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Originally Posted by TROLL
this throws me a little bit: "overall length tolerance of 10cm"
are they saying that the length of the shaft is 10cm and it varies by .2cm? or do they mean otherwise? assuming the above statement is true, here is what i came up with:
1) total range: 10.2 - 9.8 = .4
range in question: 10.03 - 9.97 = .06
answer: .06 / .4 = .15
2) total range: 10.2 - 9.8 = .4
Q1 = 9.8 + (.25*.4) = 9.9
Q3 = 9.8 + (.75*.4) = 10.1
3) total range: 10.2 - 9.8 - .4
upper 60th percentile: 9.8 + (.4*.6) = 10.04
(or what about the lower 60th percentile? that would be 10.2 - (.4*.6) = 9.96)
or i could be totally wrong, although i used to be good at this tuff, its been a real long time since i've done statistics.
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Thanks, I think you got it. I followed what you did and found it in the next chapter of the book (we didn't cover it yet) for estimating to find the probibability so I had a place to start. I used your numbers to track down the exact nunbers using the z chart in the book (the book has it but doesn't show us how to use it). Using that I figured out the standard deviation is for both sides not split between the 2 (how I got my answer for part b) After I got those 2 answers I was able to make sense of the books explination of how to find part c and got he same answer.
Thanks again
