Bolts are produced in large numbers and it is expected that there is a 4% rejection rate due to some form of defect. A batch of 40 bolts are randomly selected for inspection.
Ten such batches are randomly selected. If it is found that at least 2 bathes have at least 4 defective, the total output is considered for the scrap heap to be recycled. Find the probability that the total output is sent to the scrap heap.
Working steps please,
answer is 0.1835
Ten such batches are randomly selected. If it is found that at least 2 bathes have at least 4 defective, the total output is considered for the scrap heap to be recycled. Find the probability that the total output is sent to the scrap heap.
Working steps please,
answer is 0.1835
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Strictly speaking this should all be done by the Binomial distribution. Remember the conditions for a binomial distribution are
a) fixed number (n) of trials each with only two outcomes
b) constant probability of success (p)
c) independence of results
These all apply to both the testing of an individual batch and the batches taken as a group. However, the Poisson distribution is used as an approximation to the Binomial if n is large, p is small and you are only interested in a small number of successes. If you do use the Poisson distribution for the first part then the calculation goes like this.
For an individual batch mean = m = 40*0.04 = 1.6
P(x >= 4) = 1 - P(x = 0 or 1 or 2 or 3)
= 1 - (1 + 1.6 + (1/2)*1.6^2 + (1/6)*1.6^3)*e^(-1.6) = 0.07881 (5 dec. pl.)
[ N.B. Binomial answer to this part would be 0.07484 (5 dec. pl.) ]
For the batches as a group I think you must use the Binomial distribution because n is not large enough to justify the use of Poisson.
It will have p = 0.07881, q = 0.92119, n = 10
P(x >= 2) = 1 - P(x = 0 or 1)
= 1 - 0.92119^10 - 10*(0.92119^9)*0.07881 = 0.1835
This is not that close to the answer when the Binomial distribution is used for both parts 0.1690 (4 dec. pl.) because in the first part the conditions for using Poisson were dubious.
a) fixed number (n) of trials each with only two outcomes
b) constant probability of success (p)
c) independence of results
These all apply to both the testing of an individual batch and the batches taken as a group. However, the Poisson distribution is used as an approximation to the Binomial if n is large, p is small and you are only interested in a small number of successes. If you do use the Poisson distribution for the first part then the calculation goes like this.
For an individual batch mean = m = 40*0.04 = 1.6
P(x >= 4) = 1 - P(x = 0 or 1 or 2 or 3)
= 1 - (1 + 1.6 + (1/2)*1.6^2 + (1/6)*1.6^3)*e^(-1.6) = 0.07881 (5 dec. pl.)
[ N.B. Binomial answer to this part would be 0.07484 (5 dec. pl.) ]
For the batches as a group I think you must use the Binomial distribution because n is not large enough to justify the use of Poisson.
It will have p = 0.07881, q = 0.92119, n = 10
P(x >= 2) = 1 - P(x = 0 or 1)
= 1 - 0.92119^10 - 10*(0.92119^9)*0.07881 = 0.1835
This is not that close to the answer when the Binomial distribution is used for both parts 0.1690 (4 dec. pl.) because in the first part the conditions for using Poisson were dubious.