If the p.m.f. of a random variable X is given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.Calculating the probabilities:$P(X=0)

Vedclass · Accepted Answer

$P(X \leq 3) \leq P(X \geq 3)$

Vedclass · Answer

(C) The given probability mass function is $P(X=x) = \frac{\binom{5}{x}}{2^5}$ for $x \in \{0, 1, 2, 3, 4, 5\}$.Calculating the probabilities:$P(X=0) = \frac{\binom{5}{0}}{32} = \frac{1}{32}$,$P(X=1) = \frac{\binom{5}{1}}{32} = \frac{5}{32}$,$P(X=2) = \frac{\binom{5}{2}}{32} = \frac{10}{32}$,$P(X=3) = \frac{\binom{5}{3}}{32} = \frac{10}{32}$,$P(X=4) = \frac{\binom{5}{4}}{32} = \frac{5}{32}$,$P(X=5) = \frac{\binom{5}{5}}{32} = \frac{1}{32}$.Now evaluate the options:$A$. $P(X \leq 1) = P(0) + P(1) = \frac{1+5}{32} = \frac{6}{32}$ and $P(X \geq 4) = P(4) + P(5) = \frac{5+1}{32} = \frac{6}{32}$. Thus,$P(X \leq 1) = P(X \geq 4)$ is true.$B$. $P(X \leq 2) = P(0) + P(1) + P(2) = \frac{1+5+10}{32} = \frac{16}{32} = 0.5$. $P(X \geq 4) = \frac{6}{32} = 0.1875$. Since $0.5 \geq 0.1875$,this is true.$C$. $P(X \leq 3) = P(0) + P(1) + P(2) + P(3) = \frac{1+5+10+10}{32} = \frac{26}{32} = 0.8125$. $P(X \geq 3) = P(3) + P(4) + P(5) = \frac{10+5+1}{32} = \frac{16}{32} = 0.5$. Since $0.8125 \leq 0.5$ is false,option $C$ is not true.$D$. $P(X \leq 2) = \frac{16}{32} = 0.5$ and $P(X \geq 3) = \frac{16}{32} = 0.5$. Thus,$P(X \leq 2) = P(X \geq 3)$ is true.

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

If the p.m.f. of a random variable $X$ is given by $P(X=x) = \frac{\binom{5}{x}}{2^{5}}$ for $x = 0, 1, 2, \ldots, 5$ and $0$ otherwise,then which of the following is not true?

Similar Questions

$A$ player tosses two fair coins. He wins $Rs. 5$ if two heads appear, $Rs. 3$ if one head appears, and $Rs. 2$ if no head appears. The variance of the winning amount is:

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

$A$ random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4)$. If $\sigma^2$ is the variance and $\mu$ is the mean of $X$, then $\sigma^2 + \mu^2 =$

$A$ random variable $X$ takes the values $0, 1, 2$. Its mean is $1.2$. If $P(X=0)=0.3$,then $P(X=1)=$

$A$ random variable $X$ has the probability distribution given below. Its variance is:
$X$ $1$ $2$ $3$ $4$ $5$
$P(X=x)$ $K$ $2K$ $3K$ $2K$ $K$

Vedclass Test Series

Exam Paper Generator

Online Exam Module