A random variable $X$ has the following probability mass function.
$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P (X = x)$ $k$ $3 k$ $5 k$ $7 k$ $9 k$ $11 k$ $13 k$
(a) Find $k$ .
(b) Evaluate $P (X < 4), P (X$ $\geq$ $5)$ and $P (3 < X$ $\leq$ $6)$ .
(c) What is the smallest value of $x$ for which $P (X$ $\leq$ $x) >$ $\frac{1}{2}$ ?

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Solution

(a) $P (X = x)$ is a probability mass function.
$6 \sum x = 0 P (X = x) = 1$
(i.e.,) $P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6) = 1$ .
$\Rightarrow k + 3 k + 5 k + 7 k + 9 k + 11 k + 13 k = 1$
$\Rightarrow 49 k = 1 \Rightarrow k = \frac{1}{49}$
(b) $P (X < 4) = P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)$
$= \frac{1}{49} + \frac{3}{49} + \frac{5}{49} + \frac{7}{49} = \frac{16}{49}$
$P (X \geq 5) = P (X = 5) + P (X = 6) = \frac{11}{49} + \frac{13}{49} = \frac{24}{49}$
$P (3 < x \leq 6) = P (X = 4) + P (X = 5) + P (X = 6)$
$= \frac{9}{49} + \frac{11}{49} + \frac{13}{49} = \frac{33}{49}$
(c) The minimum value of $x$ may be determined by trial and error method.
$P (X \leq 0) = \frac{1}{49} < \frac{1}{2}; P (X \leq 1) = \frac{4}{49} < \frac{1}{2}$
$P (X \leq 2) = \frac{9}{49} < \frac{1}{2}; P (X \leq 3) = \frac{16}{49} < \frac{1}{2}$
$P (X \leq 4) = \frac{25}{49} > \frac{1}{2}$
$∴$ the smallest value of $x$ for which $P (X \leq x) > \frac{1}{2}$ is $4$ .

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