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Question
The discrete random variable X has the probability function.
| Value of X = x |
0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| P(x) | 0 | k | 2k | 2k | 3k | k2 | 2k2 | 7k2 + k |
Find k
Solution
Since the condition of Probability Mass function
`sum_("i" = 2)^oo` P(xi) = 1
`sum_("i" = 0)^7` P(xi) = 1
P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4) + P
P(x = 6) + P(x = 7) = 1
0 + k + 2k + 2k + 3k + k2 + 2k2 + 7k2 + k = 1
10k2 + 9k – 1 = 0
10k2 + 10k – k – 1 = 0
10k(k + 1) – 1(k + 1) = 0
(k + 1)(10k – 1) = 0
k + 1 = 0, 10k – 1 = 0
k = – 1, 10k = 1
k = `1/10`
k = – 1 is not possible
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