If X is N(mu, sigma squared), show that for "a" does not equal 0 holds aX + b is N(a* mu + b; a^2 sigma^2)
Last edited by mr fantastic; November 2nd 2009 at 01:55 PM. Reason: Added post title to main text.
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Originally Posted by affelix If X is N(mu, sigma squared), show that for "a" does not equal 0 holds aX + b is N(a* mu + b; a^2 sigma^2) Surely there is a proof to be found on the world wide web rather than getting someone to re-invent the wheel. Have you tried using Google?
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