The number of flaws in a given area of aluminum foil follows a Poisson distribution with a mean of 3 per nr. Let X represent the number of flaws in a 1 m2sample of foil.

a. P(X = 5)

b. P(X = 0)

c. P(X<2)

d. P(X> 1)

e. μX

f. σX

Step 1 of 7</p>

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

Step 2 of 7</p>

a) P(X=5)=

=0.101

Hence P(X=5)=0.101

Step 3 of 7</p>

b)P(X=0)=

=0.0498

Hence P(X=5)=0.0498

Step 4 of 7</p>

c) P(X<2)=P(0)+P(1)

P(1) =

=0.1494

Hence P(X<2)=0.0498+0.1494

=0.1992