added articles

stat challenge day 5
normal distrbution
poisson distribution
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Day 5 - Poisson and Normal distributions
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12 novembre 2018
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<h2>Poisson Distribution</h2>
<h3>Problem 1</h3>
<p>A random variable, <span class="math">\(X\)</span>, follows Poisson distribution with mean of 2.5. Find the probability with which the random variable <span class="math">\(X\)</span> is equal to 5.</p>
<h3>Mathematical explanation</h3>
<p>In this case, the answer is straightforward, we just need to compute the value of the Poisson distribution of mean 2.5 at 5:</p>
<div class="math">$$P(\lambda = 2.5, x=5)=\frac{\lambda^ke^{-\lambda}}{k!}$$</div>
<div class="math">$$P(\lambda = 2.5, x=5)=\frac{2.5^5e^{-2.5}}{5!}$$</div>
<div class="highlight"><pre><span></span><span class="k">def</span> <span class="nf">factorial</span><span class="p">(</span><span class="n">k</span><span class="p">):</span>
<span class="k">return</span> <span class="mi">1</span> <span class="k">if</span> <span class="n">k</span> <span class="o">==</span> <span class="mi">1</span> <span class="k">else</span> <span class="n">k</span> <span class="o">*</span> <span class="n">factorial</span><span class="p">(</span><span class="n">k</span><span class="o">-</span><span class="mi">1</span><span class="p">)</span>
<span class="kn">from</span> <span class="nn">math</span> <span class="kn">import</span> <span class="n">exp</span>
<span class="k">def</span> <span class="nf">poisson</span><span class="p">(</span><span class="n">l</span><span class="p">,</span><span class="n">k</span><span class="p">):</span>
<span class="k">return</span> <span class="p">(</span><span class="n">l</span><span class="o">**</span><span class="n">k</span> <span class="o">*</span> <span class="n">exp</span><span class="p">(</span><span class="o">-</span><span class="n">l</span><span class="p">))</span> <span class="o">/</span> <span class="n">factorial</span><span class="p">(</span><span class="n">k</span><span class="p">)</span>
<span class="n">l</span> <span class="o">=</span> <span class="mf">2.5</span>
<span class="n">k</span> <span class="o">=</span> <span class="mi">5</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that a random variable X following a Poisson distribution of mean {l} equals {k} : {round(poisson(l,k),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that a random variable X following a Poisson distribution of mean 2.5 equals 5 : 0.067
</pre></div>
<h3>Problem 2</h3>
<p>The manager of a industrial plant is planning to buy a machine of either type <span class="math">\(A\)</span> or type <span class="math">\(B\)</span>. For each days operation:</p>
<ul>
<li>The number of repairs, <span class="math">\(X\)</span>, that machine <span class="math">\(A\)</span> needs is a Poisson random variable with mean 0.88. The daily cost of operating <span class="math">\(A\)</span> is <span class="math">\(C_A=160+40X^2\)</span>.</li>
<li>The number of repairs, <span class="math">\(Y\)</span>, that machine <span class="math">\(B\)</span> needs is a Poisson random variable with mean 1.55. The daily cost of operating <span class="math">\(B\)</span> is <span class="math">\(C_B=128+40Y^2\)</span>.</li>
</ul>
<p>Assume that the repairs take a negligible amount of time and the machines are maintained nightly to ensure that they operate like new at the start of each day. What is the expected daily cost for each machine.</p>
<h3>Mathematical explanation</h3>
<p>The cost for each machine follows a law that is the square of a Poisson distribution.</p>
<div class="math">$$C_Z = a + b*Z^2$$</div>
<p>
Since the expectation is a linear operator :
</p>
<div class="math">$$E[C_Z] = aE[1] + bE[Z^2]$$</div>
<p>Knowing that <span class="math">\(Z\)</span> follows a Poisson distribution of mean <span class="math">\(\lambda\)</span> we have :
</p>
<div class="math">$$E[C_Z] = a+ b(\lambda + \lambda^2)$$</div>
<div class="highlight"><pre><span></span><span class="n">averageX</span> <span class="o">=</span> <span class="mf">0.88</span>
<span class="n">averageY</span> <span class="o">=</span> <span class="mf">1.55</span>
<span class="n">CostX</span> <span class="o">=</span> <span class="mi">160</span> <span class="o">+</span> <span class="mi">40</span><span class="o">*</span><span class="p">(</span><span class="n">averageX</span> <span class="o">+</span> <span class="n">averageX</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<span class="n">CostY</span> <span class="o">=</span> <span class="mi">128</span> <span class="o">+</span> <span class="mi">40</span><span class="o">*</span><span class="p">(</span><span class="n">averageY</span> <span class="o">+</span> <span class="n">averageY</span><span class="o">**</span><span class="mi">2</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Expected cost to run machine A : {round(CostX, 3)}&#39;</span><span class="p">)</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Expected cost to run machine A : {round(CostY, 3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Expected cost to run machine A : 226.176
