Lean Manufacturing class Book: Manufacturing Systems Modeling and Analysis PLEAS

Question

Lean Manufacturing class

Book: Manufacturing Systems Modeling and Analysis

PLEASE TRY TO DO IT STEP BY STEP so I can understand the procedure.

1.15. A manufacturing company produces parts, 97% of which are within specifi- cations and 3% are defective (outside specifications). There is apparently no pattern to the production of defective parts; thus, we assume that whether or not a part is defective is independent of other parts. (a) What is the probability that there will be no defective parts in a box of 5? (b) What is the probability that there will be exactly 2 defective parts in a box of 5? (c) What is the probability that there will be 2 or more defective parts in a box of 5? (d) Use the Poisson distribution to approximate the probability that there will be 4 or more defective parts in a box of 40. (e) Use the normal distribution to approximate the probability that there will be 20 or more defective parts in a box of 400.

Explanation / Answer

Ans:

Use Binomial distribution:

p=0.03,n=5

Let x be the number of defectives.

P(x=r)=5Cr*0.03r*0.975-r

a)P(x=0)=5C0*0.030*0.975=0.8587

b)P(x=2)=5C2*0.032*0.973=0.0082

c)P(x>=2)=1-P(x=0)-P(x=1)=1-0.8587-5C1*0.03*0.974=1-0.8587-0.1328=0.0085

d)Normal approximation:

mean=np=40*0.03=1.2

std. dev=sqrt(40*0.03*0.97)=1.079

z=(4-1.2)/1.079=2.595

P(z>=2.595)=1-P(z<=2.595)=1-0.9953=0.0047

e)mean=400*0.03=12

std dev=sqrt(400*0.03*0.97)=3.412

z=(20-12)/3.412=2.345

P(z>=2.345)=1-P(z<=2.345)=1-0.9905=0.0095

x p(x) 0 0.8587 1 0.1328 2 0.0082 3 0.0003 4 0.0000 5 0.0000

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