Question 577998
The scores on an exam are normally distributed, with a mean of 77 and a standard deviation of 10.  What percent of the scores are greater than 87??
<pre>
Since 87 is 10, exactly 1 standard deviation, namely 10, above the mean, 
its z-score is 1.  Or we can calulate the z-score by formula:

Calculate the z-score

z = {{{(x-mu)/sigma}}} = {{{(87-77)/10}}} = {{{10/10}}} = 1.

Anyway we want to find the percentage of area indicated by the
shaded portion below to the right of z=1, which 1 standard deviation
above the mean. 

{{{drawing(400,200,-3,3,-.02,.2,
graph(400,200,-3,3,-.02,.2, ( 1/(2pi) )e^(-x^2)   ),

line(1.0,0,1.0,(1/(2pi))e^(-(1.0)^2) ),
line(1.1,0,1.1,(1/(2pi))e^(-(1.1)^2) ),
line(1.2,0,1.2,(1/(2pi))e^(-(1.2)^2) ),
line(1.3,0,1.3,(1/(2pi))e^(-(1.3)^2) ),
line(1.4,0,1.4,(1/(2pi))e^(-(1.4)^2) ),
line(1.5,0,1.5,(1/(2pi))e^(-(1.5)^2) ),
line(1.6,0,1.6,(1/(2pi))e^(-(1.6)^2) ),
line(1.7,0,1.7,(1/(2pi))e^(-(1.7)^2) ),
line(1.8,0,1.8,(1/(2pi))e^(-(1.8)^2) ),
line(1.9,0,1.9,(1/(2pi))e^(-(1.9)^2) ),
line(2.0,0,2.0,(1/(2pi))e^(-(2.0)^2) ) )}}}

We are told that the middle region shaded below: 

{{{drawing(400,200,-3,3,-.02,.2,
graph(400,200,-3,3,-.02,.2, ( 1/(2pi) )e^(-x^2)   ),

line(.0,0,.0,(1/(2pi))e^(-(.0)^2) ),
line(.1,0,.1,(1/(2pi))e^(-(.1)^2) ),
line(.2,0,.2,(1/(2pi))e^(-(.2)^2) ),
line(.3,0,.3,(1/(2pi))e^(-(.3)^2) ),
line(.4,0,.4,(1/(2pi))e^(-(.4)^2) ),
line(.5,0,.5,(1/(2pi))e^(-(.5)^2) ),
line(.6,0,.6,(1/(2pi))e^(-(.6)^2) ),
line(.7,0,.7,(1/(2pi))e^(-(.7)^2) ),
line(.8,0,.8,(1/(2pi))e^(-(.8)^2) ),
line(.9,0,.9,(1/(2pi))e^(-(.9)^2) ),
line(1.0,0,1.0,(1/(2pi))e^(-(1.0)^2) ), 

line(-.0,0,-.0,(1/(2pi))e^(-(.0)^2) ),
line(-.1,0,-.1,(1/(2pi))e^(-(.1)^2) ),
line(-.2,0,-.2,(1/(2pi))e^(-(.2)^2) ),
line(-.3,0,-.3,(1/(2pi))e^(-(.3)^2) ),
line(-.4,0,-.4,(1/(2pi))e^(-(.4)^2) ),
line(-.5,0,-.5,(1/(2pi))e^(-(.5)^2) ),
line(-.6,0,-.6,(1/(2pi))e^(-(.6)^2) ),
line(-.7,0,-.7,(1/(2pi))e^(-(.7)^2) ),
line(-.8,0,-.8,(1/(2pi))e^(-(.8)^2) ),
line(-.9,0,-.9,(1/(2pi))e^(-(.9)^2) ),
line(-1.0,0,-1.0,(1/(2pi))e^(-(1.0)^2) )



)}}}
 
between z=-1 and z=+1 contains about 68.3% of the total
area.  So the rest of the shaded area, which is this,

{{{drawing(400,200,-3,3,-.02,.2,
graph(400,200,-3,3,-.02,.2, ( 1/(2pi) )e^(-x^2)   ),

line(1.0,0,1.0,(1/(2pi))e^(-(1.0)^2) ),
line(1.1,0,1.1,(1/(2pi))e^(-(1.1)^2) ),
line(1.2,0,1.2,(1/(2pi))e^(-(1.2)^2) ),
line(1.3,0,1.3,(1/(2pi))e^(-(1.3)^2) ),
line(1.4,0,1.4,(1/(2pi))e^(-(1.4)^2) ),
line(1.5,0,1.5,(1/(2pi))e^(-(1.5)^2) ),
line(1.6,0,1.6,(1/(2pi))e^(-(1.6)^2) ),
line(1.7,0,1.7,(1/(2pi))e^(-(1.7)^2) ),
line(1.8,0,1.8,(1/(2pi))e^(-(1.8)^2) ),
line(1.9,0,1.9,(1/(2pi))e^(-(1.9)^2) ),
line(2.0,0,2.0,(1/(2pi))e^(-(2.0)^2) ),

line(-1.0,0,-1.0,(1/(2pi))e^(-(1.0)^2) ),
line(-1.1,0,-1.1,(1/(2pi))e^(-(1.1)^2) ),
line(-1.2,0,-1.2,(1/(2pi))e^(-(1.2)^2) ),
line(-1.3,0,-1.3,(1/(2pi))e^(-(1.3)^2) ),
line(-1.4,0,-1.4,(1/(2pi))e^(-(1.4)^2) ),
line(-1.5,0,-1.5,(1/(2pi))e^(-(1.5)^2) ),
line(-1.6,0,-1.6,(1/(2pi))e^(-(1.6)^2) ),
line(-1.7,0,-1.7,(1/(2pi))e^(-(1.7)^2) ),
line(-1.8,0,-1.8,(1/(2pi))e^(-(1.8)^2) ),
line(-1.9,0,-1.9,(1/(2pi))e^(-(1.9)^2) ),
line(-2.0,0,-2.0,(1/(2pi))e^(-(2.0)^2) )

 )}}}

is 100% - 68.3% = 31.7% of the area, and therefore 
the desired percentage of area, which is this, 
 
{{{drawing(400,200,-3,3,-.02,.2,
graph(400,200,-3,3,-.02,.2, ( 1/(2pi) )e^(-x^2)   ),

line(1.0,0,1.0,(1/(2pi))e^(-(1.0)^2) ),
line(1.1,0,1.1,(1/(2pi))e^(-(1.1)^2) ),
line(1.2,0,1.2,(1/(2pi))e^(-(1.2)^2) ),
line(1.3,0,1.3,(1/(2pi))e^(-(1.3)^2) ),
line(1.4,0,1.4,(1/(2pi))e^(-(1.4)^2) ),
line(1.5,0,1.5,(1/(2pi))e^(-(1.5)^2) ),
line(1.6,0,1.6,(1/(2pi))e^(-(1.6)^2) ),
line(1.7,0,1.7,(1/(2pi))e^(-(1.7)^2) ),
line(1.8,0,1.8,(1/(2pi))e^(-(1.8)^2) ),
line(1.9,0,1.9,(1/(2pi))e^(-(1.9)^2) ),
line(2.0,0,2.0,(1/(2pi))e^(-(2.0)^2) ) )}}}

is half of 31.7%, and therefore about

15.9% 

Edwin</pre>