I am a begining statistics student and cannot figure it out.
121710
It applies to Histograms:\nThere is more to be said about the widths of the class intervals, sometimes called bin widths. Your choice of bin width determines the number of class intervals. This decision, along with the choice of starting point for the first interval, affects the shape of the histogram.There are some "rules of thumb" that can help you choose an appropriate width. (But keep in mind that none of the rules is perfect.) Sturgis's rule is to set the number of intervals as close as possible to 1 + Log2(N), where Log2(N) is the base 2 log of the number of observations. The formula can also be written as 1 + 3.3 Log10(N) where Log10(N) is the log base 10 of the number of observations. According to Sturgis' rule, 1000 observations would be graphed with 11 class intervals since 10 is the closest integer to Log2(1000).
421519
The key point is that, since the two normal vbls x1 and x2 are independent, their difference \n\ny = x1 - x2 \n\nis also normally distributed with mean \n\nm = m1 - m2 \n\nand variance \n\ns^2 = (s1)^2 + (s2)^2\n\nor std dev \n\ns = sqrt[(s1)^2 + (s2)^2]. \n\n(Note that its the variances that add up, *not* the std devs.) The probability that x1 > x2 is the prob that y > 0, and we know the distr of y, so we're nearly there.\n\nTo use tables or a computer prog, standardize y to \n\nz = (y-m)/s = [y-(m1-m2)] / sqrt[(s1)^2 + (s2)^2]\n\nwhich has zero mean and unit std dev. What we want is the prob that z > -m/s which is \n\n1 - Phi(-m/s) \n\nwhere Phi is the std normal c.d.f. \n\nBut computer progs usually give you not Phi but the error fctn \n\nerf(x) = (2/sqrt(pi)) integral from 0 to x exp(-t^2)dt\n\nwhich carries the same info and gives you Phi as \n\nPhi(x) = (1/2)(1 + erf(x/sqrt(2)).\n\nSo you need to put into this \n\nx = -m/s = -(m1-m2)] / sqrt[(s1)^2 + (s2)^2]\n\nand then subtract from 1 (because what you need is 1 - Phi).
423044
Sorry to say, you better pass on this data set.  \n\nNote that each row of numbers totals 100, as these are percentages, so the variables are correlated, and have limited application.\n\nI think your professor wants two of the variables to be somewhat independent, meaning one is not a function of the other.  Such as gas mileage and price, for example, if you had a list of automobile makes.\n\nRatio variables are those in which the ratio of two of the numbers have meaning, such as miles per gallon, for example.  If car A gets 15 mpg and car B gets 20 mpg, you can take the ratio of the two: 15/20 and compute 0.75, meaning car A gets 75% of the mileage of car B.
16770
Hi! I'm a statistician and I use http://www.statsoft.com as well as cheap textbooks from Half Priced Books in my area. Both seem to help. If you have a specific question, feel free to email me. Don't know how much good I'll be because I use fancy software for my analysis, but I'll try.
329154
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)
474511
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
206835
Actually, statistics books are notorious for being written in such a way that no normal human can understand them. The trick, generally, is trying to find a way to relate the problems to something you're interested in. For example, if you like basketball: one possible sampling distribution would be the number of free throws individual players make over a season. You could take a sample of, say, 10 of the players and find the mean and standard deviation. You could compute a 95% confidence interval with that information, which would allow you to say that you're 95% confident that the average number of free throws made by NBA players in a season will be between this number and that number. You could find the probability that a given player would make at least a certain number of free throws in a season by finding the z-score and looking it up on the normal probability distribution table, etc. If you're really struggling with the terminology, I recommend visiting your math help center on campus (most of them have one), or contacting a tutor.
158550
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...
