a List all the possible sums. To calculate the relative frequency of a sum, divide the number of ways which we can obtain the sum by the total number of sums of two spins.
B
b Make a bar graph using the table from Part A. Each bar should represent a sum, and the bar's height should be proportional to the corresponding relative frequency.
C
cSimulate the spinner using a random number generator.
D
d Make a bar graph using the table from Part C. Each bar should represent a sum and the bar's height should be proportional to the corresponding relative frequency.
E
e The expected value is what you get if you add up the products of the possible sums of two spins and their corresponding theoretical probabilities.
a To construct the relative-frequency table, we need to know all the possible sums and their frequencies after two spins of the wheel. Let's look at the wheel.
To find the sums and their frequencies, we will create a table with all the possible sums. The first column of the table will represent the first spin, and the first row will represent the second spin. Values in the other cells will be the sums of the corresponding spins. Let's calculate the values in the first row!
4
5
10
2
12
8
7
6
4
4+4=8
4+5=9
4+10=14
4+2=6
4+12=16
4+8=12
4+7=11
4+6=10
5
9
10
15
7
17
13
12
11
10
14
15
20
12
22
18
17
16
2
6
7
12
4
14
10
9
8
12
16
17
22
14
24
20
19
18
8
12
13
18
10
20
16
15
14
7
11
12
17
9
19
15
14
13
6
10
11
16
8
18
14
13
12
Now that we have a table of all the sums, we can calculate their frequencies by simply counting them. For example, let's count in how many ways the sum of two spins can be equal to 17.
4
5
10
2
12
8
7
6
4
8
9
14
6
16
12
11
10
5
9
10
15
7
17
13
12
11
10
14
15
20
12
22
18
17
16
2
6
7
12
4
14
10
9
8
12
16
17
22
14
24
20
19
18
8
12
13
18
10
20
16
15
14
7
11
12
17
9
19
15
14
13
6
10
11
16
8
18
14
13
12
As we can see, 17 has a frequency of four. We can calculate the other frequencies in the same way, then list them all in a relative-frequency table. The relative frequency is frequency divided by the number of total possible outcomes, which is equal to 64.
Sum
Frequency
Relative Frequency
4
1
1/64
6
2
2/64=1/32
7
2
2/64=1/32
8
3
3/64
9
4
4/64=1/16
10
5
5/64
11
4
4/64=1/16
12
7
7/64
13
4
4/64=1/16
14
7
7/64
15
4
4/64=1/16
16
5
5/64
17
4
4/64=1/16
18
4
4/64=1/16
19
2
2/64=1/32
20
3
3/64
22
2
2/64=1/32
24
1
1/64
b
The relative-frequency table we made in Part A will be essential for creating a theoretical probability distribution graph. Relative frequencies in that table are also the theoretical probabilities of the sums. Let's look at that table again! We will only focus on the sums and their relative frequencies.
Sum
Relative Frequency (Theoretical Probability)
4
1/64
6
1/32
7
1/32
8
3/64
9
1/16
10
5/64
11
1/16
12
7/64
13
1/16
14
7/64
15
1/16
16
5/64
17
1/16
18
1/16
19
1/32
20
3/64
22
1/32
24
1/64
It will be easier for us to graph the probabilities if they are in decimal notation. Let's convert them!
Sum
Relative Frequency (Theoretical Probability)
4
1/64≈ 0.016
6
1/32≈ 0.031
7
1/32≈ 0.031
8
3/64≈ 0.049
9
1/16≈ 0.063
10
5/64≈ 0.078
11
1/16≈ 0.063
12
7/64≈ 0.109
13
1/16≈ 0.063
14
7/64≈ 0.109
15
1/16≈ 0.063
16
5/64≈ 0.078
17
1/16≈ 0.063
18
1/16≈ 0.063
19
1/32≈ 0.031
20
3/64≈ 0.047
22
1/32≈ 0.031
24
1/64≈ 0.016
Now we can draw a theoretical probability distribution graph. To do that, we will assign a bar to each sum. The bar's height will represent the probability of the sum.
c We will simulate the wheel spins using a random number generator. To make a simulation, you can generate random numbers from 1 to 8. Each number will represent one color of the wheel and its corresponding value.
Outcome of the Spin
Represented by
4
1
5
2
10
3
2
4
12
5
8
6
7
7
6
8
Let's use a graphing calculator to generate integers from 1 to 8. Start by pushing the MATH button. Then scroll right to the PRB menu and choose the randInt( option.
Since we want to simulate the sum of two spins, we will need to generate numbers from 1 to 8 twice to calculate a sum. To do that, we type 1, 8, and 2 on the graphing calculator, respectively. Then press ENTER to get two outcomes.
Let's say that we generated the numbers 3 and 5. If we look at the table we made at the beginning of this part, we can see that 3 represents the situation where we spun 10 on the wheel, and 5 represents the situation where we spun 12.
Therefore, we simulated the sum 10+12=22. Now, we have to repeat this this process 100 times to get 100 sums. We will keep track of the number of times each sum occurred in the simulation. The total number of times each sum occurred is the frequency of the sum. Let's look at the example results of the simulation.
Sum
Frequency
4
2
6
2
7
3
8
4
9
6
10
7
11
7
12
11
13
8
14
13
15
5
16
7
17
6
18
7
19
3
20
5
22
3
24
1
You might get slightly different results, as the process is random after all. Now, to find the relative-frequency table we need to calculate the relative frequencies. We can do that by dividing the frequencies by the numbers of trials, which is 100.
