I need to explain what standard is, what is standard deviation.  The audience is compose by teachers, and I want to make this in an easy way.
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.
483403
Press mode on your calculator and select SD (usually option 2 on a casio).  Now enter your observations and press "data" after each observation (usually the M+ key on a casio).\n\nOnce all the data has been entered press shift and the key that corresponds to standard deviation (usually 3 on a casio).\n\nHope this helps
471532
They give you mu the mean of the population as 7\nThey give you sigma as the std dev = 1\n\nYou are working with a sampling distribution of x bar the sample mean with the sample size of n = 9\n\nThe mean of this sampling distribution is also 7 but the std dev of the sample mean is sigma/sqrt(9) = 1/3\n\nP(6.4<sample mean<7.2)\nP(6.4-7/(1/3) < Z< 7.2-7/(1/3)), look this in a table of std normal values.\n\nYou need to understand how to use this table.\n\nRemember that when you draw from a normal population a sample of size n, the sample mean is normally distributed with a std dev of sigma/sqrt(n), also called the standard error of the mean..\n\nTo solve problems like this you convert the sample varible to a standard normal distribution by z = (sample mean - mu)/std error\n\nii) Just find the z value which is the 85 percentile and work your way back  z(85th percentile) = (x_bar - mu)/std error using algebra to find the value of x_bar, the sample mean.
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When the population standard deviation is unknown, thats when you use the t-distribution, in calculating the confidence interval. \n\nThe t-distribution (also known as Student's t-distribution) measures the significance of a difference of means of small samples when two distributions are though to have the same variance, but possibly different means.\n\nBut as the sample size increases, the t-distribution gets close to the normal distribution.\n\nHope this helps!
320891
Your administration should provide you with standards, I would think?  Or ask another teacher for examples?
495441
Good question. Simply take your data and calculate the standard deviation given your data. Remember that data with higher R-squared values will also have a lower standard deviation. With this in mind, when you calculate your standard deviation with the new data, you can simply take two standard deviations in each direction to know your 95% confidence range. For example, if you had a mean of 115 and an R-Squared of .97, this would give you a standard deviation of 1.5, which means that you know that your boundries would be 115-(1.5x2)=112 and 115-(1.5x2)=118 with 95% confidence.\n\n**The key to this is that the R-Squared measures how well the regression line is "fit" to the data. The standard deviation is what tells us where to draw our boundry lines. They are related, but different.\n\nIf you want to play with this on your own, I suggest you go into excel and put in the following data (this is what I did to answer your question):\n\n113\n113\n114\n114\n115\n115\n116\n116\n117\n117\n\nThen graph it in a scatterplot and right-click on one of the points and click to "add a trendline". Select regression and on the next tab click to show the r-squared and the equation of the line. This will plot your points, your line, give you your r-squared, and give you your line equation. For your standard deviation, click a free cell and type "=stdev(......)" and select your data for the ....'s. This will tell you your standard deviation.\n\nFun question. Best of luck!\n\nPS - you might brush up on standard deviation on wikipedia via http://en.wikipedia.org/wiki/Standard_deviation or maybe even use google to find a standard deviation calculator. ;)
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population sample size (n) = 12\nsquare root of sample size = 3.464\nmean of the data = 11.692\nstandard deviation = .777\n\nIt's too hard to type in math notation so, in words, use the formula: \nmean plus or minus critical k-value multiplied by (standard deviation divided by the square-root of n)\n\nSo for your example, it would be:\n11.692 +/- k(.777/3.464) --> take .777 divided by 3.464\n= 11.692 +/- k(.224) so solving for k -->\n11.692 + k(.224) = 13 \nk = 5.84 \n11.692 - k(.224) = 10\nk = 7.6\n\nThis formula is for the confidence interval for the mean of a normal population. It assumes that this was a simple random sample and is just one of the formulas that you could possibly use. You didn't specify if the data was Normally distributed (I'm not going to check and do everything for you). Just to put this answer in perspective, a confidence level of 99.9 % corresponds to a "k-value" of 3.291
397848
I can tell that nobody wants to just do this problem.  There's nothing interesting about it.  It's a typical problem that a stat student does to show understanding.  Get out your textbook and study up how to solve it.
