In a certain manufacturing process, 5% of the tools produced turn out to bedefective. Find the probability that in a sample of 40 tools, at most 2 will bedefective: [Given: e-2 = 0.135](a) 0.555(b) 0.932(c) 0.785(d) 0.675
Question
In a certain manufacturing process, 5% of the tools produced turn out to bedefective. Find the probability that in a sample of 40 tools, at most 2 will bedefective: Given: e-2 = 0.135 0.555(b) 0.932(c) 0.785(d) 0.675
Solution
This is a problem of binomial distribution. However, since the sample size is large (n=40), we can use the Poisson approximation to the binomial distribution.
The Poisson distribution is defined by the parameter λ, which is the mean number of successes that result from the experiment. In this case, λ is the expected number of defective tools, which is np = 40 * 0.05 = 2.
The probability of having x defective tools is given by the formula:
P(X = x) = λ^x * e^-λ / x!
We want to find the probability that at most 2 tools are defective, so we need to find P(X = 0), P(X = 1), and P(X = 2), and then add these probabilities together.
P(X = 0) = 2^0 * e^-2 / 0! = 1 * 0.135 / 1 = 0.135
P(X = 1) = 2^1 * e^-2 / 1! = 2 * 0.135 / 1 = 0.27
P(X = 2) = 2^2 * e^-2 / 2! = 4 * 0.135 / 2 = 0.27
So, the probability that at most 2 tools are defective is 0.135 + 0.27 + 0.27 = 0.675.
Therefore, the answer is (d) 0.675.
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