exp 5-7: init

Signed-off-by: Amneesh Singh <natto@weirdnatto.in>

file.org

@@ -2,9 +2,11 @@
 #+LATEX_HEADER: \usepackage[margin=0.5in]{geometry}
 #+LATEX_HEADER: \usepackage{fontspec}
 #+LATEX_HAEDER: \usepackage{graphicx}
-#+LATEX_HAEDER: \usepackage{}
+#+LATEX_HEADER: \usepackage{enumitem} 
+#+LATEX_HEADER: \setlist{noitemsep}
 #+LATEX_HEADER: \setmainfont{LiberationSerif}
 #+LATEX_HEADER: \date{}
+
 #+OPTIONS: toc:nil
 #+OPTIONS: num:nil
 
@@ -82,31 +84,31 @@ Installation of Scilab and demonstration of simple programming concepts like mat
 Installed Scilab and demonstrated simple programming concepts like matrix multiplication (scalar and vector), loop, conditional statements and plotting.
 
 #+LATEX: \clearpage
-* Experiment-2
+* Experiment 2
 
 ** Objective
 Program for demonstration of theoretical probability limits.
 
 ** Method
 1. Opened Scilab console and evaluated the following commands.
-2. Declared integer n and set it to 10000, similarly declared and set another integer head_count to 0.
+2. Declared integer $n$ and set it to 10000, similarly declared and set another integer $head\_count$ to 0.
 3. Set up a loop from 1 to 10000 and generated a random number between 0 and 1 using rand() command
-4. if the value of number is less then 0.5 then incremented head_count by 1
+4. if the value of number is less then 0.5 then incremented $head\_count$ by 1
-5. Set up a function P(i) for probability of heads in trial.
+5. Set up a function $P(i)$ for probability of heads in trial.
-6. Plotted the graph of P(i) using plot() command.
+6. Plotted the graph of $P(i)$ using plot() command.
 
 ** Code
-  #+begin_src
+#+begin_src
 n = 10000;
 head_count = 0;
 for i = 1:n
-  x = rand(1)
+x = rand(1)
 
-  if x<0.5 then
+if x<0.5 then
-    head_count = head_count + 1;
+  head_count = head_count + 1;
-  end
+end
 
-  p(i) = head_count / i;
+p(i) = head_count / i;
 end
 
 disp(p(10000))
@@ -118,10 +120,12 @@ title("Probability of getting Heads");
 #+end_src
 
 ** Output
-  #+ATTR_LATEX: :width 6cm
+#+ATTR_LATEX: :width 10cm
-  [[./2.png]]
+[[./2.png]]
 
 ** Result 
+We see that as the number of lips increase the theoretical probability of 0.5 is approached and the theoretical and practical probabilities become the same.
+
 
 #+LATEX: \clearpage
 * Experiment 3
@@ -130,37 +134,37 @@ Program to plot normal distributions and exponential distributions for various p
 
 ** a) Method: Normal Distribution
 1. Opened Scilab console and evaluated the following commands.
-2. Declared 2 arrays, m_values and s_values with various parametric values of mean and standard deviation.
+2. Declared 2 arrays, $m\_values$ and $s\_values$ with various parametric values of mean and standard deviation.
-3. Set up a loop from 1 to length of the array ‘means’ and got a pair of values of mean and standard deviation.
+3. Set up a loop from 1 to length of the array $means$ and got a pair of values of mean and standard deviation.
 4. Using those values, generated the probability density function
 
 $f(\boldsymbol{x}) = \frac{1}{s}\cdot\sqrt{2\pi}\cdot e^{\frac{-(\boldsymbol{x} - m) ^ 2}{(2 * s ^ 2)}}$
 
-5. Created a range of x values to plot the normal distribution using linspace() command.
+5. [@5] Created a range of $x$ values to plot the normal distribution using linspace() command.
 5. Correctly titled and labelled the graph.
 6. Plotted various normal distribution curves using plot() command.
 
