1. ## Normal Distribution

A warehouse makes nuts and bolts on two independent machines. The internal diameter of the nuts is normally distributed with mean 0.4cm and the external diameter of the bolts is normally distributed with mean 0.42cm. They both have same variance which is determined by rate of production which corresponds to a standard deviation 0.01cm. A third machine fits the parts together provided the diameter of the nut is strictly greater than that of the bolt, other wise it rejects both.
(a) Find the probability a pair is rejected.
(b) If successive pairs of nut and bolt are produced independently, find the probability that in 20 pairs at least 1 is rejected.
(c) What is the largest value of standard deviation, that would achieve a reduction in the probability that a pair would be rejected to 0.01.

Was struggling with this topic but after working through examples have gotten much better, however not sure how to do this at all. Any help would be greatly appreciated.

2. ## Re: Normal Distribution

Originally Posted by Bloodzeed
A warehouse makes nuts and bolts on two independent machines. The internal diameter of the nuts is normally distributed with mean 0.4cm and the external diameter of the bolts is normally distributed with mean 0.42cm. They both have same variance which is determined by rate of production which corresponds to a standard deviation 0.01cm. A third machine fits the parts together provided the diameter of the nut is strictly greater than that of the bolt, other wise it rejects both.
(a) Find the probability a pair is rejected.
(b) If successive pairs of nut and bolt are produced independently, find the probability that in 20 pairs at least 1 is rejected.
(c) What is the largest value of standard deviation, that would achieve a reduction in the probability that a pair would be rejected to 0.01.

Was struggling with this topic but after working through examples have gotten much better, however not sure how to do this at all. Any help would be greatly appreciated.
The difference in diameters $(B-N)$ is normally distributed with a mean of $0.02$cm and a standard deviation of $\sqrt{2}\times 0.01$

For (a) you are asked to find the probability $\text{Pr}((B-N) \le 0)$

CB

3. ## Re: Normal Distribution

Originally Posted by CaptainBlack
The difference in diameters $(B-N)$ is normally distributed with a mean of $0.02$cm and a standard deviation of $\sqrt{2}\times 0.01$

For (a) you are asked to find the probability $\text{Pr}((B-N) \le 0)$

CB
Thank you that has helped me understand a lot. For (b) i was thinking if i work out the probability in (a) and take it away from 1 to work out the probability of success. If I then did that probability to the power 20 and took that answer away from 1 would that give me the probability that at least one is rejected? As I was thinking I might have to use a binomial distribution with successes and failures but if that other way works it would be much faster?

4. ## Re: Normal Distribution

Also worked through question (a) and got 0.0793 but not massively confident in my working (as rusty with statistical tables) so if anyone could confirm or suggest where i may have gone wrong that would be extremely helpful?

5. ## Re: Normal Distribution

Originally Posted by bloodzeed
also worked through question (a) and got 0.0793 but not massively confident in my working (as rusty with statistical tables) so if anyone could confirm or suggest where i may have gone wrong that would be extremely helpful?
~=0.0786

cb

6. ## Re: Normal Distribution

Originally Posted by Bloodzeed
Thank you that has helped me understand a lot. For (b) i was thinking if i work out the probability in (a) and take it away from 1 to work out the probability of success. If I then did that probability to the power 20 and took that answer away from 1 would that give me the probability that at least one is rejected? As I was thinking I might have to use a binomial distribution with successes and failures but if that other way works it would be much faster?
Prob that at least one is rejected is 1 - prob that none are rejected = 1-(1-0.0786)^20, which is what the binomial distribution would give.

CB