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(C) The sum of all probabilities in a probability distribution must be $1$. Thus,$\sum P(X=x) = 1$. Substituting the given values: $P(X=0) + P(X=1) +

Vedclass · Accepted Answer

$0.75$

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(C) The sum of all probabilities in a probability distribution must be $1$. Thus,$\sum P(X=x) = 1$. Substituting the given values: $P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 1$. $0.1 + k(1) + k(2) + k(5-3) + k(5-4) = 1$. $0.1 + k + 2k + 2k + k = 1$. $0.1 + 6k = 1$. $6k = 0.9$,so $k = 0.15$. We need to find the probability that you study at least two hours,which is $P(X \ge 2) = P(X=2) + P(X=3) + P(X=4)$. $P(X=2) = k(2) = 2(0.15) = 0.3$. $P(X=3) = k(5-3) = 2(0.15) = 0.3$. $P(X=4) = k(5-4) = 1(0.15) = 0.15$. $P(X \ge 2) = 0.3 + 0.3 + 0.15 = 0.75$.

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

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