Thread: If X is N(mu, sigma squared),

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)

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?