Let X be a binomially distributed random variable with mean 4 and variance (4/3). Then 54 P ( X ≤ 2) is equal to

Question

Let X be a binomially distributed random variable with mean 4 and variance (4/3). Then 54 P ( X ≤ 2) is equal to (A) (73/27) (B) (146/27) (C) (146/81) (D) (126/81)

Tardigrade Admin · Accepted Answer

np =4 n pq =4 / 3 n =6, p =2 / 3, q =1 / 3 54( P ( X =2)+ P ( X =1)+ P ( X =0)) 54( 6 C 2((2/3))2((1/3))4+ 6 C 1((2/3))1((1/3))5+ 6 C 0((2/3))0((1/3))6) =(146/27)

Q. Let $X$ be a binomially distributed random variable with mean 4 and variance $\frac{4}{3}$ . Then $54 P (X \leq 2)$ is equal to

$\frac{73}{27}$

$\frac{146}{27}$

$\frac{146}{81}$

$\frac{126}{81}$

$n p = 4$
$n pq = 4/3$
$n = 6, p = 2/3, q = 1/3$
$54 (P (X = 2) + P (X = 1) + P (X = 0))$
$54 (^{6} C_{2} (\frac{2}{3})^{2} (\frac{1}{3})^{4} +^{6} C_{1} (\frac{2}{3})^{1} (\frac{1}{3})^{5} +^{6} C_{0} (\frac{2}{3})^{0} (\frac{1}{3})^{6})$
$= \frac{146}{27}$

Q. Let X be a binomially distributed random variable with mean 4 and variance 34​. Then 54P(X≤2) is equal to

2773​

27146​

81146​

81126​