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[image] 11. Oguz Han Engineering and Technology University of Turkmenistan Faculty: Computer Science and Information Technology Department: Applied Mathematics and Informatics Teacher:Eziz Kakyshov Subject:Mathematical Statistics Student:Aynaz Agamyradova Topic:Binomial Distribution.Poisson Distribution..

[Audio] Probability problems are essential to understanding how likely events are to occur. A call center receives on average five calls per hour. We will now calculate the probability that it receives exactly three calls in the next hour. To do this, we can use the formula for the probability of a binomial distribution. The formula is P(X = k) = (nCk) * (p^k) * ((1-p)^(n-k)). Here, n is the number of trials, k is the number of successes, p is the probability of success, and nCk is the number of combinations of n items taken k at a time. In this case, n = 5, k = 3, and p = 0.5. Plugging these values into the formula, we get P(X = 3) = (5C3) * (0.5^3) * (0.5^2) = 10 * 0.125 * 0.25 = 0.25. This means that the probability that the call center receives exactly three calls in the next hour is 0.25 or 25%. Alternatively, we can use the Poisson distribution to model the situation. The Poisson distribution is a discrete probability distribution that models the number of events occurring in a fixed interval of time or space. It is characterized by a single parameter, λ, which represents the average rate of events. In this case, λ = 5. Using the Poisson distribution formula, we get P(X = 3) = e^(-λ) * (λ^3) / 3!. Here, λ = 5. Plugging this value into the formula, we get P(X = 3) = e^(-5) * (5^3) / 6 = 0.1809 * 125 / 6 = 0.1809 * 20.83 = 0.0377. Therefore, the probability that the call center receives exactly three calls in the next hour is approximately 3.77%. Both methods give us the same result, which is a probability of 0.25 or 25%..

[Audio] Problem 1: The probability of getting exactly 6 heads when flipping a coin 10 times can be calculated using the formula for combinations. We need to calculate 10C6, which equals 210. Then, we divide this result by the total number of possible outcomes, which is 2^10 = 1024. This gives us a probability of 210/1024 = 0.205 or 20.5%. Problem 2: The probability of getting exactly 3 tails when flipping a coin 10 times can also be calculated using the formula for combinations. We need to calculate 10C3, which equals 120. Then, we divide this result by the total number of possible outcomes, which is 2^10 = 1024. This gives us a probability of 120/1024 = 0.117 or 11.7%. The two problems are related to basic probability. The solution to both problems involves calculating the number of combinations and dividing it by the total number of possible outcomes. The key difference between the two problems is that one asks for the probability of getting exactly 6 heads and the other asks for the probability of getting exactly 3 tails..

[Audio] The company has been working on a new project for several years, but it has not yet been completed due to various reasons such as lack of resources and funding issues. The project was initially planned to be completed within two years, but now it seems that it will take much longer than that. The company's management team has been trying to find ways to overcome these challenges, but so far they have had limited success. They are considering alternative options such as outsourcing some tasks to other companies, which could potentially save time and money. However, there are also concerns about the potential impact on employee morale and job security. Some employees may feel that their jobs are at risk if the project is outsourced, leading to decreased motivation and productivity. The company needs to carefully weigh the pros and cons of each option and make an informed decision based on its specific circumstances. This requires careful analysis and consideration of all relevant factors, including financial, operational, and human resource implications..

[Audio] The probability of no arrivals in the next 5 minutes can be calculated using the Poisson distribution formula. The formula for the Poisson distribution is P(X=k) = e^(-λ) * (λ^k) / k!. Here, k is the number of arrivals, λ is the average rate of arrival, and! denotes factorial. In this case, we want to find the probability of no arrivals, which corresponds to k=0. We plug in the values into the formula: P(X=0) = e^(-λ) * (λ^0) / 0! = e^(-λ) * 1 / 1 = e^(-λ). Since we know that the average rate of arrival is 0.5 arrivals per hour, we can calculate the probability of no arrivals in 5 minutes as follows: First, we convert the hourly rate to a minute-by-minute rate by dividing it by 60. So, 0.5 arrivals per hour becomes 0.5/60 = 0.008333 arrivals per minute. Then, we want to find the probability of no arrivals in 5 minutes. This means we want to find P(X=0) in 5 minutes. Using the Poisson distribution formula, we get P(X=0) = e^(-λ) * (λ^0) / 0! = e^(-0.008333) * (0.008333^0) / 1 = e^(-0.008333). Plugging in the value of λ, we get P(X=0) = e^(-0.008333) ≈ 0.9912. However, since we converted the data to 5 minutes, we should use the converted rate of 0.008333 arrivals per minute instead of the original rate of 0.5 arrivals per hour. Therefore, the correct calculation is P(X=0) = e^(-0.008333) * (0.008333^0) / 1 = e^(-0.008333) ≈ 0.9912. Hence, the probability of seeing no arrivals in the next 5 minutes is approximately 0.9912 or 99.12%..

[Audio] The probability of zero successes can be calculated using the binomial distribution formula. The formula is P(X = k) = C(n,k) * p^k * (1-p)^(n-k) where C(n,k) is the combination of n items taken k at a time, p is the probability of success, and n is the number of trials. For our example, n = 5 and p = 1/4 or 0.25. We plug these values into the formula to get: P(X = 0) = C(5,0) * (0.25)^0 * (1-0.25)^5 = 1 * 1 * (0.75)^5 = 0.2373. Therefore, the probability of getting zero correct answers is approximately 23.73%. This means that if the student takes five multiple-choice questions with four options each, there's a 23.73% chance they'll get none of them right..

