Binomial distribution Help?

Question

1) Major Electronics sells transistor to the US govt. in lots of 1000. The govt. takes a random sample of 10 from each lot, puts them through a rigorous test, and accepts the lot if no more than 3 out of the 10 break down. Major electronics feels certain that at least three-fourths of its transistors will pass the govt. tests. If they are correct, what is the probability that any given lot will be accepted by the govt?
2) In a certain country, 104 boys are born for every 100 girls. In that country what percentage of families with 8 children should have 4 girls and 4 boys?

Do you know how to do it? please help

Answer 1

1. Let's start from the end. If 3/4 of transistors are good that means probability of good transistor p=3/4 and probability of bad transistor is q=1/4. If you draw 10 transistors probability that 3 are bad and 7 are good = (10 over 3)*q^3*p^7. If gov finds 3 or less bad it will accept a lot. Probability that in a group of 10 3 or less are bad is: (10 over 3)*q^3*p^7 + (10 over 2)*q^2*p^8 + (10 over 1)*q^1*p^9 + (10 over 0)*q^0*p^10. For q=1/4 and p=3/4 this amounts to 0.775875. Therefore, the probability that a lot of 1000 transistors in which every 4-th is bad will be accepted is 0.775875 meaning that out of 100 lots 77.59 will pass.

2. Probability that boy is born p = 104/204 = 0.5098. Probability that girl is born = (1 - p) = q = 0.4902. Probability of having 4 boys and 4 girls is (8 over 4)*0.5098^4*0.4902^4 = 0.273.

Answer 2

a million.) n :- sort of trials p :- probability of fulfillment on a given trial q = a million-p :- probability of failure on a given trial bear in techniques, for a sufficiently super pattern length, a binomial distribution could be approximated by potential of a classic distribution. we are able to apply here rule: If np > 5 and nq > 5, then the pattern length is satisfactorily super, and we are able to approximate by potential of a classic distribution. So calculate np and nq in each and each case, and spot whether or no longer they're extra suitable than 5. 2.) First attempt whether np > 5 and nq > 5. If this holds, then a binomial distribution could be approximated by potential of a classic distribution: N(pq, npq) or N(pq, np(a million-p) ) the place np is the propose (?) and npq is the variance (?^2) of the traditional distribution. To calculate a z-score, use the formulation: z = (x - ?)/? Plug on your values for x (given interior the question), ? = pq, and ? = squareroot(npq) to get your answer. 3.) There are sixty 5 questions, so the variety of trials n = sixty 5 Mike has a a million in 5 probability of answering each and each question properly, so p = a million/5 = 0.2 x = 10 questions we would desire to discover the probability of answering below 10 questions properly: P(X<10) you will would desire to transform x = 10 right into a z-score using an identical formulation as before: z = (x - ?)/? the place ? = pq, and ? = squareroot(npq) in case you probably did it properly, you may get: z = -0.93026 Then using your z-tables, discover P(Z

Answer 3

1) The prob. that an individual transistor will fail is 0.25
The prob. that it won't is 0.75

pob of none failing out of ten is:
(0.75)^10 = 0.0563

prob of exactly one failing is:
(0.25) (0.75)^9 x 10C9= 0.1877

prob of exactly 2 failing is;
(0.25)^2 x(0.75)^8 x 10C8 = 0.28157

prob of exactly 3 failing is:
(0.25)^3 x(0.75)^7 x 10C7 = 0.25028

The sum of all these is 0.7759 and this is the probability of 3 or less failing. The probability of a given lot passing is 77.6%


2) prob(B) = 104 /204 = 26/51
Prob(G) = 25/51

if there are 8 children:

Probability a single event of 4 boys and 4 girls is:

(26/51)^4 x (25/51)^4 = 0.0039

The number of ways this can happen is 8C4 = 70

Therefore the total probability =

70 x 0.0039 = 0.273 = 27.3%

Answer 4

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