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Absolute Value Function

X=x 01234567...

Question

$X = x$ $0$ 1 2 3 4 5 6 7
$P (X = x)$ $0$ $k$ $2 k$ $2 k$ $3 k$ $K^{2}$ $2 k^{2}$ $7 k^{2} + k$
then $P (0 < X < 5) =$

\frac{1}{10}

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\frac{3}{10}

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\frac{8}{10}

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\frac{7}{10}

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Solution

The correct option is D $\frac{8}{10}$
$\sum P (X = x_{1}) = 1$
$9 K + 10 k^{2} = 1$
$10 K^{2} + 9 K - 1 = 0$
$10 K^{2} + 10 K - K - 1 = 0$
$10 K (k + 1) - 1 (k + 1) = 0$
$K = \frac{1}{10}$
$P (0 < X < 5) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) = 8 K = \frac{8}{10}$

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