ok I DONT KNOW ANYTHING ABOUT STANDARD DEVIATIONS SO I CANT UNDERSTANT THE RESULTS ON THIS ONE ARTICLE IM READING CAN YOU HELP?
126309
A statistic used as a measure of the dispersion or variation in a distribution. It is equal to the square root of the arithmetic mean of the squares of the deviations from the arithmetic mean. Simply put, standard deviation measures how spread out the values in a data set are. More precisely, it is a measure of the average difference between the values of the data in the set. If the data points are all similar, then the standard deviation will be low (closer to zero). If the data points are highly variable, then the standard variation is high (further from zero). the greater the standard deviation, the higher the risk. it is always positive, and NEVER negative. the formula you use is....well i can't write it because i don't have the symbols on my keyboard, but i'll try:\n\n\nσ= √(Σ[D-E]^2P)\n\nwhere:\nσ=sigma=standard deviation\nΣ=capital sigma= the sum of\nD=outcome\nE=expected value\n^2= squared\nP= Probability\n√=square root\n\nhere is an example:\n\n(Assumptions)(Outcome probability)(Outcomes)\n(Pessimistic)(0.2)                     (300)\n(Successful)(0.6)                     (600)\n(Optimistic)(0.2)                     (900)\n\n(sorry, the table won't keep the spaces, so i put brackets in)\n\nok, you need the expected value, which is the sum of all of the outcomes, times their probabilities, like so:\nexpected value = 300×0.2+600×0.6+900×0.2 = 600\n\n\nthen standard deviation would be calculated like so:\nstandard deviation = {(300-600)2×0.2+(600-600)2×0.6+(900-600)2×0.2}½ = 190\n\nhope that helped! oh, and if you don't understand, the "to the power of 1/2" part is the same as taking the square root.
560442
This is taken from statistics.  Chebyshev's Theorem (Pafnuty Chebyshev, 1821-1894; sometimes spelled Tchebycheff), states that for any population or sample, the proportion of observations, whose z score has an absolute value less than or equal to k, is no less than (1 - (1 / k^2 )).  So, if you have a graph that has two standard deviations, using the theorem, would give you this equation where k=2: (1 - (1 / 2^2 )) = .7500  \n\nThis means that 75% of the data will fall between the range of data.  There is a down side to using Chebysev and that is it doesn't give any results for standard deviations less than or equal to one. Here is a small table of some of the Chebyshev results for k greater than 1. \n\n k             p\n1.25      .36\n1.50      .56\n1.75      .67\n2.00      .75\n2.25      .80\n2.50      .84\n2.75      .87\n3.00      .89\n3.25      .91\n3.50      .92\n3.75      .93\n4.00      .94
550084
We turn this into a standard normal distribution:\n\nz = (score - mean score)/std dev = 1.96\n\nsince we know P(Z>1.96) = 0.025\n\n (score -80)/5 = 1.96\n\nsolve for score\n\nscore = 89.8, please check for errors
121770
Standard deviation is often called standard error.  Say you take a sample of 100 people to know their averga ages.  You calculate the standard deviation by first knowing what the average is.  Let´s say the average age is 35 years old.  By definition, the standard deviation would be how many years are the people on your sample older or younger than the average. It could be about 3 years.  That means that most people in your sample has ages between 32 (35-3) and 38 (35+3) years.\nby now you should have the formula, and some example.  Notice that if you draw a graph made by columns with your data, the standard deviation measures exactly a proportion of the width of a column in your graph.
45555
3.56 is the correct answer.\n\n1. Find the mean of your data: 27.076\n\n2. Subtract the mean from each piece of data to find the deviations\n\n27.076 - 24.18 = 2.9\n27.076 - 31.32 = -4.24\n27.076 - 23.59 = 3.49\n27.076 - 31.54 = -4.46\n27.076 - 31.69 = -4.61\n27.076 - 24.88 = 2.2\n27.076 - 26.14 = 0.94\n27.076 - 29.2 = -2.12\n27.076 - 22.35 = 4.73\n27.076 - 25.87 = 1.21\n\n3. Square each of the deviations\n\n2.9^2 = 8.41\n-4.24^2 = 17.98\n3.49^2 = 12.18\n-4.46^2 = 19.89\n-4.61^2 = 21.25\n2.2^2 = 4.84\n0.94^2 = 0.88\n-2.12^2 = 4.49\n4.73^2 = 22.37\n1.21^2 = 1.46\n\n4. Find the mean of the squared deviations (this is the variance)\n\n11.38\n\n5. Find the square root of the mean of the squared deviations (this is the standard deviation)\n\nSQRT(11.38) = 3.37\n\nFor some reason Excel giving 3.56 for the result. Steps showing uses a lot of rounding offs.\n\nHope this helps.
536306
in Statistics in theory there should zero measurements falling beyond the three standard deviations from the mean on either side. though in practical some values do fall in this range. I don't remember the exact statistical names of these values but i think they are referred to as "outliers".
361964
(A) s = 0.2+0.4+0.4 = 1; Pm=s/3=0.33333 – mean value for P; \nsigma=sqrt{(0.3333-0.2)^2 + (0.3333-0.4)^2 + (0.3333-0.4)^2}/3; whereas sigma is standard deviation.\n(B) s = 0.2+0.2+0.5+0.2 = 1.1; Pm=s/4=0.275;\nsigma=sqrt{(0.275-0.2)^2 + (0.275-0.2)^2 + (0.275-0.5)^2 + (0.275-0.2)^2}/4 \ndo I answer your question correctly?
