Statistics - normal distribution

[cohkka]

Homework Statement

A plant manufactures 500 components a day with the diameter being random variable:
N(8.02, 0.1^2)mm

What is the probability of two randomly picked components differing by more than 0.3mm?2. Solution
I know that the solution is 0.966

The Attempt at a Solution

I thought that I needed to find the probability of the diameter differing from the mean by more than 0.3mm:
P(X < 7.72) + P(X > 8.32)
= 1 - P(7.72 < X < 8.32)
= 1 - { Phi[(8.32-8.02)/0.1] - Phi[(7.72-8.02)/0.1] }
= 1 - { Phi[3] - Phi[-3] }
= 1 - { Phi[3] - (1 - Phi[3]) }
= 1 - {0.99865 - (1-0.99865)}
= 1 - 0.9973
= 0.0027

This obviously is far from the correct answer and so I realize I must be completely on the wrong track but don't know how else one might approach this problem. I thought that the question might instead be asking for the probability of the difference of the two diameters being greater than 0.3mm so
P(a < X < a+0.3) but have no idea how you would go about finding the probability of an unknown value.

As you can probably tell, I am really confused by this so any help would be much appreciated! Thank you!

 

[vela]

You have two random variables: X₁, the diameter of the first ball, and X₂, the diameter of the second ball. Now form a third random variable Y=X₁-X₂, which is the difference in their diameters. Do you see how to solve the problem now?

 

[cohkka]

I think I might have got it now...

So for this new variable Y=X₁-X₂, I have to also calculate a new standard deviation which is sqrt(0.1² + 0.1²) = 0.1*sqrt(2)

So P(X>0.3) = 1 - P(X<0.3)
= 1 - Phi[0.3/(0.1*sqrt(2))]
= 1 - Phi[2.1213...]
= 1 - 0.983
= 0.0169
P(X<-0.3) = 0.0169
P(-0.3<X<0.3) = 1 - 2*0.0169
= 0.9661 (I guess the solution must have just been given for the difference being less than 0.3mm??)


Is this right? I understand why this would be correct but then when I apply this to the next part of the question, I do not get the right answer.
Now another plant B produces components with diameter being random variable N(7.95,0.08²)

What now is the probability of the diameter of two randomly picked components differing by more than 0.3mm if one is produced by plant B and the other by the first plant?

I went about this in the same way, calculating a standard deviation for this distribution:
sqrt(0.1² + 0.08²) = [sqrt(41)]/50

Using the mean as zero and this standard deviation, I get P(-0.3<X<0.3)=0.9808, whereas the solution I am given is 0.9662...where am I going wrong

Sorry for the long post!

 

[vela]

This is where you're going wrong:

Using the mean as zero and this standard deviation

 

[cohkka]

Ahhhhhh THANK YOU SO, SO MUCH! I have been struggling with this question for so long!