Respuesta :
Answer:
1) [tex]\frac{2}{5}[/tex]
2) 49.225
3) [tex]\frac{7}{2}[/tex]
Step-by-step explanation:
1) To find the expected value of the dice we can use the following equation:
[tex]E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})[/tex]
So in our problem the values x will be: 1/1, 1/2, 1/3, 1/4, 1/5 and 1/6 and the probavility for all values is 1/6 so the expected values will be:[tex]E(x)=(\frac{1}{1} *\frac{1}{6}) +(\frac{1}{2} *\frac{1}{6}) +(\frac{1}{3} *\frac{1}{6})+(\frac{1}{4} *\frac{1}{6})+(\frac{1}{5} *\frac{1}{6})+(\frac{1}{6} *\frac{1}{6})[/tex]
[tex]E(x)=0.167+0.083+0.056+0.042+0.033+0.028=0.409\approx \frac{2}{5}[/tex]
2) To find the variance of the expected values we can use the equation:
[tex]Var(x)=\frac{\sum_{i=1}^{n}(x_{i}-\overline{x})^{2} }{n}[/tex]
So for our problem will be:
[tex]Var(x)=\frac{(3-10.5)^2+(4-10.5)^2+(17-10.5)^2+(18-10.5)^2}{4}[/tex]
[tex]Var(x)=\frac{56.25+42.25+42.25+56.25}{4}[/tex][tex]Var(x)=\frac{196.9}{4}=49.225[/tex]
3) To find the expected value of the dice we can use the following equation:
[tex]E(x)=x_{1}*P(x_{1})+x_{2}*P(x_{2})+...+x_{n}*P(x_{n})[/tex]
So in our problem the values x will be: 1, 2, 3, 4, 5 and 6 and the probavility for all values is 1/6 so the expected values will be:[tex]E(x)=(1*\frac{1}{6}) +(2 *\frac{1}{6}) +(3 *\frac{1}{6})+(4 *\frac{1}{6})+(5 *\frac{1}{6})+(6 *\frac{1}{6})[/tex]
[tex]E(x)=0.17+0.33+0.5+0.67+0.83+1=3.5\approx \frac{7}{2}[/tex]