# Documentation

### This is machine translation

Translated by
Mouseover text to see original. Click the button below to return to the English verison of the page.

To view all translated materals including this page, select Japan from the country navigator on the bottom of this page.

# poisscdf

Poisson cumulative distribution function

## Syntax

```p = poisscdf(x,lambda) p = poisscdf(x,lambda,'upper') ```

## Description

`p = poisscdf(x,lambda)` returns the Poisson cdf at each value in `x` using the corresponding mean parameters in `lambda`. `x` and `lambda` can be vectors, matrices, or multidimensional arrays that have the same size. A scalar input is expanded to a constant array with the same dimensions as the other input. The parameters in `lambda` must be positive.

`p = poisscdf(x,lambda,'upper')` returns the complement of the Poisson cdf at each value in `x`, using an algorithm that more accurately computes the extreme upper tail probabilities.

The Poisson cdf is

`$p=F\left(x|\lambda \right)={e}^{-\lambda }\sum _{i=0}^{floor\left(x\right)}\frac{{\lambda }^{i}}{i!}$`

## Examples

collapse all

For example, consider a Quality Assurance department that performs random tests of individual hard disks. Their policy is to shut down the manufacturing process if an inspector finds more than four bad sectors on a disk. What is the probability of shutting down the process if the mean number of bad sectors () is two?

```probability = 1-poisscdf(4,2) ```
```probability = 0.0527 ```

About 5% of the time, a normally functioning manufacturing process produces more than four flaws on a hard disk.

Suppose the average number of flaws () increases to four. What is the probability of finding fewer than five flaws on a hard drive?

```probability = poisscdf(4,4) ```
```probability = 0.6288 ```

This means that this faulty manufacturing process continues to operate after this first inspection almost 63% of the time.