495271
first fin the mean with the sample of 280 \n\nmultiply 280 to .184 to get the mean \n\nmultiply 280 to .184 and to .816 (from 1-.184) then take the \n\nsquare root that will be your standard deviation \n\n\ntake the z score you will get   ( 56 - 51.52 )/6.483851\n\n=.6909   find teh valu in the standard normal distribution table \n\nyou will get 0.2549    subtract it from .5  because you need only \n\nthe probability higher than of 56  you will get \n\n24.51%
110153
I've just answered this in your previous question: "Statistics: Variability & SRS Samples of Population?"\n \nIf you look there, you'll find the formula for the s.d. of p_hat. So that increasind n by multiplying it by "m", reduces (devides) sd (or se) by sqrt(m).\n \nFor example, lets say the sd is 0.2 and the sample size is 100. If you increase the sample to 400 (multiply it by 4), the sd is devided by sqrt(4) = 2, i.e., sd = 0.1.
411737
What's your problem - your in college - do your own homework (this son of a statistician is unimpressed).  If you don't know how to construct a confidence interval look at http://facweb.bhc.edu/EastCampus/leeb/ma108/06_01w/tsld032.htm\n\nGood Luck!!!
388555
You need the standard deviation of the wieght of samples of 50 people.\n\nThe mean weight of samples of 50 people is 150 * 50, or 7500 pounds. Four tons is 8000 pounds.\n\nWith such a large sample as 50, you would expect that your sample means will each be quite close to 150. In fact, the standard deviation of the means of your samples is\n\n35 / sqrt(50)  = 35 / 5sqrt(2) = 7 / sqrt(2) = 7sqrt(2) / 2\n\nor around 5 pounds. That means that 2/3 of the time, the weight of 50 people will lie between 7495 and 7505 pounds, and 95% of the time, their weight will lie between 7490 and 7510 pounds.\n\nThe design difference between 8000 pounds and 7500 pounds represents 100 standard deviations of the sample means. That's off the charts. There is essentially "zero" probability that a sample as large as 50 will be so far from the mean as that.\n\nModulo does some calculations, but upon finding the result "enormous" abandons reason and uses some other (wrong) method that gives a more "reasonable" result. He was almost right the first time. The problem supposes an enormous design margin.
460908
Wal can't be right. The standard deviation of the means of samples is always lower than the standard deviation of the population. The formula is\n\ns / sqrt(n)\n\nwhere s is the standard deviation of the population.\n\nfor a), your answer is 15 / sqrt(10)
411814
10% > is z score of 1.281551\n7% < is z score of -1.475791\n\nz = (x - μ) / σ\n\n1.281551 = (16.3 - μ) / σ\n\n-1.475791 = 15.5 - μ) / σ\n\nMulitply by σ:\n\n1.281551σ = 16.3 - μ\n-1.475791σ = 15.5 - μ\n\nSubtract one from the other:\n1.281551σ + 1.475791σ = .8\n\n2.757342σ = .8\nσ = 0.29013448\n\nPlug that back into the equation:\n1.281551 (0.29013448) = 16.3 - μ\nμ = 15.9281779\n\nstandard deviation is ~.29 oz. and mean is ~15.93 oz.\n\nIf you go back and do that common sense check, you will see this fits in the bell curve between the percentages stated.  Only 10% is more than 16.3 oz and only 7% is less than 15.5 oz.  Since this is a normal distribution, the number has to be between these two, and should be closer to the 16.3 than the 15.5, which it is.  Clearly, the first answer submitted was wrong.  Brian made both z scores negative...
506428
1) X/25\n2) X/25\n\nExplanation:\n1) If you divide a set of values by a constant, then both the mean and the standard deviation are divided by that constant.\nSince your question is about the variance, which is the SQUARE of the standard deviation, the variance is divided by the SQUARE of the constant (5, in this case).\n\n2) When you add a constant to a set of values, the mean is increased by the amount of that constant, but the standard deviation is unchanged. This is true because the group of values has all moved up by a constant amount, but their relation to each other is unchanged, and their variation from each other is unchanged. Since the standard deviation is unchanged, the variance (the square of the standard deviation) is also unchanged.
635347
Find the midpoint of each interval.\nFor 1-3, the midpoint is (3+1)/2 = 4/2 = 2\nFor 4-6, the midpoint is (6+4)/2 = 10/2 = 5\nFor 7-9, the midpoint is (9+7)/2 = 16/2 = 8\n2(1) + 5(6) + 8(3)\n= 2+30+24\n= 56\nn = number of data points = 1+6+3 = 10\nvariance = [1/(10-1)][((2^2)(1) +(5^2)(6) +(8^2)(3)) - (56^2)/10]\n= (1/9)((4+150+192) - 3136/10)\n= (1/9)(346 - 313.6)\n= (1/9)(32.4)\n= 3.6\nstandard deviation = sqrt(variance)\n= sqrt(3.6) \n= 1.9
616296
The previous answer is correct.  Also, if the standard deviation is greater than your mean, then that may indicate your data has a large range, and most likely have an outlier.  For example, if you data is {1,1,1,100}.\nThe mean is 25.75\nThe standard deviation is 49.5.\nThe range of data is 100-1=99, with the outlier being 100.\n\nSometimes, experimentalist would throw out the outlier data and recompute statistics for the new set of data.  So if we throw out the outlier 100 and can rationalize that 100 is a corrupted data point because our previous trials were all 1's, then our new data set is {1,1,1} and\nThe mean is 1\nThe standard deviation is 0.\nThe range of data is 1-1=0.
404878
Here is what i found.  I'm not sure how right this is, but you can check it out.  Good Luck with your testing.