# Problems from 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.
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### Development:

 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 $\rightarrow \sigma=38,56$

### Solution:

 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 $\rightarrow \sigma=38,56$

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