation of \(\bar X\), \(\sigma/\sqrt n\), is its standard error. It measures the variability of the sample mean across repeated samples; it is not the standard deviation \(\sigma\) of one observation. If the observations are themselves normally distributed, the sample mean is exactly normal for every \(n\). Otherwise, the CLT supplies an approximation whose quality generally improves as \(n\) grows, subject to the population distribution and the event being estimated.

A Reliable Calculation Sequence

For each problem, state the model and its assumptions before applying a formula. Then put the event in a form that matches that model. A count is not a sample mean until it has been divided by the number of observations, and a sample mean is not a count. Keeping those quantities distinct prevents many errors.

1
Name the random quantity.
Decide whether the problem concerns a sample mean, a binomial count, or a Poisson count. State the distributional model and its parameters.
2
Find the mean and standard deviation.
For a sample mean, use \(\mu\) and \(\sigma/\sqrt n\). For a binomial count use \(np\) and \(\sqrt{np(1-p)}\). For a Poisson count use \(\lambda\) and \(\sqrt{\lambda}\).
3
Translate the event.
Express a sum as a sample-mean event if helpful. For a count, identify whether the event is an interval, lower tail, upper tail, or exact count.
4
Correct integer boundaries and standardize.
For a count event, move each integer boundary by one-half in the direction that includes the desired count values. Convert the resulting normal boundaries to \(z\)-scores.
5
Evaluate and interpret.
Use the standard normal cumulative distribution function \(\Phi(z)=P(Z\le z)\), where \(Z\sim N(0,1)\). Report the probability in the context of the event.

Why the Standard Error and Continuity Correction Appear

The sample-mean standard error follows directly from independence. Each \(X_i\) has variance \(\sigma^2\), so the variance of their sum is \(n\sigma^2\). Dividing that sum by \(n\) divides its variance by \(n^2\):

$$ \operatorname{Var}(\bar X) =\operatorname{Var}\left(\frac{1}{n}\sum_{i=1}^n X_i\right) =\frac{1}{n^2}\sum_{i=1}^n\operatorname{Var}(X_i) =\frac{n\sigma^2}{n^2} =\frac{\sigma^2}{n}. $$

Taking the square root gives \(\sigma/\sqrt n\). The CLT then justifies approximating the standardized sample mean by a standard normal variable when the sample is sufficiently large under the stated assumptions.

A count, however, takes only integer values while the approximating normal distribution is continuous. The continuity correction represents each integer \(k\) by the interval from \(k-0.5\) to \(k+0.5\). Thus, for example, the event \(X\le k\) is approximated using the continuous boundary \(k+0.5\), while \(X\ge k\) uses \(k-0.5\). For an inclusive interval \(a\le X\le b\), the corresponding normal interval runs from \(a-0.5\) to \(b+0.5\).

Worked Problems

Worked Example: A Sample Mean Between Two Limits

A laboratory measures the output of 64 independently selected components from a stable production process. Assume the component outputs are identically distributed, with mean 50 units and standard deviation 12 units, and that no exact normal population model is given. Approximate the probability that the average output of the 64 components is between 47 and 53 units.

Method and model. The observations are independent and identically distributed with finite mean \(\mu=50\) and standard deviation \(\sigma=12\). Use the CLT for their sample mean. Since the population is not specified as normal, this is an approximation, not an exact normal probability.

The sample size is \(n=64\). The sample mean has approximate mean 50 and standard error

$$ \frac{\sigma}{\sqrt n} =\frac{12}{\sqrt{64}} =\frac{12}{8} =1.5. $$

Standardize the two limits using this standard error:

$$ z_{\text{lower}}=\frac{47-50}{1.5}=-2, \qquad z_{\text{upper}}=\frac{53-50}{1.5}=2. $$

Therefore,

$$ P(47\le \bar X\le 53) \approx \Phi(2)-\Phi(-2) =0.9772-0.0228 =0.9544. $$

Using standard normal table values rounded to four decimal places, the approximate probability is 0.9544. The symmetric limits give a useful check: the probability is the central area within two standard errors of the mean. The common mistake avoided here is dividing 12 by 64; the standard error divides by \(\sqrt{64}=8\), not by 64.

Worked Example: A Continuity-Corrected Binomial Interval

A hypothetical quality audit samples 400 independent devices. Each device passes a test with probability 0.50, independently of the others. Approximate the probability that between 181 and 219 devices, inclusive, pass.

