Mathematics: 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.

v= 80, t=100Step-by-step explanation:yea...

Abox contains four bowling balls numbered 2, 4, 6,

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zay9357

02.01.2022

Mathematics

E(M) = 23/3

Var(M) = 53/9.

Step-by-step explanation:

There are three possible values for M: 4, 6, and 10.

$\begin{array}{|l|l|}\cline{1-2}\\[-1em]\text{Larger Number} & \text{Smaller Number}\\ \cline{1-2}\\[-1em]4 & 2 \\\cline{1-2}\\[-1em]6 & 2, ~4\\\cline{1-2}\\[-1em] 10 & 2, ~4, ~6\\\cline{1-2} \end{array}$ .

That's six unique combinations in total. If the balls are drawn randomly, the probability for getting each combination shall be equal. That is:

$\begin{array}{|c|c|}\cline{1-2}\\[-1em]m & P(\text{M} = m)\\\cline{1-2}\\[-1em] 2 & 0\\\cline{1-2}\\[-1em] 4 & 1/6 \\\cline{1-2}\\[-1em] 6 & 2/6\\\cline{1-2}\\[-1em] 10 & 3/6\\\cline{1-2}\end{array}$ .

Consider the formula for the expected value of a discrete random variable:

$\begin{aligned} E({\rm M})& = \sum{[m \cdot P(\mathrm{M} = m)]}\\ &= 4 \times \frac{1}{6} + 6\times \frac{2}{6} + 10 \times \frac{3}{6}\\ &= \frac{23}{3}\end{aligned}$ .

Formula for the variance of a discrete random variable (note that this formula can take many other forms):

$\begin{aligned}Var(\mathrm{M}) &= \sum{[m^{2}\cdot P(\mathrm{M}= m)]} - \mu^{2}\\&= 4^{2}\times \frac{1}{6} + 6^{2} \times \frac{2}{6} + 10^{2} \times \frac{3}{6} - \left(\frac{23}{3}\right)^{2}\\&= \frac{53}{9}\end{aligned}$ .

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