Create a simulation of the number of upsets using a random number generator.
D
d
Make a bar graph using the information from Part C. Each bar's height should be proportional to the corresponding relative frequency.
A
a
4.34, see solution.
B
b
1.90
C
c
Example Solution:
Upsets
Frequency
Relative Frequency
0
0
0/50=0
1
3
3/50=0.06
2
6
6/50=0.12
3
5
5/50=0.1
4
9
9/50=0.18
5
15
15/50=0.3
6
5
5/50=0.1
7
6
6/50=0.12
8
1
1/50=0.02
D
d
Example Graph:
Practice makes perfect
a
We can think of the numbers of upsets in a basketball tournament as the values of a random variable X. Then, the expected number of upsets is the expected value of the random variable X. Let's recall the formula for the expected value of a discrete random variable.
E(X)= ∑ [X* P(X)]The expected number of upsets is what we get if we add up all products of the possible numbers of upsets and their corresponding probabilities. We can calculate the products using the given table.
Number of Upsets Per Year
Upsets, X
0
1
2
3
4
5
6
7
8
Probability, P(X)
1/32
1/16
3/32
1/8
1/8
5/16
1/8
3/32
1/32
X* P(X)
0* 1/32=0
1* 1/16=0.0625
2* 3/32= 0.188
3* 1/8= 0.375
4* 1/8=0.5
5* 5/16= 1.5625
6* 1/8=0.75
7* 3/32=0.65625
8* 1/32=0.25
To calculate the expected value, we will sum all the values from the last row. Let's do it!
E(X)=
0+0.0625+0.188+ 0.375+ 0.5+ 1.5625+ 0.75+ 0.65625+0.25≈
4.34
The expected value of upsets per year is 4.34. Since the number of upsets is a whole number, there cannot be a year with exactly 4.34 upsets. The interpretation of this result is that the number of upsets per year is on average around 4.
b
Let's recall the formula for the standard deviation of a discrete random variable X.
σ=sqrt(∑ [[X-E(X)]^2* P(X)])
In our case, the value of X will represent the number of major upsets in the first round of a basketball tournament. 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 A we found that the expected number of upsets E(X) is equal to 4.34. We can substitute that value into the equation for σ^2.
σ^2=∑ [[X- 4.34]^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 numbers of upsets X and their corresponding probabilities P(X) from the given table.
Number of Upsets, X
P(X)
[X-4.34]^2
[X-4.34]^2* P(X)
0
1/32
[0-4.34]^2≈8.8682
8.8682*1/32≈0.5896
1
1/16
[1-4.34]^2≈11.1807
11.1807*1/16≈0.6988
2
3/32
[2-4.34]^2≈5.4932
5.4932* 3/32≈0.515
3
1/8
[3-4.34]^2≈1.8057
1.8057*1/8≈0.2257
4
1/8
[4-4.34]^2≈0.1182
0.1182*1/8≈0.0148
5
5/16
[5-4.34]^2≈0.4307
0.4307*5/16≈0.1346
6
1/8
[6-4.34]^2≈2.7432
2.7432*1/8≈ 0.3429
7
3/32
[7-4.34]^2≈7.0557
7.0557*3/32≈0.6615
8
1/32
[8-4.34]^2≈ 13.3682
13.3682*1/32≈0.4178
Now, we can find the variance σ^2 by summing all the elements from the last column. Let's do it!
σ^2=0.5896+0.6988+...≈ 3.6
Finally, we can calculate the standard deviation σ by calculating the square root of the variance σ^2.
To construct a relative-frequency table for 50 trials, we need to simulate the number of upsets in the basketball tournament. We can do that by generating random integers. To find out how many integers we should generate, let's find a common denominator for all the probabilities.
Upsets
Probability
0
1/32
1
1/16=2/32
2
3/32
3
1/8=4/32
4
1/8=4/32
5
5/16=10/32
6
1/8=4/32
7
3/32
8
1/32
We found that the common denominator of the probabilities is 32. Therefore, to simulate the numbers of upsets we can generate 32 random integers. Let's set up a proportion based on the theoretical probabilities.
Upsets
Probability
Represented by
0
1/32
1 integer
1
2/32
2 integers
2
3/32
3 integers
3
4/32
4 integers
4
4/32
4 integers
5
10/32
10 integers
6
4/32
4 integers
7
3/32
3 integers
8
1/32
1 integer
Now, we can choose exactly which integers will represent which numbers of upsets. The simplest way is to assign the first available integers to each number of upsets. 1 will represent zero upsets, 2 and 3 will represent one upset, and so on.
Upsets
Probability
Represented by
0
1/32
1
1
2/32
2 and 3
2
3/32
4 -6
3
4/32
7-10
4
4/32
11-14
5
10/32
15-24
6
4/32
25-28
7
3/32
29-31
8
1/32
32
We need to make fifty trials, so let's use a graphing calculator to generate integers from 1 to 32 fifty times. Start by pushing the MATH button. Then scroll right to the PRB menu and choose the randInt( option.
Since we want to generate integers from 1 to 32 fifty times, we type 1, 32, and 50 on a graphing calculator, respectively. Then press ENTER to get fifty outcomes.
Let's say that we generated 13. This integer is in the interval 11-14, which represents the situation where there are four upsets. We can up the frequency of this situation by one and repeat the process for all the other randomly generated integers. Let's look at example results of the simulation.
Upsets
Frequency
0
0
1
3
2
6
3
5
4
9
5
15
6
5
7
6
8
1
Note that this is a random process, so your result might be different. Finally, to finish making the relative frequency table, we will calculate the relative frequencies. We can do that by dividing the frequencies by the total number of trials, 50.
Upsets
Frequency
Relative Frequency
0
0
0/50=0
1
3
3/50=0.06
2
6
6/50=0.12
3
5
5/50=0.1
4
9
9/50=0.18
5
15
15/50=0.3
6
5
5/50=0.1
7
6
6/50=0.12
8
1
1/50=0.02
d
To make a graph of the experimental probability distribution, we will use the table we made in Part C. We will only need to look at the upsets and their relative frequencies.
Upsets
Relative Frequency
0
0
1
0.06
2
0.12
3
0.1
4
0.18
5
0.3
6
0.1
7
0.12
8
0.02
Now, we can make a bar graph. The height of a bar should be proportional to corresponding relative frequency of the number of upsets.
Again, since the process of simulating the numbers of upsets is random, your results might differ.
