Question

If x is a binomial variable with p = .4 and n = 25, write the probability that you would be looking for after you have applied the continuity correction factor for the following conditions for x: 2) x = 5 3) x > 5 4) x ≤ 5

Answer 1

`P(x=5) =0.0199\\\\P(X>5)=0.9706\\\\P(X\leq 5)=0.0294`

Explanation:

The binomial distribution formula :-

`P(x)=^nC_xp^x(1-p)^(n-x)`, here P(x) is the probability of getting success at x trial , n is the total number of trails, p is the probability of getting success in each trail.

Given : x is a binomial variable with p = 0.4 and n = 25

Then , For x=5

`P(5)=^(25)C_(5)(0.4)^5(1-0.4)^(25-5)\\\\=(25!)/(5!(25-5)!)(0.4)^5(0.6)^20=0.0198913738671\approx0.0199`

`P(x\leq5)=P(0)+P(1)+P(2)+P(3)+P(4)+P(5)\\\\=^(25)C_(0)(0.4)^0(0.6)^(25-0)+^(25)C_(1)(0.4)^1(0.6)^(25-1)+^(25)C_(2)(0.4)^2(0.6)^(25-2)+^(25)C_(3)(0.4)^3(0.6)^(25-3)+^(25)C_(4)(0.4)^4(0.6)^(25-4)++^(25)C_(5)(0.4)^5(0.6)^(25-5)\\\\=(0.6)^(25)+25(0.4)(0.6)^(24)+300(0.4)^2(0.6)^(23)+2300(0.4)^3(0.6)^(22)+12650(0.4)^4(0.6)^(21)+53130(0.4)^5(0.6)^(20)=0.02936220\approx0.0294`

Now,
`P(x>5)=1-P(x\leq5)=1-0.0294=0.9706`