For the probability distribution of a discret | vedclass

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

Question

(B) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = \frac{1}{10} + (K + \frac{2}{10}) + (K + \frac{3}{10}) + (K

Vedclass · Accepted Answer

$\frac{4}{5}$

Vedclass · Answer

(B) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = \frac{1}{10} + (K + \frac{2}{10}) + (K + \frac{3}{10}) + (K + \frac{3}{10}) + (K + \frac{4}{10}) + (K + \frac{2}{10}) = 1$$\Rightarrow 5K + \frac{15}{10} = 1$$\Rightarrow 5K + 1.5 = 1$$\Rightarrow 5K = -0.5$$\Rightarrow K = -0.1 = -\frac{1}{10}$Now,substitute $K = -\frac{1}{10}$ into the table:For $x = -2, P = \frac{1}{10}$For $x = -1, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$For $x = 0, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$For $x = 1, P = -\frac{1}{10} + \frac{3}{10} = \frac{2}{10}$For $x = 2, P = -\frac{1}{10} + \frac{4}{10} = \frac{3}{10}$For $x = 3, P = -\frac{1}{10} + \frac{2}{10} = \frac{1}{10}$The mean $\mu = \sum x P(x) = (-2)(\frac{1}{10}) + (-1)(\frac{1}{10}) + (0)(\frac{2}{10}) + (1)(\frac{2}{10}) + (2)(\frac{3}{10}) + (3)(\frac{1}{10})$$\mu = \frac{-2 - 1 + 0 + 2 + 6 + 3}{10} = \frac{8}{10} = \frac{4}{5}$

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

For the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $K + \frac{2}{10}$ $K + \frac{3}{10}$ $K + \frac{3}{10}$ $K + \frac{4}{10}$ $K + \frac{2}{10}$

Similar Questions

$A$ player tosses two coins. He wins $Rs. 1$ if $1$ head appears,$Rs. 2$ if $2$ heads appear. But he loses $Rs. 3$ if no head appears. The mean of the prize money is

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(X)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,find $P[X=3]$.

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to

If the mean of a Poisson distribution is $\frac{1}{2}$, then the ratio of $P(X=3)$ to $P(X=2)$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module

For the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:$X = x$$-2$$-1$$0$$1$$2$$3$$P(X = x)$$\frac{1}{10}$$K + \frac{2}{10}$$K + \frac{3}{10}$$K + \frac{3}{10}$$K + \frac{4}{10}$$K + \frac{2}{10}$