1. ## factory metal

Any help offered in solving the below question will be much appreciated.
Thanks
Kingman

In a factory of metal plates, small dent occurs at random at an average of 1 dent in every 2 metal plates.
The number of dent detected in a metal plate is denoted by X and follows a Poisson distribution.
a) Show that, in a randomly chosen metal plate, the probability that there are at least 3 dents is 0.0144, correct to 3 Significant figures. (0.0144)
b) A box contains 15 such metal plates. Inspections are carried out at random to ensure quality. A box is rejected if it contains at least 2 metal plates with at least 3 dents each. Find the probability that a randomly chosen box is rejected. (0.0192)
c) 100 randomly chosen boxes are inspected. Find, by using a suitable approximation, the probability that more than 98 boxes are not rejected. (0.428) (Use Poisson Distribution to approximate Binomial Distribution)

2. You need only the fact that the Poisson Distribution is wonderfully scalable.

We have 1 dent every 2 plates.

This translates smoothly to

1/2 dent every 1 plate or

15/2 dents every 15 plates or

50 dents every 100 plates.

Of course, at some point, often around 30 or so, it may become burdensome to calculate what you want using the Poisson directly. That's where the Normal approximation comes in.

Let's see what you get.

3. ## Thanks

Thanks

4. Did you show any work on the first two?

5. ## here are the working for a and b

here are the working for a and b but part c need help
thanks

6. Poisson distribution - Wikipedia, the free encyclopedia

Check out that link of an explanation of how to use a poisson distribution to approximate a binomial distribution.

Your answer from part b was the probability that a box of metal plates would be rejected. (0.0192)

For part c, we're looking for the probability that more than 98 boxes are not rejected, which is the same as the probability of less than 2 boxes being rejected.

So we have p=0.0192 and n=100

For a poisson approximation of of a binomial distribution

X~poisson( $\lambda$=np=1.92)

From there just solve:

$pr(X<2)=\frac{(e^{-1.92})(1.92^{x})}{x!}$