Question

Suppose that we model the occurrence of defects on a fabric manufacturing line as a Poisson process with rate 0.01 per square foot. Use the central limit theorem (both with and without the correction for continuity) to approximate the probability that one would find at least 15 defects in 2000 square feet of fabric.

Short answer

Approximate value for the probability that one would find at least 15 defects in 2000 square feet of fabric is 0.8951.

Step-by-step solution

Given information

Need to model the occurrence of defects on a fabric manufacturing line as a Poisson process with rate 0.01 per square foot.

Calculating approximate value of the probability that one would find at least 15 defects in 2000 square feet of fabric.

The number of defects in 2000 square feet follows has the Poisson distribution with mean is

\(2000 \times 0.01 = 20\)

As mean and variance of Poisson distribution is same.

The central limit theorem approximation is the normal distribution with mean 20 and variance 20.

Without correction for continuity, the approximate probability of at least 15 defects is:

\(\begin{align}1 - \phi \left( {\frac{{15 - 20}}{{\sqrt {20} }}} \right) &= 1 - \phi \left( {\frac{{ - 5}}{{\sqrt {20} }}} \right)\\ &= 1 - \phi \left( { - 1.1180} \right)\end{align}\)

\(1 - \phi \left( {\frac{{15 - 20}}{{\sqrt {20} }}} \right) = 0.8682\)

With the continuity correction, we get

\(\begin{align}1 - \phi \left( {\frac{{14.5 - 20}}{{\sqrt {20} }}} \right) &= 1 - \phi \left( {\frac{{ - 5.5}}{{\sqrt {20} }}} \right)\\ &= 1 - \phi \left( { - 102295} \right)\end{align}\)

\(1 - \phi \left( {\frac{{14.5 - 20}}{{\sqrt {20} }}} \right) = 0.8906\)

The actual Poisson probability is 0.8951.