Method and model. Let \(X\) be the number of devices that pass. Under the stated independent, constant-probability assumptions, \(X\sim\operatorname{Binomial}(n=400,p=0.50)\). Its mean and variance are

$$ E[X]=np=400(0.50)=200, \qquad \operatorname{Var}(X)=np(1-p)=400(0.50)(0.50)=100. $$

Thus the standard deviation is \(\sqrt{100}=10\). Both \(np=200\) and \(n(1-p)=200\) are large, so a normal approximation is reasonable. For the inclusive integer interval, apply the continuity correction: \(181\le X\le219\) becomes \(180.5\le Y\le219.5\), where \(Y\) is the approximating normal variable with mean 200 and standard deviation 10.

$$ z_{\text{lower}}=\frac{180.5-200}{10}=-1.95, \qquad z_{\text{upper}}=\frac{219.5-200}{10}=1.95. $$

The required probability is therefore

$$ P(181\le X\le219) \approx \Phi(1.95)-\Phi(-1.95) =0.9744-0.0256 =0.9488. $$

The approximation is 0.9488. As a check, the corrected interval is symmetric about the mean, so its probability is twice the area from zero to \(1.95\).

Common mistake. Omitting the correction would use boundaries 181 and 219, producing \(z=-1.9\) and \(z=1.9\), and an estimated probability of \(0.9426\). That calculation excludes part of the continuous area representing the endpoint counts. The corrected boundaries give the appropriate approximation for an inclusive count interval.

Worked Example: A Poisson Upper-Tail Probability

Suppose a hypothetical monitoring system records events according to a Poisson process at a constant rate of 25 events per day, with independent counts over disjoint time intervals. Approximate the probability of at least 110 events during a four-day monitoring period.

Method and model. For a Poisson process with rate 25 per day over four days, the count \(X\) has a Poisson distribution with parameter

$$ \lambda=(25\text{ events/day})(4\text{ days})=100\text{ events}. $$

The time units cancel, leaving the expected count over the entire exposure. For a Poisson random variable, the mean and variance are both \(\lambda\), so \(E[X]=100\) and the standard deviation is \(\sqrt{100}=10\). Since \(\lambda=100\) is large, use a normal approximation.

The event \(X\ge110\) is an upper tail. The continuity-corrected event is \(Y\ge109.5\), not \(Y\ge110.5\), because the count value 110 is included. Its standardized boundary is

$$ z=\frac{109.5-100}{10}=0.95. $$

Thus,

$$ P(X\ge110) \approx P(Z\ge0.95) =1-\Phi(0.95) =1-0.8289 =0.1711. $$

The estimated probability is 0.1711. As a check, the requested event is above the mean, so the result should be less than one-half. A frequent boundary error is using 110.5, which corresponds to excluding 110 and approximating \(X\ge111\), a different event. Another error is using \(\lambda=25\) rather than multiplying the daily rate by four days.

Choosing and Checking the Approximation

The calculation is only as useful as the model and approximation behind it. For a sample mean, verify independence, a common mean and variance where the identically distributed model is being used, and finite variance. The CLT describes behavior as sample size increases; it does not guarantee an accurate approximation for every small sample or every unusually skewed population. If the population is exactly normal, the sample mean is exactly normal under independent sampling, regardless of sample size.

For a binomial count, check that the trials are independent and share the same success probability. A common practical diagnostic for using a normal approximation is that both \(np\) and \(n(1-p)\) are sufficiently large; a frequently used rule of thumb is that each is at least 10. This is a guideline, not a guarantee of a specified error. For a Poisson count, the normal approximation is generally more credible when \(\lambda\) is large. Small rates and far-tail events can make normal approximations poor.

Continuity correction applies to discrete integer counts, not to a sample mean merely because the original observations happen to be recorded as whole numbers. The sample mean can take fractional values, and its CLT standardization uses the event’s stated boundaries directly. For counts, make the event’s inclusiveness explicit before moving a boundary. For example, \(X<k\) means \(X\le k-1\) for an integer-valued count, so its corrected upper boundary is \(k-0.5\); \(X\le k\) instead uses \(k+0.5\).

Key takeaway: First identify the model and the random quantity. Use \(\sigma/\sqrt n\) for a sample mean, \(\sqrt{np(1-p)}\) for a binomial count, and \(\sqrt{\lambda}\) for a Poisson count. Correct integer boundaries before standardizing, then check that the approximation is plausible for the model and event.

Check Your Understanding

For each problem, state the relevant normal approximation and apply a continuity correction when the random quantity is an integer count.

  1. Independent measurements have mean 72 and standard deviation 16. For a sample of 64 measurements, use the CLT to approximate the probability that the sample mean is between 70 and 75. (Answer: 0.7745)
  2. A binomial count has \(n=400\) and \(p=0.25\). Approximate the probability that the count is between 90 and 110, inclusive. (Answer: 0.7747)
  3. A Poisson process has rate 16 events per hour over a four-hour period. Approximate the probability of at least 73 events. (Answer: 0.1440)
  4. Independent observations have mean 4 and standard deviation 1. For \(n=100\), approximate the probability that their sum exceeds 420. (Answer: 0.0228)
  5. A Poisson count has mean \(\lambda=64\). Approximate the probability that the count is at most 70. (Answer: 0.7917)

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