Mathematics, 04.01.2021 21:10 abdullaketbi71

You roll 2 dice. If the sum of the dice is: 2, 3, 4 you win $20. If the sum is 5, 6, 7, 8 you win $10. If the sum is 9, 10, 11 you lose $20. If the sum is 12 you lose $25. It costs $5 to play. Create a probability distribution of the random variable, "winnings" including the cost to play, for this game. What is the expected value of the "winnings" for this game? What is the variance for the "winnings" for this game? What is the standard deviation? Is this game a "fair game"? Why or why not?

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Answer from: quece233

1) The expected value of the "winnings" is $(-0.97 $\overline 2$ )

2) The variance for the "winnings" is $0.57966

3) The standard deviation for the "winnings" is$0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

Step-by-step explanation:

1) The probability of having a sum of 2 = 1/6×1/6 = 1/36

The probability of having a sum of 3 = 1/6×1/6 = 1/36

The probability of having a sum of 4 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 5 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 6 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 7 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 8 = 1/6×1/6 + 1/6×1/6 + 1/6×1/6 = 1/12

The probability of having a sum of 9 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 10 = 1/6×1/6 + 1/6×1/6 = 1/18

The probability of having a sum of 11 = 1/6×1/6 = 1/36

The probability of having a sum of 12 = 1/6×1/6 = 1/36

The values are;

For 2, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 3, we have 1/36 × (20 - 5) = 0.41 $\overline 6$

For 4, we have 1/18 × (20 - 5) = 0.8 $\overline 3$

For 5, we have 1/18 × (10 - 5) = 0.2 $\overline 7$

For 6, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 7, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 8, we have 1/12 × (10 - 5) = 0.41 $\overline 6$

For 9, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 10, we have 1/18 × (-20 - 5) = -1.3 $\overline 8$

For 11, we have 1/36 × (-20 - 5) = -0.69 $\overline 4$

For 12, we have 1/36 × (-25 - 5) = -0.69 $\overline 4$

The expected value of the winnings is given as follows;

0.41 $\overline 6$ + 0.41 $\overline 6$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.8 $\overline 3$ + 0.41 $\overline 6$ + 0.41 $\overline 6$ + -1.3 $\overline 8$ -1.3 - 0.69 $\overline 4$ - 0.69 $\overline 4$ = -0.97 $\overline 2$

Therefore, the expected value = $-0.97 $\overline 2$ , which is one is expected to lose $0.97 $\overline 2$

2) Using Microsoft Excel, we have;

The variance for the "winnings", σ² = $0.57966

3) The standard deviation for the "winnings" = √σ² = √(0.57966) ≈ $0.761354

4) The game is not a fair game because one is expected to lose $0.97 $\overline 2$

Answer from: Quest

(0,-3) see attached graph

step-by-step explanation:

a. to solve the system, graph the two equations.

x = y+3 becomes y=x-3 has m=1/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move up 1 unit and 1 unit to the right. plot a point and connect.

y=-4x+3 has m =-4/1 and b=-3. start at (0,-3) on the y-axis and plot a point. then move down four units and to the right 1 unit. plot a point and connect.

the solution is the point (x,y) where the two lines intersect. by the graph below it is (0,-3).

b. to solve by substitution, substitute y = -4x-3 into x= y+3

x= (-4x-3) +3

x=-4x

5x=0

x=0

now substitute x = 0 back in to find y.

y= -4(0)-3

y=0-3

y=-3

solution is (0,-3)

c. use elimination by adding the equations together.