Please include working out , thanks A manufacturing company has three lines pro-
ID: 3135408 • Letter: P
Question
Please include working out , thanks
A manufacturing company has three lines pro- ducing firps. Line 1 manufactures 45% of the flirps, line 2 manufactures 35% and line 3 manufactures 20%. However some flirps produced are faulty. 3% produced in line 1, 1% in line 2 and 0.05% in line 3 are defective. (a) Because of its lower rate of defectives, a special order for 10,000 flirps has been assigned to production line 3. Use a sensible approximation to obtain an estimate of the probability that no more than 4 firps in this batch will be defective. (b) Before being transported to retail outlets, the firps are sealed into packets of 12 and sent to a central storage depot. A customer received a packet produced in March, which contained 1 defective firp. What is the probability that this packet was manufactured on production line 1? (c) In one year, line 2 produces a total of 580,000 flirps. What is the probability that between 5700 and 5750 (inclusive) defective flirps are produced in this yearExplanation / Answer
answer a) n=10000,p=0.005 , mean=np=10000*0.0005=5
we use poisson distribution to calculte the required probability in this situation by using the MS-EXCEL COMMANd
=POISSON(4,5,TRUE)=0.4405
answer b) here P(line1)=0.45, P(faulty|line1)=0.03
we want P(line1|faulty)=P(line1)* P(faulty|line1)=0.45*0.03=0.0135
answer c) in this case we use poisson distribution
n=580000,p=0.01, mean=variance=n*p=5800
sd=sqrt(var)=sqrt(5800)=76.15
now we use standard normal approximation
for 5700, z-value z1=(5700-5800)/76.15=-1.3131
for 5750 ,z-value z2=(5750-5800)/76.15=-0.6565
P(-1.3131<z<-0.6565)=P(z<-0.6565)-P(z<-1.3131)=0.2557-0.0946=0.1611
required proabability =0.1611