## PROBBNML Function

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 . 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 . In terms of
acceptance sampling, *n* is the number of items in the
sample. |

*m* | is the number of successes, where . 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

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 Language Reference: Dictionary*.

Copyright © 1999 by SAS Institute Inc., Cary, NC, USA. All rights reserved.