The probability distribution of a random variable X is given below: X 0 1 2 3 P(X) k k 2 k 4 k 8 1. Determine the value of k. 2. Determine P(X 2) and P(X 2) 3. Find P(X 2)+P(X 2)

We have, X 0 1 2 3 P(X) k k 2 k 4 k 8 1. Since, _ i=1 ^nP_ i =1,i=1,2, ....,n and P_ i 0 + k 2 + k 4 + k 8 =1 8k+4k+2k+k=8 =8 15 2. P(X 2)=P(0)+P(1)+P(2) =k+ k 2 + k 4 = (4k+2k+k) 4 = 7k 4 =7 4 8 15 =14 15 And P(X 2)=P(3)= k 8 =1 8 8 15 =1 15 3. P(X 2)+P(X 2) =14 15 +1 15 =1

The probability distribution of a random variable X is given belo…

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Question

The probability distribution of a random variable X is given below:

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Determine the value of k.
Determine $\text{P}(\text{X}\leq2)$ and $\text{P}(\text{X}\geq2)$
Find $\text{P}(\text{X}\leq2)+\text{P}(\text{X}\geq2)$

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Answer

We have,

$\text{X}$	$0$	$1$	$2$	$3$
$\text{P}(\text{X})$	$\text{k}$	$\frac{\text{k}}{2}$	$\frac{\text{k}}{4}$	$\frac{\text{k}}{8}$

Since, $\sum_\limits{\text{i}=1}^\text{n}\text{P}_{\text{i}}=1,\text{i}=1,2, ....,\text{n}$ and $\text{P}_{\text{i}}\geq0$

$\therefore\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}+\frac{\text{k}}{8}=1$
$\Rightarrow8\text{k}+4\text{k}+2\text{k}+\text{k}=8$
$\therefore\text{k}=\frac{8}{15}$

$\text{P}(\text{X}\leq2)=\text{P}(0)+\text{P}(1)+\text{P}(2)$

$=\text{k}+\frac{\text{k}}{2}+\frac{\text{k}}{4}$
$=\frac{(4\text{k}+2\text{k}+\text{k})}{4}=\frac{7\text{k}}{4}$
$=\frac{7}{4}\cdot\frac{8}{15}=\frac{14}{15}$
And $\text{P}(\text{X}\geq2)=\text{P}(3)=\frac{\text{k}}{8}$
$=\frac{1}{8}\cdot\frac{8}{15}=\frac{1}{15}$