Question

A manufacturer who produces medicines bottles finds that 0.1% of the bottles are defective. The bottles are packed in boxes containing 500 bottles. A drug manufacturer buys 100 boxes from producer of bottles. Using possion distribution, how many bottles will contain

(1) no defectives (2) at least two defectives

Answer 1

We have:

$N=100, n=500$

Probability of a defective bottle:

$p=0.1\%=0.001$

$\lambda=np=500\cdot0.001=0.5$

If $X$ denotes the number of defective bottles in a pack of 500, then by Poisson distribution:

$P(X=x)=e^{-0.5}\frac{0.5^x}{x!}=\frac{0.6065\cdot0.5^x}{x!}$ ; $x=0,1,2,...$

Hence in a lot of 100 boxes the frequency of boxes with $x$ defective bottles is given by

$f(x)=NP(X=x)=\frac{100\cdot0.6065\cdot0.5^x}{x!}$

1) Number of bottles with no defectives:

$=100P(X=0)=100\cdot0.6065=61$

2) Number of bottles with at least two defectives:

$=100P(X\geq2)=100(1-P(X=0)-P(X=1))=$

$=100(1-0.6065-0.6065\cdot0.5)=9$