HELP! STANDARD DEVIATION kills!?
453647
93.16667 =- 3*10.63 \n61.28 to 105.04, so all of the values fit in this range.\n\nThe previous answerers have been wrong about Chebyshev Inequality.  According to that, there should be 1-(1/k)^2 values within k standard deviations of the mean.  So, according to the theorem, there should be 1-(1/3)^2 = 8/9 = 88.89% within 3 SD's of the mean.
860672
Use Statistics\n\ntake your numbers and find the standard deviation this is your uncertainty (±.23). Average all the numbers this is your beaker mass (106.49). Beaker mass plus uncertainty is best estimate with uncertainty (106.49±0.23)
323296
First you need to ask yourself if you have any reason to believe that measurements of nitrogen oxide [chemical formula -- Just say "NO"] emissions across different cars should be normally distributed.  Since this is a statistics problem, we assume that the answer is yes.\n\nThen you compute the sample mean (the sum of those seven measurements, divided by seven) and the sample variance (the sum of (x - sample mean)^2 for those seven measurements x, all divided by six).  The sample standard deviation is then the square root of the sample variance.\n\nFinally, you look up on your table or with your calculator or wherever what "Z score" corresponds to 98% of the area under the normal pdf curve being between\n\n(mean - Z standard deviations, mean + Z standard deviations)\n\nThey want to know if that upper bound is less than 0.165 grams / mile [a peculiar mix of metric and non-metric units].\n\nThat's probably how they want you to solve it.  Compare that to the Bayesian approach (Google for it).
101856
Standard deviation is a measure of the variability in a sample.  A small standard deviation means that most observations are clost to the mean (average).  The standard deviation is the square root of the variance.  The variance is calculated by taking the difference between each observation and the mean, squaring it, summing for all observations, then dividing by the number of observations less one.\n\nwith m = mean and X1 ... Xn being the observations:\n\nvariance: V = [(X1-m)^2 + (X2-m)^2 ... + (Xn-m)^2] / (n-1)\nstandard deviation: sigma = sqrt(V)
280200
If a random sample size of 25 corresponds to a standard deviation of 30, then you can expect a random sample of size 25N to roughly correspond to a standard deviation of 30 / sqrt(N).\n\nIf 99% of the area under the bell curve is between -Z and +Z standard deviations, then you want 30 Z / sqrt(N) to be less than or equal to 0.5.\n\nMaybe a little more because of the variation of the estimated standard deviation from the true standard deviation.\n\nAlso check that the theory and the measurements both suggest it is appropriate to use the normal or Gaussian distribution.
1038416
1 standard deviation range is\n\n74-13=61\nto\n74+13=87\n\nSo any score between 61-87 is 1 standard deviation of the mean 74. So a 62 is in that range.
305208
oh puke, what a pain in the rear.  i don't envy you.  like the other dude said use excel, but if you have to do it on paper, here's the formula:\n \nIn other words, the standard deviation of a discrete uniform random variable X can be calculated as follows:\n\n1.  find the mean of the set of numbers\n2.  subtract the mean from the first number\n3.  square the difference between that number and the mean\n4.  repeat steps 2 and 3 for all numbers\n5.  add all of the differences from step 3\n6.  dvide this number by the number of numbers you had in the set\n7.  take the square root of this\n\n\nthat's the standard deviation.
387360
The problem is incompletely or improperly described.\n\nEach member of the population has some characteristic value.  The mean is the average of this value taken over the entire population, which, assuming each member is an equal contributor, is M = sum(values)/N, where N is the population.  If each member has the same value, say 515, then mean would be 515 and the standard deviation would be zero.  If each even-numbered member has value 1030, and each odd-number member has value 0, the mean is 515 and the standard deviation is (approximately) 515.\n\nSo, you have provided enough information to solve the problem satisfactorily.
108492
Assuming you are talking about standard deviation...\n\nSigma is the square root of variance.  If you need to add sigma's together, they are added by taking the square room of the sum of squares.
40612
Sigma\none sigma is one standard deviation from the mean, etc.
