Let X be a discrete random variable whose probability distribution is defined as follows:
$P(\mathrm{X}=x)= \begin{cases}k(x+1) & \text { for } x=1,2,3,4 \\ 2 k x & \text { for } x=5,6,7 \\ 0 & \text { otherwise }\end{cases}$
where k is a constant. Calculate
(i) the value of k
(ii) E (X)
(iii) Standard deviation of X.
(i) Given, P(X = x) = k(x + 1) for x = 1, 2, 3, 4
So, P(X = 1) = k(1 + 1) = 2k
P(X = 2) = k(2 + 1) = 3k
P(X = 3) = k(3 + 1) = 4k
P(X = 4) = k(4 + 1) = 5k
Also, P(X = x) = 2kx for x = 5, 6, 7
P(X = 5) = 2k(5) = 10k
P(X = 6) = 2k(6) = 12k
P(X = 7) = 2k(7) = 14k
And, for otherwise it is 0.
Thus, the probability distribution is given by
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