STA301 - Assignment Solution

Answer (a):
We know that:
E(X) = np = mean of a binomial distribution = 40
Var(X) = npq = variance of a binomial distribution =?
S.D(X) = √npq = standard deviation of a binomial distribution = 6
So,
Var(X) = [S.D(X)] 2
Var(X) = [√npq] 2
Var(X) = [6] 2
Var(X) = 36
Var(X) = npq
36 = (40)q
q = 36/40
q = 0.9
p = 1 – q
p = 1 – 0.9
p = 0.1
E(X) = np
40 = n(0.1)
n = 40/0.1
n = 400
OR
Var(X) = npq
36 = n(0.1)(0.9)
n(0.09) = 36
n = 36/0.09
n = 400
Answer (b):
We know that:
P(X=x) = (μx e-μ)/x! for x = 0, 1, 2, …, ∞
Here;
e = 2.71828
n = 500
p = 0.1% or 0.001
μ = np
μ = 500 * 0.001
μ = 0.5
Our Poisson formula becomes:
p(x; 0.5) = (0.5x e-0.5)/x!
i. No defective
p(x; 0.5) = (0.5x e-50)/x!
p(0; 0.5) = (0.50 (2.71828)-0.5)/0!
p(0; 0.5) = (1 * 0.606531)
p(0; 0.5) = 0.606531 or 60.65% chances that there will be no defective.
ii. At least two defective
P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4) + ….
P(X ≥ 2) = 1 – P(X=1) – P(X=0)
So, first find P(X=1):
p(x; 0.5) = (0.5x e-0.5)/x!
p(1; 0.5) = (0.51 (2.71828)-0.5)/1!
p(1; 0.5) = (0.5 * 0.606531)/1
p(1; 0.5) = 0.303265 or 30.33% chances that there will be exactly one defective
P(X ≥ 2) = 1 – P(X=1) – P(X=0)
P(X ≥ 2) = 1 – 303265 – 606531
P(X ≥ 2) = 0.090204 or 9.02% chances that at least 2 will be defective.