CS 4953 Experimentation in Computer Science Assignment 2 Comments: Spring 2003

• For Problem 6, I was looking for an explanation, not just the answer. Since I did not make this clear, I gave credit if the answer was correct. Just putting in two number is not sufficient. If you cannot prove it analytically, then at leas write a program to show a table. Here is the one I generated. n and m start at 2 and increase by one as you go across and down.
  0.333 0.400 0.467 0.524 0.571 0.611 0.644 0.673
  0.400 0.400 0.429 0.464 0.500 0.533 0.564 0.591
  0.467 0.429 0.429 0.444 0.467 0.491 0.515 0.538
  0.524 0.464 0.444 0.444 0.455 0.470 0.487 0.505
  0.571 0.500 0.467 0.455 0.455 0.462 0.473 0.486
  0.611 0.533 0.491 0.470 0.462 0.462 0.467 0.475
  0.644 0.564 0.515 0.487 0.473 0.467 0.467 0.471
  0.673 0.591 0.538 0.505 0.486 0.475 0.471 0.471
  
• For Problem 7, I meant 1K to mean 1K bits but some interpreted it to mean 1K bytes. That was OK.
• There does not seem to be a simple answer for Problem 8.
• For problems 9 and 10, you needed to remember a bit of calculus.
• Problem 14 gave several people trouble. Make sure you understand how to do it.
• Below is the program I wrote to do number 15. It uses the Trapezoidal Rule to integrate the pdf from a to b for various values of a and b. For a=10 and b=20 we get the answer to the problem. For a=1 and b=20 we get P(z>1) which can be checked with the normal distribution tables. To simplify the the check, tval = .5 - val = P(0 < z < a) is displayed. This can be directly read from the normal table.

  #include <stdio.h>
  #include <math.h>
  
  static double f(double x) {
     return exp(-x*x/2.0);
  }
  
  static double integrate(double a, double b, int N) {
     double delta;
     double sum = 0.0;
     int i;
  
     delta = (b -a)/N;
     sum = f(a) + f(b);
     sum = sum/2.0;
     for (i=1;i<N;i++)
        sum = sum + f(a + delta*i);
     sum = delta * sum;
     sum = sum/sqrt(2*M_PI);
     return sum;
  }
  
  int main(void) {
     int N = 1000;
     double a = 1.0;
     double b = 20.0;
     double val;
     double tval;
     int i;
  
     for (i=0; i<10; i++) {
        val = integrate(a,b,N);
        tval = 0.5 - val;
        fprintf(stderr,"a = %f, b = %f, N = %d, tval = %f, val = %E\n",
              a, b, N, tval, val);
        a = a + 1;
     }
     return 0;
  }
  

  Output:

  a = 1.000000, b = 20.000000, N = 1000, tval = 0.341337, val = 1.586625E-01
  a = 2.000000, b = 20.000000, N = 1000, tval = 0.477247, val = 2.275305E-02
  a = 3.000000, b = 20.000000, N = 1000, tval = 0.498650, val = 1.350218E-03
  a = 4.000000, b = 20.000000, N = 1000, tval = 0.499968, val = 3.168266E-05
  a = 5.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 2.867909E-07
  a = 6.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 9.871830E-10
  a = 7.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 1.280713E-12
  a = 8.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 6.225810E-16
  a = 9.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 1.129521E-19
  a = 10.000000, b = 20.000000, N = 1000, tval = 0.500000, val = 7.626264E-24