Expected cost to run machine A : 286.1
</pre></div>
<h2>Normal Distribution</h2>
<h3>Problem 1</h3>
<p>In a certain plant, the time taken to assemble a car is a random variable, <span class="math">\(X\)</span>, having a normal distribution with a mean of 20 hours and a standard deviation of 2 hours. What is the probability that a car can be assembled at this plant in:</p>
<p>Less than 19.5 hours?
Between 20 and 22 hours?</p>
<h3>Mathematical explanation</h3>
<p><span class="math">\(X\)</span> is a real-valued random variable following a normal distribution : the probability of assembly the car in less than 19.5 hours is the cumulative distribution function of X evaluated at 19.5:</p>
<div class="math">$$P(X\leq 19.5)=F_X(19.5)$$</div>
<p>For a normal distribution, the cumulative distribution function is :
</p>
<div class="math">$$\Phi(x)=\frac{1}{2}\left(1+erf\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right)$$</div>
<div class="highlight"><pre><span></span><span class="kn">import</span> <span class="nn">math</span>
<span class="k">def</span> <span class="nf">cumulative</span><span class="p">(</span><span class="n">x</span><span class="p">,</span><span class="n">mean</span><span class="p">,</span><span class="n">sd</span><span class="p">):</span>
<span class="k">return</span> <span class="mf">0.5</span><span class="o">*</span><span class="p">(</span><span class="mi">1</span><span class="o">+</span><span class="n">math</span><span class="o">.</span><span class="n">erf</span><span class="p">((</span><span class="n">x</span><span class="o">-</span><span class="n">mean</span><span class="p">)</span><span class="o">/</span><span class="p">(</span><span class="n">sd</span><span class="o">*</span><span class="n">math</span><span class="o">.</span><span class="n">sqrt</span><span class="p">(</span><span class="mi">2</span><span class="p">))))</span>
<span class="n">mean</span> <span class="o">=</span> <span class="mi">20</span>
<span class="n">sd</span> <span class="o">=</span> <span class="mi">2</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that the car is built in less than 19.5 hours : {round(cumulative(19.5,mean,sd),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that the car is built in less than 19.5 hours : 0.401
</pre></div>
<p>Similarly, the probability that a car is built between 20 and 22hours can be computed thanks to the cumulative density function:</p>
<div class="math">$$P(20\leq x\leq 22) = F_X(22)-F_X(20)$$</div>
<div class="highlight"><pre><span></span><span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that the car is built between 20 and 22 hours : {round(cumulative(22,mean,sd)-cumulative(20,mean,sd),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that the car is built between 20 and 22 hours : 0.341
</pre></div>
<h3>Problem 2</h3>
<p>The final grades for a Physics exam taken by a large group of students have a mean of <span class="math">\(\mu=70\)</span> and a standard deviation of <span class="math">\(\sigma=10\)</span>. If we can approximate the distribution of these grades by a normal distribution, what percentage of the students:
<em> Scored higher than 80 (i.e., have a <span class="math">\(grade \gt 80\)</span>))?
</em> Passed the test (i.e., have a <span class="math">\(grade \gt 60\)</span>)?
* Failed the test (i.e., have a <span class="math">\(grade \lt 60\)</span>)?</p>
<h3>Mathematical explanation</h3>
<p>Here again, we need to appy the cumulative density function to get the probabilities :</p>
<p>Probability that they scored higher than 80 :
</p>
<div class="math">$$P(X\gt80) = 1- P(X\lt80)$$</div>
<div class="math">$$P(X\gt80) = 1- F_X(80)$$</div>
<div class="highlight"><pre><span></span><span class="n">mean</span> <span class="o">=</span> <span class="mi">70</span>
<span class="n">sd</span> <span class="o">=</span> <span class="mi">10</span>
<span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that the the student scored higher than 80 : {round(1- cumulative(80,mean,sd),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that the the student scored higher than 80 : 0.159
</pre></div>
<p>Probability that they passed the test :
</p>
<div class="math">$$P(X\gt60) = 1- P(X\lt60)$$</div>
<div class="math">$$P(X\gt80) = 1- F_X(60)$$</div>
<div class="highlight"><pre><span></span><span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that the the student passed the test : {round(1- cumulative(60,mean,sd),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that the the student passed the test : 0.841
</pre></div>
<p>Probability that they failed : </p>
<div class="math">$$P(X\lt60) = F_X(60)$$</div>
<div class="highlight"><pre><span></span><span class="k">print</span><span class="p">(</span><span class="n">f</span><span class="s1">&#39;Probability that the student failed the test: {round(cumulative(60,mean,sd),3)}&#39;</span><span class="p">)</span>
</pre></div>
<div class="highlight"><pre><span></span>Probability that the student failed the test: 0.159
