Question

Question: An octahedral die has eight faces that are numbered \(1\) through \({\bf{8}}\).

a) What is the expected value of the number that comes up when a fair octahedral die is rolled?

b) What is the variance of the number that comes up when a fair octahedral die is rolled?

Short answer

Answer

a) The Expected value of the number that comes up when a fair octahedral die is rolled is \(\frac{9}{2}\).

b) The variance of the numbers that comes up when a fair octahedral die is rolled is \(\frac{{21}}{4}\).

Step-by-step solution

Given data

A fair octahedral die is rolled.

Concept of Probability 

Probability is simply how likely something is to happen. Whenever we’re unsure about the outcome of an event, we can talk about the probabilities of certain outcomes—how likely they are. The analysis of events governed by probability is called statistics.

Calculation for the expected value

a)

Let us assume\(X\)is the number that comes up when a fair octahedral die is rolled. The variable\(X\)takes the random values\(1\;,\;2\;,\;3\;,\;4\)or\(8\)with the probability\(1/8\).

Expected value is given as:

\(\begin{aligned}{}E(X) &= \frac{1}{8}(1 + 2 + \ldots .. + 8)\\E(X) &= \frac{9}{2}\end{aligned}\)

Therefore, the Expected value of the number that comes up is \(\frac{9}{2}\).

Calculation for the variance

b)

A fair octahedral die is rolled.

\(\begin{aligned}{}E\left( {{X^2}} \right) &= \frac{1}{8}\left( {{1^2} + {2^2} + {3^2} + {4^2} + {5^2} + \ldots \ldots .. + {8^2}} \right)\\E\left( {{X^2}} \right) &= \frac{{51}}{2}\end{aligned}\)

Variance is given as:

\(\begin{aligned}{}V(X) &= E\left( {{X^2}} \right) - E{()^X}\\V(X) &= \frac{{51}}{2} - {()^{\frac{9}{2}}}\\V(X) &= \frac{{21}}{4}\end{aligned}\)

Therefore, the variance of the numbers that comes up is \(\frac{{21}}{4}\).