A robot is used to prepare case of peanut butter for shipment. As the 12 cases a

Question

A robot is used to prepare case of peanut butter for shipment. As the 12 cases are being loaded two of the cases are dropped. Before the operator can isolate the cases, they are mixed in with the other cases. (This is the end of production run, so there are no replacement cases.) What is the probability that the operator will find the two broken cases in the first four randomly chosen cases (from 12 cases)? Pay attention: there is no replacement. What's the other alternative distribution to solve this problem? Is this distribution useful Why no or why yes?

Explanation / Answer

Ans:

Binomial distribution:

Probability that a case is broken=2/12=0.167

Probability that a case is not broken=1-0.167=0.833

n=4,p=0.167

Probability that the operator will find the two broken cases in the first four randomly chosen cases=P(2)

=4C2 (0.167)2*(0.833)2

=0.1161

We can use Normal approximation to the binomial distribution:

Mean=np=4*0.167=0.668<=5

For Normal approximation,np>=5 and np(1-p)>=5

So,it is not valid in this case.

F

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.