I have to pass a test on-line, but I really don't get it. I need to pass it at least 3 times without mistakes (knowing it does change every time i log on...).

here is an example of test:


Faults occur at random on lengths of cable at a rate of 2 per kilometre.
The probability of no faults on a 200 metre section is (3 d.p.)?
The probability of 7 or more faults on a 4 kilometre section is (3 d.p.)?

I've found:
Poisson distribution (0.4)
0.4 = 2/5 as 1000/200 = 5
P(0)=e^(-0.4)=0.6703
So the probability of no fault on a 200 metre section is 0.6703, right?

but now do i have to change to Poisson (8), as it's on 4 kilometres?
the answer was 0.687...

Caora,

I sent you a private message.

On both exercises, I have the 1st correct but not the second... I got confused after...

the other excercise

A paint manufacturer produces cans of paint of nominal volume 1 litre. In fact the volume of paint in a can is a Normal random variable with mean 1.08 litres and standard deviation 0.05 litres.

A single can is examined; the probability that it contains less than 1 litre is (3 d.p.)? the answer is 0.055
There are symbols I can't reproduce on screen, but I understand how to find the answer.

The volume, x, which 10% of cans exceed is (in litres) (2 d.p.)?
I found 1.08 but the answer was 1.14
What I did:
I looked on the statistical table which is given to help for 0.1 (the 10%) = 0.5398
then I took the negative of it (the teacher did that in the tutorial)
-0.5398*0.05+1.08

I guess it wasn't how to solve the problem



I think I know why I got it wrong, I looked at the wrong table.....Shame on me

Caora,

Yes you'll use Poisson(8) for the second part.

P(X=7) = 8^7*exp(-8)/7! = 0.140
P(X=8) = ...
P(X=9) = ...
You'll have to go on till the probability is close to zero, then add up all the calculated probabilities and it should be 0.687.

For the paint can problem, if you let X be normally distributed with mean=1.08 and sd=0.05,

P(X < 1)
=P[(X-mean)/sd < (1-mean)/sd]
=P[Z < (1-1.08)/0.05]
=P(Z< -1.6)
=1-P(Z<1.6)
=1-0.9452
=0.0548
=0.055 approx.