Question 1173518: Find the z-score that corresponds to the following areas.
Area to the left is 1.24
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! with normal distributions, the total area is 1 and anything else has to be between 0 and 1.
if we assume you are talking in percents, then the total area is 100% and 1.24% / 100% would be equal to .0124.
the z-score with an area of .0124 to the left of it would be equal to -2.244503871.
round this to 3 decimal places and it becomes -2.245.
i used my ti-84 plus scientific calculator to get the more detailed answer.
the tables and the calculators assume that the area is a ratio of the area to the left of the z-score divided by the whole area under the normal distribution curve.
the whole area under the normal distribution curve is assumed to be equal to 1.
that's why they show the area to the left of a z-score as a ratio between 0 and 1.
here's a reference on z-score tables and how to find the area to the left of the z-score.
https://mathbitsnotebook.com/Algebra2/Statistics/STzScores.html#:~:text=To%20find%20a%20specific%20area,on%20a%20standard%20normal%20distribution.&text=You%20need%20both%20tables!
most tables are full tables, so concentrate on that.
the table i normally reference can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
using that table, i found that the z-score of -2.24 gave an area of .01255 to the left of it and the z-score of -2.25 gave an area of .01222 to the left of it.
.0124 was in between z-score of -2.24 and -2.25.
i manually interpolated to get a z-score of -2.2445 which is equal to -2.245 when rounded to 3 decimal places.
whether i used my ti-84 plus calculator, or used the table and manually interpolated, i got a z-score of -2.245 rounded to 3 decimal places.
that's usually close enough.
a calculator you can use online can be found at https://stattrek.com/online-calculator/normal.aspx
using that calculator, i got a z-score of -2.245 with a lot less work than manual interpolation.
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