349
file.org
349
file.org
@@ -2,9 +2,11 @@
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#+LATEX_HEADER: \usepackage[margin=0.5in]{geometry}
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||||
#+LATEX_HEADER: \usepackage{fontspec}
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||||
#+LATEX_HAEDER: \usepackage{graphicx}
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||||
#+LATEX_HAEDER: \usepackage{}
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||||
#+LATEX_HEADER: \usepackage{enumitem}
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||||
#+LATEX_HEADER: \setlist{noitemsep}
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#+LATEX_HEADER: \setmainfont{LiberationSerif}
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||||
#+LATEX_HEADER: \date{}
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||||
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#+OPTIONS: toc:nil
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#+OPTIONS: num:nil
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@@ -82,31 +84,31 @@ Installation of Scilab and demonstration of simple programming concepts like mat
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Installed Scilab and demonstrated simple programming concepts like matrix multiplication (scalar and vector), loop, conditional statements and plotting.
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#+LATEX: \clearpage
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* Experiment-2
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* Experiment 2
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||||
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||||
** Objective
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Program for demonstration of theoretical probability limits.
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** Method
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1. Opened Scilab console and evaluated the following commands.
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2. Declared integer n and set it to 10000, similarly declared and set another integer head_count to 0.
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2. Declared integer $n$ and set it to 10000, similarly declared and set another integer $head\_count$ to 0.
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||||
3. Set up a loop from 1 to 10000 and generated a random number between 0 and 1 using rand() command
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4. if the value of number is less then 0.5 then incremented head_count by 1
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5. Set up a function P(i) for probability of heads in trial.
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6. Plotted the graph of P(i) using plot() command.
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4. if the value of number is less then 0.5 then incremented $head\_count$ by 1
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5. Set up a function $P(i)$ for probability of heads in trial.
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6. Plotted the graph of $P(i)$ using plot() command.
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** Code
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#+begin_src
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||||
#+begin_src
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n = 10000;
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head_count = 0;
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for i = 1:n
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x = rand(1)
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||||
x = rand(1)
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if x<0.5 then
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if x<0.5 then
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head_count = head_count + 1;
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||||
end
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||||
end
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p(i) = head_count / i;
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p(i) = head_count / i;
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end
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disp(p(10000))
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@@ -118,10 +120,12 @@ title("Probability of getting Heads");
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#+end_src
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** Output
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#+ATTR_LATEX: :width 6cm
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||||
[[./2.png]]
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#+ATTR_LATEX: :width 10cm
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[[./2.png]]
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** Result
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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.
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#+LATEX: \clearpage
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* Experiment 3
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@@ -130,23 +134,23 @@ Program to plot normal distributions and exponential distributions for various p
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** a) Method: Normal Distribution
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1. Opened Scilab console and evaluated the following commands.
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2. Declared 2 arrays, m_values and s_values with various parametric values of mean and standard deviation.
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3. Set up a loop from 1 to length of the array ‘means’ and got a pair of values of mean and standard deviation.
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2. Declared 2 arrays, $m\_values$ and $s\_values$ with various parametric values of mean and standard deviation.
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3. Set up a loop from 1 to length of the array $means$ and got a pair of values of mean and standard deviation.
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4. Using those values, generated the probability density function
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|
||||
$f(\boldsymbol{x}) = \frac{1}{s}\cdot\sqrt{2\pi}\cdot e^{\frac{-(\boldsymbol{x} - m) ^ 2}{(2 * s ^ 2)}}$
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5. Created a range of x values to plot the normal distribution using linspace() command.
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5. [@5] Created a range of $x$ values to plot the normal distribution using linspace() command.
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5. Correctly titled and labelled the graph.
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6. Plotted various normal distribution curves using plot() command.
