A single measurement varies with the population standard deviation σ, but the average of n measurements varies much less. When the population is normal – or n is large enough for the central limit theorem to apply – the sample mean x̄ follows a normal distribution with mean μ and standard deviation σ/√n, called the standard error.
That lets you answer questions such as “what is the chance that the average height of 7 randomly chosen women is below 160 cm?” or “how likely is the average fill of 25 bottles to be between 498 and 502 ml?”. The method is the same as for a single value, except that the z-score divides by the standard error instead of σ.
Choose the type of probability, enter one or two bounds, and the calculator returns the probability with the standard error, the z-scores and a chart of the sampling distribution with the area shaded. For comparison it also shows the probability for a single observation, which is always less extreme – averages cluster more tightly around μ.
How to use the Normal Probability for Sampling Distributions Calculator
- Population mean (μ)
- Population standard deviation (σ)
- Sample size (n)
- What probability do you want?
- X₁ (bound for the sample mean)
- X₂ (upper bound)
Everything is calculated in your browser, so results appear instantly and nothing you type is sent to a server.
How it works
The standard error σ/√n is computed first; each bound is converted to z = (X − μ)/(σ/√n) and the normal CDF gives the probability. Upper tails use 1 − Φ(z), ranges a difference of Φ values.
Worked example
Heights of adult women have μ = 161.3 cm and σ = 7.1 cm. For a random sample of 7 women the standard error is 7.1/√7 = 2.684 cm. The probability that their average height is below 160 cm is P(z < (160 − 161.3)/2.684) = Φ(−0.484) = 31.40%. For a single woman it would be Φ(−0.183) = 42.74%, and for a sample of 25 (standard error 1.42 cm) it drops to Φ(−0.915) = 18.00%.
When to use it
Use this calculator for questions about averages: quality control of mean fill weights, the average score of a class, the mean waiting time over many customers. For a single observation use the normal distribution calculator; for proportions (percent of a sample) use the sampling distribution of the sample proportion calculator.
Formula
σx̄ = σ ÷ √n · z = (X − μ) ÷ (σ/√n) · P(x̄ < X) = Φ(z) · P(X₁ < x̄ < X₂) = Φ(z₂) − Φ(z₁)
Worked example
Inputs — Population mean (μ): 161.3; Population standard deviation (σ): 7.1; Sample size (n): 7; What probability do you want?: P(x̄ < X₁) – below; X₁ (bound for the sample mean): 160.
Results — Probability = 31.4039 %; Result = P(x̄ < 160) = 0.314039 (31.4039%); Standard error of the mean σ/√n = 2.6835; z-score of X₁ = -0.4844; Complement (1 − probability) = 68.5961 %; Same probability for one observation (n = 1) = 42.736 %.
Key terms
- Sample mean (x̄)
- The average of the n values in a sample.
- Standard error
- σ/√n, the standard deviation of the sample mean.
- Sampling distribution
- The distribution of x̄ over all possible samples of size n.
Frequently asked questions
What is the mean of the sampling distribution of x̄?
It equals the population mean μ – the sample mean is an unbiased estimator.
What is the standard deviation of x̄?
σ/√n, the standard error. With σ = 7.1 and n = 7 it is 2.684.
What is the chance that a sample mean exceeds μ?
Exactly 50% when x̄ is normal, because the distribution is symmetric about μ.
When can I use this if the population is not normal?
When n is large (n ≥ 30 is the usual guide); the central limit theorem then makes x̄ approximately normal.
What if I do not know σ?
With a sample standard deviation s and a small sample, use the t distribution with n − 1 degrees of freedom; for large n, using s here is a fair approximation.
Why is the probability for one person different?
A single value has standard deviation σ, the mean of n values only σ/√n, so extreme averages are rarer than extreme individuals.