The number of flaws in a given area of aluminum foil follows a Poisson distribution with a mean of 3 per \(\mathrm{m}^{2}\). Let \(X\) represent the number of flaws in a 1 \(\mathrm{m}^{2}\) sample of foil.

a. P(X = 5)

b. P(X = 0)

c. P(X < 2)

d. P(X > 1)

e. \(\mu_{X}\)

f. \(\sigma_{X}\)

Equation Transcription:

Text Transcription:

m^2

X

mu_X

sigma_X

Solution

Step 1 of 7

Let X represents the no.of flaws in a sample

Here Xpoisson()

The pmf of Poisson distribution is p(x)=,x=0,1,2,............

Given mean =3

The mean of poisson distribution is

So =3