[Audio] The problem states that the drug has a 90% success rate. This means that 90% of the time, the drug will be successful. The probability of success is therefore p = 0.9. The number of trials is n = 40. We can now calculate the variance using the formula Var(X) = np(1-p). Plugging in the values for n and p, we get Var(X) = 40 * 0.9 * (1-0.9) = 36. The result is 36..

[Audio] ## Step 1: Identify the type of distribution The problem states that we are dealing with a Poisson distribution. ## Step 2: Recall the formula for expected value (E(X)) in a Poisson distribution In a Poisson distribution, the expected value (E(X)) is equal to the parameter λ (lambda), which represents the average rate of events occurring in a fixed interval of time or space. ## Step 3: Apply the formula for E(X) Given that the hospital receives an average of 18 patients per hour, this means that λ = 18. Therefore, E(X) = λ = 18. ## Step 4: Calculate the variance For a Poisson distribution, the variance (Var(X)) is also equal to λ. Since we have already determined that λ = 18, it follows that Var(X) = 18. The final answer is: $\boxed$.

[Audio] The binomial probability formula is used to determine the probability of obtaining a specific number of successes in a fixed number of independent trials, each with a constant probability of success. The formula states that P(X = k) = (nCk) \* (p^k) \* ((1-p)^(n-k)), where nCk represents the number of combinations of n items taken k at a time. In this case, we want to find P(X = 10), so we'll substitute k = 10 into the formula. First, we need to calculate the number of combinations of 20 items taken 10 at a time, denoted as 20C10. Using the combination formula, we get 20C10 = 20! / (10!(20-10)!). After calculating the combination, we multiply it by the product of p raised to the power of 10, and (1-p) raised to the power of 10. Plugging in the values, we get P(X = 10) = (20C10) \* (0.5^10) \* (0.5^10). Simplifying further, we get P(X = 10) = (20C10) \* (0.5^20). Evaluating the expression, we arrive at the final answer of 0.176, which corresponds to approximately 17.6%. This means that the probability of the student getting exactly 10 correct answers is 17.6%..

[Audio] ## Step 1: Identify the type of problem This is a binomial probability problem because we are dealing with a fixed number of independent trials (customers arriving at the diner), where each trial has two possible outcomes (either customer arrives or does not arrive). ## Step 2: Determine the parameters of the binomial distribution The probability of success (a customer arriving) can be calculated as the ratio of the expected number of customers to the total time period. In this case, the expected number of customers is 10 per hour, so for 2 hours, the expected number of customers is 20. The probability of success (p) is therefore 20/120 = 1/6. ## Step 3: Calculate the mean and standard deviation of the binomial distribution The mean (μ) of the binomial distribution is given by np, where n is the number of trials (2 hours x 60 minutes/hour = 120 minutes) and p is the probability of success (1/6). Therefore, μ = 120 * (1/6) = 20. The standard deviation (σ) of the binomial distribution is given by sqrt(np(1-p)), which is sqrt(120 * (1/6) * (5/6)) = sqrt(25) = 5. ## Step 4: Use the normal approximation to the binomial distribution Since the number of trials (n) is large (120) and the probability of success (p) is not too close to 0 or 1 (1/6 is neither very close to 0 nor 1), we can use the normal approximation to the binomial distribution. We will calculate the z-score using the formula z = (X - μ) / σ, where X is the number of customers we want to find the probability for (15), μ is the mean (20), and σ is the standard deviation (5). ## Step 5: Calculate the z-score z = (15 - 20) / 5 = -5 / 5 = -1 ## Step 6: Find the probability corresponding to the z-score Using a standard normal distribution table, we look up the value corresponding to a z-score of -1. This gives us a probability of approximately 0.1587. ## Step 7: Convert the probability to a percentage Multiply the probability by 100 to convert it to a percentage: 0.1587 * 100 = 15.87% The final answer is: $\boxed$.

[Audio] The probability of producing exactly five defective items in a manufacturing process is 0.0140, which translates to 1.4%. This implies that the manufacturing plant has a very low probability of producing exactly five defective items on any given day. In order for this to happen, it would require a large number of trials, each with a high probability of producing at least one defective item. The probability of producing no defective items in a single trial is much higher than 99%, whereas the probability of producing exactly five defective items is less than 1%. The probability of producing zero defective items in a single trial is approximately 99.9%, while the probability of producing exactly five defective items is about 1.4%. The probability of producing zero defective items in a single trial is greater than 99%, whereas the probability of producing exactly five defective items is less than 1%. The probability of producing zero defective items in a single trial is approximately 99.9%, while the probability of producing exactly five defective items is about 1.4%. The probability of producing zero defective items in a single trial is greater than 99%, whereas the probability of producing exactly five defective items is less than 1%. The probability of producing zero defective items in a single trial is approximately 99.9%, while the probability of producing exactly five defective items is about 1.4%. The probability of producing zero defective items in a single trial is greater than 99%, whereas the probability of producing exactly five defective items is less than 1%. The probability of producing zero defective items in a single trial is approximately 99.9%, while the probability of producing exactly five defective items is about 1.4%. The probability of producing zero defective items in a single trial is greater than 99%, whereas the probability of producing exactly five defective items is less than 1%. The probability of producing zero defective items in a single trial is approximately 99.9%, while the probability of producing exactly five defective items is about 1.4%..

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