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(A) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = 0.1 + k + 2k + 2k + k = 1$$0.1 + 6k = 1 \implies 6k = 0.9 \i

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$0.75, 0.3, 0.55$

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(A) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = 0.1 + k + 2k + 2k + k = 1$$0.1 + 6k = 1 \implies 6k = 0.9 \implies k = 0.15$The probability distribution is:$X$$0$$1$$2$$3$$4$$P(X)$$0.1$$0.15$$0.3$$0.3$$0.15$$1$. Probability of studying at least two hours $(X \geq 2)$:$P(X=2) + P(X=3) + P(X=4) = 0.3 + 0.3 + 0.15 = 0.75$$2$. Probability of studying exactly two hours $(X=2)$:$P(X=2) = 0.3$$3$. Probability of studying at most two hours $(X \leq 2)$:$P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.15 + 0.3 = 0.55$

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

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$K$	$3K$	$5K$	$7K$	$8K$	$K$

Similar Questions

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

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$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$