Answer to Question #281777 in Statistics and Probability for Pankaj

Question

In a lot of Tv manufactured,each tv checked and found Avg.defect per tv =4,w hat is the possibilities of maximum defect in any tv?

Answer 1

using Poisson distribution, probability that there are no defects in tv:

$P(x=0)=e^{-\lambda}=e^{-4}=0.0183$

then probability that there are defects in tv:

$P(x\ge0)=1-0.0183=0.9817$

mean:

$\mu=\lambda=4$

standard deviation:

$\sigma=\sqrt{\lambda}=2$

for normal distribution:

for $P=0.0183$ :

$z=\frac{x-\mu}{\sigma}=\frac{x-4}{2}=2.09$

possibilities of maximum defect in any tv:

$x=2\cdot2.09+4=8.18\approx8$ defects