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I am new in this, and I am hoping for some assistance.

I have the table (attached) and need to calculate the value corresponding to Probability (-2.026). Can someone help me how to look at it from the table below? (If we have a negative Z score, how to calculate P value in the table with just positive Z scores?)

PS: I know we can look it up in negative Z score table but I wish to understand, how we can do it in this specific table?

Z Score

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  • $\begingroup$ You might find the explanations at stats.stackexchange.com/questions/250241/… to be helpful, because it's essentially the same problem. $\endgroup$
    – whuber
    Apr 15 '18 at 19:26
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    $\begingroup$ Also, -2.026 is not a probability score as they are always between 0 and 1. This is a z-score $ z = \frac{x-\mu}{\sigma}$ . Simply put, a z-score is the number of standard deviations from the mean a data point is. $\endgroup$
    – COOLBEANS
    Apr 15 '18 at 20:30
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In short, just subtract the values in this table above from 1.

So for a z-score of -2, with a p-value of 0.5, that gives (1- 0.9798) = 0.0202

You will get a negative z-score when $ x $ is less than $ \mu $ in the following equation:

$ z = \frac{x-\mu}{\sigma}$

As z-scores move from negative to positive they are moving from left to right on the bell curve. The z score is zero in the center. Imagine moving a vertical line across the image and asking yourself what proportion of the area under this curve is to the left of this line. At the far left it will be zero, in the middle 50% and at the far right 100%.

enter image description here

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    $\begingroup$ I think the answer is not clear because the OP did not state whether this is a one-sided or two-sided test. $\endgroup$ Apr 15 '18 at 23:32
  • $\begingroup$ As written, this answer isn't quite correct. for a z-score of -2, with a p-value of 0.5 doesn't make sense here, and is reading the table incorrectly. $\endgroup$ Nov 24 '18 at 13:31
  • $\begingroup$ The example in the answer is also incorrect. A z value of –2 yields a probability of 0.0228, not the value given in the answer. $\endgroup$ Nov 24 '18 at 13:39

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