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2026-01-01
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<p>Last updated on<strong>November 24, 2025</strong></p>
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<p>Last updated on<strong>November 24, 2025</strong></p>
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<p>The binomial distribution is used to measure how much the probabilities differ from the expected value (mean). This value shows the difference between the sampled observations and the expected value. In this topic, we are going to learn more about the variance of binomial distribution.</p>
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<p>The binomial distribution is used to measure how much the probabilities differ from the expected value (mean). This value shows the difference between the sampled observations and the expected value. In this topic, we are going to learn more about the variance of binomial distribution.</p>
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<h2>What is Binomial Distribution?</h2>
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<h2>What is Binomial Distribution?</h2>
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<p>A<a>binomial distribution</a>is a type<a>of</a><a>probability</a>model that counts the<a>number</a>of “successes” in a given<a>set</a>of independent trials. It is used only when there are only two possible outcomes per trial (e.g., heads vs. tails or pass vs. fail) and the probability of success is<a>constant</a>. Essentially, it calculates the likelihood of receiving a specific result in a repeated “yes or no” experiment. </p>
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<p>A<a>binomial distribution</a>is a type<a>of</a><a>probability</a>model that counts the<a>number</a>of “successes” in a given<a>set</a>of independent trials. It is used only when there are only two possible outcomes per trial (e.g., heads vs. tails or pass vs. fail) and the probability of success is<a>constant</a>. Essentially, it calculates the likelihood of receiving a specific result in a repeated “yes or no” experiment. </p>
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<p>\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) </p>
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<p>\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\) </p>
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<p>Consider a student guessing randomly on a 10-<a>question</a>multiple-choice test, where each question has four answer options. This scenario fits a binomial distribution because: </p>
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<p>Consider a student guessing randomly on a 10-<a>question</a>multiple-choice test, where each question has four answer options. This scenario fits a binomial distribution because: </p>
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<ul><li>Fixed Trials: There are exactly 10 questions. </li>
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<ul><li>Fixed Trials: There are exactly 10 questions. </li>
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<li>Binary Outcome: Each guess is either “Correct” or “Incorrect.” </li>
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<li>Binary Outcome: Each guess is either “Correct” or “Incorrect.” </li>
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<li>Constant Probability: Since the student is guessing blindly, the chance of getting it right is always 25% (1 in 4) for every question. </li>
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<li>Constant Probability: Since the student is guessing blindly, the chance of getting it right is always 25% (1 in 4) for every question. </li>
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<li>Independence: Guessing on question #1 doesn't change the odds for question #2. </li>
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<li>Independence: Guessing on question #1 doesn't change the odds for question #2. </li>
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</ul><p>You could use the binomial formula to calculate the exact odds of the student passing (getting 6 or more correct) by pure luck.</p>
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</ul><p>You could use the binomial formula to calculate the exact odds of the student passing (getting 6 or more correct) by pure luck.</p>
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<h2>What is Variance of Binomial Distribution?</h2>
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<h2>What is Variance of Binomial Distribution?</h2>
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<p>The<a>variance</a>of a<a>binomial</a>distribution measures how much the number of successes in your trials is likely to differ from the<a>average</a>(<a>mean</a>) number of successes.</p>
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<p>The<a>variance</a>of a<a>binomial</a>distribution measures how much the number of successes in your trials is likely to differ from the<a>average</a>(<a>mean</a>) number of successes.</p>
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<p>In simpler<a>terms</a>, it tells you how “spread out” your results are.</p>
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<p>In simpler<a>terms</a>, it tells you how “spread out” your results are.</p>
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<p><strong>Low Variance:</strong>Results are tightly clustered around the average. You can be confident the outcome will be close to the mean.</p>
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<p><strong>Low Variance:</strong>Results are tightly clustered around the average. You can be confident the outcome will be close to the mean.</p>
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<p><strong>High Variance:</strong>Results are widely scattered.3 predicting the exact outcome is harder because the possibilities are more spread out.</p>
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<p><strong>High Variance:</strong>Results are widely scattered.3 predicting the exact outcome is harder because the possibilities are more spread out.</p>
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<p>If you were to graph this:</p>
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<p>If you were to graph this:</p>
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<ul><li>Large Variance makes the graph look flatter and wider (like a low hill). </li>
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<ul><li>Large Variance makes the graph look flatter and wider (like a low hill). </li>
