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The weighted sample mean, ¯, is itself a random variable. Its expected value and standard deviation are related to the expected values and standard deviations of the observations, as follows. For simplicity, we assume normalized weights (weights summing to one).
These values are used to calculate an E value for the estimate and a standard deviation (SD) as L-estimators, where: E = (a + 4m + b) / 6 SD = (b − a) / 6. E is a weighted average which takes into account both the most optimistic and most pessimistic estimates provided. SD measures the variability or uncertainty in the estimate.
The expected value of a random variable is the weighted average of the possible values it might take on, with the weights being the respective probabilities. More generally, the expected value of a function of a random variable is the probability-weighted average of the values the function takes on for each possible value of the random variable.
Since the probabilities must satisfy p 1 + ⋅⋅⋅ + p k = 1, it is natural to interpret E[X] as a weighted average of the x i values, with weights given by their probabilities p i. In the special case that all possible outcomes are equiprobable (that is, p 1 = ⋅⋅⋅ = p k), the weighted average is given by the standard average. In the ...
The mathematics of the distribution resulted from the authors' desire to make the standard deviation equal to about 1/6 of the range. [ 2 ] [ 3 ] The PERT distribution is widely used in risk analysis [ 4 ] to represent the uncertainty of the value of some quantity where one is relying on subjective estimates, because the three parameters ...
It is a measure used to evaluate the performance of regression or forecasting models. It is a variant of MAPE in which the mean absolute percent errors is treated as a weighted arithmetic mean. Most commonly the absolute percent errors are weighted by the actuals (e.g. in case of sales forecasting, errors are weighted by sales volume). [3]
This algorithm can easily be adapted to compute the variance of a finite population: simply divide by n instead of n − 1 on the last line.. Because SumSq and (Sum×Sum)/n can be very similar numbers, cancellation can lead to the precision of the result to be much less than the inherent precision of the floating-point arithmetic used to perform the computation.
One way of seeing that this is a biased estimator of the standard deviation of the population is to start from the result that s 2 is an unbiased estimator for the variance σ 2 of the underlying population if that variance exists and the sample values are drawn independently with replacement. The square root is a nonlinear function, and only ...