Mathematics, 05.09.2019 18:30 renee9913
Abox contains four bowling balls numbered 2, 4, 6, and 10 according to their weight. two are drawn randomly without replacement. let the random variable m denote the max of the two numbers on the balls drawn. calculate e(m) and var(m). hint: start by specifying the joint distribution of the number on the balls.
Answers: 1
![\begin{array}{|l|l|}\cline{1-2}\\[-1em]\text{Larger Number} & \text{Smaller Number}\\ \cline{1-2}\\[-1em]4 & 2 \\\cline{1-2}\\[-1em]6 & 2, ~4\\\cline{1-2}\\[-1em] 10 & 2, ~4, ~6\\\cline{1-2} \end{array}](/tpl/images/0223/6083/e4a85.png)
.![\begin{array}{|c|c|}\cline{1-2}\\[-1em]m & P(\text{M} = m)\\\cline{1-2}\\[-1em] 2 & 0\\\cline{1-2}\\[-1em] 4 & 1/6 \\\cline{1-2}\\[-1em] 6 & 2/6\\\cline{1-2}\\[-1em] 10 & 3/6\\\cline{1-2}\end{array}](/tpl/images/0223/6083/6f25a.png)
.![\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 \times \frac{1}{6} + 6\times \frac{2}{6} + 10 \times \frac{3}{6}\\ &= \frac{23}{3}\end{aligned}](/tpl/images/0223/6083/a4a0f.png)
.![\begin{aligned}Var(\mathrm{M}) &= \sum{[m^{2}\cdot P(\mathrm{M}= m)]} - \mu^{2}\\&= 4^{2}\times \frac{1}{6} + 6^{2} \times \frac{2}{6} + 10^{2} \times \frac{3}{6} - \left(\frac{23}{3}\right)^{2}\\&= \frac{53}{9}\end{aligned}](/tpl/images/0223/6083/bbe82.png)
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Abox contains four bowling balls numbered 2, 4, 6, and 10 according to their weight. two are drawn r...
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