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(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

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$\frac{4}{5}$

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(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}$

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

$X$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X)$	$0.1$	$0.2$	$0.2$	$0.3$	$0.15$	$0.05$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$k-1$	$3k$	$k$	$3k$	$3k^2$	$k^2$	$k^2+k$

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

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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}$