Question

Suppose you have a binomial distribution with n = 20 and p = 0.4. find p(9 ≤ x ≤ 10).

Answer 1

`\mathbb P(9\le X\le 10)=\mathbb P(X=9)+\mathbb P(X=10)`

Since


`\mathbb P(X=x;n,p)=\dbinom nxp^x(1-p)^(n-x)`

you have


`\mathbb P(9\le X\le10)=\dbinom{20}90.4^9(1-0.4)^(20-9)+\dbinom{20}{10}0.4^(10)(1-0.4)^(20-10)`

`\mathbb P(9\le X\le10)=\dbinom{20}90.4^90.6^(11)+\dbinom{20}{10}0.4^(10)0.6^(10)`

`\mathbb P(9\le X\le10)\approx0.2769`