Question 869536
<pre><font face = "consolas" color = "indigo" size = 3><b>
Hi
Verbiage is for a binomial Distribution. 
p = .15,  n = 8
P(x = 4) = binompdf(8, .15, 4) = .0185
P(X &#8804; 4) = binomcdf(8, .15, 4) = .9971 0r 99.71% **(Common sense tells us 4 is unusual)
mean = .15*8 = 1.2. sd = sqrt(1.2*.85) = 1.010
 4 nearly 3 SD to the right of mean 1.2, very unusual result
For the normal distribution: 
one  standard deviation from the mean accounts for about 68% of the set 
two standard deviations from the mean account for about 95%
and three standard deviations from the mean account for about 99.7%.
Below:  z = 0, z = ± 1, z= ±2 , z= ±3 are plotted.  
Note: z = 0 (x value the mean) 50% of the area under the curve is to the left and %50 to the right
{{{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))}}}