2 (20 points) Items produced on an assembly line meet the dimensional tolerances

Question

2 (20 points) Items produced on an assembly line meet the dimensional tolerances 95% of the time and each trial is independent. 2.1 If eight items are sampled, what is the probability that all meet tolerances? What is the probability that at least seven meet the tolerances? 2.2 Sampling is performed until an item does not meet tolerance. What is the mean number of samples needed until an item does not meet tolerance? What is the standard deviation of the number of trials needed until the first such item is detected?

Explanation / Answer

Ans:

2.1)Use binomial distribution with n=8 and p=0.95

P(all meet tolerances)=P(x=8)=0.958=

P(atleast 7 meet tolerances)=P(x=7)+P(x=8)

=8C7*0.957*0.051+0.958

=0.2793+0.6634=0.9428

2.2)We can use Geometric distribution for number of trials until we meet success

So,here,p=1-0.95=0.05

*(as here success means that we find a item that does not tolerance)

Mean=1/p=1/0.05=20

Variance=(1-p)/p2=(1-0.05)/0.052=380

Standard deviation=sqrt(380)=19.5

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