1. In a process that manufactures bearings, 90% of the bearings meet a thickness

Question

1. In a process that manufactures bearings, 90% of the bearings meet a thickness specification A shipment contains 500 bearings. A shipment is acceptable if at least 440 of the 500 bearings meet the specification. Assume that each shipment contains a random sample of bearings. Use the normal approximation to the binomial to solve the following questions: c. What is the probability that a given shipment has less than 450 bearings that eet the specification? (5 points) a. What is the probability that a given shipment is acceptable (5 points b. What is the probability that more than 285 out of 300 shipments are acceptable? (5 points) d. What proportion of bearings must meet the specification in order that 99% of the shipments are acceptable? (5 points)

Explanation / Answer

Ans:

Normal approximation to the binomial:

n=500,p=0.9

mean=np=500*0.9=450

standard dev=sqrt(500*0.9*0.1)=6.71

a)

z=(450-450)/6.71=0

P(z<0)=0.5

b)P(acceptable)=P(x>=440)

z=(440-450)/6.71=-10/6.71=-1.49

P(z>=-1.49)=P(z<=1.49)=0.9319

P(acceptale)=0.9319

c)Now,use binomial distribution with n=300,p=0.9319

mean=300*0.9319=279.57

std dev=sqrt(300*0.9319*(1-0.9319))=4.36

z=(285-279.57)/4.36=1.25

P(z>1.25)=1-P(z<=1.25)=1-0.8944=0.1056

d)

P(acceptable)=P(x>=440)

P(x>=440)=1-BINOMDIST(439,500,0.9,TRUE)=0.9382

now use p=0.91

P(x>=440)=1-BINOMDIST(439,500,0.91,TRUE)=0.9902

using normal approximation:

mean=500*0.91=455

std. dev=SQRT(0.91*0.09*500)=6.4

z=(440-455)/6.4=-2.34

P(z>=-2.34)=P(z<=2.34)=0.9904

Hence,p=0.91 for 99% shipments to be acceptable.

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