A random variable $X$ has the following probability distribution.
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P (X = x)$ $0$ $k$ $2 k$ $2 k$ $3 k$ $k^{2}$ $2 k^{2}$ $7 k^{2} + k$
Find (i) $k$ (ii) The mean and (iii) $P (0 < X < 5)$

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Solution

(i) The total probability should add up to $1$ .
$∴ 0 + k + 2 k + 2 k + 3 k + k^{2} + 2 k^{2} + 7 k^{2} + k = 1$
$∴ 10 k^{2} + 9 k - 1 = 0$
$∴ 10 k^{2} + 10 k - k - 1 = 0$
$∴ (10 k - 1) (k + 1) = 0$
Since $k$ cannot be negative, $k = \frac{1}{10}$

(ii) Mean $= \sum x_{i} P_{i}$
$= 0 \times 0 + 1 \times k + 2 \times 2 k + 3 \times 2 k + 4 \times 3 k + 5 \times k^{2} + 6 \times 2 k^{2} + 7 \times (7 k^{2} + k)$
$= k + 4 k + 6 k + 12 k + 5 k^{2} + 12 k^{2} + 49 k^{2} + 7 k$
$= 30 k + 66 k^{2}$
$= 3 + 0.66$
$= 3.66$

(iii) $P (0 < X < 5) = P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)$
$∴ P (0 < X < 5) = k + 2 k + 2 k + 3 k = 9 k = 0.9$

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