Can you calculate a statistics question for me?
552270
would calculate your chance of me calculating a statistics question for you at 1234.367:1
963387
You add up the total number you got right - (1/4) * the total you got wrong. \n\nSo it depends on how many questions you had on the reading and how many you had on the math.
64925
This is not the best place to ask for help. Meet your faculty advisor, or best - meet your Stats professor. Personally I think the best approach is to just take a book, and solve the problems in it to ace Stats.
804399
I just took Intro to Statistics at Hunter College, and got an A+.  It's not really difficult.  There's no real calculation in it, except to find the average, and range, and such.  They emphasize understanding the terminology (normal distribution, variance, etc.), understanding the advantages of using one method over another, which test to use under certain situations, and such. \n\nHope I helped!
861788
There is no accurate way to do this because too many variables are involved. You can do like I do though and run a series of tests and then average them.\n\nhttp://www.toast.net/
729266
Yes.  This would require the use of a t confidence interval.  That is because the sample is small, the population is normal, and the population standard deviation is unknown.  You would have to compute the mean and the standard deviation of the six values to determine the confidence limits.\n\nedit: BTW, the mean is 81.98 and the standard deviation is 4.18.  The confidence interval I get is (77.18, 86.87)
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%
1071099
Here is a link to a users manual-specifically the statistics section. You can go to the full contents.\nhttp://www.infj.ulst.ac.uk/NI-Maths/Guides/TexTI-85/Ch3/THBStats.html
1014917
You must  write your question more understandable.\n\nFor the test results, you can calculate the average, variance, standard deviation, min and max grades. You can also find the frequency of the marks.
408817
X = sample mean, U = theoretical mean, S = standard deviation, and N = number of data...\n\ni assume u know how to calculate X and S from above. \n\n.95 = P(1.96< (X-U)/(S/sqrt N) < 1.96)...solve for U..U will be in between to values, which means that you are 95% confident that U will lie in that range.
185718
I don't know of any people who tutor online.\n\nThere are a number of good free statistics textbooks.  See the link.\n\nYou may also find it helpful to visit the library and read what different textbooks have to say about things that are stumping you.
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.
808142
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.
1018969
Just google it dear..very simple..calculate mean and average the sum of square of variation of each score from the mean.\nhttp://davidmlane.com/hyperstat/A16252.html
343921
http://www.statistics.gov.uk/
624712
This problem is solvable.\nProbability of event A is written P(A).\n\nP(A)= Total # of outcomes that correspond to event A (divided by -or- over) Total # of outcomes in sample\n\nAnswer:\nP(A) = 12/20 = 3/5
365087
You can expect mind numblingly boring calculations for most of your summer.  The mathematics behind statistics is fairly elementary, it's the staying awake and alert part that's difficult...\nGood Luck, I had to take Statistics in college, and I have to say... It was the best one hour nap 3 times a week a person could possibly ask for...