## Solved – Finding the mean and standard deviation of an unknown distribution from a sample

I am interested in finding the mean and standard deviation of the whole distribution by looking only at a random sample. I don't know anything else about the distribution (for example I don't know if the distribution is normal or not). Is what I'm asking even possible? Best Answer Sure, your best guess of the … Read more

## Solved – Are two sets of 1-D data are equal, if they have same mean and variance and same size

I have two datasets of size n. They have same mean and variance. Is it possible that they have the same entries as their data? What if I considered variance using L1 norm? EDIT: By L1 norm I mean that, the mean of the deviation with the mean of the data. Best Answer Is it … Read more

## Solved – Is it allowed to use averages on a dataset to improve correlation

I have a dataset with a dependent and an independent variable. Both are not a time series. I have 120 observations. The correlation coefficient is 0.43 After this calculation, I have added a column for both variables with the average for every 12 observations, resulting in 2 new columns with 108 observations (pairs). The correlation … Read more

## Solved – Expected number of dice rolls require to make a sum greater than or equal to K

A 6 sided die is rolled iteratively. What is the expected number of rolls required to make a sum greater than or equal to K? Before Edit P(Sum>=1 in exactly 1 roll)=1 P(Sum>=2 in exactly 1 roll)=5/6 P(Sum>=2 in exactly 2 rolls)=1/6 P(Sum>=3 in exactly 1 roll)=5/6 P(Sum>=3 in exactly 2 rolls)=2/6 P(Sum>=3 in exactly … Read more

## Solved – Expected number of dice rolls require to make a sum greater than or equal to K

A 6 sided die is rolled iteratively. What is the expected number of rolls required to make a sum greater than or equal to K? Before Edit P(Sum>=1 in exactly 1 roll)=1 P(Sum>=2 in exactly 1 roll)=5/6 P(Sum>=2 in exactly 2 rolls)=1/6 P(Sum>=3 in exactly 1 roll)=5/6 P(Sum>=3 in exactly 2 rolls)=2/6 P(Sum>=3 in exactly … Read more

## Solved – Standard deviation of revenue where numbers are only positive

Is there a way for standard deviation to ignore negative numbers? In other words, how can you make it so that 2 or 3 deviations do not extend into negative territory? When looking at a set of only positive numbers, within one or two standard deviations, I end up with negative numbers. It's essentially impossible … Read more

## Solved – Standard deviation of revenue where numbers are only positive

Is there a way for standard deviation to ignore negative numbers? In other words, how can you make it so that 2 or 3 deviations do not extend into negative territory? When looking at a set of only positive numbers, within one or two standard deviations, I end up with negative numbers. It's essentially impossible … Read more

## Solved – Mean centering interaction terms

I want to mean center my interaction terms in a regression model (i.e., make the mean zero for each variable). I understand that I am supposed to mean center my variables first and then multiply them together to create my interaction term. But is it a problem that when I multiply two negative scores, I … Read more

## Solved – Why does mean tend be more stable in different samples than median

Section 1.7.2 of Discovering Statistics Using R by Andy Field, et al., while listing virtues of mean vs median, states: … the mean tends to be stable in different samples. This after explaining median's many virtues, e.g. … The median is relatively unaffected by extreme scores at either end of the distribution … Given that … Read more

## Solved – Is the sample mean a better point estimate of the population median than the sample median

A beginner's question to check I've understood correctly. A basic stats textbook says: "The variance of the sampling distribution of the median is greater than that of the sampling distribution of the mean. It follows that sample mean is likely to be closer to the population mean than the sample median. Therefore, the sample mean … Read more