Suppose that 60% of conductors in Arbaminch use xy wire what is probability of a rundom sample of 120 contractors use 0.5 or less xy wire?
This is a binomial probability problem with n = 120 trials and p = 0.60 probability of success (using xy wire).
The formula for the probability of x success in n trials with probability of success p is:
P(x) = (nCx) * p^x * (1-p)^(n-x)
where nCx is the number of ways to choose x objects from n.
To find the probability of 0.5 or less contractors using xy wire, we need to add up the probabilities of x = 0, 1, 2, ..., 60 successes:
P(0.5 or less) = P(0) + P(1) + P(2) + ... + P(60)
We can use a binomial distribution table or calculator to find each individual probability, or we can use a normal approximation since n is large and p is not too close to 0 or 1.
Using the normal approximation, we can convert the binomial distribution to a normal distribution with mean np = 120*0.6 = 72 and standard deviation sqrt(np(1-p)) = sqrt(120*0.6*0.4) = 5.48. Then we can use the normal distribution to find the probability of x being less than or equal to 60.5 (since we want the probability of 0.5 or less, which is equivalent to 60.5 or fewer). We have:
Z = (60.5 - 72) / 5.48 = -2.15
Looking up the area to the left of Z = -2.15 in a standard normal distribution table or calculator, we find:
P(Z < -2.15) = 0.0158
This is the probability of observing 60.5 or fewer contractors using xy wire in a random sample of 120. Therefore, the probability of 0.5 or less contractors using xy wire is approximately 0.0158 (or 1.58%).