Discussion
Suppose that 12 independent observations on a random variable X are given by 2, 2, 2, 2, 4, 4, 5, 5, 6, 9, 9, 10 and that Y is a random variable with c.d.f. \(F_{12}\), where \(F_{12}\) is the empirical distribution function of the above data. Then :
Answer: B
The true statement is \(P[Y \le 6] = 0.75\).
A variable with the empirical distribution function \(F_{12}\) puts mass 1/12 on each of the 12 observations. Nine of them (2, 2, 2, 2, 4, 4, 5, 5, 6) are at most 6, so \(P[Y \le 6] = 9/12 = 0.75\).
The others fail. The mean is the sample mean:
\[ \begin{aligned} E(Y) &= \frac{8 + 8 + 10 + 6 + 18 + 10}{12} \\ &= \frac{60}{12} = 5 \end{aligned} \]
not 4.5. \(P(Y = 7) = 0\) because 7 was not observed, while \(P(Y = 6) = 1/12\). And \(P(Y = 2) = 4/12\) is larger than \(P(Y = 9) = 2/12\), not smaller.
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