Question

1. A machine produces screws, 1% of which are defective. Find the probability that in a box of 200 screws there are no defectives. Use both binomial method and Poisson method.

a) From above say the machine produces 1 box every 10 minutes. What is the probability the machine can go 15 minutes without producing a defective screw.

b) From above, how many defects can we expect in a week? What is the standard deviation of this number? Using the normal approximation what is the bar for the number of defects per week to be considered 10% the most?

Answer 1

Ans:

n=200

p=0.01

Using binomial:

P(no defectives)=(1-0.01)^200=0.1340

Using Poisson:

lambda=200*0.01=2

P(no defective)=P(x=0)=e^-2*(2⁰/0!)=0.1353

a)lambda rate=1/10

P(t>15)=e^-15/10=e^-1.5=0.2231

b)1 box in 10 min

So,in a week,number of boxes=7*24*60/10=1008

Number of screws=200*1008=201600

Expected number of defects=201600*0.01=2016

standard deviation=sqrt(201600*0.01*0.99)=44.675

P(Z<=z)=0.1

z=normsinv(0.1)=-1.282

At most number of defects=2016-1.282*44.675=1958.75