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Q: Does variance and standard deviation assume nominal data?

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The answer will depend on the underlying distribution for the variable. You may not simply assume that the distribution is normal.

z =0 and P(X< x) = 0.5 Explanation: z = (x-xbar)/sd, where xbar is the estimated mean or average of the sample, sd is the standard deviation, and x is the value of the particular outcome. We change x to z so that we can use the normal distribution or t-tables tables, which are based on a zero mean and 1 standard deviation. For example: What is the probability that the mean value of the distribution is 5 or less, given the sample average is 5 and the sd is 2? The z-score would be (5-5)/2 which is equal to 0. The probability, if we assume the normal or t-distribution, is 0.50. (see normal distribution tables) I hope this makes sense to you. The normal distribution is symmetrical. Per the example, a sample average of 5 tells you there is equal chance of the population mean being above and below 5.

I assume you refer to the distance between the points.I assume you refer to the distance between the points.I assume you refer to the distance between the points.I assume you refer to the distance between the points.

Assume that you are correlating two variables x and y. If there is an increasing relationship between x and y, (that is , the graph of y=a+bx, slopes upward), the correlation coefficient is positive. Similarly, if there is a decreasing relationship, the correlation coefficient is negative. The correlation coefficient can assume values only between -1 and 1.

The answer is 9^4 (9 to the 4th power.) OR Standard form may also refer to factoring, in which case it is (9x9)(9x9); which, through FOIL (First, Inner, Outer, Last), would still give you the simplified answer of 9^4. Standard form in Algebra, or graphing, is much different and involves moving all variables to one side and all constants to the other...but you don't have variables so I'm going to assume you aren't asking for that kind of standard form.

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Since this is regarding statistics I assume you mean lower case sigma (σ) which, in statistics, is the symbol used for standard deviation, and σ2 is known as the variance.

square (25/36) = 5/6 = .833

No. It is defined to be the positive square root of ((the sum squared deviation divided by (the number of observations less one))

A standard deviation is a statistical measure of the variation there in a population or group. A standard deviation of 1 means that 68% of the members of the population are withing plus or minus the value of the standard deviation from the average. For example: assume the average height of men is 5 feet 9 inches, and the standard deviation is three inches. Then 68% of all men are between 5' 6" and 6' which is 5'9" plus or minus 3 inches. [Note: this is only to illustrate and is not intended to be a real/correct statistic of men's heights.]

The purpose of obtaining the standard deviation is to measure the dispersion data has from the mean. Data sets can be widely dispersed, or narrowly dispersed. The standard deviation measures the degree of dispersion. Each standard deviation has a percentage probability that a single datum will fall within that distance from the mean. One standard deviation of a normal distribution contains 66.67% of all data in a particular data set. Therefore, any single datum in the data has a 66.67% chance of falling within one standard deviation from the mean. 95% of all data in the data set will fall within two standard deviations of the mean. So, how does this help us in the real world? Well, I will use the world of finance/investments to illustrate real world application. In finance, we use the standard deviation and variance to measure risk of a particular investment. Assume the mean is 15%. That would indicate that we expect to earn a 15% return on an investment. However, we never earn what we expect, so we use the standard deviation to measure the likelihood the expected return will fall away from that expected return (or mean). If the standard deviation is 2%, we have a 66.67% chance the return will actually be between 13% and 17%. We expect a 95% chance that the return on the investment will yield an 11% to 19% return. The larger the standard deviation, the greater the risk involved with a particular investment. That is a real world example of how we use the standard deviation to measure risk, and expected return on an investment.

Usually, industrial use of standard deviation is involved in quality control and testing. A product such as cement, is produced in batches, and I assume, requires periodic testing to ensure consistent properties. The sample test variations can be evaluated using standard deviation. If the standard deviation is high, it is likely that inferior product could be shipped. Probability analysis can determine the chance that product below certain standards would be shipped.

The answer will depend on the underlying distribution for the variable. You may not simply assume that the distribution is normal.

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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

Standard deviation is a measure of the dispersion of the data. When the standard deviation is greater than the mean, a coefficient of variation is greater than one. See: http://en.wikipedia.org/wiki/Coefficient_of_variation If you assume the data is normally distributed, then the lower limit of the interval of the mean +/- one standard deviation (68% confidence interval) will be a negative value. If it is not realistic to have negative values, then the assumption of a normal distribution may be in error and you should consider other distributions. Common distributions with no negative values are gamma, log normal and exponential.

In general you cannot. You will need to know more about the distribution of the variable - you cannot assume that the distribution is uniform or Normal.

If it is possible to assume normality, simply convert the desired score to a z-score, and look up the probability for that.

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