83169
Figure out how many standard deviations from the mean those values are:  (13500-10000)/2000 = 1.75; (8000-10000)/2000 = -1; 11500-10000)/2000 = 0.75.\n\nNow look up the areas under the curve for those z-values and those are the answers.  While you're at it, find the value 0.10 in the table and see how many z-values (standard deviations) that is from the mean.  Multiply by 2000.\n\nNote that you have to add the values in part b - they're on opposite sides of the mean.
475940
Yes, you need to convert relative standard derivation to standard derivation first. Use relative standard derivation times the mean.\n\nIn the question, I didnt see anything about number of measurements, so there's no way we can tell n. Since we dont know the number of measurements, we cant tell which distribution, Students' T  or normal distribution, we should use. Look into the questions again, with this much info, we cant do nothing here.
177686
Quick Definitions\n\nAll these definitions use the set of numbers  3, 5, 7, 3, 11, 1.\n\nAverages--\n\nMEAN\nThe mean is all of the numbers added up (3+5+7+3+11+1=30) divided by how many numbers there is (30/6) so the mean in this case is |5|\n\nMEDIAN\nTo find the median you have to arrange the numbers in order of size (1,3,3,5,7,11) and then it is the median is the middle number. If there is an even amount of numbers, then you take the mean of the two middle numbers (3+5=8; 8/2=4) so in this case the mean would be |4|\n\nMODE\nThe mode is simply the most common number. In this case 3 appears more than any of the other numbers, so 3 is the mode.\n\nHope this all helps with your math classes!
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A.  Since the values range from 5 to 55, and if we believe that almost all of the values are within 2 sd's of the mean, then that would mean that there are a total of 4 sd's between 5 and 55.  So we can approximate this standard deviation by saying\n\n(55-5)/4 = 12.5\n\nB.  Since you are taking a sample of 400, then the standard error of the mean is 12.5/sqrt(400) = 0.625.  Since the sample size is large, then the sample mean is approximately normal.  To be within 2 units, then you would have to be within mu-2 and mu+2.  So we can standardize.\n\nz = (mu-2-mu)/0.625 = -3.2\nz = mu+2-mu)/0.625 = 3.2\n\nThe area between these two values would then be approximately 0.9986.\n\nC. In this case, our hypothesis test would look like this.\n\nH0: mu = 35\nHA: mu < 35\n\nOur test statistic would be (31.75-35)/0.625 = -5.2\n\nWe would then reject the null hypothesis.  So we would conclude that the scientists are correct.
421757
Quickly, based on my own past learning of these things:\nThe mean is used when the distribution is normal, no lumps on either end or any place but in the middle.\nThe standard deviation is how far apart scores are from each other, another way to think of it is how alike they are.  \nThe mean and standard deviation go together in a NORMAL distribution as above... one big bell shaped hump in the middle.\nStandard error of the mean:  \n\nIt may be easier for you to check out this link:\nhttp://research.med.umkc.edu/tlwbiostats/stnderrmean.html
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Mean is the average, when adding all numbers then dividing by the amount of numbers given.\n\nMedian is the numbers in the middle average - 8
275020
I have written a math book that explains math in plain English.  It is called Mauk Math.  I teach math and I could not find any math book that I liked so I decided to write my own.  Math does NOT have to be difficult.  It CAM be explained in simple terms and it CAN be easy and even FUN.  I use music to learn certain concepts and help retain it.  I use songs to help break it down and make it easier.  For instance, when I teach how to make negative exponents positive I use Will Smith's song, SWITCH. We play the song in class and sing it.  Because that's how you make a negative exponent positive.... you switch it !  If it is on the top of a fraction... you switch it to the bottom.  If it is on the bottom...you switch it to the top...EASY!  In typical math books, they make it difficult by explaining it like this: a number taken to the -nth power in the denominator can be make positive by rewriting the fraction with the denominator as the numerator taken to the nth power.  Ridiculous...especially for the struggling student.  MAUK MATH uses M.C. Hammer's Hammertime to teach the order of operations.  You always do everything in parentheses first.  If something is outside the parentheses....CAN'T TOUCH THAT na na na na. break it down.  Math can be easy and fun with the right book.... my book.  LOL
