Is that all parts in Q1?
This isn't right. It's always good to do a common sense check: is it really likely that 8.53% of components are going to fail in 600 hours, when the mean is considerably less than 600? I think you're looking at the wrong side of the distribution; you need the percentage of cases that fall under a z score of 1.37, not the percentage above it.
Looks right. You can check these quite easily yourself using this tool.b) percentage that fail between 471 and 525 hrs (My answer: 33.72%)
I think this is right.c) lifetime that components can exceed with 95% confidence (My answer 613.64 hrs)
Is the answer 91.47%?This isn't right. It's always good to do a common sense check: is it really likely that 8.53% of components are going to fail in 600 hours, when the mean is considerably less than 600? I think you're looking at the wrong side of the distribution; you need the percentage of cases that fall under a z score of 1.37, not the percentage above it.
Ok, thanks. Please can you check the final question.
It's definitely wrong. Are you sure you used the right numbers for the mean and sd?
here is my working for Q3
Well. Um. That does shed some light on the problem. When you're deciding whether or not you're looking at a one-tail or a two-tail problem it doesn't change what the mean and standard deviation are. The stuff you have labeled '2way' is just completely wrong. Completely forget anything you were thinking when you decided to go that route. Do the problem using the 1way numbers.
I have kept the mean and SD the same, however I get a z value of 5.6. This value is not on my tables so what should I do?
I still don't understand how to do this last question
I think your first attempt is very close.
Let be the travelling time to and from work respectively.
Then
Moreover we assume are independent.
Then the total working time is also
normally distributed.
To determine the distribution of , you just need to compute
Note that since they are independent,
and thus
Now you want to evaluate the probability
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