it is shaped roughly like a bell... a bell curve.
17.7 and 20.9
The standard deviation for the Woodcock-Johnson III Tests of Achievement is typically set at 15. This is consistent with many standardized tests, which use a mean of 100 and a standard deviation of 15 to represent scores on a normal distribution. This allows for the interpretation of individual test scores in relation to the broader population.
The answer depends on the degrees of freedom (df). If the df > 1 then the mean is 0, and the standard deviation, for df > 2, is sqrt[df/(df - 2)].
Z-scores standardize data from various distributions by transforming individual data points into a common scale based on their mean and standard deviation. This process involves subtracting the mean from each data point and dividing by the standard deviation, resulting in a distribution with a mean of 0 and a standard deviation of 1. This transformation enables comparisons across different datasets by converting them to the standard normal distribution, facilitating statistical analysis and interpretation.
To determine the percentage of scores between 61 and 82, you would need to know the distribution of the scores (e.g., normal distribution) and the total number of scores. If the data is normally distributed, you can use the mean and standard deviation to find the percentage of scores in that range using a z-score table. Without specific data, it isn't possible to provide an exact percentage.
17.7 and 20.9
It is 68.3%
If the standard deviation of 10 scores is zero, then all scores are the same.
Assuming a normal distribution 68 % of the data samples will be with 1 standard deviation of the mean.
All the scores are equal
mean
T-scores and z-scores measure the deviation from normal. The normal for T-score is 50 with standard deviation of 10. if the score on t-score is more than 50, it means that the person scored above normal (average), and vise versa. The normal for Z-score is 0. If Z-score is above 0, then it means that person scored above normal (average), and vise versa.
The standard deviation for the Woodcock-Johnson III Tests of Achievement is typically set at 15. This is consistent with many standardized tests, which use a mean of 100 and a standard deviation of 15 to represent scores on a normal distribution. This allows for the interpretation of individual test scores in relation to the broader population.
None.z-scores are linear transformations that are used to convert an "ordinary" Normal variable - with mean, m, and standard deviation, s, to a normal variable with mean = 0 and st dev = 1 : the Standard Normal distribution.
No, it is called the absolute deviation.
In general, you cannot. If the distribution can be assumed to be Gaussian [Normal] then you could use z-scores.
The cumulative probability up to the mean plus 1 standard deviation for a Normal distribution - not any distribution - is 84%. The reference is any table (or on-line version) of z-scores for the standard normal distribution.