Library / Probability distributions / PRD-43
Probability distributions
Sampling without replacement: expected count
Expected number of items of the given type in a sample of n from a lot of N containing K: the sample size times the fraction of the lot that is of that type.
- Ref
- PRD-43
- Rev
- v1
- Field
- Probability distributions
- Result
- E[X]
The expected count is nK/N: the sample size times the share of the lot that is of the given type. It is the same as for sampling with replacement; only the variance changes, through the finite-population factor.
Expected number of items of the given type in a sample of n from a lot of N containing K: the sample size times the fraction of the lot that is of that type.
Calculator
- Share of the lot that is of the type, K/N0.16
- Sampling fraction, n/N0.1
- Sample size needed to expect one, N/K6.25
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- A sample of 5 from a lot where 16 % are of the type should contain about 0.8 of them.
- To expect one such item on average the sample needs about 6.25 items.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing () | Local step | Change in result |
|---|---|---|---|---|---|---|
| n Sample size n | 1 to 20 | 0.16 | 3.2 | 3.04 | +1 | +20 % |
| K Items of the type in the lot K | 0 to 20 | 0 | 2 | 2 | +1 | +12.5 % |
| N Lot size N | 20 to 200 | 0.2 | 2 | 1.8 | +1 | −1.96 % |
Sweep of Sample size n
| Sample size n | Expected count |
|---|---|
| 1 | 0.16 |
| 3.375 | 0.54 |
| 5.75 | 0.92 |
| 8.125 | 1.3 |
| 10.5 | 1.68 |
| 12.88 | 2.06 |
| 15.25 | 2.44 |
| 17.63 | 2.82 |
| 20 | 3.2 |
Sweep one variable
Variables and how each one contributes
Expected count
What the equation gives you. Everything else feeds it.
Lot size N
The lot size (a count): the total number of items. A larger lot behaves more like sampling with replacement. Raising it by 10 (to 60) moves the result by −16.7 %, from 0.8 to 0.6667.
Items of the type in the lot K
The number of items of the specified type in the lot (a count), such as defectives. More items of the type raise the chance of finding them. Raising it by 1 (to 9) moves the result by +12.5 %, from 0.8 to 0.9.
Sample size n
The sample size (a count): the number of distinct items drawn. A larger sample finds more of the type and reduces the variance. Raising it by 1 (to 6) moves the result by +20 %, from 0.8 to 0.96.
Worked example with the values above
- Share of the lot: K/N = 8 / 50 = 0.16.
- Multiply by the sample size 5.
- E[X] = 0.8.
Assumptions and limits
- Items are drawn at random without replacement from a finite lot, and every item is equally likely to be chosen.
- N, K, n and x are whole numbers with 0 ≤ K, n ≤ N.
Common mistakes
- Using the binomial formula for a large share of a small lot: the finite-population correction matters when n/N is above about 10 %.
- Sampling convenient items instead of choosing at random.
- Entering an x outside the possible range: the probability is then exactly zero.
Statistics and uncertainty
A mean, variance or percentile is calculated from fitted parameters, and the uncertainty in those parameters passes straight through. A distribution fitted to a small sample can have parameters that are far from the truth, especially in the tails and for the variance, so quote the result together with the sample size it rests on.
Where it is used
- Acceptance sampling and inspection of finite lots.
- Quality audits, card-style problems and capture–recapture estimates.
Maths you need
- Combinations: ways of choosing items, C(n, k)
- Sampling without replacement: draws that remove the item chosen