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## 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 realise 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!