When an unfair dice is thrown,the probability | vedclass

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(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$. Since the sum of all probabilities must be $1$,we have: $\sum_{k=1}^6 P(X=k) = 1$ $P(1^2) + P(2

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$\frac{441}{91}$

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(D) Given $P(X=k) = k^2 P$ for $k = 1, 2, 3, 4, 5, 6$. Since the sum of all probabilities must be $1$,we have: $\sum_{k=1}^6 P(X=k) = 1$ $P(1^2) + P(2^2) + P(3^2) + P(4^2) + P(5^2) + P(6^2) = 1$ $P(1 + 4 + 9 + 16 + 25 + 36) = 1$ $91P = 1 \Rightarrow P = \frac{1}{91}$. The mean of $X$ is given by $E(X) = \sum_{k=1}^6 k \cdot P(X=k)$. $E(X) = \sum_{k=1}^6 k \cdot (k^2 P) = P \sum_{k=1}^6 k^3$. $E(X) = \frac{1}{91} (1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3)$. $E(X) = \frac{1}{91} (1 + 8 + 27 + 64 + 125 + 216) = \frac{441}{91}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

When an unfair dice is thrown,the probability of getting a number $k$ on it is $P(X=k)=k^2 P$,where $k=1, 2, 3, 4, 5, 6$ and $X$ is the random variable denoting a number on the dice,then the mean of $X$ is

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Then $E(x) = ?$

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