find the mean & standard deviation for the number of left-handed students in such classes of 25 students?
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in order to do statistics you need data\n\nie.  we need number to do this
111022
7\n\norder them; if you have odd number of entries, take the middle one; if even - take average of two middle ones.\n\n\n\nand it's a statistics question
220083
I am pondering this question a bit. I might have a better answer for you, but here is a start:\n- - - - - - - - - - - -  - \n\nYour data is not exactly amenable to standard quantitative analysis. You can't really form an average, for example. \n\nYou can find the mode. The most common answer is, diet. So the mode of your data is "diet". \n\nThe data categories, as they stand, are not represented by ordinal numbers. So it is a bit hard to even describe the distribution, let alone the dispersion of the distribution. What you could do, is say that 35% preferred diet, and 65% did not. Then you could try to estimate the error bounds on these figures, given some assumptions. Or do a robust estimate using a jackknife or bootstrap.\n\n- - - - - -  - - \n\nHere is another idea. You might want to know the average number of people that are in a given category. That is, you might want to know how many people like a given type of soda. You have 5 categories, so the average is \n\n(20)/5=4\n\nThere are an average of 4 people in each category.\n\nNow you can find the standard deviation.\n\nFirst find an estimate of the variance V using the standard formula:\n\nV=(1/4)*[(1-4)^2 + (3-4)^2 +(3-4)^2 +(7-4)^2 +(6-4)^2 ]\nV=(1/4)*[9 + 1 + 1 + 9 +4 ]\nV=(1/4)*[24]\nV=6\n\nThen take the square root\n\nS=sqrt(V)= sqrt(6)\nS=sqrt(6) = 2.44948974\n\nThe average soft drink is liked by 4 people, but there is a standard deviation of about 2.4.\n\nSo one can say a soft drink is liked by 4 += 2.4 people (if we set the error bounds to be 1 standard deviation, but some people set it at 2).\n\nYou could even convert this to percentages. Then you could say that a soft drink is liked by 20% of the people, +- 12%.\n\nThe standard deviation assumes that the distribution is normal, that is a bell-shaped curve (aka Gaussian). However, it probably is not. A robust estimate of the dispersion of the dispersion can be formed just resampling methods, like the bootstrap or the jackknife. I have placed a link to these if you are interested, but I think it might be too advanced for your needs.\n\nAnother thing to do is to look at the median, and the median average deviation, or the median average deviation away from the median. A reference discusses these and other measures of dispersion if you are interested.\n\nI still feel that one might be able to do more with this. I will think some more and if I have any good ideas I will post them. I am not a statistician, so it is likely that one of them might have a much better suggestion right away.
898911
Take the average of the absolute value of each of the numbers minus the mean...\n\n( |2.44-2.23| + |3.54-2.23| + |1.37 - 2.23|  + |1.5 - 2.23| + |2.3-2.23| )/5
1118574
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
906033
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%
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There are 20 students in the class. There are 4 possibilities when you choose the 3 random students. I've included the probabilities of these 4 possibilities. 1) and 3) are the answers to your questions.\n\n1)   3 boys = (4/20)(3/19)(2/18) = 0.0035\n\n2)    2 b, 1 g = (4/20)(3/19)(16/18)x3(=3C2) = 0.0842\n\n3)    2 g, 1 b = (16/20)(15/19)(4/18)x3(=3C2) = 0.4211\n\n4)     3 g = (16/20)(15/19)(14/18) =  0.4912\n\nFor the probabilities to be correct they must sum to 1:\n\n0.0035+0.0842+0.4211+0.4912 = 1
937201
First find the mean (average).  Then subtract each data number from the mean, and square the difference.  Add up all these squared differences and find the average of that sum.  Then take the square root of that average.  That's the standard deviation.
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If the claim is correct, then the mean of samples of size 25 has a normal distribution with a mean of 10 (the population mean) ounces and a standard deviation of 0.5/sqrt(25) = 0.1.\n\nz = (10.15-10)/0.1 = 1.5\n\nAccording to the standard normal table, the probability is 0.0669.\n\nThis is due to the fact that the population is normal.\n\nedit: You guys, it is not three sd's above.  You have to think in terms of standard error.  It is 1.5 standad errors above.  Do the math.
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Standard deviation tells how far away a number in a set is from the mean. Like in data where the numbers are 2,4,6,8,10, the mean of this data is 6. So you have to find the difference of each number and the mean, then square each answer and add together(Sum of squared deviations). Divide this answer by the same number you divided the mean by - 1. So in this case you divided the the total of the numbers above by five, so divide the sum of squared deviation/5-1, this gives you the number they call the variance, take the square root of the answer, and you will get the standard deviation.  So, the sum of the squares for this example is 40, by taking the the square root, you get 6.32
125699
take out the mid-class (by adding the upper class limit and lower class limit and dividing it by 2)of each class interval (in a separate column) and then multiply frequency with x (in a separate column) take out the sum of the fx column.\n\nThe formula for mean is\n              summation of fx divided by summation of f\ni.e\n    divide the sum of fx column by the sum of frequency
127297
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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1.  Using your Normal Distribution table (sorry, mine is buried under my dirty laundry), find the Std Devs that represents .18 (.50 - .32).  Multiply by 30 (Std Dev).  This number represents the number of pages short of 150 (the mean).  It's probably around 140.\n\n2.  The second group (.18 below 150 pages + .25 above 150).  Repeat the process for 1 above. (about 140 to 165)\n\n3.  The third group is from .25 to .49 above the mean (about 165 to 240)\n\nGood luck.
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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.
1156519
If you are using a sample set of data you can press stat/edit - then press enter/enter a list of numbers/ press stat again/move the arrow to calc/ select 1-Var stats/ press enter/press 2nd and then 1 (or L1)/ press enter.\n\nx bar = the sample mean\nSx = the sample standard deviation\nVariance = Sx squared
745794
If 22.5% of 1500 students will not graduate:\n\n0.225*1500 = 337.5 students will not graduate.\n\nAll of the students left will graduate, which is:\n\n1500 - 337.5 = 1162.5 ~ 1163 students will graduate.\n\nI hope this helps!
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Your data set would look like this:\n\n10 10 10 10 10 10 10 10 10 10 20 20 20 20 20 20 20 20 235 235\n\nThe mean is 730/20 = 36.5.\nThe median is (10+20)/2 = 15.\nThe mode is 10.\n\nSo certainly what was stated by the dean isn't a lie.\n\nNow, if I were to calculate, say, the average number of classmates for each person in the statistics classes, there would be 100 9's, 160 19's, and 470 234's.  So the average number of classmates for people taking these classes is approximately 113,920/730 = 156.05.  The median is 234 and the mode is also 234.  So the newspaper isn't really lying either.\n\nI guess you can say that they are looking at this from two different perspecitves, so in that sense, there are two truths.