it is shaped roughly like a bell... a bell curve.
17.7 and 20.9
T-scores have a mean of 50 and a standard deviation of 10. These values are fixed and do not change regardless of the distribution of T-scores.
T score is usually used when the sample size is below 30 and/or when the population standard deviation is unknown.
They should be.
68% of the scores are within 1 standard deviation of the mean -80, 120 95% of the scores are within 2 standard deviations of the mean -60, 140 99.7% of the scores are within 3 standard deviations of the mean -40, 180
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.
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.
This question cannot be answered unless the bowling scores are provided.