The sum of two random variables that are normally distributed will be also be normally distributed. Use the link and check out the article. It'll save a cut and paste.
Yes, and the new distribution has a mean equal to the sum of the means of the two distribution and a variance equal to the sum of the variances of the two distributions. The proof of this is found in Probability and Statistics by DeGroot, Third Edition, page 275.
Given "n" random variables, normally distributed, and the squared values of these RV are summed, the resultant random variable is chi-squared distributed, with degrees of freedom, k = n-1. As k goes to infinity, the resulant RV becomes normally distributed. See link.
The Central Limit Theorem (abbreviated as CLT) states that random variables that are independent of each other will have a normally distributed mean.
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The sum of two random variables that are normally distributed will be also be normally distributed. Use the link and check out the article. It'll save a cut and paste.
Yes, and the new distribution has a mean equal to the sum of the means of the two distribution and a variance equal to the sum of the variances of the two distributions. The proof of this is found in Probability and Statistics by DeGroot, Third Edition, page 275.
Given "n" random variables, normally distributed, and the squared values of these RV are summed, the resultant random variable is chi-squared distributed, with degrees of freedom, k = n-1. As k goes to infinity, the resulant RV becomes normally distributed. See link.
The Central Limit Theorem (abbreviated as CLT) states that random variables that are independent of each other will have a normally distributed mean.
Yes it is. That is actually true for all random vars, assuming the covariance of the two random vars is zero (they are uncorrelated).
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Stochastic processes are families of random variables. Real-valued (i.e., continuous) random variables are often defined by their (cumulative) distribution function.
Yes, to approximately standard normal.If the random variable X is approximately normal with mean m and standard deviation s, then(X - m)/sis approximately standard normal.
Discrete variables must be countable and not negative. So no a negative number must be a continuous variable.
The answer depends on how the sample is selected. If it is a simple random sample, of size n, then it is distributed approximately normally with the same mean as the population mean.The answer depends on how the sample is selected. If it is a simple random sample, of size n, then it is distributed approximately normally with the same mean as the population mean.The answer depends on how the sample is selected. If it is a simple random sample, of size n, then it is distributed approximately normally with the same mean as the population mean.The answer depends on how the sample is selected. If it is a simple random sample, of size n, then it is distributed approximately normally with the same mean as the population mean.
For the purpose of analyses, they should be independent, identically distributed random variables. But the ideal is that they are all 0.
According to the Central Limit Theorem, the arithmetic mean of a sufficiently large number of iterates of independent random variables at a given condition is normally distributed. This is based on the condition that each random variable has well defined-variance and expected value.