</pre></div>
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<h1>
Normal Distribution
</h1>
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<time class="published" datetime="2018-11-12T10:21:00+01:00">
12 novembre 2018
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<li>Mathematics</li>
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<h2>Normal Distribution</h2>
<p>The probability density of normal distribution is:
</p>
<div class="math">$$\mathcal{N}(\mu,\sigma^2)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$</div>
<p>where,
<em> <span class="math">\(\mu\)</span> is the mean (or expectation) of the distribution. It is also equal to median and mode of the distribution.
</em> <span class="math">\(\sigma^2\)</span> is the variance.
* <span class="math">\(\sigma\)</span> is the standard deviation. </p>
<h2>Standard Normal Distribution</h2>
<p>If <span class="math">\(\mu=0\)</span> and <span class="math">\(\sigma=1\)</span>, then the normal distribution is known as standard normal distribution:
</p>
<div class="math">$$\phi(x)=\frac{e^{-\frac{x^2}{2}}}{\sigma\sqrt{2\pi}}$$</div>
<p>Every normal distribution can be represented as standard normal distribution:</p>
<div class="math">$$\mathcal{N}(\mu,\sigma^2)=\frac{1}{\sigma}\phi(\frac{x-\mu}{\sigma})$$</div>
<h2>Cumulative Probability</h2>
<p>Consider a real-valued random variable, <span class="math">\(X\)</span>. The cumulative distribution function of <span class="math">\(X\)</span> (or just the distribution function of <span class="math">\(X\)</span>) evaluated at <span class="math">\(x\)</span> is the probability that <span class="math">\(X\)</span> will take a value less than or equal to <span class="math">\(x\)</span>:</p>
<div class="math">$$F_X(x)=P(X\leq x)$$</div>
<p>
also,
</p>
<div class="math">$$P(a\leq X\leq b)=P(a\lt X\lt b)=F_X(b)-F_X(a)$$</div>
<p>the cumulative distribution function for a function with normal distribution is:
</p>
<div class="math">$$\Phi(x)=\frac{1}{2}\left(1+erf\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right)$$</div>
<p>
where <span class="math">\(erf\)</span> is the error function:
</p>
<div class="math">$$erf(z)=\frac{2}{\sqrt{\pi}}\int_0^ze^{-x^2}dx$$</div>
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<h1>
Poisson Distribution
</h1>
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<time class="published" datetime="2018-11-12T09:02:00+01:00">
12 novembre 2018
</time>
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<li>Mathematics</li>
<li>Statistics</li>
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<h2>Poisson Experiment</h2>
<p>Poisson experiment is a statistical experiment that has the following properties:
<em> The outcome of each trial is either success or failure.
</em> The average number of successes (<span class="math">\(\lambda\)</span>) that occurs in a specified region is known.
<em> The probability that a success will occur is proportional to the size of the region.
</em> The probability that a success will occur in an extremely small region is virtually zero. </p>
<h2>Poisson Distribution</h2>
<p>A Poisson random variable is the number of successes that result from a Poisson experiment. The probability distribution of a Poisson random variable is called a Poisson distribution:</p>
<div class="math">$$P(k,\lambda)=\frac{\lambda^ke^{-\lambda}}{k!}$$</div>
<p>where :
<em> <span class="math">\(\lambda\)</span> is the average number of successes that occur in a specified region.
</em> <span class="math">\(k\)</span> is the actual number of successes that occur in a specified region.
* <span class="math">\(P(k,\lambda)\)</span> is the Poisson probability, which is the probability of getting exactly <span class="math">\(k\)</span> successes when the average number of successes is <span class="math">\(\lambda\)</span>. </p>
<h2>Example</h2>
<p>The average number of goals in the soccer world cup is 2.5. The probability that 4 goals are scored is then:
</p>
<div class="math">$$p(\lambda=2.5,k=4)=\frac{2.5^4e^{-2.5}}{4!}=0.133$$</div>
<h2>Expectation for the Poisson distribution</h2>
<p>Consider some Poisson random variable, <span class="math">\(X\)</span>. Let <span class="math">\(E[X]\)</span> be the expectation of <span class="math">\(X\)</span>. Find the value of <span class="math">\(E[X^2]\)</span>. </p>
<p>Let <span class="math">\(Var(X)\)</span> be the variance of <span class="math">\(X\)</span>. Recall that if a random variable has a Poisson distribution, then:
<em> <span class="math">\(E[X]=\lambda\)</span>
</em> <span class="math">\(Var[X]=\lambda\)</span></p>
<p>Now, we'll use the following property of expectation and variance for any random variable, <span class="math">\(X\)</span>:
</p>
<div class="math">$$Var(X)=E[X^2]-(E[X])^2$$</div>
<div class="math">$$E[X^2]=Var(X)+(E[X])^2$$</div>
<p>So, for any random variable having a Poisson distribution, the above result can be rewritten as:
</p>
<div class="math">$$E[X^2]=\lambda + \lambda^2$$</div>
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<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

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@ -171,7 +171,7 @@ Numpy array with unique values : [10 20 30 40 50 60]
<footer class="footer">
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<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

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@ -151,7 +151,7 @@
<footer class="footer">
<div class="container">
<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

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@ -112,7 +112,7 @@
<footer class="footer">
<div class="container">
<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

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@ -5,7 +5,7 @@ xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">
<url>
<loc>redoules.github.io/</loc>
<lastmod>2018-11-11T21:55:09-00:00</lastmod>
<lastmod>2018-11-12T20:49:03-00:00</lastmod>
<changefreq>daily</changefreq>
<priority>0.5</priority>
</url>
@ -17,6 +17,27 @@ xmlns="http://www.sitemaps.org/schemas/sitemap/0.9">
<priority>0.5</priority>
</url>
<url>
<loc>redoules.github.io/blog/Statistics_10days-day5.html</loc>
<lastmod>2018-11-12T13:07:00+01:00</lastmod>
<changefreq>monthly</changefreq>
<priority>0.5</priority>
</url>
<url>
<loc>redoules.github.io/mathematics/normal.html</loc>
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<changefreq>monthly</changefreq>
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</url>
<url>
<loc>redoules.github.io/mathematics/poisson.html</loc>
<lastmod>2018-11-12T09:02:00+01:00</lastmod>
<changefreq>monthly</changefreq>
<priority>0.5</priority>
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<url>
<loc>redoules.github.io/mathematics/Geometric.html</loc>
<lastmod>2018-11-11T21:19:00+01:00</lastmod>

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@ -344,7 +344,7 @@ Done.
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<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

View File

@ -631,7 +631,7 @@ Done.
<footer class="footer">
<div class="container">
<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

View File

@ -216,7 +216,7 @@ Done.
<footer class="footer">
<div class="container">
<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

View File

@ -346,7 +346,7 @@ Done.
<footer class="footer">
<div class="container">
<p class="text-muted">
<center>This project contains 56 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<center>This project contains 59 pages and is available on <a href="https://github.com/redoules/redoules.github.io">GitHub</a>.
<br/>
Copyright &copy; Guillaume Redoulès,
<time datetime="2018">2018</time>.

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