Answer to Question #320981 in Statistics and Probability for Calvin

Question

 A discrete random variable X has probability distribution function f(x) = 12! /x!(12−x)!p x (1 − p) 12−x x = 0, 1, 2, .., 12 0, elsewhere

(i) if p = 0.3, find Pr(X > 3).

(ii) find possible values of p if Var[X] is equal to 1.92. 

Answer 1

$f\left( x \right) =\frac{12!}{x!\left( 12-x \right) !}p^x\left( 1-p \right) ^{12-x},x=0,1,...,12\Rightarrow X\sim Bin\left( 12,p \right) \\i:P\left( X>3 \right) =1-P\left( X=0 \right) -P\left( X=1 \right) -P\left( X=2 \right) =\\=1-\sum_{i=0}^2{C_{12}^{i}p^i\left( 1-p \right) ^{12-i}}=1-0.7^{12}-12\cdot 0.3\cdot 0.7^{11}-\frac{12\cdot 11}{2}\cdot 0.3^2\cdot 0.7^{10}=0.747185\\ii: Var\left( X \right) =np\left( 1-p \right) \\12p\left( 1-p \right) =1.92\Rightarrow p^2-p+0.16=0\Rightarrow p\in \left\{ 0.2,0.8 \right\}$