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A table gives P(x=1)=0.15, P(x=2)=0.25, P(x=3)=0.30, P(x=4)=0.20, P(x=5)=0.15. What is wrong with this probability distribution table?
AThe x values are not listed in order
BThe values of P(x) do not sum to one
CEvery P(x) value is too small to use
DThe random variable X is not discrete here
Answer & Solution
Correct answer: B. The values of P(x) do not sum to one
1. Add the five listed probabilities: 0.15 + 0.25 + 0.30 + 0.20 + 0.15 = 1.05.
2. A discrete PDF requires the column of probabilities to sum to exactly one.
3. 1.05 does not equal one, so the table breaks the second characteristic of a discrete PDF.
4. The x values are already in order and X is discrete, ruling out options B, C, and D.
_Source: OpenStax Introductory Statistics (CC BY 4.0), Ch 4 "Discrete Random Variables", section 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable_
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