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The simplest case of a normal distribution is known as the standard normal distribution or unit normal distribution. This is a special case when μ = 0 {\textstyle \mu =0} and σ 2 = 1 {\textstyle \sigma ^{2}=1} , and it is described by this probability density function (or density): φ ( z ) = e − z 2 2 2 π . {\displaystyle \varphi (z ...
If a data distribution is approximately normal then about 68 percent of the data values are within one standard deviation of the mean (mathematically, μ ± σ, where μ is the arithmetic mean), about 95 percent are within two standard deviations (μ ± 2σ), and about 99.7 percent lie within three standard deviations (μ ± 3σ).
There are three parameters: the mean of the normal distribution (μ), the standard deviation of the normal distribution (σ) and the exponential decay parameter (τ = 1 / λ). The shape K = τ / σ is also sometimes used to characterise the distribution. Depending on the values of the parameters, the distribution may vary in shape from almost ...
The Template:Infobox probability distribution generates a right-hand side infobox, based on the specified parameters. To use this template, copy the following code in ...
Diagram showing the cumulative distribution function for the normal distribution with mean (μ) 0 and variance (σ 2) 1. These numerical values "68%, 95%, 99.7%" come from the cumulative distribution function of the normal distribution. The prediction interval for any standard score z corresponds numerically to (1 − (1 − Φ μ,σ 2 (z)) · 2).
[citation needed] In the first two methods the interest is in estimating the parameters of the distribution, ν and σ, from a sample of data. This can be done using the method of moments, e.g., the sample mean and the sample standard deviation. The sample mean is an estimate of μ 1 ' and the sample standard deviation is an estimate of μ 2 1/2.
Let and be respectively the cumulative probability distribution function and the probability density function of the ( , ) standard normal distribution, then we have that [2] [4] the probability density function of the log-normal distribution is given by:
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.