Question

If binomial distribution X~Bin(50,0.4) coincides with normal distribution X*~N(μ,σ²) , then find the value of μ, σ and

P[X ≤ 5] , P[X ≤ 15] , P[5 ≤ X ≤ 15] , P[5 < X < 15].

Answer 1

$$\mu=np=50(0.4)=20$$ $$\sigma^2=npq=50(0.4)(1-0.4)=12$$

$$\sigma=\sqrt{12}=2\sqrt{3}$$

Continuity Correction Factor 

Using Normal Distribution with Continuity Correction $P(X\leq 5)$

Using Normal Distribution with Continuity Correction

$$P(X<5.5)=P(Z<\dfrac{5.5-20}{2\sqrt{3}})$$ $$\approx P(Z<-4.18579)\approx0.00001421$$

Using Normal Distribution with Continuity Correction $P(X\leq 15)$

Using Normal Distribution with Continuity Correction

$$P(X<15.5)=P(Z<\dfrac{15.5-20}{2\sqrt{3}})$$ $$\approx P(Z<-1.299038)\approx0.096965$$

Using Normal Distribution with Continuity Correction $P(5\leq X\leq 15)=P(X\leq 15)-P(X<5)$

Using Normal Distribution with Continuity Correction

$$P(X<15.5)-P(X<4.5)$$ $$=P(Z<\dfrac{15.5-20}{2\sqrt{3}})-P(Z<\dfrac{4.5-20}{2\sqrt{3}})$$ $$\approx P(Z<-1.299038)-P(Z<-4.474465)$$ $$\approx0.096965-0.00000383$$ $$\approx0.096961$$

Using Normal Distribution with Continuity Correction $P(5< X<15)=P(X< 15)-P(X\leq5)$

Using Normal Distribution with Continuity Correction

$$P(X<14.5)-P(X<5.5)$$ $$=P(Z<\dfrac{14.5-20}{2\sqrt{3}})-P(Z<\dfrac{5.5-20}{2\sqrt{3}})$$ $$\approx P(Z<-1.587713)-P(Z<-4.18579)$$ $$\approx0.05617565-0.00001421$$ $$\approx0.05616144$$