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(D) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(x_i) = 0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1

Vedclass · Accepted Answer

$0.33$

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(D) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(x_i) = 0 + 2k + k + 3k + 2k^2 + 2k + (k^2 + k) + 7k^2 = 1$Combining the terms,we get: $10k^2 + 9k = 1$Rearranging into a quadratic equation: $10k^2 + 9k - 1 = 0$Factoring the equation: $(10k - 1)(k + 1) = 0$Since $k$ must be positive for probabilities to be non-negative,we have $k = \frac{1}{10} = 0.1$.We need to find $P(3 $P(3 Substituting $k = 0.1$:$P(3 < x \leq 6) = 3(0.1)^2 + 3(0.1) = 3(0.01) + 0.3 = 0.03 + 0.3 = 0.33$.

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

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

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:

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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: