The second moment of a Poisson-distributed random variable is 2. The mean of the random variable is
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- If the random variable x has a Poisson Distribution with mean μ = 3.95, find the probability that x = 16.Austin decided to give Taylor Swift another chance. So, he listened through all her songs from 2009. There were 8 in total. Let the random variable X be the number of songs in which she complained about some boy. The probability of complaining about a boy is 56%. (a) Is X a discrete or continuous random variable? How do you know? (b) What are the possible values of X? (c) Does X follow a distribution that we covered in class? If so, how do we know? Properly label the distribution.Find the mean and standard deviation for each binomial random variable
- Two coins are selected from a pocket that contains 3 nickels and 4 quarters. The random variable is the total value in cents of the 2 selected coins. Find the probability distribution. Find the expected value, variance, and standard deviation.Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 4.0 minutes and standard deviation of 2.0 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n₁ = 46 customers in the first line and n₂ = 52 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X₁ and the mean service time for the longer one X₂ is more than 0.3 minutes. Assume that the service times for each customer can be regarded as independent random variables. Round your answer to two decimal places (e.g. 98.76). P = !After careful research you observe that the number of hackers active on each day is a random variable Bin(3,0.5) distribution. If no hackers are active, the probability of the website failure is 0.15; If one hacker is active the probability of the website failure is 0.3; if two or more hackers are active the probability of the website failure is 0.5. You conclude that the total probability of website failure on a given day is two decimal place). Assuming there is a failure, the probability no hackers are active is (to (to two decimal places).
- Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 3.5 minutes and standard deviation of 3.5 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n₁ = 2 customers in the first line and n₂ = 13 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X₁ and the mean service time for the longer one X₂ is more than 0.1 minutes. Assume that the service times for each customer can be regarded as independent random variables. Round your answer to two decimal places (e.g. 98.76). P =Suppose 2 dice are rolled, and suppose the discrete variable x counts the number of 3’s that show up. Plot the probability distribution for x using a bar graph. What is the expected value for the random variable x? What is the variance for the random variable x?Five percent of U.S. employees who are late for work blame oversleeping. You randomly select four U.S. employees who are late for work and ask them whether they blame oversleeping. The random variable represents the number of U.S. employees who are late for work and blame oversleeping. Find the mean of the binomial distribution. μ= (Round to the nearest hundredth as needed.)
- 78% of U.S. adults think that political correctness is a problem in America today. You randomly select six U.S. adults and ask them whether they think that political correctness is a problem in America today. The random variable represents the number of U.S. adults who think that political correctness is a problem in America Find the variance of the binomial distribution. σ2=nothing (Round to the nearest tenth as needed.)21% of U.S. employees who are late for work blame oversleeping. You randomly select four U.S. employees who are late for work and ask them whether they blame oversleeping. The random variable represents the number of U.S. employees who are late for work and blame oversleeping. Find the mean of the binomial distribution.Service time for a customer coming through a checkout counter in a retail store is a random variable with the mean of 1.0 minutes and standard deviation of 0.5 minutes. Suppose that the distribution of service time is fairly close to a normal distribution. Suppose there are two counters in a store, n = 11 customers in the first line and n2 = 30 customers in the second line. Find the probability that the difference between the mean service time for the shorter line X1 and the mean service time for the longer one X2 is more than 0.5 minutes. Assume that the service times for each customer can be regarded as independent random variables. Round your answer to two decimal places (e.g. 98.76). P = i
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