Question 1174090
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Hi,
{{{drawing(400,200,-5,5,-.5,1.5, graph(400,200,-5,5,-.5,1.5, exp(-x^2/2)), green(line(1,0,1,exp(-1^2/2)),line(-1,0,-1,exp(-1^2/2))),green(line(2,0,2,exp(-2^2/2)),line(-2,0,-2,exp(-2^2/2))),green(line(3,0,3,exp(-3^2/2)),line(-3,0,-3,exp(-3^2/2))),green(line( 0,0, 0,exp(0^2/2))),locate(4.8,-.01,z),locate(4.8,.2,z))}}}

Above is the Standard Normal Curve:  μ = 0  and  σ = 1  
Included are various z-scores demonstrating the AREA 
under the Standard Normal distribution Curve  according to the
value of a particular z-score.  

PROBABILITY of a particular x-value is an AREA as defined by {{{z =blue (x - mu)/blue(sigma)}}} 
to the left of that z-score and is commonly written as P(z ≤ its value)
Whether one uses a z-score /table 0r a calculator, Probability will be
computed as representing an Area under the Standard Normal Curve.

In Your case:  μ = 1000  and  σ = 100 
(a) P(x > 1000) is the Area  to the right of  μ = 1000  
(According to the Above, obviously it is 50% of the Area under the curve 0r P = .50
    P (x > 1000) can be written as 1- P(x ≤ 1000) = .50

(b) P(1000 ≤ x < 1000) = P(1000 ≤ x ≤ 2000) = normalcdf(1000,2000,1000,100)
(c) P(1000 < x < 1000) = P(1000 ≤ x  ≤ 2000) = normalcdf(1000,2000,1000,100)

(d) P(x < 1000) =  P(x ≤ 1000) = .5
  
Wish You the Best in your Studies.
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