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Can anyone explain when should I add 0.5 to the z-score?

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Could you give some context? It's going to be hard to answer your question without more information. –  Mike Spivey May 6 '11 at 17:35
    
example Pr(z <= 1.35) = 0.5 + Pr(0 <= Z <= 1.35) = 0.5 + 0.4115 –  optimus May 6 '11 at 17:37
    
and why is Pr(0 <= z <= 1.5) = 0.4332 without adding 0.5? –  optimus May 6 '11 at 17:42
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Well, if the distribution is symmetric and continuous then $Pr(z\leq 0)=0.5$. –  Rasmus May 6 '11 at 17:44
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2 Answers

up vote 3 down vote accepted

Based on the example you give in your comment above, I think what's being used is that $P(z\lt 0)=0.5$ (and $P(z\gt 0)=0.5$, too), so that $P(z\le 1.35)=P((z\lt0)\text{ or }(0\le z\le 1.35))=0.5+P(0\le z\le 1.35)$.

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well, can you further elaborate? –  optimus May 6 '11 at 17:46
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$z$-scores are typically on datasets that are expected to be normally distributed and the $z$-score is the number of standard deviations from the mean. So, the probability of being below (or above) the mean should be 0.5. –  Isaac May 6 '11 at 17:47
    
I think Isaac has it right. Some statistics tables give the $z$ score probability corresponding to $P(0 \leq Z \leq z)$, while some give it for $P(Z \leq z)$. I'm guessing OP is looking at one of the former. (See, for example, mathsisfun.com/data/standard-normal-distribution-table.html) –  Mike Spivey May 7 '11 at 3:17
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If your problem involves binomial probabilities, and you wish to use the normal approximation to the binomial, you would add .5 to the z formula (not the score) as a continuity correction factor.

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