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<li>Small Variance makes the graph look taller and thinner (like a spike).</li>
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<li>Small Variance makes the graph look taller and thinner (like a spike).</li>
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</ul><h2>Formula for Variance of Binomial Distribution</h2>
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</ul><h2>Formula for Variance of Binomial Distribution</h2>
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<p>The<a>formula</a>for the variance (\(\sigma^2\)) of a binomial distribution is:</p>
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<p>The<a>formula</a>for the variance (\(\sigma^2\)) of a binomial distribution is:</p>
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<p>\(Var(X) = \sigma^2 = n \cdot p \cdot (1 - p)\)</p>
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<p>\(Var(X) = \sigma^2 = n \cdot p \cdot (1 - p)\)</p>
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<p>Sometimes written as:</p>
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<p>Sometimes written as:</p>
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<p>\(Var(X) = npq\)</p>
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<p>\(Var(X) = npq\)</p>
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<p>Where,</p>
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<p>Where,</p>
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<p>n is the Total number of trials,</p>
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<p>n is the Total number of trials,</p>
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<p>p is the Probability of success in a single trial, and</p>
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<p>p is the Probability of success in a single trial, and</p>
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<p>q is the Probability of failure (1-p).</p>
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<p>q is the Probability of failure (1-p).</p>
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<p>Example Calculation:</p>
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<p>Example Calculation:</p>
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<p>Imagine you flip a fair coin (p=0.5) 100 times (n=100).</p>
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<p>Imagine you flip a fair coin (p=0.5) 100 times (n=100).</p>
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<ul><li>Find the Mean (\(\mu\)): n \times p = \(100 \times 0.5 = 50\ heads.\) </li>
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<ul><li>Find the Mean (\(\mu\)): n \times p = \(100 \times 0.5 = 50\ heads.\) </li>
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<li>Calculate Variance (\(\sigma^2\)): \(100 \times 0.5 \times (1 - 0.5) = 25\) </li>
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<li>Calculate Variance (\(\sigma^2\)): \(100 \times 0.5 \times (1 - 0.5) = 25\) </li>
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<li>Standard Deviation (\(\sigma\)): \(\sqrt{25} = 5.\) </li>
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<li>Standard Deviation (\(\sigma\)): \(\sqrt{25} = 5.\) </li>
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</ul><p>This means that while you expect 50 heads, it is very normal for the result to vary by about 5 heads (getting 45 to 55 heads). </p>
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</ul><p>This means that while you expect 50 heads, it is very normal for the result to vary by about 5 heads (getting 45 to 55 heads). </p>
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<h2>How to Calculate the Variance of Binomial Distribution?</h2>
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<h2>How to Calculate the Variance of Binomial Distribution?</h2>
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<p>Calculating the variance of a binomial distribution is a straightforward process once you have your<a>variables</a>defined. It essentially measures the “spread” or volatility of your<a>data</a>around the average. Here is the step-by-step guide to performing this calculation manually.</p>
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<p>Calculating the variance of a binomial distribution is a straightforward process once you have your<a>variables</a>defined. It essentially measures the “spread” or volatility of your<a>data</a>around the average. Here is the step-by-step guide to performing this calculation manually.</p>
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<p><strong>Step 1:</strong>Identify the Total Trials (n)</p>
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<p><strong>Step 1:</strong>Identify the Total Trials (n)</p>
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<p>Determine how many times the experiment or trial is being performed.</p>
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<p>Determine how many times the experiment or trial is being performed.</p>
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<ul><li>Example: You are flipping a coin 50 times. (n = 50) </li>
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<ul><li>Example: You are flipping a coin 50 times. (n = 50) </li>
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</ul><p><strong>Step 2:</strong>Identify the Probability of Success (p)</p>
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</ul><p><strong>Step 2:</strong>Identify the Probability of Success (p)</p>
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<p>Determine the likelihood of getting the “successful” outcome in a single, individual trial.</p>
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<p>Determine the likelihood of getting the “successful” outcome in a single, individual trial.</p>
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<ul><li>Example: The probability of getting “Heads” in one flip is 0.5. (p = 0.5) </li>
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<ul><li>Example: The probability of getting “Heads” in one flip is 0.5. (p = 0.5) </li>
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</ul><p><strong>Step 3:</strong>Calculate the Probability of Failure (q)</p>
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</ul><p><strong>Step 3:</strong>Calculate the Probability of Failure (q)</p>
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<p>Subtract the probability of success from 1. This gives you q, or (1 - p).</p>
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<p>Subtract the probability of success from 1. This gives you q, or (1 - p).</p>
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<ul><li>Example: 1 - 0.5 = 0.5. (q = 0.5) </li>
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<ul><li>Example: 1 - 0.5 = 0.5. (q = 0.5) </li>
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</ul><p>(Note: If p was 0.2, then q would be 0.8)</p>
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</ul><p>(Note: If p was 0.2, then q would be 0.8)</p>
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<p><strong>Step 4:</strong>Multiply Them All Together</p>
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<p><strong>Step 4:</strong>Multiply Them All Together</p>
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<p>Multiply n by p by q.</p>
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<p>Multiply n by p by q.</p>
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<ul><li>Calculation: \(50 \times 0.5 \times 0.5\) </li>
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<ul><li>Calculation: \(50 \times 0.5 \times 0.5\) </li>
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<li>\(50 \times 0.5 = 25\) </li>
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<li>\(50 \times 0.5 = 25\) </li>
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<li>\(25 \times 0.5 = 12.5\) </li>
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<li>\(25 \times 0.5 = 12.5\) </li>
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</ul><p>Answer: The variance is 12.5.</p>
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</ul><p>Answer: The variance is 12.5.</p>
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<h2>Derivation of Variance of Binomial Distribution</h2>
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<h2>Derivation of Variance of Binomial Distribution</h2>
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<p>The binomial distribution represents the probability of getting a specific number of successes in independent trials. The possible outcomes of each trial are success and failure. In every trail the probability of success remains the same. Let X be the number of successes in the n trials. Then the variance of X can be calculated as:</p>
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<p>The binomial distribution represents the probability of getting a specific number of successes in independent trials. The possible outcomes of each trial are success and failure. In every trail the probability of success remains the same. Let X be the number of successes in the n trials. Then the variance of X can be calculated as:</p>
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<p>σ2 = E (X2) - (E (X))2</p>
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<p>σ2 = E (X2) - (E (X))2</p>
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<p>Now, let’s find the E(X), then the mean of X, which is np, here n is the number of trials and p is the probability of success. </p>
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<p>Now, let’s find the E(X), then the mean of X, which is np, here n is the number of trials and p is the probability of success. </p>
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<p>Then, we have to identify the E (X2). This refers to the squared values of X. Also, we need to find the expected value of X2. X2 has a distribution where each outcome is squared because X follows a binomial distribution. </p>
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<p>Then, we have to identify the E (X2). This refers to the squared values of X. Also, we need to find the expected value of X2. X2 has a distribution where each outcome is squared because X follows a binomial distribution. </p>
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<p> \(E (X^2) = \displaystyle\sum_{k=0}^{n}k^2\). P (X = k) Next, using the<a>probability mass function</a>(PMF) of the binomial distribution, we can find the probability of getting k successes in n trials. P (X = k) = \(\binom{n}{k}\) pk (1 - p) n - k </p>
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<p> \(E (X^2) = \displaystyle\sum_{k=0}^{n}k^2\). P (X = k) Next, using the<a>probability mass function</a>(PMF) of the binomial distribution, we can find the probability of getting k successes in n trials. P (X = k) = \(\binom{n}{k}\) pk (1 - p) n - k </p>
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<p>We can substitute this formula into an<a>equation</a>for E (X2) and then analyze the<a>sum</a>. Finally, add the values of E (X) and E (X2) into the formula of variance. Then simplify it to get the variance of the binomial distribution as (σ2) = np (1 - p)</p>
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<p>We can substitute this formula into an<a>equation</a>for E (X2) and then analyze the<a>sum</a>. Finally, add the values of E (X) and E (X2) into the formula of variance. Then simplify it to get the variance of the binomial distribution as (σ2) = np (1 - p)</p>
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<h2>Real-life applications of the Variance of Binomial Distribution</h2>
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<h2>Real-life applications of the Variance of Binomial Distribution</h2>
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<p>The variance of a binomial distribution measures how much the number of successes deviates from the expected mean. In the fields of medical research, finance, sports analysis, and manufacturing, the role of the variance of the binomial distribution is vital. </p>
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<p>The variance of a binomial distribution measures how much the number of successes deviates from the expected mean. In the fields of medical research, finance, sports analysis, and manufacturing, the role of the variance of the binomial distribution is vital. </p>
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<ul><li><strong>Medical:</strong>To predict how many patients may have side effects from new drugs, the binomial distribution is used by medical professionals such as doctors and researchers. </li>
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<ul><li><strong>Medical:</strong>To predict how many patients may have side effects from new drugs, the binomial distribution is used by medical professionals such as doctors and researchers. </li>
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<li><strong>Banking:</strong>Banks use the binomial distribution to model the likelihood of a specific number of credit card transactions being fraudulent. </li>
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<li><strong>Banking:</strong>Banks use the binomial distribution to model the likelihood of a specific number of credit card transactions being fraudulent. </li>
