You use the z-transformation.
For any variable X, with mean m and standard error s,
Z = (X - m)/s is distributed as N(0, 1).
You use the z-transformation.
For any variable X, with mean m and standard error s,
Z = (X - m)/s is distributed as N(0, 1).
You use the z-transformation.
For any variable X, with mean m and standard error s,
Z = (X - m)/s is distributed as N(0, 1).
You use the z-transformation.
For any variable X, with mean m and standard error s,
Z = (X - m)/s is distributed as N(0, 1).
It's the same as a z-Transformation. for all xi: (xi-mean(x)) / std(x)
The normal distribution would be a standard normal distribution if it had a mean of 0 and standard deviation of 1.
The standard normal curve is symmetrical.
The standard normal distribution is a normal distribution with mean 0 and variance 1.
A mathematical definition of a standard normal distribution is given in the related link. A standard normal distribution is a normal distribution with a mean of 0 and a variance of 1.
About half the time.
It's the same as a z-Transformation. for all xi: (xi-mean(x)) / std(x)
A chi square is square of standard normal variate, so all values are positive
There is no simple formula to calculate probabilities for the normal distribution. Those for the standard normal have been calculated by numerical methods and then tabulated. As a result, probabilities for the standard normal can be looked up easily.
The F-variate, named after the statistician Ronald Fisher, crops up in statistics in the analysis of variance (amongst other things). Suppose you have a bivariate normal distribution. You calculate the sums of squares of the dependent variable that can be explained by regression and a residual sum of squares. Under the null hypothesis that there is no linear regression between the two variables (of the bivariate distribution), the ratio of the regression sum of squares divided by the residual sum of squares is distributed as an F-variate. There is a lot more to it, but not something that is easy to explain in this manner - particularly when I do not know your knowledge level.
Only the mean, because a normal distribution has a standard deviation equal to the square root of the mean.
You calculate standard deviation the same way as always. You find the mean, and then you sum the squares of the deviations of the samples from the means, divide by N-1, and then take the square root. This has nothing to do with whether you have a normal distribution or not. This is how you calculate sample standard deviation, where the mean is determined along with the standard deviation, and the N-1 factor represents the loss of a degree of freedom in doing so. If you knew the mean a priori, you could calculate standard deviation of the sample, and only use N, instead of N-1.
The standard deviation in a standard normal distribution is 1.
The standard deviation in a standard normal distribution is 1.
The normal distribution would be a standard normal distribution if it had a mean of 0 and standard deviation of 1.
The standard normal distribution has a mean of 0 and a standard deviation of 1.
The standard normal curve is symmetrical.