Question

A company manufactures bulbs. The probability of getting a defective bulb is 0.045. A sample of 100 bulbs were selected. Use Poisson approximation to binomial distribution to find the probability of finding at most 3 non-defective bulbs

Answer 1

The Poisson distribution is

$P\left(X=x\right)=\frac{e^{-\lambda }\lambda ^x}{x!}$

Let $x$ denote the non - defective fuse

$P=1-0.045=0.955$

$n=100$

$mean=\lambda =np=100\times0.955=95.5$

$P\left(atmost\:3\:non-defective\:fuses\right)= \sum_{x=0}^3 \frac{e^{-95.5}95.5^x}{x!}$

$P\left(atmost\:3\:non-defective\:fuses\right)=$ $0.01528$