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

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    $\begingroup$ Could you give some context? It's going to be hard to answer your question without more information. $\endgroup$ May 6, 2011 at 17:35
  • $\begingroup$ example Pr(z <= 1.35) = 0.5 + Pr(0 <= Z <= 1.35) = 0.5 + 0.4115 $\endgroup$ May 6, 2011 at 17:37
  • $\begingroup$ and why is Pr(0 <= z <= 1.5) = 0.4332 without adding 0.5? $\endgroup$ May 6, 2011 at 17:42
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    $\begingroup$ Well, if the distribution is symmetric and continuous then $Pr(z\leq 0)=0.5$. $\endgroup$
    – Rasmus
    May 6, 2011 at 17:44

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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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  • $\begingroup$ well, can you further elaborate? $\endgroup$ May 6, 2011 at 17:46
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    $\begingroup$ $z$-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. $\endgroup$
    – Isaac
    May 6, 2011 at 17:47
  • $\begingroup$ 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) $\endgroup$ May 7, 2011 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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