Suppose that in a plant that manufactures integrated circuit chips, 15% of chips

Q: Suppose that in a plant that manufactures integrated circuit chips, 15% of chips

c) By Negative Binomial Distribution, then rth success in n trials is P(n, r) = C(n-1, r-1) p^r (1-p)^(n-r) As n = 19, r = 3, p = 0.15, then P(19, 3) = C(18,2) (0.15^3) (0.85^16) = 0.038341405 [ANSWER] ******************** d) E(n) = 1/p = 1/0.15 = 6.666666667 [ANSWER] ***************** e) E(x) = n p = 10*0.15 = 1.5 [ANSWER]

ID: 3127789 • Letter: S

Question

Suppose that in a plant that manufactures integrated circuit chips, 15% of chips are defective, and let us select chips randomly from the output of the plant. If 10 chips are selected, what is the probability that at most one chip is defective? If 8 chips are selected, what is the probability that exactly two chips work (i.e. are not defective)? What is the probability that the third defective chip will be the nineteenth chip selected? How many chips would you expect to select when encountering the first defective chip? How many defective chips would you expect to find in a selection of 10 chips?

Explanation / Answer

By Negative Binomial Distribution, then rth success in n trials is

P(n, r) = C(n-1, r-1) p^r (1-p)^(n-r)

As n = 19, r = 3, p = 0.15, then

P(19, 3) = C(18,2) (0.15^3) (0.85^16) = 0.038341405 [ANSWER]

********************

E(n) = 1/p = 1/0.15 = 6.666666667 [ANSWER]

*****************

E(x) = n p = 10*0.15 = 1.5 [ANSWER]

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Suppose that in a plant that manufactures integrated circuit chips, 15% of chips

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