what are the mean and standard deviation of the mean number of strikes per square kilometers?
825061
i think you need a set of numbers to deviate from first, and then you can easily google both of these. then it is just a bunch of calculator work
114212
First you have to find the mean (average) of the set of numbers. Standard deviation measures how spread out the values in a data set are. To find standard deviation, take the difference of each number in the data set and the mean, square them individually and then add them together. Then divide that by the number of values. And then take the square root of all of that. It's really hard to write that equation in just plain text, but the link shows you what it looks like.\n\nIt's easier to see with an example.\nFind the standard deviation of 45.7, 43.8, 42.6, and 48.1.\n\nThe mean is the sum of all the numbers divided by the amount.\n(45.7+43.8+42.6+48.1)/4= 180.2/4= 45.05\n\nNow find standard deviation, sd:\nsd^2= [(45.7-45.05)^2 + (43.8-45.05)^2 + (42.6-45.05)^2 + (48.1-45.05)^2]/4\n\nsd^2= [(.65)^2 + (-1.25)^2 + (-2.45)^2 + (3.05)^2]/4\n\nsd^2= (.4425 + 1.5625 + 6.0025 + 9.3025)/4\n\nsd^2= 17.29/4=4.3225\n\nsd=sqrt(4.3226)= 2.08
878193
EDIT: This answer has been corrected.  See note below.\n\n\nIn the case of the dice, you have discrete values in the probability distribution.  It is not a "normal" (bell-curve) distribution.  But you can calculate the std dev. as follows:\n\nFor every possible value v of a throw i, compute P(vi).  In a large number of throws N, the expected number of times a value vi occurs, nvi = N*P(vi).  The mean value would then be m = (1/N)∑nv*v = (1/N)∑N*v*P(v) = ∑v*P(v).  The standard deviation would then be\n\nd^2 = [1/(N-1)]∑(xi - m)^2 =  [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑v*P(vi) +N*m^2}\n\nd = √ [1/(N-1)]*{∑[N*vi*P(vi)]^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\n\nYou can't eliminate N (the total no of triails) but you can compute the result for a large N without actually rolling the dice.\n\nEDIT:  There is an error in the formula: the terms N and P(vi) should not be squared in the first term of the sum. The formula should be \n\nd = √ [1/(N-1)]*{∑N*P(vi)*vi^2 - 2*N*m*∑vi*P(vi) +N*m^2}\n\nThen, for large N, N-1≈1 and the N's cancel out to give\n\nd = √ [∑P(vi)*vi^2 - 2*m*∑vi*P(vi) +m^2]\n\nSo for all possible values of the dice (vi), compute the probabiity of that value P(vi), the mean value m = ∑vi*P(vi),  and sum over all value according to the above formula.\n\nPlugging the mean value into the formula gives the simplification:\n\nd = √ [∑P(vi)*vi^2 - 2*m^2 +m^2] = √ [∑P(vi)*vi^2 - m^2]\n\nFINAL:  Made final correction.  Above formula is now correct; the m^2 is not under the summation sign.\n\nCalculated for an ordinary pair of dice: m = 7, d = 2.415\nYou can see the calculation here http://img452.imageshack.us/img452/1734/dicestddevlk8.png
677379
If the population has a mean of 58 and a sd of 8, then the sample means of size 100 will have a mean of 58 and a sd of 8/sqrt(100) = 0.8\n\na) Within 3 of 58 means that the sample mean will be between 58-3 = 55 and 58+3 = 61\n\nz = (55-58)/0.8 = -3.75\nz = (61-58)/0.8 = 3.75\n\nFrom the table, the area between the two is almost 1.\n\nb) The mean weight is less than 2.5 of 58 or more means less than 58-2.5 = 55.5\n\nz = (55.5-58)/0.8 = -3.13\n\nFrom the table, the are greater than that is almost 0.\n\nc) z = (57.5-58)/0.8 = -0.63\nz = (58.7-58)/0.8 = 0.88\n\nFrom the table, the area between is 0.811 - 0.264 = 0.547.\n\nd) 56 +- (1.96)(0.8)\n56 +- 1.568\n(54.432,57.568)
869354
standard deviation is one of measure of variability use in statistics which has the same unit as the given data and is equivalent ot the square root of the variance.\n\nIt measures how all the data varies. It indicates how the data in a distribution scatter or spread.\n\nThe lower the standard deviation, the closer are all the data around the mean.\nThat is, the data are not scattered and they are less heterogenuous, less variable.\n\nThe greater its value, the more the spreadness of the data around the mean.\n\nFor example,\nGiven:\nA: 45,46,40,43,48,46\nB:10,45,28,100,158,468\n\nThe standard deviation of the data A is 2.804757863\nThe standard deviation of the data B is 171.9400089\n\n]since the statndard deviation of the data A is less that in B,\nthe data in A are closer to each other, compare to the data B.
