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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.2011-05-06
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    example Pr(z <= 1.35) = 0.5 + Pr(0 <= Z <= 1.35) = 0.5 + 0.41152011-05-06
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    and why is Pr(0 <= z <= 1.5) = 0.4332 without adding 0.5?2011-05-06
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    Well, if the distribution is symmetric and continuous then $Pr(z\leq 0)=0.5$.2011-05-06

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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?2011-05-06
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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 0.5.2011-05-06
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    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, http://www.mathsisfun.com/data/standard-normal-distribution-table.html)2011-05-07
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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.