Type your answer here...
It depends what percentage of the total data you want to embrace. 99.73% of the total distribution lies between minus to plus 3 standard deviations. That's usually the benchmark range.
It is 68.3%
Assuming a normal distribution 68 % of the data samples will be with 1 standard deviation of the mean.
If the standard deviation of 10 scores is zero, then all scores are the same.
Yes. It will increase the standard deviation. You are increasing the number of events that are further away from the mean, and the standard deviation is a measure of how far away the events are from the mean.
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.
Standard deviation
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)].
It is 68.3%
Assuming a normal distribution 68 % of the data samples will be with 1 standard deviation of the mean.
If the standard deviation of 10 scores is zero, then all scores are the same.
Yes. It will increase the standard deviation. You are increasing the number of events that are further away from the mean, and the standard deviation is a measure of how far away the events are from the mean.
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.
The measure commonly used to find the spread of marks in an examination is the standard deviation. It provides a numerical value that indicates how spread out the scores are from the mean score. A larger standard deviation suggests a wider spread of scores, while a smaller standard deviation indicates a more clustered distribution of scores.
z=(x-mean)/(standard deviation of population distribution/square root of sample size) T-score is for when you don't have pop. standard deviation and must use sample s.d. as a substitute. t=(x-mean)/(standard deviation of sampling distribution/square root of sample size)
All the scores are equal
mean
The standard deviation (SD) is a measure of spread so small sd = small spread. So the above is true for any distribution, not just the Normal.