This is a binomial probability problem, since we are interested in the number of successes (contractors who use 0.5 or less xy wire) out of a fixed number of trials (120 contractors).
The probability of success for each trial is 0.6 (since 60% of contractors use xy wire). The number of trials is 120.
We can use the binomial probability formula:
P(X ≤ 60) = Σ(k=0 to 60) (120 choose k) * 0.6^k * 0.4^(120-k)
where P(X ≤ 60) is the probability of 60 or fewer successes (i.e. contractors who use 0.5 or less xy wire), and (120 choose k) is the binomial coefficient that gives the number of ways to choose k successes out of 120 trials.
We can use software or a calculator to evaluate this sum:
P(X ≤ 60) = 0.012
Therefore, the probability of a random sample of 120 contractors using 0.5 or less xy wire is 0.012 or about 1.2%.
Suppose that 60% of contractors in Arbaminch use xy wire. What is the probability of rondum sample of 120 contractors use 0.5 or less xy wire?
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