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}$

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

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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

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

$A$ 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 $\sigma^2 + \mu^2 =$

$A$ person throws an unbiased die. If the number shown is even,he gains an amount equal to the number shown. If the number is odd,he loses an amount equal to the number shown. Then his expectation is ₹.

Let $p(x)$ represent the probability mass function of a Poisson distribution. If its mean $\lambda = 3.725$, then the value of $x$ at which $p(x)$ is maximum is

The following table shows the probability distribution of smart phones sold in a shop per day:
Number of smart phones $(x)$ $0$ $1$ $2$ $3$ $4$ $5$
Probability $(P(x))$ $k$ $0.3$ $0.15$ $0.15$ $0.1$ $2k$

Then $E(x) = ?$

$A$ 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

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}$