Sum
Frequency
Relative Frequency
4
2
2/100=0.02
6
2
2/100=0.02
7
3
3/100=0.03
8
4
4/100=0.04
9
6
6/100=0.06
10
7
7/100=0.07
11
7
7/100=0.07
12
11
11/100=0.11
13
8
8/100=0.08
14
13
13/100=0.13
15
5
5/100=0.05
16
7
7/100=0.07
17
6
6/100=0.06
18
7
7/100=0.07
19
3
3/100=0.03
20
5
5/100=0.05
22
3
3/100=0.03
24
1
1/100=0.01
d For this part we will need the relative-frequency table we made in Part C. We only need to look at the sums and their relative frequencies.
Sum
Relative Frequency
4
0.02
6
0.02
7
0.03
8
0.04
9
0.06
10
0.07
11
0.07
12
0.11
13
0.08
14
0.13
15
0.05
16
0.07
17
0.06
18
0.07
19
0.03
20
0.05
22
0.03
24
0.01
Now, we can draw a experimental probability distribution graph. To do that, we will assign a bar to each sum. The heights of the bars should be proportional to the relative frequencies of the sums they represent.
This is only an example graph. Creating this graph was affected by the random process, so your result might be different.
e The expected value of a discrete random variable X is the weighted sum of its values. In our case, the values of the random variable X represent the possible sums of two spins.
E(X)= ∑ [X* P(X)]
In other words, the expected value is what you get if you add up all products of the possible sums of two spins and their corresponding probabilities. Now, let's use the table we made in Part A to find the products.
Sum, X
Theoretical Probability, P(X)
X* P(X)
4
1/64
4* 1/64≈ 0.06
6
1/32
6* 1/32≈ 0.19
7
1/32
7* 1/32≈ 0.22
8
3/64
8* 3/64≈ 0.38
9
1/16
9* 1/16≈ 0.56
10
5/64
≈ 0.78
11
1/16
≈ 0.69
12
7/64
≈ 1.31
13
1/16
≈ 0.81
14
7/64
≈ 1.53
15
1/16
≈ 0.94
16
5/64
≈ 1.25
17
1/16
≈ 1.06
18
1/16
≈ 1.13
19
1/32
≈ 0.59
20
3/64
≈ 0.94
22
1/32
≈ 0.69
24
1/64
≈ 0.38
Finally, we can calculate the expected value by summing all values from the the third column. Let's do it!
E(X)=
0.06+0.19+0.22+0.38+... ≈ 13.5
The expected value of the sum of two spins is 13.5.
f Let's recall the formula for the standard deviation σ of a discrete random variable X. In our case, X represents the sum of the values of two spins of the wheel.
σ = sqrt(∑ [X-E(X)]^2* P(X))
To calculate the standard deviation, we will first calculate the expression inside the radical, which is called the variance of a random variable X.
σ^2 = ∑ [X-E(X)]^2* P(X)
In Part B we found that E(X)≈ 13.5, we can substitute that value into the equation.
σ^2 = ∑ [X- 13.5]^2* P(X)
The variance σ^2 is a sum of elements. Let's calculate each element of the sum one at a time. We can substitute the possible values of the sums X and their corresponding probabilities P(X) from the table we made in Part A.
Sum, X
P(X)
[X-13.5]^2
[X-13.5]^2* P(X)
4
1/64
[4-13.5]^2=90.25
90.25* 1/64≈ 1.410
6
1/32
[6-13.5]^2= 56.25
56.25* 1/32≈ 1.758
7
1/32
[7-13.5]^2=42.25
42.25 * 1/32≈ 1.320
8
3/64
[8-13.5]^2=30.25
30.25* 3/64≈ 1.418
9
1/16
[9-13.5]^2=20.25
20.25* 1/16≈ 1.266
10
5/64
[10-13.5]^2=12.25
12.25* 5/64≈ 0.957
11
1/16
[11-13.5]^2=6.25
6.25* 1/16≈ 0.391
12
7/64
[12-13.5]^2=2.25
2.25* 7/64≈ 0.246
13
1/16
[13-13.5]^2=0.25
0.25*1/16≈ 0.0156
14
7/64
[14-13.5]^2=0.25
0.25* 7/64≈ 0.027
15
1/16
[15-13.5]^2=2.25
2.25* 1/16≈ 0.141
16
5/64
[16-13.5]^2=6.25
6.25* 5/64≈ 0.488
17
1/16
[17-13.5]^2=12.25
12.251/16≈ 0.766
18
1/16
[18-13.5]^2=20.25
20.25*1/16≈ 1.266
19
1/32
[19-13.5]^2=30.25
30.25 * 1/32≈ 0.945
20
3/64
[20-13.5]^2=42.25
42.25* 3/64≈ 1.980
22
1/32
[22-13.5]^2=72.25
72.25* 1/32≈ 2.258
24
1/64
[24-13.5]^2=110.25
110.25* 1/64≈ 1.723
Now, we can calculate the variance by summing all the elements from the last column. Let's do it!
σ^2=1.410+1.758+1.320... ≈ 18.38
Finally, we can calculate the standard deviation by calculating the square root of the variance σ^2.