 ** a) Code
-  #+begin_src
+#+begin_src
 m_values = [0, 1, -1];
 s_values =[0.5, 1, 1.5];
 
 for m = m_values
-  for s =s_values
+for s =s_values
-    t =grand(1, 1000, "nor", m, s);
+  t =grand(1, 1000, "nor", m, s);
 
-    x= linspace(m-4*s, m+4*s, 1000)
+  x= linspace(m-4*s, m+4*s, 1000)
-    y = (1/s*sqrt(2*%pi))*exp(-(x - m).^2/(2*s^2));
+  y = (1/s*sqrt(2*%pi))*exp(-(x - m).^2/(2*s^2));
 
-    plot(x, y);
+  plot(x, y);
-    hold on;
+  hold on;
-    xgrid();
+  xgrid();
-  end
+end
 
 end
 legend("m=0, s=0.5", "m=1, s=0.5", "m=-1, s=0.5",...
-       "m=0, s=1"  , "m=1, s=1"  , "m=-1, s=1", ...
+      "m=0, s=1"  , "m=1, s=1"  , "m=-1, s=1", ...
-       "m=0, s=1.5", "m=1, s=1.5", "m=-1, s=1.5");
+      "m=0, s=1.5", "m=1, s=1.5", "m=-1, s=1.5");
 
 xlabel("x")
 ylabel("Probability density function");
@@ -168,34 +172,34 @@ title("Normal distributions for various parametric values");
 #+end_src
 
 ** a) Output
-  #+ATTR_LATEX: :width 6cm
+#+ATTR_LATEX: :width 10cm
-  [[./3a.png]]
+[[./3a.png]]
 
 ** b) Method: Exponential Distribution
 1. Opened Scilab console and evaluated the following commands.
-2. Declared an array, Lambda_values with various values of lambda
+2. Declared an array, $lambda\_values$ with various values of $lambda$
-3. Set up a loop from 1 to length of the array Lambda_values and the function grand() is used to generate 1000 random numbers from the exponential distribution.
+3. Set up a loop from 1 to length of the array $lambda\_values$ and the function grand() is used to generate 1000 random numbers from the exponential distribution.
 4. Using those values, generated the probability density function
 
 $f(x) = Y = lambda \cdot e^{-lambda \cdot x}$
 
-5. Created a range of x values to plot the exponential distribution using linspace() command.
+5. [@5] Created a range of $x$ values to plot the exponential distribution using linspace() command.
 5. Correctly titled and labelled the graph.
 6. Plotted various exponential distribution curves using plot() command.
 
 ** b) Code
-  #+begin_src
+#+begin_src
 lambda_values = [0.5, 1, 2];
 for lambda = lambda_values
-  t = grand(1, 1000, "exp", lambda);
+t = grand(1, 1000, "exp", lambda);
-  x = linspace(0, 8/lambda, 1000);
+x = linspace(0, 8/lambda, 1000);
-  y = lambda * exp(-lambda * x);
+y = lambda * exp(-lambda * x);
 
-  plot(x, y);
+plot(x, y);
 
-  xgrid();
+xgrid();
 
-  hold on;
+hold on;
 end
 
 xlabel("x");
@@ -206,8 +210,8 @@ legend(["lambda=0.5", "lambda=1", "lambda=2"]);
 #+end_src
 
 ** b) Output
-  #+ATTR_LATEX: :width 6cm
+#+ATTR_LATEX: :width 10cm
-  [[./3b.png]]
+[[./3b.png]]
 
 #+LATEX: \clearpage
 * Experiment 4
@@ -217,13 +221,13 @@ Program to plot normal distributions and exponential distributions for various p
 ** Theory 
 Binomial Distribution: A probability distribution that summarizes the likelihood that a variable will take one of two independent values under a given set of parameters. The distribution is obtained by performing a number of Bernoulli trials. A Bernoulli trial is assumed to meet each of these criteria: 
 1. There must be only 2 possible outcomes. 
-2. Each outcome has a fixed probability of occurring. A success has the probability of p, and a failure has the probability of 1 – p. 
+2. Each outcome has a fixed probability of occurring. A success has the probability of $p$, and a failure has the probability of $1 – p$.
-3. Each trial is completely independent of all others. 
+3. Each trial is completely independent of all others.
 4. To calculate the binomial distribution values, we can use the binomial distribution formula:
 
 $P(X = x) = {}^{n}C_{x} \cdot p^x \cdot (1 - p)^{n - x}$
 
-where `n` is the total number of trials, `p` is the probability of success, and `x` is the number of successes. We can calculate the binomial distribution values for each possible value of `x` using this formula and the values of `n` and `p` given above. 
+where $n$ is the total number of trials, $p$ is the probability of success, and $x$ is the number of successes. We can calculate the binomial distribution values for each possible value of $x$ using this formula and the values of $n$ and $p$ given above. 
  