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<li><strong>Emails:</strong>Email providers apply the binomial distribution to estimate the chance of receiving a certain number of spam emails per day. </li>
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<li><strong>Emails:</strong>Email providers apply the binomial distribution to estimate the chance of receiving a certain number of spam emails per day. </li>
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<li><strong>Park Systems:</strong>Park systems use the binomial distribution to estimate the probability of the overflow of rivers in a year due to heavy rainfall. </li>
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<li><strong>Park Systems:</strong>Park systems use the binomial distribution to estimate the probability of the overflow of rivers in a year due to heavy rainfall. </li>
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<li><strong>Retail:</strong>To calculate the probability of receiving a specific number of shopping returns each week, retail stores use the binomial distribution. </li>
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<li><strong>Retail:</strong>To calculate the probability of receiving a specific number of shopping returns each week, retail stores use the binomial distribution. </li>
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</ul><h2>Common Mistakes and How to Avoid Them on Variance of Binomial Distribution</h2>
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</ul><h2>Common Mistakes and How to Avoid Them on Variance of Binomial Distribution</h2>
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<p>The variance of the binomial distribution tells us how much our actual results differ from the expected value on average. However, some mistakes can lead to incorrect calculations and interpretations. By understanding the common mistakes of the variance of the binomial distribution, students can improve their statistical skills and practical knowledge. </p>
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<p>The variance of the binomial distribution tells us how much our actual results differ from the expected value on average. However, some mistakes can lead to incorrect calculations and interpretations. By understanding the common mistakes of the variance of the binomial distribution, students can improve their statistical skills and practical knowledge. </p>
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<h3>Tips and Tricks of Variance of Binomial Distribution</h3>
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<h3>Tips and Tricks of Variance of Binomial Distribution</h3>
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<p>The variance of the binomial distribution helps us to understand how much the results fluctuate around the mean. Understanding the concepts of variance of binomial distribution is useful in working with<a>statistics</a>, probability, and risk assessment. Here are some of the tricks and tips that help us to effectively work with the fundamental concept. </p>
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<p>The variance of the binomial distribution helps us to understand how much the results fluctuate around the mean. Understanding the concepts of variance of binomial distribution is useful in working with<a>statistics</a>, probability, and risk assessment. Here are some of the tricks and tips that help us to effectively work with the fundamental concept. </p>
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<ul><li>Use the correct formula for the variance of the binomial distribution. Before concluding, double-check the structure of the formula. </li>
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<ul><li>Use the correct formula for the variance of the binomial distribution. Before concluding, double-check the structure of the formula. </li>
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<li>The probability of success (p) must be between 0 and 1. Here, p must satisfy 0 ≤ p ≤ 1. </li>
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<li>The probability of success (p) must be between 0 and 1. Here, p must satisfy 0 ≤ p ≤ 1. </li>
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<li>The variance decreases when p is near 0 or 1. </li>
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<li>The variance decreases when p is near 0 or 1. </li>
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<li>The value of variance depends on the value of n. But in these cases, the value of p should be constant. The variance increases with a larger n. </li>
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<li>The value of variance depends on the value of n. But in these cases, the value of p should be constant. The variance increases with a larger n. </li>
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<li>Mean and variance are related. The variance denotes how much the values in a<a>probability distribution</a>differ from the mean. The mean in a binomial distribution is denoted as (μ) = np. Also, variance (σ2) = np (1 - p).</li>
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<li>Mean and variance are related. The variance denotes how much the values in a<a>probability distribution</a>differ from the mean. The mean in a binomial distribution is denoted as (μ) = np. Also, variance (σ2) = np (1 - p).</li>
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</ul><h3>Problem 1</h3>
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</ul><h3>Problem 1</h3>
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<p>Find the variance of the binomial distribution having 15 trials and a probability of success of 0.6.</p>
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<p>Find the variance of the binomial distribution having 15 trials and a probability of success of 0.6.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>3.6</p>
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<p>3.6</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We can use the formula for the variance of a binomial distribution: Variance (σ2) = np (1 - p)</p>
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<p>We can use the formula for the variance of a binomial distribution: Variance (σ2) = np (1 - p)</p>
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<p>Here, n is the number of trials = 15</p>
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<p>Here, n is the number of trials = 15</p>
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<p>p is the probability of success = 0.6</p>
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<p>p is the probability of success = 0.6</p>
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<p>Hence, the prob failure = 1 - p = 1 - 0.6 = 0.4</p>
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<p>Hence, the prob failure = 1 - p = 1 - 0.6 = 0.4</p>
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<p>Now, we can substitute the values to the formula: (σ2) = np (1 - p)</p>
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<p>Now, we can substitute the values to the formula: (σ2) = np (1 - p)</p>
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<p>15 × 0.6 × 0.4 </p>
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<p>15 × 0.6 × 0.4 </p>
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<p>15 × 0.24 = 3.6</p>
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<p>15 × 0.24 = 3.6</p>
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<p>The variance is 3.6.</p>
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<p>The variance is 3.6.</p>
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<p>It means that the number of successes will fluctuate around the mean with a variance of 3.6 </p>
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<p>It means that the number of successes will fluctuate around the mean with a variance of 3.6 </p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A factory produces 10 bulbs daily. The probability of a defective bulb is 0.2. Find the variance of the defective bulbs per day.</p>
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<p>A factory produces 10 bulbs daily. The probability of a defective bulb is 0.2. Find the variance of the defective bulbs per day.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>1.6</p>
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<p>1.6</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>To find the variance of the defective bulbs per day, we can apply the binomial variance formula. Here, </p>
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<p>To find the variance of the defective bulbs per day, we can apply the binomial variance formula. Here, </p>
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<p>n = 10 </p>
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<p>n = 10 </p>
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<p>p = 0.2 </p>
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<p>p = 0.2 </p>
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<p>1 - p = 1 - 0.2 = 0.8 </p>
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<p>1 - p = 1 - 0.2 = 0.8 </p>
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<p>The variance formula is: </p>
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<p>The variance formula is: </p>
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<p>Variance (σ2) = np (1 - p)</p>
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<p>Variance (σ2) = np (1 - p)</p>
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<p>σ2 = 10 × 0.2 × 0.8 = 1.6</p>
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<p>σ2 = 10 × 0.2 × 0.8 = 1.6</p>
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<p>Hence, the variance of defective bulbs per day is 1.6</p>
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<p>Hence, the variance of defective bulbs per day is 1.6</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Felix takes 20 quizzes. The probability of passing each quiz is 0.8. Find the variance of the number of quizzes passed.</p>
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<p>Felix takes 20 quizzes. The probability of passing each quiz is 0.8. Find the variance of the number of quizzes passed.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>3.2</p>
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<p>3.2</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>Variance (σ2) = np (1 - p) is the formula for the variance of the binomial distribution.</p>
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<p>Variance (σ2) = np (1 - p) is the formula for the variance of the binomial distribution.</p>
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<p>Here, n = 20</p>
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<p>Here, n = 20</p>
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<p>p = 0.8 </p>
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<p>p = 0.8 </p>
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<p>1 - p = 1 - 0.8 = 0.2</p>
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<p>1 - p = 1 - 0.8 = 0.2</p>
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<p>Now, we can substitute the values.</p>
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<p>Now, we can substitute the values.</p>
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<p> σ2 = 20 × 0.8 × 0.2 </p>
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<p> σ2 = 20 × 0.8 × 0.2 </p>
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<p> σ2 = 20 × 0.16 = 3.2</p>
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<p> σ2 = 20 × 0.16 = 3.2</p>
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<p>The number of quizzes Felix passes fluctuates around the mean with a variance of 3.2</p>
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<p>The number of quizzes Felix passes fluctuates around the mean with a variance of 3.2</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>A basketball player takes 30 free throws. The probability of making a basket is 0.6. Find the variance of successful shots.</p>
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<p>A basketball player takes 30 free throws. The probability of making a basket is 0.6. Find the variance of successful shots.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>7.2</p>
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<p>7.2</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>To find the answer, we can use the formula, Variance (σ2) = np (1 - p)</p>
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<p>To find the answer, we can use the formula, Variance (σ2) = np (1 - p)</p>
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<p>Where, n = 30 </p>
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<p>Where, n = 30 </p>
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<p>p = 0.6</p>
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<p>p = 0.6</p>
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<p>1 - p = 1 - 0.6 = 0.4</p>
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<p>1 - p = 1 - 0.6 = 0.4</p>
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<p>So the formula will be: σ2 = 30 × 0.6 × 0.4 </p>
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<p>So the formula will be: σ2 = 30 × 0.6 × 0.4 </p>
