Sampling Distribution Calculator

Probabilities for a sample mean тАФ standard error, the Central Limit Theorem, and the normal CDF, step by step
P(sample mean > 103) with mu = 100, sigma = 15, n = 25
P(67 < sample mean < 69) with mu = 68, sigma = 3, n = 36
Standard error of the mean for sigma = 12 and n = 64
Normal approximation to the binomial for n = 200, p = 0.4

The Distribution of Sample Means

Take repeated samples of size nn and record x╦Й\bar{x} each time. Those means form their own distribution тАФ the sampling distribution of the mean тАФ and it is narrower than the population:

╬╝x╦Й=╬╝,╧Гx╦Й=╧Гn\mu_{\bar{x}} = \mu, \qquad \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}

That second quantity is the standard error (SE). It is not the population standard deviation, and substituting ╧Г\sigma for it is the single most common error on this topic.

When is x╦Й\bar{x} normal? Either condition suffices:

  1. The population is normal тАФ then x╦Й\bar{x} is exactly normal for any nn, even n=2n = 2.
  2. The Central Limit Theorem тАФ for any population with finite variance, x╦Й\bar{x} is approximately normal once nn is large. nтЙе30n \ge 30 is the usual rule of thumb; strongly skewed populations need more.

Also assumed: observations are independent. If sampling without replacement from a finite population, that holds well enough when nтЙд0.05Nn \le 0.05N; otherwise apply the finite population correction (NтИТn)/(NтИТ1)\sqrt{(N-n)/(N-1)}.

Computing the Probability

Standardise with the standard error, then read the cumulative distribution function ╬ж(z)=P(ZтЙдz)\Phi(z) = P(Z \le z):

z=x╦ЙтИТ╬╝╧Г/nz = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}

zz╬ж(z)\Phi(z)
тИТ2-20.02280.0228
тИТ1-10.15870.1587
000.50000.5000
110.84130.8413
1.6451.6450.95000.9500
1.961.960.97500.9750
220.97720.9772

Then P(x╦Й>a)=1тИТ╬ж(za)P(\bar{x} > a) = 1 - \Phi(z_a) and P(a<x╦Й<b)=╬ж(zb)тИТ╬ж(za)P(a < \bar{x} < b) = \Phi(z_b) - \Phi(z_a).

Is ╧Г\sigma known? If you only have the sample standard deviation ss, the statistic follows a tt distribution with nтИТ1n-1 degrees of freedom, not zz. For nтЙе30n \ge 30 the two are close enough that many textbooks use zz anyway.

Normal approximation to the binomial. A count XтИ╝Bin(n,p)X \sim \text{Bin}(n,p) is approximately N(np,тАЕтАКnp(1тИТp))N(np,\; np(1-p)) when npтЙе10np \ge 10 and n(1тИТp)тЙе10n(1-p) \ge 10. Apply a continuity correction of ┬▒0.5\pm 0.5.

Common Mistakes to Avoid

  • Dividing by ╧Г\sigma instead of ╧Г/n\sigma/\sqrt{n}. That answers a question about one individual, not about a mean of nn.
  • Dividing by nn instead of n\sqrt{n}. The standard error shrinks with the square root: quadrupling nn only halves it.
  • Assuming the CLT fixes everything. It says nothing about outliers, dependent observations, or a biased sampling method. A large nn of a bad sample is still a bad sample.
  • Using zz when ╧Г\sigma is unknown and nn is small. Use tnтИТ1t_{n-1}.
  • Skipping the continuity correction in the binomial approximation тАФ with nn near the npтЙе10np \ge 10 boundary it shifts the answer noticeably.
  • Confusing P(X>a)P(X > a) with P(x╦Й>a)P(\bar{x} > a). The second is far smaller, because means vary less than individuals.

Examples

Step 1: The population is normal, so x╦Й\bar{x} is normal for any nn
Step 2: Standard error: ╧Г/n=15/25=15/5=3\sigma/\sqrt{n} = 15/\sqrt{25} = 15/5 = 3
Step 3: z=103тИТ1003=1.00z = \dfrac{103 - 100}{3} = 1.00
Step 4: P(x╦Й>103)=1тИТ╬ж(1)=1тИТ0.8413P(\bar{x} > 103) = 1 - \Phi(1) = 1 - 0.8413
Answer: PтЙИ0.1587P \approx 0.1587

Step 1: n=36тЙе30n = 36 \ge 30, so the CLT applies
Step 2: Standard error: 3/36=3/6=0.53/\sqrt{36} = 3/6 = 0.5
Step 3: z1=67тИТ680.5=тИТ2z_1 = \dfrac{67 - 68}{0.5} = -2 and z2=69тИТ680.5=2z_2 = \dfrac{69 - 68}{0.5} = 2
Step 4: P=╬ж(2)тИТ╬ж(тИТ2)=0.9772тИТ0.0228P = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228
Answer: PтЙИ0.9544P \approx 0.9544

Step 1: Check: np=80тЙе10np = 80 \ge 10 and n(1тИТp)=120тЙе10n(1-p) = 120 \ge 10, so the approximation is valid
Step 2: ╬╝=np=80\mu = np = 80, ╧Г=np(1тИТp)=48тЙИ6.9282\sigma = \sqrt{np(1-p)} = \sqrt{48} \approx 6.9282
Step 3: Continuity correction: P(XтЙе90)тЙИP(Y>89.5)P(X \ge 90) \approx P(Y > 89.5)
Step 4: z=89.5тИТ806.9282тЙИ1.3712z = \dfrac{89.5 - 80}{6.9282} \approx 1.3712
Step 5: P=1тИТ╬ж(1.3712)тЙИ1тИТ0.9148P = 1 - \Phi(1.3712) \approx 1 - 0.9148
Answer: PтЙИ0.085P \approx 0.085

Frequently Asked Questions

It is the standard deviation of the sampling distribution of the mean, sigma / sqrt(n). It measures how much a sample mean typically varies from the population mean, and it is always smaller than the population standard deviation for n > 1.

Averaging cancels random error: high and low values in a sample offset each other. The cancellation improves with the square root of n, so the spread of sample means is sigma/sqrt(n). Quadrupling the sample size halves the standard error.

Not if n is large. If the population is normal, x-bar is exactly normal at any n. If it is not, the Central Limit Theorem makes x-bar approximately normal for n of about 30 or more тАФ larger if the population is heavily skewed or has outliers.

Use t with n-1 degrees of freedom whenever the population standard deviation is unknown and you substitute the sample standard deviation s. The t distribution has heavier tails, which matters most for small n; by n = 30 the difference is under 3%.

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