Question 1136607
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This function models the <a href="https://en.wikipedia.org/wiki/Normal_distribution">normal distribution</a> curve when mu = 0 and sigma = 1, aka this is the standard Z distribution



The general template is 
*[Tex \LARGE f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}]
Plug in mu = 0 and sigma = 1 to get this slightly simpler equation
*[Tex \LARGE f(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-(x-\mu)^2}{2\sigma^2}]
*[Tex \LARGE f(x) = \frac{1}{1*\sqrt{2\pi}}e^{\frac{-(x-0)^2}{2*1^2}]
*[Tex \LARGE f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}]
which is exactly what you were given to start with. 



Find the area under the curve that is to the left of z = -1.00 using a table such as <a href="http://users.stat.ufl.edu/~athienit/Tables/Ztable.pdf">this one</a>. You should find that P(Z < -1.00) = 0.1587


Using that same table, you should also find P(Z < 1.00) = 0.8413


Subtract the areas to get the region between the proper z values we want
P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)
P(-1 < Z < 1) = 0.8413 - 0.1587
P(-1 < Z < 1) = 0.6826



Answer: <font color=red size=4>0.6826</font> (which is approximate)
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