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Question
State if the following is not the probability mass function of a random variable. Give reasons for your answer
| Z | 3 | 2 | 1 | 0 | −1 |
| P(Z) | 0.3 | 0.2 | 0.4 | 0 | 0.05 |
Solution 1
P.m.f. of random variable should satisfy the following conditions:
- 0 ≤ pi ≤ 1
- ∑pi = 1
| Z | 3 | 2 | 1 | 0 | −1 |
| P(Z) | 0.3 | 0.2 | 0.4 | 0 | 0.05 |
Here ∑pi = 0.3 + 0.2 + 0.4 + 0 + 0.05
= 0.95 ≠ 1
Hence, P(Z) cannot be regarded as p.m.f. of the random variable Z.
Solution 2
Here, pi ≥ 0, `AA` i = 1, 2, ...., 5
Now consider,
`sum_("i" = 1)^5 "P"_"i"` = 0.3 + 0.2 + 0.4 + 0 + 0.05
= 0.95 ≠ 1
∴ Given distribution is not p.m.f.
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