Question 623216
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Hi,
A population has a mean of 300 and a standard deviation of 18. A sample of 144 observations will be taken. 
The probability that the sample mean will be between 297 to 303 is:
Note: graph of Standard Normal Curve showing z = 0 and z = ±1, ±2, ±3 
{{{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))}}}
In your Ex: P(297 < <span style="text-decoration: overline">X</span> < 303)
   P ( -3/18/12) < z < 3 /18/12)) = P ( 2 < z < 2) = .9545  
Agree with You.  
to get .8864, that would be P( -1.5 < z < 1.5) 
Using z = ± 1.5 would be incorrect thing to do in this case.
For the normal distribution: 
one  standard deviation from the mean accounts for about 68.2% of the set 
two standard deviations from the mean account for about 95.5%
and three standard deviations from the mean account for about 99.7%.