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