In an effort to cut costs and improve profits, any US companies have been turning to outsourcing. In fact, according to Purchasing magazine, 54\% of companies surveyed outsourced some part of their manufacturing process in the past two to three years. Suppose 555 of these companies are contacted.
What is the probability percentage that 285 or more companies outsourced some part of their manufacturing process in the past two or three years? Round the percent to two decimal plac
Answer
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 \(\boxed{89.47\%}\).
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 \(\boxed{89.47\%}\).
