Question

In a sompling the mean number of defective
about bolt manufactured by a machine in a
sample of 20
is 2. Determine the expected
number of sample of such 500 to contain
2 defective bolts
at least​

Answer 1

Answer:

Using binomial distribution

Probability of defective (P)=2/10=0.1

Probability of non defective (q)=1−0.1=0.9

n=20 and N=100

P(x=X)= nCx×px×qn−x

To expect 3 defective

P(x=3)= 20C3(0.1)3(0.1)17=0.19

∴ No. of packets containing 3 defective

=N×P(X=3)

=100×0.19=19...packets

2) By poisons distribution

we have m=2.,n=20,p=0.1P(X=x)=e−mmxx!

for 3 defective

P(X=3)=e−22x3!=0.135×86=0.18

∴ No. of packets containing 3 defective is =N×P(X=3)=100×0.18=18packets.