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A hospital researcher records X = the number of times a post-op patient rings the nurse in a 12-hour shift, for a random sample of 50 patients, with P(x=0)=4/50, P(x=1)=8/50, P(x=2)=16/50, P(x=3)=14/50, P(x=4)=6/50, and P(x=5)=2/50. Do these six probabilities form a valid discrete PDF?
ANo, because P(x=2) exceeds one half
BNo, because x should start at one
CYes, because every x value is an integer
DYes, because the six values sum to one
Answer & Solution
Correct answer: D. Yes, because the six values sum to one
1. Add the six fractions: 4 + 8 + 16 + 14 + 6 + 2 = 50, all over 50.
2. 50/50 = 1, so the probabilities sum to one, and every individual value lies between zero and one.
3. Both characteristics of a discrete PDF hold, so the table is valid.
4. P(x=2) = 16/50 = 0.32, well under one half, so option B is false, and option C wrongly demands x start at one, which the definition never requires.
_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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