50256
There are lots of ways to do this. Most of them assume a distribution to the data, that is, sort of a histogram of the scores. So for example, a bell curve (also called a normal distribution or a Gaussian distribution) is a very common curve to grade on. It is not that fair in some ways, because it is symmetric, and students on the bottom of the curve might feel short changed. If you have a group of very good students, or a group of very bad students, things can be a bit distorted.  \n\n\nSome of the methods are pretty technical. But there might be some easy ways to do this. \n\nThe first method sort of automates the procedure described by the previous answers. It has the advantage that if anyone asks, you can tell them you really DID use a normal distribution curve and here are the parameters involved etc. I do not know if this is an important consideration in this situation for you, however, or not. \n \nIn comment 2., a number of other methods are mentioned, particularly in the references, but I suspect these might be too technical for your needs. \n\n1. Try just sorting the students into groups and then assigning grades to each group. Sort the student's grades, smallest to largest. Then suppose you want to give out grades A, B, C, D, and F. This method depends on knowing where the middle is, and what standard deviation is. \n\nFor the middle: try taking an arithmetic average. Just add up all the scores Xj, and divide by the number of students N. Something like : \n\nM=Sum Xj/N\n\nFor the standard deviation: This is a bit trickier, but not too bad with a computer or calculator. Take each score, subtract off the arithmetic average from the score, square the resulting number and add all these numbers, dividing by N-1. Something like:\n\nSD=Sum [(Xj-M)^2]/(N-1)\n\n\nIf you can't do this yourself easily, there are calculators to do this on the internet. For example, look at web sites in references 1 and 2.\n\nThen go to the third web site in the references (ref 3) and type in your values for mean and SD. \n\n\nSuppose that you find mean =60 and SD=10. Then type it in the spaces in the web calculator. \n\nEntering different values in the calculator will tell you what proportion of students are in a category. So suppose you decide that all students with grades less than 40 will get an F. Type in 40 on the calculator, and then you find that this corresponds to a shaded area on the curve of 0.022750. This means that 2.275% of the students will be assigned an F. \n\nTake the bottom 2.3% of the students and give them an F.\n\nThen suppose you decide that scores from 40 to 50 should get a D. Type in the spaces in the calculator, that you want the area between 40 and 50. The calculator tells you that this area is 0.135905. \n\nTake the next 13.6 % and give them a D. \n\nSuppose you decide that the scores between 50 and 60 will get a C. Type into the website calculator, you want the area between 50 and 60. The calculator tells you the area is 0.341345. \n\nTake the next 34% and give them a C. \n\nSuppose you decide that the students with scores between 60 (the average remember) and say 70 should get a B. Into the web calculator, look for the area beween 60 and 70. The calculator tells you the area is .341345 again. \n\nSo take the next 34% and give them a B. \n\nSuppose you decide that all students with a score above 70 should get an A. In the web calculator, look for the area above 70 and you find that the area is 0.158655. \n\n\nThe top 16% get an A. \n\nNotice this is still sort of arbitrary, since you are deciding what is an F, a D, a C, a B and an A. But you are using the normal distribution to help you a bit. \n\nYou can play around with this quite a bit. You can make it so the same number of students who get an F get an A, and same number who get a D get a B. However, even though it is done on a "curve", it is all still quite arbitrary. You really do not need the web calculator except to help you get a feel for
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Standard deviation = sqrt(4) =2\nWe want to calculate :\nP(70 - 4 < n < 70 + 4 )\ni.e P(66 < n < 74 ) = P ( (66 - 70)/2 < s < (74 - 70)/2)\nSo P ( -2 < s < 2) = 2 P( o < s < 2 )\nUsing the table of areas under the standard distribution curve\nthen the percentage = 2* 0.4772 = 0.9544 = 95.44%\nWhere s denotes standard normal distribution\nand n denotes normal distribution