520808
Documented strikes average 300 a year.  67 people die per year on average. Undocumented strikes likely push the injury list even higher.\n\nOn a side note it is Estimated that there are 25 MILLION cloud to ground strikes in the U.S. alone each year!\n\nThese are just U.S. stats there would be no telling how many there are worldwide.
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.
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The mean is the sum of the numbers divided by the count.\n\nSo, when you add up the numbers above, you get 57.5.  There are 10 shoe sizes, so you divide by 10 to get 5.75.  That is your mean.\n\nWhen you add Susan's shoe, the mean becomes 6.  The count goes from 10 to 11.  With 11 items, you get a mean of 6 only if they add up to 66.  You currently have 57.5, so you need to figure out what shoe size brings the total to 66.\n\n66 – 57.5 = 8.5\n\nSo, Susan's shoe size mus be 8.5 to bring the total up to 66 and thus the mean of 6.
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)
59697
typically the measures of central tendency are mean, median, mode\n\nMean--essentially the average\nMedian--essentially the middle number, you find this by listing all numbers in the set in order, then start to cross off from each end equally, if the set has an odd number of items you will end on the median, if it does not (thus even) there will be two numbers which you must take the average of.  if you have recurring numbers then you cant cross out all of them in one cross-out, you need to count each individually\nMode--this is the last measure of central tendency, it is the most recurring number in a set,it is possible to have more than one Mode\n\nhope this helps!!
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You have all the information you need.\n\nThe Mean is the same as the average. It's just another term for a simple average. So it's $7500\n\nThe standard error of the mean is the standard deviation divided by the square root of the number of samples. So it's just $650 divided by 10, or 65.
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http://www.windows.ucar.edu/tour/link=/earth/statistics.html
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Standard Deviation is a simple measure of how much a list of numbers varies. It quickly tells you how much deviation there is from the average.\n\nThink of the familiar bell curve, the so-called normal curve. It might show how tall students are in your school\n\nAbout 68% of the students will have heights between the 1 standard deviation mark on either side of the central peak -- 34% on the left, and 34% on the right.\n\n68.26% of the area under the curve is within one standard deviation of the mean.\n\n95.45% of the area is within two standard deviations.\n\n99.73% of the area is within three standard deviations.\n\n99.99% of the area is within four standard deviations.\n\nIt is a quick way to explain a large hunk of data!\n\nSo for instance if one standard deviation in your school height list is 1 inch, then you'll know that 68 out of hundred kids are within 1 inch of the central value.
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Try the Bureau of Labor Statistics :)
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110,860 sq kilometers (that's 42,803 square miles)\n\nStraight from the CIA Factbook (link below)
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Not sure what exactly type of statistics you are looking for, but here are two pages\nhttp://www.mdeda.org/info.asp\nhttp://www.unos.org/
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Standard deviation of a random variable X is defined as: \nό=√E((X-E(X))2) = √E(X2)-(E(X))2\n(the 2's of course are squares)