A production process historically produces 95% items that are considered acceptable quality-wise and 5% that are defective.A sample of 50 items is taken. Determine the following probablities:
Probability that exactly 47 items will be good.
Try a binomial probability table or use a binomial probability function, which is this:
P(x) = (nCx)(p^x)[q^(n-x)]
x = 47
n = 50
p = .95
q = .05 (q is 1-p)
I'll let you take it from here.
To determine the probability that exactly 47 items will be good, we can use the binomial probability formula. The formula is:
P(X = k) = (nCk) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability of getting exactly k successes (in this case, exactly 47 good items)
n is the sample size (in this case, 50)
k is the number of successes (in this case, 47)
p is the probability of success (in this case, the probability of getting a good item, which is 0.95)
(1 - p) is the probability of failure (in this case, the probability of getting a defective item, which is 0.05)
nCk is the binomial coefficient (the number of ways to choose k items out of n)
Plugging in the values into the formula, we get:
P(X = 47) = (50C47) * (0.95)^47 * (1 - 0.95)^(50 - 47)
To calculate the binomial coefficient, we use the formula:
nCk = n! / (k! * (n-k)!)
Calculating the binomial coefficient, we get:
50C47 = 50! / (47! * (50-47)!)
= 50! / (47! * 3!)
Calculating the factorial values, we have:
50! = 50 * 49 * 48 * ... * 3 * 2 * 1
47! = 47 * 46 * 45 * ... * 3 * 2 * 1
3! = 3 * 2 * 1
Substituting these values into the binomial coefficient, we get:
50C47 = (50 * 49 * 48 * ... * 3 * 2 * 1) / ((47 * 46 * 45 * ... * 3 * 2 * 1) * (3 * 2 * 1))
Finally, plugging in the values into the probability formula, we can calculate the probability that exactly 47 items will be good.