 ** Problem statement
 6 fair dice are tossed 1458 times. Getting a 2 or a 3 is counted as success. Fit a binomial distribution and calculate expected frequencies. 
@@ -251,28 +255,271 @@ Calculate it.
 |  6  |       312.50       |            0.002132             |
 
 ** Steps
-1. Find the number of cases, here, we take it as n. 
+1. Find the number of cases, here, we take it as $n$. 
-2. Find the probability of success. Here, we take it as s (=2/6). 
+2. Find the probability of success. Here, we take it as $s =2/6$.
-3. Define the number of times the process is repeated, and mark it as N. Here, acc to question, it’s 1458. 
+3. Define the number of times the process is repeated, and mark it as $N$. Here, acc to question, it’s 1458. 
-4. Take a variable x that varies from 0 to number of cases. 
+4. Take a variable $x$ that varies from 0 to number of cases. 
 5. Apply the Formula for Binomial Probability Distribution.
 6. Apply the formula for the Frequency. 
 7. Plot the Graph.
 
 ** Code
-  #+begin_src
+#+begin_src
-  n=6;
+n=6;
-  s=1/3;
+s=1/3;
-  N=1458;
+N=1458;
-  x=0:n;
+x=0:n;
-  P= (1-s).^(n-x).*s.^x.*factorial(n)./(factorial(x).*factorial(n-x));
+P= (1-s).^(n-x).*s.^x.*factorial(n)./(factorial(x).*factorial(n-x));
-  E= P*N;
+E= P*N;
-  clf();
+clf();
-  plot(x,E,"b.-"); 
+plot(x,E,"b.-"); 
-  #+end_src
+#+end_src
 
