Hi
Verbiage is for a binomial Distribution.
p = .15, n = 8
P(x = 4) = binompdf(8, .15, 4) = .0185
P(X ≤ 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