Tarjei,
The key is that if U is any random variable with mean 0 and variance
sigma^2 and a>0, then
Pr(abs(U)>a) = Pr(U > a) + Pr(U <- a) = 2P(U>a) because of the symmetry of
the normal distribution.
Pr(U>a) = 1-Pr(U<a) = 1-Pr(U/sigma < a/sigma) = 1-Pr(Z<a/sigma) where Z is
standard normal so that Pr(Z<x) is tabulated.
You can apply this to X1-X2 since it has mean 2-2=0
Ellen Hertz
Tarjei Knapstad wrote:
> I recently ran into this problem when assisting my girlfriend with her
> studies for an exam in basic statistics, but I've been unable to solve
> it:
>
> I have a random variable X normally distributed with known mean and
> standard deviation, say, X~N(2, 0.5). If I make two samples from this
> distribution, how do I find the probability of the absolute difference
> between these samples being larger than a given number, say 1? (I think
> the textbook example involved retrieving two cabbages from a field and
> weighing them, X being the weight)
>
> Any help would be much appreciated.
>
> Regards,
> --
> Tarjei Knapstad
> siv.ing.
>
> The Norwegian University of Science and Technology
> Dept. of Chemistry and Biology
> Institute of Physical Chemistry, Group of Chemometrics
> Trondheim, Norway.
>
> ([EMAIL PROTECTED] --- phone: 47 73 59 41 70)
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