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(D) The sum of probabilities in a probability distribution is $1$.$3K^2 + K + K^2 + 2K = 1$ $4K^2 + 3K - 1 = 0$ $(4K - 1)(K + 1) = 0$ Since $P(X=x) \g

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$\frac{15}{16}$

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(D) The sum of probabilities in a probability distribution is $1$.$3K^2 + K + K^2 + 2K = 1$ $4K^2 + 3K - 1 = 0$ $(4K - 1)(K + 1) = 0$ Since $P(X=x) \geq 0$, $K$ must be positive, so $K = \frac{1}{4}$. The distribution is: $X=x$$1$$3$$5$$2$$P(X=x)$$\frac{3}{16}$$\frac{1}{4}$$\frac{1}{16}$$\frac{1}{2}$ The mean $\mu = E(X) = \sum x_i P(x_i) = 1(\frac{3}{16}) + 3(\frac{1}{4}) + 5(\frac{1}{16}) + 2(\frac{1}{2}) = \frac{3}{16} + \frac{12}{16} + \frac{5}{16} + \frac{16}{16} = \frac{36}{16} = \frac{9}{4}$. $E(X^2) = \sum x_i^2 P(x_i) = 1^2(\frac{3}{16}) + 3^2(\frac{1}{4}) + 5^2(\frac{1}{16}) + 2^2(\frac{1}{2}) = \frac{3}{16} + \frac{36}{16} + \frac{25}{16} + \frac{32}{16} = \frac{96}{16} = 6$. Variance $\sigma^2 = E(X^2) - [E(X)]^2 = 6 - (\frac{9}{4})^2 = 6 - \frac{81}{16} = \frac{96 - 81}{16} = \frac{15}{16}$.

If a random variable $X$ has the following probability distribution, then its variance is
$X=x$ $1$ $3$ $5$ $2$
$P(X=x)$ $3 K^2$ $K$ $K^2$ $2 K$

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If a random variable $X$ has the following probability distribution, then its variance is $X=x$$1$$3$$5$$2$$P(X=x)$$3 K^2$$K$$K^2$$2 K$