A binomial distribution has a probability of success = 0.8
Calculate the probability of exactly 3 successes in 8 trials:
f(k;n,p) = | n! * pkqn - k | |
k!(n - k)! |
q = 1 - p
q = 1 - 0.8
q = 0.2
n! = 8!
8! = 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
8! = 40320
k! = 3!
3! = 3 * 2 * 1
3! = 6
(n - k)! = (8 - 3)!
(n - k)! = 5!
5! = 5 * 4 * 3 * 2 * 1
5! = 120
P(X = 3) = | 8! * 0.830.2(8 - 3) | |
3!(8 - 3)! |
P(X = 3) = | 40320 * 0.512 * 0.25 | |
6 * 120 |
P(X = 3) = | 40320 * 0.512 * 0.00032 | |
720 |
P(X = 3) = | 6.6060288 | |
720 |
Since q = 1 - p, we have n(1 - p) = 8(1 - 0.8)
nq = 8(0.2)
nq = 1.6
μ = np
μ = 8 x 0.8
μ = 6.4
σ2 = np(1 - p)
σ2 = 8 x 0.8 x (1 - 0.8)
σ2 = 6.4 x 0.2
σ2 = 1.28
σ = √σ2 = √np(1 - p)
σ = √1.28
σ = 1.1314
Skewness = | 1 - 2p |
√np(1 - p) |
Skewness = | 1 - 2(0.8) |
√8(0.8)(1 - 0.8) |
Skewness = | 1 - 1.6) |
√8(0.8)(0.2) |
Skewness = | -0.6 |
√1.28 |
Skewness = -0.46875
Kurtosis = | 1 - 6p(1 - p) |
np(1 - p) |
Kurtosis = | 1 - 6(0.8)(1 - 0.8) |
8(0.8)(1 - 0.8) |
Kurtosis = | 1 - (4.8)(0.2) |
8(0.8)(0.2) |
Kurtosis = | 1 - 0.96 |
1.28 |
Kurtosis = 0.03125