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checking normal distribution thingy

Bryn
in Applied Mathematics

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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.

  • Author

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) ?

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