Help! Statistics problem

[mostasteless]

Thanks to not having a license I was forced to miss one of my stat classes last week, this week the teacher gave us 10 problems to do and have in tomorrow to be graded. 9 of the problems I had no trouble with but the last one has me stumped and the book is no help. If anyone here can help me out with the answer (and how you came up wit it if you have the time) I would be eternally grateful .

Question:
A manufactured gearbox input shaft has an overall length tolerance of 10cm with a standard deviation of 0.2cm. The overall lengths are normally distributed. The QA department periodically selects parts for measurement of the overall length. The following questions concern the overall length measurement.

a) What is the probability that a randomly selected shaft would have a length between 9.97 and 10.03cm? (I came up with .88)

b) What is the first and third quartiles for the length of the shafts? (no clue)

c) Shafts with lengths above the 60th percentile will be rejected. What length represents the 60th percentile? (came up with 10.05067)


Thanks
John

[03sLoMaX]

???? lol...i would help you if i could man...sadly i failed out of basic algebra in highschool and again in college....numbers or anything like that arent my thing i guess, lol

[mostasteless]

OK I have been playing with the numbers some more and here is what I came up with, still don't think I am correct but maybe I did figure it out
a) .56
b) Q1=9.95. Q3=10.05
c) 10.02

[Intercooled T]

shouldn't you be working? geez.... lol

[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.

[03sLoMaX]

i feel so stupid when i look at this stuff, lol

[300zTT]

ummmmmmmmmmmmmmmmmmmmmmmmmmmm

[Kerplunk105]

wow, I didn't know you guys spoke greek!

[mostasteless]

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.

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