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
Accepted Solution
A:
Answer: [tex]P(x=5) =0.0199\\\\P(X>5)=0.9706\\\\P(X\leq 5)=0.0294[/tex]Step-by-step explanation:The binomial distribution formula :-[tex]P(x)=^nC_xp^x(1-p)^{n-x}[/tex], 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 = 25Then , For x=5[tex]P(5)=^{25}C_{5}(0.4)^5(1-0.4)^{25-5}\\\\=\dfrac{25!}{5!(25-5)!}(0.4)^5(0.6)^20=0.0198913738671\approx0.0199[/tex][tex]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[/tex]Now, [tex]P(x>5)=1-P(x\leq5)=1-0.0294=0.9706[/tex]