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Original
2026-01-01
Modified
2026-02-28
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<p>A Poisson distribution is used to predict or estimate the<a>number</a><a>of</a>times an event might occur within a given period of time. This type of distribution method is specifically used when the<a>variables</a>are discrete count variables. We usually use Poisson distribution when dealing with variables, such as economic and financial<a>data</a>.</p>
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<p>A Poisson distribution is used to predict or estimate the<a>number</a><a>of</a>times an event might occur within a given period of time. This type of distribution method is specifically used when the<a>variables</a>are discrete count variables. We usually use Poisson distribution when dealing with variables, such as economic and financial<a>data</a>.</p>
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<p>The<a>formula</a>used to calculate the<a>probability</a>of an event occurring discreetly over a given period of time is:</p>
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<p>The<a>formula</a>used to calculate the<a>probability</a>of an event occurring discreetly over a given period of time is:</p>
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<p>\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)</p>
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<p>\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)</p>
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<p>Where:</p>
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<p>Where:</p>
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<p>e is approximately 2.718 (Euler’s number),</p>
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<p>e is approximately 2.718 (Euler’s number),</p>
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<p>λ is the<a>average</a>number of events in the interval</p>
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<p>λ is the<a>average</a>number of events in the interval</p>
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<p>k! is the factorial of k</p>
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<p>k! is the factorial of k</p>
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<p>k is the actual number of occurrences.</p>
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<p>k is the actual number of occurrences.</p>
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<p>In a Poisson distribution, both the mean and variance are equal to λ. Here, λ is greater than 0.</p>
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<p>In a Poisson distribution, both the mean and variance are equal to λ. Here, λ is greater than 0.</p>
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<h3><strong>Example:</strong></h3>
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<h3><strong>Example:</strong></h3>
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<p>Imagine you are waiting for text messages from your friends. Based on your history, you know that on average, you receive three text messages per hour. Even though the average is 3, you know that in any specific hour, the actual number could vary. You might get 0, 3, or 5 texts. The text messages arrive independently of one another (receiving one doesn't change the chance of receiving another).</p>
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<p>Imagine you are waiting for text messages from your friends. Based on your history, you know that on average, you receive three text messages per hour. Even though the average is 3, you know that in any specific hour, the actual number could vary. You might get 0, 3, or 5 texts. The text messages arrive independently of one another (receiving one doesn't change the chance of receiving another).</p>
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<p>This is a classic Poisson Distribution problem because:</p>
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<p>This is a classic Poisson Distribution problem because:</p>
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<ul><li>We know the average rate (\lambda = 3). </li>
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<ul><li>We know the average rate (\lambda = 3). </li>
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<li>We are looking at a fixed time interval (1 hour). </li>
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<li>We are looking at a fixed time interval (1 hour). </li>
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<li>Events are independent.</li>
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<li>Events are independent.</li>
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</ul><p><strong>How the Probability Looks:</strong></p>
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</ul><p><strong>How the Probability Looks:</strong></p>
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<p>Using the Poisson distribution, we can calculate the likelihood of receiving a specific number of texts in the next hour.</p>
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<p>Using the Poisson distribution, we can calculate the likelihood of receiving a specific number of texts in the next hour.</p>
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<ul><li>0 Texts: There is about a 5% chance you receive absolute silence. </li>
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<ul><li>0 Texts: There is about a 5% chance you receive absolute silence. </li>
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<li>1 Text: There is about a 15% chance. </li>
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<li>1 Text: There is about a 15% chance. </li>
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<li>2 Texts (The Average): There is about a 22% chance. </li>
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<li>2 Texts (The Average): There is about a 22% chance. </li>
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<li>3 Texts: There is about a 22% chance. </li>
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<li>3 Texts: There is about a 22% chance. </li>
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<li>5 Texts: There is only about a 10% chance.</li>
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<li>5 Texts: There is only about a 10% chance.</li>
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</ul><p><strong>Calculation:</strong></p>
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</ul><p><strong>Calculation:</strong></p>
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<p>\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)</p>
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<p>\(P(X=k) = \frac{\lambda^k e^{-\lambda}}{k!}\)</p>
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<p>Where:</p>
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<p>Where:</p>
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<ul><li>\(P(X=k)\) is the probability of getting exactly k texts. </li>
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<ul><li>\(P(X=k)\) is the probability of getting exactly k texts. </li>
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<li>\(\lambda\) (Lambda) is the average rate (3 texts). </li>
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<li>\(\lambda\) (Lambda) is the average rate (3 texts). </li>
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<li>e is Euler's number (\approx 2.718). </li>
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<li>e is Euler's number (\approx 2.718). </li>
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<li>k! is the factorial of k.</li>
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<li>k! is the factorial of k.</li>
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</ul><p>Example Calculation for three texts:</p>
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</ul><p>Example Calculation for three texts:</p>
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<p>\(P(X=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27 \cdot 0.0498}{6} \approx 0.22\)</p>
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<p>\(P(X=3) = \frac{3^3 \cdot e^{-3}}{3!} = \frac{27 \cdot 0.0498}{6} \approx 0.22\)</p>
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<p>Use the Poisson distribution calculator to check. </p>
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<p>Use the Poisson distribution calculator to check. </p>
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