Answer:
(a) P(X<=3) = 0.9658, P(X<3) = 0.8728
(b) P(X>=4) = 0.0342
(c) P(1<= X <=3) = 0.6884
(d) E(X) = 1.25, σ(X) = 1.089
(e) P(X=0) = 0.0769
Step-by-step explanation:
We are given that X ~ Bin(25, .05) which means that this can be approximated as a binomial distribution. The formula for calculating probability using the binomial distribution is:
P(X=x) = ⁿCₓ pˣ qⁿ⁻ˣ
where n = total no. of trials
x = no. of successful trials
p = probability of success
q = probability of failure = 1 - p
We have n = 25, p = 0.05 and q = 0.95
(a) P(X<=3) = P(X=0) + P(X=1) + P(X=2) + P(X=3)
= ²⁵C₀ (0.05)⁰ (0.95)²⁵⁻⁰ + ²⁵C₁ (0.05)¹ (0.95)²⁵⁻¹ + ²⁵C₂ (0.05)² (0.95)²⁵⁻² + ²⁵C₃ (0.05)³ (0.95)²⁵⁻³
= 0.2774 + 0.3649 + 0.2305 + 0.09302
P(X<=3) = 0.9658
P(X<3) = P(X=0) + P(X=1) + P(X=2)
= ²⁵C₀ (0.05)⁰ (0.95)²⁵⁻⁰ + ²⁵C₁ (0.05)¹ (0.95)²⁵⁻¹ + ²⁵C₂ (0.05)² (0.95)²⁵⁻²
= 0.2774 + 0.3649 + 0.2305
P(X<3) = 0.8728
(b) P(X>=4) = 1 - P(X<4)
= 1 - (P(X=0) + P(X=1) + P(X=2) + P(X=3))
= 1 - 0.9658
P(X>=4) = 0.0342
(c) P(1<= X <=3) = P(X=1) + P(X=2) + P(X=3)
= ²⁵C₁ (0.05)¹ (0.95)²⁵⁻¹ + ²⁵C₂ (0.05)² (0.95)²⁵⁻² + ²⁵C₃ (0.05)³ (0.95)²⁵⁻³
= 0.3649 + 0.2305 + 0.09302
P(1<= X <=3) = 0.6884
(d) E(X) = np
= (25)(0.05)
E(X) = 1.25
σ(X) = √npq
= √(25)(0.05)(0.95)
σ(X) = 1.089
(e) n=50 and p= 0.05. We need P(X=0) so,
P(X=0) = ⁵⁰C₀ (0.05)⁰ (0.95)⁵⁰⁻⁰
= (1)*(1)*(0.0769)
P(X=0) = 0.0769
