3over9 is the answer to the question
It is mean + 2*standard deviation.
Yes, depending on the data being studied. Standard deviation can be thought of as the magnitude of the average distance between the data points and their mean.
The mean alone is not enough to provide an answer.
The answer is about 16% Using the z-score formula(z = (x-u)/sd) the z score is 1. This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
The answer depends on the value of the standard deviation. Without that information, the question cannot be answered.
It is mean + 2*standard deviation.
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 (SD) is a measure of spread so small sd = small spread. So the above is true for any distribution, not just the Normal.
Yes, depending on the data being studied. Standard deviation can be thought of as the magnitude of the average distance between the data points and their mean.
The mean alone is not enough to provide an answer.
The answer is about 16% Using the z-score formula(z = (x-u)/sd) the z score is 1. This means that we want the percentage above 1 standard deviation. We know from the 68-95-99.7 rule that 68 percent of all the data fall between -1 and 1 standard deviation so there must be about 16% that falls above 1 standard deviation.
There are two points of infection (the points where the curvature changes its direction) which lie at a distance of one standard deviation above mean and one standard deviation below mean.
The answer depends on the value of the standard deviation. Without that information, the question cannot be answered.
The answer will depend on what the distribution is. Non-statisticians often assum that the variable that they are interested in follows the Standard Normal distribution. This assumption must be justified. If that is the case then the answer is 81.9%
You can calculate standard deviation by addin the numbers of data that are together and dividing that number by the amount pieces of data.THAT IS TOTALLY INCORRECT.What was answered above was the calculation for getting an (mean) average.If you take five numbers for example 1, 2, 3, 4, 5 then the (mean) average is 3.But the standard deviation between them is 1.58814 and the variance is 2.5Also the population std. deviation will be 1.41421 and the population variance will be 2.see standard-deviation.appspot.com/
The "z-score" is derived by subtracting the population mean from the measurement and dividing by the population standard deviation. It measures how many standard deviations the measurement is above or below the mean. If the population mean and standard deviation are unknown the "t-distribution" can be used instead using the sample mean and sample deviation.
You also know that x is 1.036 times the standard deviation of the variable above its mean. Anything more than that would require further information about the mean and/or the variance of the variable.