Exercise 149 Page 457

a To find the expected value for a spin, we first have to calculate the expected value for each region. This is the product of the probability of spinning it and its value. From the diagram, we know that one region has a central angle of 135^(∘) and another has a central angle of 90^(∘).

The region that pays 4, will occupy the remaining angle of the spinner. With this information, we can write and solve an equation. θ+90^(∘)+135^(∘)=360^(∘) ⇔ θ =135^(∘) Let's add this information to the diagram.

Let's find the expected value from each region.

Winning	Probability	Winnings * Probability	Expected winning
- 12	90/360	- 12(90/360)	- 3
16	135/360	16(135/360)	6
4	135/360	4(135/360)	1.5

The expected value is the sum of the last column in the table above. - 3+6+1.5=4.5

b The expected value from the two regions that are not x, are unchanged. Therefore, the only calculation we have to redo is that of region x. Instead of earning 4, you earn - 8.

- 8(135/360)=- 3 Finally, we will add the expected value from all three regions to get the expected value from one spin when x=- 8. - 3+6+(- 3)=0

c Let's label the expected value from region x as e. Since the expected value of spinning the wheel once should be 6, we can write and solve the following equation.

- 3+6+e=6 ⇔ e=3The expected value from region x should be 3. By equating this number with the expression describing the expected value from region x, we can determine what this region's value must be. We get the following equation. x(135/360)=3 Let's solve this equation for x.

x(135/360)=3

CalcQuot

Calculate quotient

x(0.375)=3

DivEqn

.LHS /0.375.=.RHS /0.375.

x=8

If you win 8 when you get region x, the expected value from one spin on the wheel is 6.