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Question
A random variable X has the following probability mass function.
| x | 1 | 2 | 3 | 4 | 5 |
| F(x) | k2 | 2k2 | 3k2 | 2k | 3k |
Find the value of k
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Solution
Given f(x) in a probability mass function
`sum_x` f(x) = 1
k2 + 2k2 + 3k2 + 2k + 3k = 1
6k2 + 5k = 1
6k2 + 5k – 1 = 0
(k + 1)(6k – 1) = 0
k = `1/6`
k ≠ –1 neglecting negative terms
Probability mass function
| x | 1 | 2 | 3 | 4 | 5 |
| F(x) | `1/36` | `2/36` | `3/36` | `2/6` | `3/6` |
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