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** a) Code
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#+begin_src
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#+begin_src
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m_values = [0, 1, -1];
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s_values =[0.5, 1, 1.5];
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for m = m_values
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for s =s_values
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for s =s_values
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t =grand(1, 1000, "nor", m, s);
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x= linspace(m-4*s, m+4*s, 1000)
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@@ -155,7 +159,7 @@ for m = m_values
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plot(x, y);
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hold on;
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xgrid();
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end
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end
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end
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legend("m=0, s=0.5", "m=1, s=0.5", "m=-1, s=0.5",...
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@@ -168,34 +172,34 @@ title("Normal distributions for various parametric values");
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#+end_src
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** a) Output
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#+ATTR_LATEX: :width 6cm
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[[./3a.png]]
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#+ATTR_LATEX: :width 10cm
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[[./3a.png]]
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** b) Method: Exponential Distribution
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1. Opened Scilab console and evaluated the following commands.
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2. Declared an array, Lambda_values with various values of lambda
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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.
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2. Declared an array, $lambda\_values$ with various values of $lambda$
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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.
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4. Using those values, generated the probability density function
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$f(x) = Y = lambda \cdot e^{-lambda \cdot x}$
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5. Created a range of x values to plot the exponential distribution using linspace() command.
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5. [@5] Created a range of $x$ values to plot the exponential distribution using linspace() command.
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5. Correctly titled and labelled the graph.
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6. Plotted various exponential distribution curves using plot() command.
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** b) Code
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#+begin_src
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#+begin_src
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lambda_values = [0.5, 1, 2];
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for lambda = lambda_values
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t = grand(1, 1000, "exp", lambda);
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x = linspace(0, 8/lambda, 1000);
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y = lambda * exp(-lambda * x);
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t = grand(1, 1000, "exp", lambda);
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x = linspace(0, 8/lambda, 1000);
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y = lambda * exp(-lambda * x);
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plot(x, y);
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plot(x, y);
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xgrid();
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xgrid();
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hold on;
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hold on;
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end
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xlabel("x");
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@@ -206,8 +210,8 @@ legend(["lambda=0.5", "lambda=1", "lambda=2"]);
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#+end_src
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** b) Output
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#+ATTR_LATEX: :width 6cm
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[[./3b.png]]
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#+ATTR_LATEX: :width 10cm
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[[./3b.png]]
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#+LATEX: \clearpage
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* Experiment 4
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@@ -217,13 +221,13 @@ Program to plot normal distributions and exponential distributions for various p
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** Theory
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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:
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1. There must be only 2 possible outcomes.
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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.
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2. Each outcome has a fixed probability of occurring. A success has the probability of $p$, and a failure has the probability of $1 – p$.
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3. Each trial is completely independent of all others.
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4. To calculate the binomial distribution values, we can use the binomial distribution formula:
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$P(X = x) = {}^{n}C_{x} \cdot p^x \cdot (1 - p)^{n - x}$
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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.
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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.
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** Problem statement
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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.
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@@ -251,28 +255,271 @@ Calculate it.
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| 6 | 312.50 | 0.002132 |
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** Steps
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1. Find the number of cases, here, we take it as n.
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2. Find the probability of success. Here, we take it as s (=2/6).
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3. Define the number of times the process is repeated, and mark it as N. Here, acc to question, it’s 1458.
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4. Take a variable x that varies from 0 to number of cases.
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1. Find the number of cases, here, we take it as $n$.
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2. Find the probability of success. Here, we take it as $s =2/6$.
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3. Define the number of times the process is repeated, and mark it as $N$. Here, acc to question, it’s 1458.
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4. Take a variable $x$ that varies from 0 to number of cases.
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5. Apply the Formula for Binomial Probability Distribution.
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6. Apply the formula for the Frequency.
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7. Plot the Graph.
