A sample of 8 parts is drawn from a lot of 6.000 parts. It is known that 20.0% o
ID: 3130343 • Letter: A
Question
A sample of 8 parts is drawn from a lot of 6.000 parts. It is known that 20.0% of the parts carry at least one defect. Let X be the random variable representing the number of parts in the sample that carry at least one defect What is the sample space of X? Because this is a finite population that contains items of two types (successes and failures), what is the notation of the distribution that we use to model X? (In other words, you should write X~Bin(n,p), but replace n and p with their values for this problem.) What is the probability that exactly three parts are defective? Use the formula and verify using the binomial table in your book. What is the probability that less than two parts are defective? Use the formula and verify using the binomial table in your book. What is the probability that more than 6 parts are defective? Use the formula and verify using the binomial table in your book. What problems are there in analyzing a sample of 350 parts?Explanation / Answer
a)
Note that the number of parts with defect is at least 0 and at most 8 [as we sampled 8 parts]
S = {0,1,2,3,4,5,6,7,8} [ANSWER]
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b)
Here, n = 8, p = 0.20, so the distribution is
X~Bin(8, 0.20). [ANSWER]
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c)
Note that the probability of x successes out of n trials is
P(n, x) = nCx p^x (1 - p)^(n - x)
where
n = number of trials = 8
p = the probability of a success = 0.2
x = the number of successes = 3
Thus, the probability is
P ( 3 ) = 0.14680064 [ANSWER]
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d)
Note that P(fewer than x) = P(at most x - 1).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 8
p = the probability of a success = 0.2
x = our critical value of successes = 2
Then the cumulative probability of P(at most x - 1) from a table/technology is
P(at most 1 ) = 0.50331648
Which is also
P(fewer than 2 ) = 0.50331648 [ANSWER]
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e)
Note that P(more than x) = 1 - P(at most x).
Using a cumulative binomial distribution table or technology, matching
n = number of trials = 8
p = the probability of a success = 0.2
x = our critical value of successes = 6
Then the cumulative probability of P(at most x) from a table/technology is
P(at most 6 ) = 0.99991552
Thus, the probability of at least 7 successes is
P(more than 6 ) = 8.448*10^-5 [ANSWER]
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f)
If there are 350 parts, then this is a significant part of the lot of 6000 already. Hence, we cannot approximate the distribution to be binomial anymore. It becomes a hypergeometric distribution already.