Sampling Distribution Calculator

Probabilities for a sample mean — 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 xˉ\bar{x} each time. Those means form their own distribution — the sampling distribution of the mean — and it is narrower than the population:

Ξxˉ=Ξ,σxˉ=σn\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 σ\sigma for it is the single most common error on this topic.

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

  1. The population is normal — then xˉ\bar{x} is exactly normal for any nn, even n=2n = 2.
  2. The Central Limit Theorem — for any population with finite variance, xˉ\bar{x} is approximately normal once nn is large. nâ‰Ĩ30n \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 nâ‰Ī0.05Nn \le 0.05N; otherwise apply the finite population correction (N−n)/(N−1)\sqrt{(N-n)/(N-1)}.

Computing the Probability

Standardise with the standard error, then read the cumulative distribution function ÎĶ(z)=P(Zâ‰Īz)\Phi(z) = P(Z \le z):

z=xˉ−Ξσ/nz = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}}

zzÎĶ(z)\Phi(z)
−2-20.02280.0228
−1-10.15870.1587
000.50000.5000
110.84130.8413
1.6451.6450.95000.9500
1.961.960.97500.9750
220.97720.9772

Then P(xˉ>a)=1−ÎĶ(za)P(\bar{x} > a) = 1 - \Phi(z_a) and P(a<xˉ<b)=ÎĶ(zb)−ÎĶ(za)P(a < \bar{x} < b) = \Phi(z_b) - \Phi(z_a).

Is σ\sigma known? If you only have the sample standard deviation ss, the statistic follows a tt distribution with n−1n-1 degrees of freedom, not zz. For nâ‰Ĩ30n \ge 30 the two are close enough that many textbooks use zz anyway.

Normal approximation to the binomial. A count X∞Bin(n,p)X \sim \text{Bin}(n,p) is approximately N(np,  np(1−p))N(np,\; np(1-p)) when npâ‰Ĩ10np \ge 10 and n(1−p)â‰Ĩ10n(1-p) \ge 10. Apply a continuity correction of Âą0.5\pm 0.5.

Common Mistakes to Avoid

  • Dividing by σ\sigma instead of σ/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 σ\sigma is unknown and nn is small. Use tn−1t_{n-1}.
  • Skipping the continuity correction in the binomial approximation — with nn near the npâ‰Ĩ10np \ge 10 boundary it shifts the answer noticeably.
  • Confusing P(X>a)P(X > a) with P(xˉ>a)P(\bar{x} > a). The second is far smaller, because means vary less than individuals.

Examples

Step 1: The population is normal, so xˉ\bar{x} is normal for any nn
Step 2: Standard error: σ/n=15/25=15/5=3\sigma/\sqrt{n} = 15/\sqrt{25} = 15/5 = 3
Step 3: z=103−1003=1.00z = \dfrac{103 - 100}{3} = 1.00
Step 4: P(xˉ>103)=1−ÎĶ(1)=1−0.8413P(\bar{x} > 103) = 1 - \Phi(1) = 1 - 0.8413
Answer: P≈0.1587P \approx 0.1587

Step 1: n=36â‰Ĩ30n = 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=67−680.5=−2z_1 = \dfrac{67 - 68}{0.5} = -2 and z2=69−680.5=2z_2 = \dfrac{69 - 68}{0.5} = 2
Step 4: P=ÎĶ(2)−ÎĶ(−2)=0.9772−0.0228P = \Phi(2) - \Phi(-2) = 0.9772 - 0.0228
Answer: P≈0.9544P \approx 0.9544

Step 1: Check: np=80â‰Ĩ10np = 80 \ge 10 and n(1−p)=120â‰Ĩ10n(1-p) = 120 \ge 10, so the approximation is valid
Step 2: Ξ=np=80\mu = np = 80, σ=np(1−p)=48≈6.9282\sigma = \sqrt{np(1-p)} = \sqrt{48} \approx 6.9282
Step 3: Continuity correction: P(Xâ‰Ĩ90)≈P(Y>89.5)P(X \ge 90) \approx P(Y > 89.5)
Step 4: z=89.5−806.9282≈1.3712z = \dfrac{89.5 - 80}{6.9282} \approx 1.3712
Step 5: P=1−ÎĶ(1.3712)≈1−0.9148P = 1 - \Phi(1.3712) \approx 1 - 0.9148
Answer: P≈0.085P \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 — 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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