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** Code
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#+begin_src
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n=6;
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s=1/3;
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N=1458;
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x=0:n;
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P= (1-s).^(n-x).*s.^x.*factorial(n)./(factorial(x).*factorial(n-x));
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E= P*N;
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clf();
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plot(x,E,"b.-");
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#+end_src
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#+begin_src
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n=6;
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s=1/3;
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N=1458;
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x=0:n;
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P= (1-s).^(n-x).*s.^x.*factorial(n)./(factorial(x).*factorial(n-x));
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E= P*N;
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clf();
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plot(x,E,"b.-");
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#+end_src
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** Output
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#+ATTR_LATEX: :width 6cm
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[[./4.png]]
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#+ATTR_LATEX: :width 10cm
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[[./4.png]]
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#+LATEX: \clearpage
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* Experiment 5
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|
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** Objective
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Program for fitting of binomial distributions after computing mean and variance.
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||||
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** Theory
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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:
|
||||
|
||||
$P(X = x) = {}^{n}C_{x} \cdot p^x \cdot (1 - p)^{n - x}$
|
||||
|
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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.
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** Problem statement
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A set of three similar coins are tossed 100 ($N$) times with the following results
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| <c> | <c> | <c> | <c> | <c> |
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| No. of Heads ($x$) | 0 | 1 | 2 | 3 |
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| Frequency ($f$) | 36 | 40 | 22 | 2 |
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Fit a binomial distribution and calculate expected frequencies.
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** Method
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1. For given data, total number of coins($n$) = 3 and total number of trials($N$) would be = 100
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2. Find $X \cdot F$ to calculate the mean, for all corresponding values of $X$ and $F$, and calculate $\Sigma F \cdot x$
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Mean would then be = $\frac{\Sigma F \cdot x}{n}$ i.e., $\frac{90}{100} = 0.9$.
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3. Since we know for binomial distribution, $mean = n \cdot p$. So, $0.9 = 3 \cdot p$ => $p = 0.3$.
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4. Calculate variance. We already know that $var = n \cdot p \cdot (1-p)$.
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5. Now, calculate Binomial distribution, $P(X= x)$ and calculate expectes valued using $P \cdot N$.
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| | $X$ | $F$ | $X \cdot F$ |
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|---+------------+-----+-------------|
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| | <c> | <c> | <c> |
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| / | < | < | <> |
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| | 0 | 36 | 0 |
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| | 1 | 40 | 40 |
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| | 2 | 22 | 44 |
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| | 3 | 2 | 6 |
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| | TOTAL($N$) | 100 | 90 |
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| $X$ | 0 | 1 | 2 | 3 | TOTAL |
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|--------------------+-----+------+------+-----+-------|
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| <c> | <c> | <c> | <c> | <c> | <c> |
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| Observed Frequency | 26 | 40 | 2 2 | 2 | 100 |
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| Expected Frequency | 343 | 44.1 | 18.9 | 2.7 | 100 |
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** Code
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#+begin_src
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n = 3;
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N = 100;
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F = [36, 40, 22, 2];
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X = [0,1,2,3];
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FX = (36*0+40*1+22*2+2*3);
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Mean = FX/N;
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disp("Mean = ", Mean);
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p = Mean/n;
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disp("Probability of success = ",p);
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var = n*p*(1-p);
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disp("Variance = ",var);
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x = 0:3;
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P= (1-p).^(n-x).*p.^x.*((factorial(n))./((factorial(x).*factorial(n-x))));
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disp("Binomial distribution = ",P);
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E = N*P;
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disp("Expected Frequency = ", E);
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clf();
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plot(X,E,"b.-");
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xlabel('X ');
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ylabel('Expected Frequency');
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title('Binomial Distribution ');
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|
||||
Y = grand(1, 80, "bin", 3,p);
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disp(Y);
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histplot(30, Y, normalization=%f, style=5);
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#+end_src
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** Output
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#+ATTR_LATEX: :width 10cm
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[[./5.png]]
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|
||||
#+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.
|
||||
|
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#+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]]
|
||||
|
||||
2
toc.tex
2
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
|
||||
|
||||
Reference in New Issue
Block a user