
Phil C. answered 02/04/21
MPH, Biostatistics + 20 years' teaching
<< What is P(-1.23 < Z < 0)?
(What is the probability that Z is between -1.23 and 0?) >> NOTE: this is a pretty good translation of the expression; well done!
I prefer "anywhere from -1.23 to 0..." but for continuous models they are the same thing.
(1) ESTIMATE: well let's see... P(-infinity < Z < 0) = 0.50 or 1/2, because that's half of the scores in the normal distribution, yes? Half of the scores are "below the mean" or "below average?"
So my fearless estimate is... less than 0.50.
(2) Therefore, a proposed solution of 0.8907 is "dead in the water" from the get-go. It is not less than 0.50, and I am VERY confident the P(z) I am looking for is < 0.50.
(3) << Start at line 1.2....(how do I know to start at this line, why not -1.2?)
...read along until 1.23 >>
Sounds like you are using a "z-score" table. NOTE: I HAVE SEEN AT LEAST 3 DIFFERENT TYPES, and they all work differently. The most common type, if you locate the cell value at 1.2 + 0.03, is the cumulative area (or probability) from -infinity UP TO 1.23.
Your idea to find the value at -1.23 sounds better to me, IF you then recall: that value will be the area below Z = 0 that you DON'T want. subtract that amount from 0.50, and you will obtain the area you DO want. This is true if your table works as I described above. (values are cumulative areas below that upper bound).
Do you know how to check your answer with a calculator or online site? This comes in handy sometimes.
MORAL: 0.8907 cannot be the value of P(Z) in the setting you have described.
Hope this helps!
Phil C.

Phil C.
Well you could... but I teach my class that even a gross estimate is often helpful. The lower half of the SNDist. is 0.50 ... we would go with that: "it's under 50%." But even that weeds out solutions like 0.89 etc. which is very useful when using a table.02/05/21

Phil C.
Also, my students are taught to always sketch a picture of the area we want to calculate. I like sketches. :')02/05/21
Brian P.
I am not sure how you estimate .5 Couldn't I estimate .4, .6 etc...?02/05/21