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(D) The mean $E(X)$ is given by $\sum_{j=0}^{\infty} j \cdot P(X=j)$.Since $P(X=0) = 0 \cdot \frac{1}{2} = 0$,we have $E(X) = \sum_{j=1}^{\infty} j \c

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$\frac{3}{4}$ and $\frac{1}{8}$

Vedclass · Answer

(D) The mean $E(X)$ is given by $\sum_{j=0}^{\infty} j \cdot P(X=j)$.Since $P(X=0) = 0 \cdot \frac{1}{2} = 0$,we have $E(X) = \sum_{j=1}^{\infty} j \cdot \frac{1}{3^j}$.This is an arithmetico-geometric series of the form $\sum_{j=1}^{\infty} j r^j$ where $r = \frac{1}{3}$.The sum is given by $\frac{r}{(1-r)^2} = \frac{1/3}{(1-1/3)^2} = \frac{1/3}{4/9} = \frac{3}{4}$.For $P(X 	ext{ is positive and even})$,we sum $P(X=j)$ for $j \in \{2, 4, 6, \ldots\}$.$P(X 	ext{ is positive and even}) = \sum_{k=1}^{\infty} P(X=2k) = \sum_{k=1}^{\infty} \frac{1}{3^{2k}} = \sum_{k=1}^{\infty} \left(\frac{1}{9}ight)^k$.This is a geometric series with first term $a = \frac{1}{9}$ and common ratio $r = \frac{1}{9}$.The sum is $\frac{a}{1-r} = \frac{1/9}{1-1/9} = \frac{1/9}{8/9} = \frac{1}{8}$.Thus,the mean is $\frac{3}{4}$ and the probability is $\frac{1}{8}$.

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$2k$	$k$	$3k$	$2k^2$	$2k$	$k^2+k$	$7k^2$

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

Similar Questions

The distribution of a random variable $X$ is given below. The value of $k$ is:
$X = x$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $k$ $\frac{1}{5}$ $2k$ $\frac{3}{10}$ $k$

If a random variable $x$ has the probability distribution as follows:
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(x)$ $0$ $2k$ $k$ $3k$ $2k^2$ $2k$ $k^2+k$ $7k^2$

Then $P(3 < x \leq 6)$ is equal to:

The p.d.f. of a discrete random variable $X$ is defined as $f(x) = \begin{cases} kx^2, & x \in \{0, 1, 2, 3, 4, 5, 6\} \\ 0, & \text{otherwise} \end{cases}$. Then the value of $F(4)$ (c.d.f.) is:

$A$ random variable $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If $\operatorname{var}(X) : E(X) = 4 : 1$,then $n$ is equal to

$A$ random variable $X$ has the following probability distribution:

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

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