The probability distribution of a random vari | vedclass

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(C) Given the expected value $E(X) = \sum X_i P(X_i) = \frac{263}{15}$.Calculating the sum: $E(X) = (4k \cdot \frac{2}{15}) + (\frac{30}{7}k \cdot \fr

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$\frac{11}{15}$

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(C) Given the expected value $E(X) = \sum X_i P(X_i) = \frac{263}{15}$.Calculating the sum: $E(X) = (4k \cdot \frac{2}{15}) + (\frac{30}{7}k \cdot \frac{1}{15}) + (\frac{32}{7}k \cdot \frac{2}{15}) + (\frac{34}{7}k \cdot \frac{1}{5}) + (\frac{36}{7}k \cdot \frac{1}{15}) + (\frac{38}{7}k \cdot \frac{2}{15}) + (\frac{40}{7}k \cdot \frac{1}{5}) + (6k \cdot \frac{1}{15})$.$E(X) = \frac{k}{105} [ (28 \cdot 2) + (30 \cdot 1) + (32 \cdot 2) + (34 \cdot 3) + (36 \cdot 1) + (38 \cdot 2) + (40 \cdot 3) + (42 \cdot 1) ] = \frac{k}{105} [ 56 + 30 + 64 + 102 + 36 + 76 + 120 + 42 ] = \frac{526k}{105} = \frac{263}{15}$.Solving for $k$: $k = \frac{263}{15} \cdot \frac{105}{526} = \frac{263}{526} \cdot \frac{105}{15} = \frac{1}{2} \cdot 7 = \frac{7}{2}$.Substituting $k = \frac{7}{2}$ into the values of $X$: $X$ takes values ${14, 15, 16, 17, 18, 19, 20, 21}$.We need $P(X $P(X < 20) = \frac{2}{15} + \frac{1}{15} + \frac{2}{15} + \frac{1}{5} + \frac{1}{15} + \frac{2}{15} = \frac{2+1+2+3+1+2}{15} = \frac{11}{15}$.

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$X=x$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

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