The probability distribution of a discrete random variable X is given below: X 2 3 4 5 P(X) (5/k) (7/k) (9/k) (11/k) The value of k is

Question

The probability distribution of a discrete random variable X is given below: X 2 3 4 5 P(X) (5/k) (7/k) (9/k) (11/k) The value of k is (A) 8 (B) 16 (C) 32 (D) 48

Tardigrade Admin · Accepted Answer

We have Sigma P(X) = 1 ⇒ (5/k) + (7/k) +(9/ k) +(11/k) = 1 ⇒ (32/k)= 1 ⇒ k= 32

X	$2$	$3$	$4$	$5$
P(X)	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$

Q. The probability distribution of a discrete random variable $X$ is given below:

X $2$ $3$ $4$ $5$

P(X) $\frac{5}{k}$ $\frac{7}{k}$ $\frac{9}{k}$ $\frac{11}{k}$

The value of $k$ is

$8$

$16$

$32$

$48$

We have $Σ P (X) = 1$
$\Rightarrow \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1$
$\Rightarrow \frac{32}{k} = 1$
$\Rightarrow k = 32$

Q. The probability distribution of a discrete random variable X is given below: X 2 3 4 5 P(X) k5​ k7​ k9​ k11​ The value of k is

8

16

32

48

We have ΣP(X)=1 ⇒k5​+k7​+k9​+k11​=1 ⇒k32​=1 ⇒k=32

Q. The probability distribution of a discrete random variable $X$ is given below:

X $2$ $3$ $4$ $5$

P(X) $\frac{5}{k}$ $\frac{7}{k}$ $\frac{9}{k}$ $\frac{11}{k}$

The value of $k$ is

$8$

$16$

$32$

$48$

We have $Σ P (X) = 1$
$\Rightarrow \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1$
$\Rightarrow \frac{32}{k} = 1$
$\Rightarrow k = 32$