421541
1. If you know the variances you'd know the mean, because you need to calculate the mean to find the variance. Newer stat texts have finally started to get smart and quit putting stupid topics like this in there!\n\n2. If you suspect that the population variances are equal, you can combine the samples to get one estimate of that variance. That's called 'pooling.' When variances are equal use the formula that combines those variances for the calculation.\n\n3. The p-value comes from this calculation so you don't use it to determine which formula to use. Your background knowledge is what you would use to determine whether variances are likely to be equal.\n\n4. It's never wrong not to pool. Pooling when appropriate does give a test more power, but not pooling is always OK.
441593
Mean: (the same as the average) add all the numbers and divide the total by the amount of numbers added. \nExample: 1 5 7 2 9\n1+5+7+3+9=25\nDivide by 5\nMean=5\n\nMedian: the middle number when placed in numerical order low-high\nExample: 1 2 4 6 9 10 15\nFind the middle term (in this case 6 but if there is no exact middle and it is between 2 numbers like this: 1 2 4 5 it would be in between the numbers in the middle (so it would be 3) )\nSo the median is 6\n\nMode: The most repeated number\nExample: 1 5 8 1 4 9 1 5\nThe mode would be 1 because it appears three times) \n\nRange is the difference between the highest and lowest number\nExample: 1 4 5 9 3 \n9-1=8\nRange=8\nHope this helps.
645627
OH NOES THIS IS LIKE A MATH PROBLEM!!!! i can't handle any more variables :( sry.\n\n\nhahahaha.
512198
If you assume it's normal, then yeah, although I assume there is more data than you have given as you do not have the variance there, although you could treat this as a binomial distribution giving it mean 0.3, variance n0.3*0.7, although when calculating the confience interval you'll want to divide by n^2 for that variance.
225538
First you have to put the numbers in order:\n\n\n3, 4, 8, 8, 8, 10, 12, 13, 14, 16\n\nThere are 10 numbers. To find the median, you want to go to the middle. The middle is between 8 and 10, so you average those to find a median of 9.\n\nThere are two quartiles (like quarters) made up of the elements less than 9 and two quartiles made up of the elements more than 9.\n\nNow you need to find the median of each half. For the first five numbers, the median, middle number, is 8. For the second five numbers, the median is 13. \n\nSo the Q1 goes from 3 to 8, Q2 goes from 8 to 9, Q3 goes from 9 to 13, and Q4 goes from 13 to 16. These are all based on how the data is distributed (spread out) around the median in the middle.\n\nHope that helps.
623646
No. This is what we call probability. Take a statistics course.
6561
You might try DAP. You can find it here along with a lot of other free statistical software.\nhttp://members.aol.com/johnp71/javasta2.html
419700
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!
44832
There will be a table in your statistics textbook that gives probabilities for a normal (Gaussian) distribution.  Usually it is for a "standard normal" where the mean is 0, and the standard deviation is 1.  The table will show upper tail probabilities for values of x ranging from 0. to 3. (or higher), where the it gives the probability that random sample xi > x.  For general samples with a mean different than 0 and a standard deviation different than 1, you have to apply a conversion to use this table, as follows:\n\nLet m = the mean of a normal distribution\nq = the standard deviation (usually called by the greek letter sigma), yi = boundary of upper tail distribution.\nIn this case, m = 92500, q = 8400, yi = 95000.  The value of x to look up in the table is \n(1) x = (yi-m)/q.  \nThis will tell you the probability that ONE random element, xi > x.\n\nHowever, if you take a sample of N elements, if the standard deviation of the entire sample is q, the standard deviation on the MEAN of the sample is \n(2) q'=q/sqrt(N).\nSo for the sample of 64 families, the standard deviation on the mean is 8400/sqrt(64) = 1050.  This is the value to use for q in equation (1).  Equation (1) then gives x = (95000-92500)/1050 = 2.38.\n\n  So the probability is the same as the probability that x > 2.38 for a standard normal distribution.  From the table (or from an online calculator like the one at\nhttp://bayes.math.montana.edu/cgi-bin/Rweb/buildModules.cgi\nI find that the probability = 0.008656319 , less than 1%