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I am looking through an example (found here https://answers.yahoo.com/question/index?qid=20080303223718AA2T5qI) on the central limit theorem. I understand most of the best answer and am able to follow how they derive most of the numbers. What I am confused on is where the numbers for this line (= 0.8286574 - 0.1713426) come from? How are they derived. I am guessing this is basic stat principle that I am not grasping.

I have pasted a part of the solution given from the best answer on the webpost below

In this question we have 
X ~ Normal( μx = 0.49 , σx² = 0.0009996 ) 
X ~ Normal( μx = 0.49 , σx = 0.03161645 ) 

Find P( 0.46 < X < 0.52 )
= P( ( 0.46 - 0.49 ) / 0.03161645 < ( X - μ ) / σ < ( 0.52 - 0.49 ) / 0.03161645 )
= P( -0.948873 < Z < 0.948873 ) 
= P( Z < 0.948873 ) - P( Z < -0.948873 ) 
= 0.8286574 - 0.1713426    <------HOW ARE THESE NUMBERS DERIVED???
= 0.6573148
soccergal_66
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  • Using math typesetting will attract more readers and answers. Some more information: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Sycorax Mar 25 '17 at 03:04
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    They're straight out of normal tables (or equivalently straight out of normal cdf functions of computer packages). See the examples here and here which solve slightly different problems but illustrate the use of tables. – Glen_b Mar 25 '17 at 05:13
  • @Glen_b Reading through the comments you had provided on your 2nd example, do I understand this correctly, the (Z < 0.948873) is a z-value and 0.8286 is the corresponding value that goes with the z-value=0.948873? Looking at the z-table here in your post, I see that a z-value of 0.94 is 0.8264 so how did they get 0.8286? – soccergal_66 Mar 25 '17 at 20:01
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  • The "value that goes with 0.948873" is a probability, specifically the probability that Z is less than 0.948873 (it's also an area, since in the density function, area represents probability). 2. That table only has two decimal places for Z so under ideal conditions you can get about two decimal place accuracy back (more if you use interpolation). However, notice that 0.948873 is a lot closer to 0.95 than 0.94, so if you're not doing interpolation, look up 0.95 not 0.94. ... ctd
    • – Glen_b Mar 26 '17 at 00:07
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    ctd... My guess is what they've actually done is call a function on a computer (stats packages can do it, as can essentially any general purpose spreadsheet program). If I do that in R I get exactly the same value they do. If I use tables but with linear interpolation (how to do that is discussed in detail here) it would give 0.828656, which is very close to the exact 0.82865739... (linear interpolation often works pretty well in normal tables) – Glen_b Mar 26 '17 at 00:07
  • @Glen_b thanks that is an incredibly helpful link on linear interpolation and understandable. Out of curiosity how do you calculate it in R as I am also in the process of learning R? – soccergal_66 Mar 26 '17 at 00:43
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    In R ... pnorm(0.948873) gives [1] 0.8286574 ... by comparison, in Excel or LibreOffice, =NORMSDIST(0.948873) should give 0.828657396. Similar commands for the standard normal cdf are easy to find in other packages. – Glen_b Mar 26 '17 at 01:05