Step 1 :We are given a problem involving a binomial distribution, where the probability of success (p) is 0.54, the number of trials (n) is 555, and we need to find the probability of 285 or more successes.
Step 2 :The binomial distribution formula is: \(P(X = k) = C(n, k) * (p^k) * ((1-p)^(n-k))\), where \(P(X = k)\) is the probability of k successes, \(C(n, k)\) is the number of combinations of n items taken k at a time, p is the probability of success, n is the number of trials, and k is the number of successes.
Step 3 :However, calculating this directly for k = 285 to 555 would be computationally intensive. Instead, we can use the normal approximation to the binomial distribution, which is valid when both np and n(1-p) are greater than 5.
Step 4 :The normal approximation to the binomial distribution is given by: \(Z = (X - np) / \sqrt{np(1-p)}\), where Z is the standard normal random variable, X is the number of successes, np is the mean of the binomial distribution, and \(\sqrt{np(1-p)}\) is the standard deviation of the binomial distribution.
Step 5 :We can calculate the Z-score for X = 285 and then find the area to the right of this Z-score under the standard normal curve, which represents the probability of 285 or more successes.
Step 6 :Given n = 555, p = 0.54, and X = 285, we find that the mean is approximately 299.70 and the standard deviation is approximately 11.74.
Step 7 :The Z-score is then approximately -1.25, and the corresponding probability is approximately 0.8947, or 89.47%.
Step 8 :Final Answer: The probability percentage that 285 or more companies outsourced some part of their manufacturing process in the past two or three years is \(\boxed{89.47\%}\).