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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.

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What statistic is the average amount by which the scores in a distribution vary from the mean?

Standard deviation


What is the mean and standard deviation of a distribution of T-scores?

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)].


What percent of the scores in a normal distribution will fall within one standard deviation?

It is 68.3%


How many of scores will be within 1 standard deviation of the population mean?

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 0?

If the standard deviation of 10 scores is zero, then all scores are the same.


Will increasing the frequency of scores in the tails of a distribution effect the standard deviation How or Why?

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.


How much is 84 percentile equals mean plus 1 standard deviation or mean plus 1.4 standard deviation. Can you give me reference also please?

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.


What is the sampling distribution when the standard deviation is known?

When the standard deviation of a population is known, the sampling distribution of the sample mean will be normally distributed, regardless of the shape of the population distribution, due to the Central Limit Theorem. The mean of this sampling distribution will be equal to the population mean, while the standard deviation (known as the standard error) will be the population standard deviation divided by the square root of the sample size. This allows for the construction of confidence intervals and hypothesis testing using z-scores.


Which measure would you use to fine the spread of marks in an examination?

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.


What role do z scores play in this transformation of data from multiple distributions to standard normal distribution?

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.


How do you calculate Z and T 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)


If standard deviation of 10 scores is 0?

All the scores are equal

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