Bryn Posted May 28, 2004 Share Posted May 28, 2004 If X has the distribution N(102.3,1.96), calculate P(X<100) Now to get the standard normal distribution it's (100 - mew)/sigma so should be (100-102.3)/SQRt(1.96) ... P(Z<-1.64) That is correct isn't it? It's just that the book as got an answer that indicates they've used 1.96 instead of SQRT(1.96), but 1.96 is sigma squared right? btw is there anyway to do all the mathematical symbol, list sqare root, sigma, mew etc. Link to comment Share on other sites More sharing options...
wolfson Posted May 28, 2004 Share Posted May 28, 2004 No Z = X - u / sigma, not sqroot. Link to comment Share on other sites More sharing options...
Bryn Posted May 29, 2004 Author Share Posted May 29, 2004 yea but 1.96 is sigma squared isn't it? so to get sigma you need the sqrt. is it not in the format X ~ N(u , sigma^squared) ? Link to comment Share on other sites More sharing options...
Dave Posted May 29, 2004 Share Posted May 29, 2004 Yes, N ~ N(mu, sigma^2). Link to comment Share on other sites More sharing options...
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