How to Solve Discrete Random Variables — Univariate RV — moments of a tabulated distribution Questions on Exam P

Exam P Topic: Discrete Random Variables — Univariate RV — moments of a tabulated distribution Verified Procedural Question
Sample Practice Problem ID: #JQ10P
A quality score \(X\) has distribution (value:probability) 20:0.15, 30:0.10, 40:0.05, 50:0.20, 60:0.10, 70:0.10, 80:0.30. Calculate the probability a score is within one standard deviation of the mean.
(A)0.3000
(B)0.3375
(C)0.1125
(D)0.1500
(E)0.4500
📖 Worked Solution & Strategy
The score has a finite discrete distribution, so its moments are exact probability-weighted sums. After computing the mean and standard deviation, only listed support values satisfying the event may contribute probability. First calculate mean \(\mu=\sum_x x p(x)=55.0000\) and second moment \(\operatorname{E}[X^2]=\sum_x x^2 p(x)=3500.00\). The standard deviation is \(\sigma=\sqrt{\operatorname{E}[X^2]-\mu^2}=21.7945\). The interval \([\mu-\sigma, \mu+\sigma]\) contains support values \(\{40,50,60,70\}\). Summing their probabilities gives \(\Pr(\lvert X-\mu\rvert\le\sigma)=\frac{9}{20}\). Thus \(\Pr(\lvert X-\mu\rvert\leq\sigma)=\frac{9}{20}\), which matches the correct option.
Therefore, the result is \(\frac{9}{20}\approx 0.4500\), so the correct answer is option (E).

Final Answer: Option (E)

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