The number of flaws in a given area of aluminum foil

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