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<p>σ2 = 30 × 0.24 = 7.2</p>
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<p>σ2 = 30 × 0.24 = 7.2</p>
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<p>The variance of successful shots is 7.2</p>
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<p>The variance of successful shots is 7.2</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>In a shop, 40 customers visit daily. The probability that a customer makes a purchase is 0.6. Find the variance of the number of customers making a purchase.</p>
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<p>In a shop, 40 customers visit daily. The probability that a customer makes a purchase is 0.6. Find the variance of the number of customers making a purchase.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>9.6</p>
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<p>9.6</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>To find the variance of the number of customers making a purchase, we can use the formula. Here, n = 40</p>
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<p>To find the variance of the number of customers making a purchase, we can use the formula. Here, n = 40</p>
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<p>p = 0.6 </p>
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<p>p = 0.6 </p>
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<p>1 - p = 1 - 0.6 = 0.4</p>
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<p>1 - p = 1 - 0.6 = 0.4</p>
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<p>The binomial variance formula is:</p>
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<p>The binomial variance formula is:</p>
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<p> Variance (σ2) = np (1 - p)</p>
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<p> Variance (σ2) = np (1 - p)</p>
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<p>σ2 = 40 × 0.6 × 0.4 </p>
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<p>σ2 = 40 × 0.6 × 0.4 </p>
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<p>σ2 = 40 × 0.24 = 9.6</p>
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<p>σ2 = 40 × 0.24 = 9.6</p>
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<p>The variance of the number of customers making a purchase is 9.6</p>
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<p>The variance of the number of customers making a purchase is 9.6</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on Variance of Binomial Distribution</h2>
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<h2>FAQs on Variance of Binomial Distribution</h2>
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<h3>1.Define the variance of the binomial distribution.</h3>
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<h3>1.Define the variance of the binomial distribution.</h3>
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<p>The variance of the binomial distribution explains how much the values differ from the mean. A binomial distribution involves n independent trials, and the probability of success is denoted as p. Also, the probability of failure is q (1 - p). </p>
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<p>The variance of the binomial distribution explains how much the values differ from the mean. A binomial distribution involves n independent trials, and the probability of success is denoted as p. Also, the probability of failure is q (1 - p). </p>
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<h3>2.Explain the formula of variance of the binomial distribution.</h3>
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<h3>2.Explain the formula of variance of the binomial distribution.</h3>
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<p>To find the value of n and p, we can use the formula: Variance (σ2) = np (1 - p) Here n is the number of trials, then p is the probability of success, and 1 - p is the probability of failure. The<a>symbol</a>for variance is σ2 and is the<a>square</a>of the<a>standard deviation</a>. </p>
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<p>To find the value of n and p, we can use the formula: Variance (σ2) = np (1 - p) Here n is the number of trials, then p is the probability of success, and 1 - p is the probability of failure. The<a>symbol</a>for variance is σ2 and is the<a>square</a>of the<a>standard deviation</a>. </p>
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<h3>3. What is variance in a binomial distribution?</h3>
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<h3>3. What is variance in a binomial distribution?</h3>
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<p>Variance tells us how much the number of successes fluctuates from the mean. A higher variance means the result is far different from the mean, and a low variance means the results stay more consistent. </p>
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<p>Variance tells us how much the number of successes fluctuates from the mean. A higher variance means the result is far different from the mean, and a low variance means the results stay more consistent. </p>
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<h3>4.Is it possible for the variable to be negative?</h3>
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<h3>4.Is it possible for the variable to be negative?</h3>
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<p>No, variance cannot be negative. It measures how much values differ from the mean, so the value of variance will always be zero or positive. </p>
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<p>No, variance cannot be negative. It measures how much values differ from the mean, so the value of variance will always be zero or positive. </p>
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<h3>5.How are n and variance related?</h3>
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<h3>5.How are n and variance related?</h3>
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<p>As n increases, the variance also increases, provided that p remains constant. Here, n is the number of trials and p is the probability of success. Also, if n doubles, it results in the doubling of variance, only if p remains constant. </p>
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<p>As n increases, the variance also increases, provided that p remains constant. Here, n is the number of trials and p is the probability of success. Also, if n doubles, it results in the doubling of variance, only if p remains constant. </p>
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