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Suppose that in a particular geographic region, the mean and standard deviation of scores on a reading test are 100 points, and 12 points, respectively. Our interest is in the scores of 55 students in a particular school who received a mean score of 96.
Comparison of the various grading methods in a normal distribution, including: standard deviations, cumulative percentages, percentile equivalents, z-scores, T-scores. In statistics, the standard score is the number of standard deviations by which the value of a raw score (i.e., an observed value or data point) is above or below the mean value of what is being observed or measured.
Cumulative from mean gives a probability that a statistic is between 0 (mean) and Z. Example: Prob(0 ≤ Z ≤ 0.69) = 0.2549. Cumulative gives a probability that a statistic is less than Z. This equates to the area of the distribution below Z. Example: Prob(Z ≤ 0.69) = 0.7549. Complementary cumulative
It is also the continuous distribution with the maximum entropy for a specified mean and variance. [18] [19] Geary has shown, assuming that the mean and variance are finite, that the normal distribution is the only distribution where the mean and variance calculated from a set of independent draws are independent of each other. [20] [21]
The mean and the standard deviation of a set of data are descriptive statistics usually reported together. In a certain sense, the standard deviation is a "natural" measure of statistical dispersion if the center of the data is measured about the mean. This is because the standard deviation from the mean is smaller than from any other point.
For example, to calculate the 95% prediction interval for a normal distribution with a mean (μ) of 5 and a standard deviation (σ) of 1, then z is approximately 2. Therefore, the lower limit of the prediction interval is approximately 5 ‒ (2⋅1) = 3, and the upper limit is approximately 5 + (2⋅1) = 7, thus giving a prediction interval of ...
The data set [90, 100, 110] has more variability. Its standard deviation is 10 and its average is 100, giving the coefficient of variation as 10 / 100 = 0.1; The data set [1, 5, 6, 8, 10, 40, 65, 88] has still more variability. Its standard deviation is 32.9 and its average is 27.9, giving a coefficient of variation of 32.9 / 27.9 = 1.18
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).