hi, i'd also really like some help for another binomial distribution question, i have no idea where to start... the question is:
A manufacturer makes 10,000 ball point pens per day and estimates that 400 will be defective. She decides that if a random sample of 10 pens contain more than 1 defective pen, she will institute quality control meausures. Find the probability that these measures are introduced.
Here's one way to do this problem:
n = 10
p = 400/10000 = .04
q = 1 - p = 1 - .04 = .96
You will need to find P(2) through P(10). Add those values for your probability.
You can use a binomial probability table, or calculate by hand using the following formula: P(x) = (nCx)(p^x)[q^(n-x)]
I hope this will help.
You can also take 1 - [P(0) + P(1)], which is easier than finding P(2) through P(10). This way you will just need to find P(0) and P(1). Either way you can still use a binomial probability table or calculate by hand.
I hope this will also help.
To solve this question, we can use the binomial distribution formula. The binomial distribution is used to calculate the probability of a specific number of successes in a fixed number of trials, given a known probability of success for each trial.
In this case, the manufacturer wants to find the probability of selecting more than 1 defective pen in a sample of 10 pens. Let's break down the question into the components we need for the binomial distribution formula:
1. The number of trials (n): We are sampling 10 pens.
2. The probability of success (p): The manufacturer estimates that out of 10,000 pens, 400 will be defective. So the probability of selecting a defective pen is 400/10,000 = 0.04.
3. The number of successes (k): We want to find the probability of selecting more than 1 defective pen. This means we need to find the probability of selecting 2, 3, 4, 5, 6, 7, 8, 9, or 10 defective pens.
Now, let's calculate the probability using the binomial distribution formula:
P(X > 1) = 1 - P(X = 0) - P(X = 1)
P(X = k) = C(n,k) * p^k * (1-p)^(n-k)
where C(n,k) is the number of combinations of n objects taken k at a time.
P(X = 0) = C(10,0) * 0.04^0 * (1-0.04)^(10-0)
P(X = 1) = C(10,1) * 0.04^1 * (1-0.04)^(10-1)
Using a calculator or spreadsheet, you can calculate the above probabilities and substitute them into the formula to find P(X > 1).
P(X > 1) = 1 - P(X = 0) - P(X = 1)
I hope this helps you solve the problem! Let me know if you have any further questions.