The Student Room Group

normal distribution with pistons and cylinders

hello :smile:

i am unsure of how to do this question:
(all units are cm)

diameter of piston ~ N(10.42,0.03²)
diameter of cylinder ~ N(10.52, 0.04²)

piston has to fit into cylinder

"If pairs of pistons and cylinders are selected at random for assembly, for what proportion will the piston not fit into the cylinder?"

ok, so i am confused :confused:

if piston is P and cylinder is C - do we want to work out the difference -> C-P and this must be bigger than 0???

please help

i would be grateful if u could clarify the first step of the method rather than solve the question :smile:
Reply 1
can anyone help?
Reply 2
To subtract normal random variables, you subtract the means and add the variances.

In symbols, N(a, u^2) - N(b, v^2) = N(a - b, u^2 + v^2).

C - P
~ N(0.1, 0.03^2 + 0.04^2)
= N(0.1, 0.05^2)

Select the following text if you want the rest of the solution:

P(piston doesn't fit)
= P(C - P < 0)
= P(Z < -0.1/0.05) where Z ~ N(0, 1)
= P(Z < -2)
= Phi(-2)
= 0.023
Reply 3
is this s1? :redface:
Reply 4
thanks Jonny W

thats the way that i did it but i didn't finish the question because i didn't think i was doing it right :redface:

oh, and nice idea on how to not show me the answer if i don't want it - really appreciate it :smile:


by the way - this is for university stats course

not S1 - in S1 the normal distribution stuff was a lot simpler
Reply 5
i've done that part of the question and got the probability to be 0.0228

next part:
"What is the chance that in 100 pairs, selected at random: (i) every piston will fit, (ii) not more than 2 or the pistons will fail to fit?"

how do i do that?
it says to calculate the probabilities (a) exactly, and (b) using a Poisson approximation

so i am confused again - how do i calculate 'exactly'?

do i use binomial but approximate it to normal??? :frown:
Reply 6
anyone?
Reply 7
Let X be the number of the 100 pairs that don't fit.

(a)
X ~ Bin(100, 0.0228)

(i)
P(X = 0)
= 0.9772^100
= 0.100

(ii)
P(X <= 2)
= 0.9772^100 + 100*0.9772^99*0.0228 + 100C2*0.9772^98*0.0228^2
= 0.602

(b)
Approximately, X ~ Po(2.28).

(i)
P(X = 0)
\approx e^(-2.28)
= 0.103

(ii)
P(X <= 2)
\approx e^(-2.28) (1 + 2.28 + 2.28^2/2!)
= 0.603
Reply 8
wow!

thanks once again

i used the Binomial approximation in another question so i guessed it was that.

how do u know all this stuff? do u do loads of stats courses? :biggrin:
Reply 9
Stupid ST102.. grr.. :smile:
Reply 10
CharXlotte
Stupid ST102.. grr.. :smile:


ST102 :biggrin:

have u finished all of the 'Christmas Work'?
Reply 11
Just started today.. I've been trying to hold it off for as looooong as possible :frown: Why is it out to ruin my life?! Hehe, I've done a few questions. Good luck to you!! x
Reply 12
CharXlotte
Just started today.. I've been trying to hold it off for as looooong as possible :frown: Why is it out to ruin my life?! Hehe, I've done a few questions. Good luck to you!! x


i have been doing this 'fun' christmas work for ages! nearly a week and i still have some questions to do :mad:
Reply 13
Stats stats stats, tra la la la la.. is what's going on in my head. It's not pretty. Not at all.