Assuming a normal distribution, the proportion falling between the mean (of 8) and 7 with standard deviation 2 is:
z = (7 - 8) / 2 = -0.5 → 0.1915 (from normal distribution tables)
→ less than 7 is 0.5 - 0.1915 = 0.3085 = 0.3085 x 100 % = 30.85 %
(Note: the 0.5 in the second sum is because half (0.5) of a normal distribution is less than the mean, not because 7 is half a standard deviation away from the mean, and the tables give the proportion of the normal distribution between the mean and the number of standard deviations from the mean.)
T score is usually used when the sample size is below 30 and/or when the population standard deviation is unknown.
A single number, such as 478912, always has a standard deviation of 0.
The standard deviation and the arithmetic mean measure two different characteristics of a set of data. The standard deviation measures how spread out the data is, whereas the arithmetic mean measures where the data is centered. Because of this, there is no particular relation that must be satisfied because the standard deviation is greater than the mean.Actually, there IS a relationship between the mean and standard deviation. A high (large) standard deviation indicates a wide range of scores = a great deal of variance. Generally speaking, the greater the range of scores, the less representative the mean becomes (if we are using "mean" to indicate "normal"). For example, consider the following example:10 students are given a test that is worth 100 points. Only 1 student gets a 100, 2 students receive a zero, and the remaining 7 students get a score of 50.(Arithmetic mean) = 100 + 0(2) + 7(50) = 100 + 0 + 350 = 450/10 studentsSCORE = 45In statistics, the median refers to the value at the 50% percentile. That means that half of the scores fall below the median & the other half are above the median. Using the example above, the scores are: 0, 0, 50, 50, (50, 50), 50, 50, 50, 100. The median is the score that has the same number of occurrences above it and below it. For an odd number of scores, there is exactly one in the middle, and that would be the median. Using this example, we have an even number of scores, so the "middle 2" scores are averaged for the median value. These "middle" scores are bracketed by parenthesis in the list, and in this case are both equal to 50 (which average to 50, so the median is 50). In this case, the standard deviation of these scores is 26.9, which indicates a fairly wide "spread" of the numbers. For a "normal" distribution, most of the scores should center around the same value (in this case 50, which is also known as the "mode" - or the score that occurs most frequently) & as you move towards the extremes (very high or very low values), there should be fewer scores.
2 standard deviation's below the mean
The standard deviation varies from one data set to another. Indeed, 100 may not even be anywhere near the range of the dataset.
T score is usually used when the sample size is below 30 and/or when the population standard deviation is unknown.
The area between the mean and 1 standard deviation above or below the mean is about 0.3413 or 34.13%
X = 50 => Z = (50 - 70)/12 = -20/12 = -1.33 So prob(X < 50) = Prob(Z < -1.33...) = 0.091
There must be a formula, but in the mean time there is a handy site that does it for you. [See related link below for the converter]
A single number, such as 478912, always has a standard deviation of 0.
Standard deviation calculation is somewhat difficult.Please refer to the site below for more info
A single number, such as 478912, always has a standard deviation of 0.
There are approximately 16.4% of students who score below 66 on the exam.
z=-20/12 = -1.667 Assuming normal distribution, P(Z < -1.667) = 0.04779 or 4.8% of the scores should be less than 50. You can get the probabilities by looking them up on a table or use Excel, where +Normdist(50,70,12,true). My normal table has only 2 digit accuracy so for -1.67 = 0.0475.
Suppose the random variable, X, that you are studying, has a mean = m, and standard deviation (sd) = s. Then z = 1.33 is equivalent to saying that(x - m)/s = 1.33 or that your observed value is greater than the mean by 1.33 times the sd.
Yes. If the variance is less than 1, the standard deviation will be greater that the variance. For example, if the variance is 0.5, the standard deviation is sqrt(0.5) or 0.707.
34.1% of the data values fall between (mean-1sd) and the mean.