PROBBNML Function :: SAS/QC(R) 13.1 User's Guide

PROBBNML Function

Subsections:

Syntax
Description
Examples

computes the probability that an observation from a binomial $(n,p)$ distribution will be less than or equal to m.

Syntax

PROBBNML $(p,n,m)$

where

p	is the probability of success for the binomial distribution, where $0\leq p\leq 1$ . In terms of acceptance sampling, p is the probability of selecting a nonconforming item.
n	is the number of independent Bernoulli trials in the binomial distribution, where $n\geq 1$ . In terms of acceptance sampling, n is the number of items in the sample.
m	is the number of successes, where $0\leq m\leq n$ . In terms of acceptance sampling, m is the number of nonconforming items.

Description

The PROBBNML function returns the probability that an observation from a binomial distribution (with parameters n and p) is less than or equal to m. To compute the probability that an observation is equal to a given value m, compute the difference of two values for the cumulative binomial distribution.

In terms of acceptance sampling, the function returns the probability of finding m or fewer nonconforming items in a sample of n items, where the probability of a nonconforming item is p. To find the probability that the sample contains exactly m nonconforming items, compute the difference between PROBBNML $(p,n,m)$ and PROBBNML $(p,n,m-1)$ .

In addition to using the PROBBNML function to return the probability of acceptance, the function can be used in calculations for average sample number, average outgoing quality, and average total inspection in Type B single-sampling. See Evaluating Single-Sampling Plans for details.

The PROBBNML function computes

$\sum _{j=0}^{m}(\stackrel{n}{_ j})p^{j}(1-p)^{n-j}$

where m, n, and p are defined in the preceding list.

Examples

The following statements compute the probability that an observation from a binomial distribution with p = 0.05 and n = 10 is less than or equal to 4:

data;
   prob=probbnml(0.05,10,4);
   put prob;
run;

These statements result in the value 0.9999363102. In terms of acceptance sampling, for a sample of size 10 where the probability of a nonconforming item is 0.05, the probability of finding 4 or fewer nonconforming items is 0.9999363102.

The following statements compute the probability that an observation from a binomial distribution with p = 0.05 and n = 10 is exactly 4:

data;
   p=probbnml(0.05,10,4) - probbnml(0.05,10,3);
   put p;
run;

These statements result in the value 0.0009648081.

For additional information on probability functions, refer to SAS Functions and CALL Routines: Reference.