Asked by elle
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.
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.
Answers
Answered by
MathGuru
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.
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.
Answered by
MathGuru
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.
I hope this will also help.
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