Question

A manufacturer who produces medicine 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 the producer of bottles. Using Poisson distribution, the number of boxes with no defective bottle is

• A. $100\times e^{-0.1}$
• B. $100\times e^{-0.5}$
• C. $100\times e^{-0.05}$
• D. $100\times e^{-0.01}$

Answer 1

Answer: (B)

$100\times e^{-0.5}$

$p=\frac{0.1}{100}=0.001,n=500\lambda =np=500\times 0.001=0.5N=100$
We know that $p\left(r\right)=e^{-1}\frac{{\lambda }^r}{r!}$
Number of boxes containing no defective bottle
$=N.P.(r=0)=1000\times e^{0.5}\frac{{\left(0.5\right)}^0}{0!}=1000\times e^{-0.5}$