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

Sum of all probabilities must be equal to $1$
So we get $K + 2 K + 2 K + 3 K + K^{2} + 2 K^{2} + 7 K^{2} + K = 1$
$\Rightarrow 10 K^{2} + 9 K = 1$
$\Rightarrow K = - 1, \frac{1}{10}$
But probability cannot be negative , therefore the value of $K = \frac{1}{10}$
The mean is $1. K + 2.2 K + 3.2 K + 4.3 K + 5 K^{2} + 6.2 K^{2} + 7.7 K^{2} + 7 K$
$= 30 K + 66 K^{2} = 3 + 0.66 = 3.66$

$P (0 < x < 5) = P (x = 1) + P (x = 2) + P (x = 3) + P (x = 4) = K + 2 K + 2 K + 3 K = 8 K = 0.8$

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