405752
Specifically: press STAT, then ENTER. This takes you to the lists. Put in a value and press enter, and continue until all your values are entered into the list. \nThen press STAT again. This time use the arrow key to move to CALC, then press 1-Var Stats. Then press 2nd then watever list you entered the data into. List one would be 2nd then 1 key, list two is 2nd and 2 key. L1, L2, L3 L4 L5 and L6 all have their own key. \n\nPress enter. The first number is x bar, or the mean. The SD, standard deviation is Sx. \n\nHope that helps
266644
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.
874605
In a normal distribution:\n68% of the points are within 1 standard deviation\n95% of the points are within 2 standard deviations\n\nYou'll have to look at your normal distribution tables.  You want to find the point where the area is 90%.  This happens around z = 1.645 standard deviations.\n\nSo they should advertise the life of the calculators as 2,150 +/- 450 * 1.645\n\nThat works out to a range of between 1,410 and 2,890 hours, or 2,150 +/- 740 hours.
585237
Since we are not given any raw data, we must assume normality and use the population standard deviation.\n\nThe 90% CI for the population mean is given by:\n(xbar-z(.05)*sigma/sqrt(n), xbar+z(.05)*sigma/sqrt(n))\n\nHere we are told xbar = 12.4, sigma = 6, and n = 54.  Using a table of normal probabilities, we find z(.05) = 1.645.  Simple calculation shows the CI is:\n(11.0569, 13.7431)\n\nFor part (2), we want our margin or error to be 1 with a 90% confidence level.  Our population standard deviation is still 6 and z(.05) is still 1.645.  Thus, we want:\n1.645*6/sqrt(n) = 1\n9.87/sqrt(n) = 1\nsqrt(n) = 9.87\nn = 97.4169\nRound up to get n = 98.\n\nFor your additional comments:\nIncreasing the confidence level also increases the margin of error--not (a).\nThe sample standard deviation could be higher or lower than the population standard deviation, so it could increase or decrease the margin of error, but there is no guarantee--not (b).\nIncreasing the sample size to 100 makes for a better estimate of the mean and would definitely decrease the margin of error.\n\nThus, (c) is the correct choice.
979610
Well, you've got the averages. You can now find the total points my multiplying the mean with 10. You can also determine the mean deviation, standard deviation, variance, co-variance etc of the data of each group and determine the relative levels of consistancy of each group and compare the spread of the data.\nFor further stuff...look it up in wikipedia.
303094
OK.  From any given population, you can take many, many different samples of size n.  For each of those samples, you can find the mean.  So if you take 1000 samples of size 50,  Then you would have 1000 sample means.  Now each of those sample means are probably going to be different from the overall population mean, since samples are going to look different from the population.  Also, each of the 1000 sample means are going to look different from each other because the mean is coming from different samples.\n\nSo if I were to draw a histogram (hopefully you know what that is; it is a bar graph that you can use to look at the distribution of numerical data).  Anyway, if I were to draw a histogram of my 1000 sample means,  You should notice the following three things...\n\n1.  The center of your histogram should be the same as you population mean (so if the mean of the population is 20, then the histogram of my 1000 sample means should center around 20).\n\n2.  As the sample size gets larger, the spread of the histogram should get smaller (so the spread of the means of samples of size 50 would be smaller than for the means of samples of size 25).  This is because with larger samples, you would have more information about the population, and thus you should be able to get closer to the population mean.\n\n3.  As the sample size gets larger, the shape of the histogram will look more and more like a normal curve.\n\nThis is because of the Central Limit Theorem, that states that if you take a sample of size n from any population with a finite variance, the distribution of the mean of the sample would get closer and closer to a normal curve as n gets larger.  The mean of the normal would be the same as the population mean, and the standard deviation of the normal (usually called the standard error) would  be the standard deviation of the population divided by the square root of n.  Notice that as n gets larger, the standard error gets smaller, just as described before.  So typically you can approximate these histrgrams with a normal curve if n > 25.\n\nThis phenomenon is hard to describe in words alone.  Looking at pictures of these histograms that I described would help you understand it better.
999718
what was the question?\n\n\nDo you want to know terms like mean, average, mode, standard deviation, etc.?