|
AP Statistics / Mr. Hansen |
Name: _________________________ |
Solution to Pi Simulation Exercise
Your task was to estimate p by performing a Monte Carlo simulation involving the
first quadrant quarter circle of radius 2. Here are the steps:
- Select a random real number between 0 and 2 to
serve as the x coordinate and
another random real number between 0 and 2 to serve as the y coordinate. [Note: On the TI-83, this is accomplished with the function
rand*2. In Excel, the formula would be =RAND()*2. If you are using the
random digit table, you would take a large number of digits at a time, say
10, and treat them as decimal fraction with an implied leading decimal
point, and then multiply by 2 to scale them in the range of 0 to 2.]
- If y
< (4 – x2)0.5,
that means that the random (x, y) ordered pair is within the
circle. Count this as a success. Otherwise, count the event as a failure.
- Repeat steps 1 and 2 for a total of 30 trials.
- Let
. Remember, the notation 
means sample proportion. The sample
proportion will be an unbiased estimator of the true probability p. - Since » (area of quarter circle)/(area of square) = p/4, we can multiply through by 4 to conclude that
.
Here is a tabulation of
results:
|
Trial # |
x |
y |
Success?
(1=yes, 0=no) |
|
1 |
1.514869 |
0.716877 |
1 |
|
2 |
0.523154 |
1.529781 |
1 |
|
3 |
0.802115 |
1.278847 |
1 |
|
4 |
0.415451 |
0.817405 |
1 |
|
5 |
0.299436 |
0.58402 |
1 |
|
6 |
1.850538 |
0.705966 |
1 |
|
7 |
1.724866 |
0.858874 |
1 |
|
8 |
1.247713 |
1.688632 |
0 |
|
9 |
0.012287 |
0.056046 |
1 |
|
10 |
0.223077 |
1.922212 |
1 |
|
11 |
1.032865 |
0.495525 |
1 |
|
12 |
1.332576 |
1.832073 |
0 |
|
13 |
1.340788 |
0.805873 |
1 |
|
14 |
0.71197 |
1.074217 |
1 |
|
15 |
0.94439 |
1.988424 |
0 |
|
16 |
0.602156 |
0.140416 |
1 |
|
17 |
0.912871 |
1.164401 |
1 |
|
18 |
1.564564 |
0.540651 |
1 |
|
19 |
0.881133 |
0.729464 |
1 |
|
20 |
1.945962 |
0.416888 |
1 |
|
21 |
0.735593 |
1.772785 |
1 |
|
22 |
1.748613 |
0.014502 |
1 |
|
23 |
0.481978 |
0.939933 |
1 |
|
24 |
0.71744 |
0.052331 |
1 |
|
25 |
1.341016 |
1.939306 |
0 |
|
26 |
1.389113 |
1.649355 |
0 |
|
27 |
1.085715 |
0.918535 |
1 |
|
28 |
0.693089 |
0.021667 |
1 |
|
29 |
1.156587 |
0.06176 |
1 |
|
30 |
1.291855 |
1.666163 |
0 |
|
|
|
|
|
|
|
|
Success count |
24 |
|
|
0.800 |
||
Therefore, the estimate of p would be 4(0.800) = 3.2.
Using an Excel
spreadsheet and a macro, I ran 5 million trials and obtained an estimate of
3.140911 for p.