 ** Output
-  #+ATTR_LATEX: :width 6cm
+#+ATTR_LATEX: :width 10cm
-  [[./4.png]]
+[[./4.png]]
 
 #+LATEX: \clearpage
+* Experiment 5
+ 
+** Objective
+Program for fitting of binomial distributions after computing mean and variance. 
+ 
+** Theory
+Binomial Distribution: A probability distribution that summarizes the likelihood that a variable will take one of two independent values under a given set of parameters. The distribution is obtained by performing a number of Bernoulli trials. A Bernoulli trial is assumed to meet each of these criteria: 
+1. There must be only 2 possible outcomes. 
+2. Each outcome has a fixed pobability of occurring. A success has the probability of $p$, and a failure has the probability of $1 – p$. 
+3. Each trial is completely independent of all others.
+4. To calculate the binomial distribution values, we can use the binomial distribution formula:
+
+$P(X = x) = {}^{n}C_{x} \cdot p^x \cdot (1 - p)^{n - x}$
+
+where $n$ is the total number of trials, $p$ is the probability of success, and $x$ is the number of successes. We can calculate the binomial distribution values for each possible value of $x$ using this formula and the values of $n$ and $p$ given above. 
+ 
+** Problem statement
+A set of three similar coins are tossed 100 ($N$) times with the following results
+
+|        <c>         | <c> | <c> | <c> | <c> |
+| No. of Heads ($x$) |  0  |  1  |  2  |  3  |
+|  Frequency ($f$)   | 36  | 40  | 22  |  2  |
+ 
+Fit a binomial distribution and calculate expected frequencies. 
+ 
+** Method
+1. For given data, total number of coins($n$) = 3 and total number of trials($N$) would be = 100 
+2. Find $X \cdot F$ to calculate the mean, for all corresponding values of $X$ and $F$, and calculate $\Sigma F \cdot x$
+Mean would then be = $\frac{\Sigma F \cdot x}{n}$ i.e., $\frac{90}{100} = 0.9$. 
+
+3. Since we know for binomial distribution, $mean = n \cdot p$. So, $0.9 = 3 \cdot p$ => $p = 0.3$. 
+4. Calculate variance. We already know that $var = n \cdot p \cdot (1-p)$. 
+5. Now, calculate Binomial distribution, $P(X= x)$ and calculate expectes valued using $P \cdot N$. 
+
+|   |    $X$     | $F$ | $X \cdot F$ |
+|---+------------+-----+-------------|
+|   |    <c>     | <c> |     <c>     |
+| / |     <      |  <  |     <>      |
+|   |     0      | 36  |      0      |
+|   |     1      | 40  |     40      |
+|   |     2      | 22  |     44      |
+|   |     3      |  2  |      6      |
+|   | TOTAL($N$) | 100 |     90      |
+
+|        $X$         |  0  |  1   |  2   |  3  | TOTAL |
+|--------------------+-----+------+------+-----+-------|
+|        <c>         | <c> | <c>  | <c>  | <c> |  <c>  |
+| Observed Frequency | 26  |  40  | 2 2  |  2  |  100  |
+| Expected Frequency | 343 | 44.1 | 18.9 | 2.7 |  100  |
+
+** Code
+#+begin_src
+n = 3;
+N = 100;   
+F = [36, 40, 22, 2];
+X = [0,1,2,3];
+
+FX = (36*0+40*1+22*2+2*3);
+Mean = FX/N;
+disp("Mean = ", Mean);
+
+p = Mean/n;
+disp("Probability of success = ",p);
+
+var = n*p*(1-p);
+disp("Variance = ",var);
+
+x = 0:3; 
+P= (1-p).^(n-x).*p.^x.*((factorial(n))./((factorial(x).*factorial(n-x))));
+disp("Binomial distribution = ",P);
+
+E = N*P;
+disp("Expected Frequency = ", E);
+
+clf();
+plot(X,E,"b.-");
+
+xlabel('X ');
+ylabel('Expected Frequency');
+title('Binomial Distribution ');
+
+Y = grand(1, 80, "bin", 3,p);
+disp(Y);
+
+histplot(30, Y, normalization=%f, style=5); 
+#+end_src
+
+
+** Output
+#+ATTR_LATEX: :width 10cm
+[[./5.png]]
+
+#+LATEX:\clearpage
+* Experiment 6
+** Objective
+Program for fitting of Poisson distributions for given value of lambda.. 
+ 
+** Theory
+Poisson Distribution: The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, assuming that these events occur with a known constant rate and independently of the time since the last event. It is often used to model rare events. 
+
+The Poisson distribution has only one parameter, often denoted as $\lambda$, which represents the average rate of occurrence of the events over the given interval. The probability of observing exactly k events in the interval is given by the Poisson probability mass function: 
+
+$P(X=k) = (e^{-\lambda} * \lambda^k) / k!$
+
+where $X$ is the random variable representing the number of events, $e$ is the base of the natural logarithm, and $k!$ denotes the factorial of $k$.
+
+
+The mean and variance of the Poisson distribution are both equal to $\lambda$, which means that the distribution is unimodal and symmetric around $\lambda$. 
+
+The Poisson distribution is also a limiting case of the binomial distribution, when the number of trials $n$ goes to infinity and the probability of success $p$ goes to zero, but the product $n \cdot p$ remains constant and equal to $\lambda$.
+ 
+** Problem statement
