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Binomial distribution answe check

Hi, I'm very new to statistics and would like some one to check over the first part of the question i have done as well as some help on the other two parts.

1) could some one please check part a and see if i have the correct answer .

2) For the second part do i have to work out the probability of 6 machines failing then seven and eight etc all the way up to 60 and and the results together, so i will have a page full of calculations?

3) i know i have to calculate Part b first but what exactly does the numbers mean in terms of the machines failing.

Any help would be greatly appreciated.

Re: Binomial distribution answe check

Your answer to 1) is correct but $\displaystyle 3.27 \times 10^{-3}$ is 0.327%

For 2) find the chance that 6 machines will fail, and the chance 5 will fail, same for 4, 3, 2, 1, 0 machines and add up all the chances. That is the chance that 6 or less will fail. Remember that [chance 6 or less fail] + [chance more than 6 will fail] = 1

Use this fact to find the chance that more than 6 will fail.

Discuss your findings they probably just want you to talk about [chance 6 or less fail] + [chance more than 6 will fail] = 1 and how it is faster and still valid.

Re: Binomial distribution answe check

Yes, you answer for (a) is correct.

Quote:

2) For the second part do i have to work out the probability of 6 machines failing then seven and eight etc all the way up to 60 and and the results together, so i will have a page full of calculations?

It will be a lot easier to find the probability that **less than** 6 machines will fail, then subtract from 1.

I don't see any "part b" in your attachment.

Re: Binomial distribution answe check

Quote:

Originally Posted by

**djstar** Hi, I'm very new to statistics and would like some one to check over the first part of the question i have done as well as some help on the other two parts.

1) could some one please check part a and see if i have the correct answer .

2) For the second part do i have to work out the probability of 6 machines failing then seven and eight etc all the way up to 60 and and the results together, so i will have a page full of calculations?

3) i know i have to calculate Part b first but what exactly does the numbers mean in terms of the machines failing.

It looks correct to me. You can cheek it (60!/((5!)(55!))(.22)^(5)(.78)^(55) - Wolfram|Alpha.

I do not see the rest of the question.

What is the whole statement?

Re: Binomial distribution answe check

Thank you all for your kind replies, it makes sense to find the probability that less that 6 machines will fail and take it away from 1 I'll work that out in the next day and get you guys to check it please. Just one thing is how is 3.27 \times 10^{-3} 0.327% ???

ps sorry about the part 'B' bit. part b was the question about 6 or more machines failing.

Re: Binomial distribution answe check

Quote:

Originally Posted by

**djstar** Just one thing is how is [tex]3.27 \times 10^{-3}/tex] 0.327% ???.

$\displaystyle 3.27 \times 10^{-3}= 0.00327$ this is in decimal form. To convert decimal to percentage multiply by 100.