Average, variance and standard deviation
We have the following discrete random: If the result of throwing a perfect dice is a prime number, the payoff will be the result times $10$. We include in the table these payoffs. Assign payoffs to the other results from throwing the dice.
- Fill in the following table:
| Result of the dice | probability | payoff |
| $1$ | $1/6$ | $10$ |
| $2$ | ? | ? |
| $3$ | ? | $30$ |
| $4$ | ? | ? |
| $5$ | $1/6$ | ? |
| $6$ | $1/6$ | ? |
Find the average payoff if we throw the dice only once.
Find the variance and the standard deviation.
| Result of the dice | probability | payoff |
| $1$ | $1/6$ | $10$ |
| $2$ | $1/6$ | $20$ |
| $3$ | $1/6$ | $30$ |
| $4$ | $1/6$ | $8$ |
| $5$ | $1/6$ | $50$ |
| $6$ | $1/6$ | $120$ |
$$\mu=\sum_i p_i\cdot x_i=\dfrac{1}{6}\cdot10+\dfrac{1}{6}\cdot20+\dfrac{1}{6}\cdot30+\dfrac{1}{6}\cdot8+\dfrac{1}{6}\cdot50+\dfrac{1}{6}\cdot120$$ $$\mu=\dfrac{238}{6}=39,67$$
The variance is calculated first: $$\sigma^2=\sum_i x_i^2\cdot p_i - \mu^2=\dfrac{1}{6}(10^2+20^2+30^2+8^2+50^2+120^2)-39,67^2$$
variance $\rightarrow \sigma^2=1486,95$
standard deviation $\rightarrow \sigma=38,56$
| Result of the dice | probability | payoff |
| $1$ | $1/6$ | $10$ |
| $2$ | $1/6$ | $20$ |
| $3$ | $1/6$ | $30$ |
| $4$ | $1/6$ | $8$ |
| $5$ | $1/6$ | $50$ |
| $6$ | $1/6$ | $120$ |
$\mu=\dfrac{238}{6}=39,67$
variance $\rightarrow \sigma^2=1486,95$
standard deviation $\rightarrow \sigma=38,56$