+A set of three similar coins are tossed 100 ($N$) times with the following results in Table 1. Note the difference with experiment number 4 where “observed frequencies” were not given. 
+ 
+#+CAPTION: Table 1
+|        <c>         | <c> | <c> | <c> | <c> |
+|         /          |  <  |     |     |     |
+| No. of Heads ($x$) |  0  |  1  |  2  |  3  |
+|  Frequency ($f$)   | 36  | 40  | 22  |  2  |
+
+Fit a binomial distribution and calculate expected frequencies. 
+ 
+** Method
+1. For given data, create a table (Table 2) to record the values of $x$, $f$, and $xf$. Here, $x$ denotes the number of heads obtained in a single toss, $f$ is the frequency of obtaining $x$ heads, and $xf$ is the product of $x$ and $f$. 
+
+#+CAPTION: Table 2
+| <c>  | <c> | <c> | <c> | <c> |
+|  /   |  <  |     |     |     |
+| $x$  |  0  |  1  |  2  |  3  |
+| $f$  | 36  | 40  | 22  |  2  |
+| $xf$ |  0  | 40  | 44  |  6  |
+
+2. [@2] The mean of the binomial distribution can be calculated as $N \cdot p$, where $N$ is the total number of trials and $p$ is the probability of getting a head. Here, $N=100$. 
+3. Calculate the value of $p$, using the formula $p = \frac{x}{N}$, where $x$ is the number of heads obtained in the 100 tosses. So, $p = (036 + 140 + 222 + 32)/100 = 0.64$. 
+4. Using the formula for the binomial distribution, calculate the expected frequency of obtaining x heads in a single toss as: 
+$P(X = x) = {}^{n}C_{x} \cdot p^x \cdot (1 - p)^{n - x}$
+#+OPTIONS: \n:t
+
+Where $n$ is the total number of trials, $x$ is the number of successful trials, and ${}^{n}C_{x}$ is the binomial coefficient. We can then multiply this value by 100 to get the expected frequency for each value of $x$. 
+
+Tabulate this information in Table 3. 
+
+#+CAPTION: Table 3
+|                  <c>                  |  <c>  |  <c>  |  <c>  |  <c>  |        |
+|                   /                   |   <   |       |       |       |        |
+|                  $X$                  |   0   |   1   |   2   |   3   |  Total |
+|          Observed Frequency           |  36   |  40   |  22   |   2   |    100 |
+| Expected Frequency $100 \cdot P(X=x)$ | 23.65 | 37.84 | 27.51 | 10.00 | 100.00 |
+
+** Code
+#+begin_src
+observed_freq = [36, 40, 22, 2]; 
+N = sum(observed_freq); 
+x = 0:3; 
+ 
+mean = sum(x .* observed_freq) / N; 
+ 
+// Poisson distribution with lambda = mean 
+expected_freq = N * exp(-mean) * (mean .^ x) ./ factorial(x); 
+ 
+// Output tables 
+disp(["x", "Observed Freq", "Expected Freq"]); 
+disp([x', observed_freq', expected_freq']); 
+ 
+// Plot of expected frequencies 
+clf(); 
+plot(x, expected_freq, '.-', 'LineWidth', 2); 
+xlabel('Number of Heads'); 
+ylabel('Expected Frequency'); 
+title('Fitting of Poisson distribution'); 
+legend('Expected Frequencies'); 
+#+end_src
+
+** Output
+#+ATTR_LATEX: :width 10cm
+[[./6.png]]
+
+#+LATEX: \clearpage
+* Experiment 7
+
+** Objective: 
+Plotting Regression line for the given data points. 
+
+** Formulation and Method
+*** Regression line
+\begin{equation}
+  y = m \cdot x + b
+\end{equation}
+
+where, $m$ = slope, $b$ = y-intercept.  
+
+*** Method of least squares:
+\begin{equation}
+	E = \Sigma_i (Y_i – (m \cdot x +b))^2
+\end{equation}
+Find $\frac{\partial e}{\partial m}, \frac{\partial e}{\partial b} = 0$
+
+*** Problem statement
+Find the regression line for the given data points (x,y)
+
+|  X  |  Y   |
+|-----+------|
+| <c> | <c>  |
+| 20  | 0.18 |
+| 60  | 0.37 |
+| 100 | 0.35 |
+| 140 | 0.70 |
+| 160 | 0.56 |
+| 220 | 0.75 |
+| 260 | 0.18 |
+| 300 | 0.36 |
+| 340 | 1.17 |
+| 380 | 1.65 |
+
+** Method
+1. Make $x$ data and $y$ data lists.
+2. Use regline() to find $m$ and $b$ 
+3. Use scatter() to plot $x$ data and $y$ data
+4. Use plot() to draw regression line.
+
+
+** Results 
+The following are required as output:
+1. The full code.
+2. The plot for regression line and data points.
+
+** Code
+#+begin_src
+x_data = [20,60,100,140,160,220,260,300,340,380]
+y_data = [0.18,0.37,0.35,0.70,0.56,0.75,0.18,0.36,1.17,1.65]
+
+[a, b] = reglin(x_data, y_data);
+scatter(x_data,y_data,30,"x")
+plot(x_data, a*x_data+b,"red")
+
+xlabel("X")
+ylabel("Y")
+title("Simple Linear Regression")
+#+end_src
+
+** Output
+#+ATTR_LATEX: :width 10cm
+[[./7.png]]

toc.tex

@@ -22,7 +22,7 @@ Experiments according to the lab syllabus prescribed by GGSIPU
 \begin{table}[h]
  \fontsize{11}{12}\selectfont{
     \renewcommand{\arraystretch}{2.5}
-    \begin{tabular}{|p{0.6cm}|p{8cm}|p{2cm}|p{2cm}|p{1cm}|} \hline
+    \begin{tabular}{|p{0.6cm}|p{10cm}|p{2cm}|p{2cm}|p{1cm}|} \hline
       \textbf{Exp. No.} & \textbf{Experiment Name} & \textbf{Performance Date} & \textbf{Date Checked}& \textbf{Marks} \\ \hline \hline
       & & & & \\ \hline
       & & & & \\ \hline