DC average value & root mean square value ?
82973
try what that guy said
655920
The standard deviation is the square root of the variance. Now, the variance is the "average squared difference (distance) of each observation from the mean".\n \nLets think about it. You want a measure of how dispersed the data is, right? Well, suppose you have heights for your data and your trying to compare two groups. For example, you want to know if the height´s of New Yorkers vary more or less than the heights of Floridians. You are NOT trying to see which are taller.\n \nSo you can calculate the average height of each group and then see haw far away from this average each individual height is and then calculate the average distance from the average height. If everyone had the same height, the average distance would be ZERO, reflecting that the population´s height is very CONCENTRATED around its average.\n \nBut why do you use the squared difference? Because if you were to calculate the average of "x - average(x)" you would always get zero, so by squaring and taking the square, you always get a positive number (the distance is always positive).\n \nGot it?
880121
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.
546400
D=square root ((x1-x2)^2 + (y1-y2)^2)
453647
93.16667 =- 3*10.63 \n61.28 to 105.04, so all of the values fit in this range.\n\nThe previous answerers have been wrong about Chebyshev Inequality.  According to that, there should be 1-(1/k)^2 values within k standard deviations of the mean.  So, according to the theorem, there should be 1-(1/3)^2 = 8/9 = 88.89% within 3 SD's of the mean.
430930
1. let mu=true mean salary for industry\n2. Null Hyp. mu=50000\n3. Alt. Hyp. mu<50000\n4. alpha=.05\n5. test stat: t=(mu-xbar)/(standard deviation/sq.rt(sample  size))\n6. Assumptions: random sample, approx. normal distribution, pop.std.dev. unknown\n7. t= (50000-47000)/(4000/3)= -2.25\n8. p= .027\n9. Since p= .027 < alpha=.05, reject null hypothesis\n10. Based on the sample, as long as it has a normal distribution, it can be concluded that Initech pays their employeess less than the industry average.\n\nHope that helps; this 10-step format is the one that I was taught and have always used.
193948
The sq rt of x is x to the 1/2 power; 1 over sqrt x is x to the -1/2 power\n\nThen use your normal rule for exponents : d/dx (x^n) = n x^(n-1)\n\nd/dx x^-1/2 = -1/2 x^(-3/2)\n\n(This equals -1 over the square root of x cubed if you need to write it that way.)
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.
697345
I get a different answer. Surely none of these answers can be right: in no case is the interval symmetric about the mean = 10.\n\nSample size n = 4\nSample mean xbar = 40/4 = 10\nPopulation mean = mu (unknown)\nSample variance s^2 = (0 + 4 + 1 + 1)/(4-1) = 2\n(Note: dividing here by n-1 = 4-1 is more accurate than by n because it makes s^2 an exactly unbiased estimator of the population variance sigma^2. This would matter little if n were large.)\n\nStd error = s/sqrt(n) = sqrt(2)/2 = 0.707\nSo xbar-mu is normally distributed with mean 0 and std dev 0.707 (as estimated from the sample; its exact but unknown value is sigma/sqrt(n) = sigma/2).\nSo the 95% confidence interval for mu is\n\n[10 - 1.96*0.707, 10 + 1.96*0.707] = [8.614,11.386].\n\n\nRemark 1: With such a small sample, you really have to take (n-1) = 3 instead of n = 4 in calculating the sample variance s^2. The formula for s^2 "mean of squares - square of mean" under-estimates sigma^2 by the factor n/(n-1) = 4/3 here. This is not inconsiderable. The unbiased estimate of s^2 from this sample is 2, and NOT 1.5 as in Wal C's solution.\nEven the value s^2 = 2 will lead to under-estimating the population std dev. sigma itself (because sqrt is a concave fctn, s is an under-estimating estimator of sigma even though s^2 is an unbiased estimator of sigma^2).\n\nRemark 2: Wal C's solution is even more seriously wrong. She should have used the std error, the std dev of the sample mean, which is sigma/sqrt(n). But she used the std dev. of a SINGLE trial, sigma. If that were correct, there would never be any point in increasing the sample size! The point is of course that the std error sigma/sqrt(n) goes down as n increases -- but you've got to use this fact, not ignore it!\n\nRemark 3: My solution above is also faulty. The sample size is so small that you must use the Student t-distribution and not the normal distribution. (xbar-mu)/(s/sqrt(n)) is not normally distributed; it has the Student t-distribution with 4-1=3 degrees of freedom. The 97.5 percentile is 3.18 (and not 1.96 as for the normal distr), and so the 95% confidence interval for mu is\n\n[10 - 3.18*0.707, 10 + 3.18*0.707] = [7.75, 12.25].
553133
This is an example for standard deviation:\n\nSay you have these values and need to find the deviation for them\n5, 10, 15, 20, 25 \n\nMean = 15\n\nNow, calculate the difference of the values from the mean\n\n15-5 = 10\n15-10 = 5\n15-15 = 0\n20 - 15 = -5\n25 - 15 = -10\n\nNow, square the differences and add them\n\n10^2 + 5^2 + (-5)^2 + (-10)^2 = 100 + 25 + 25 + 100 = 250\n\nStandard deviation is the squared root of this sum divided by the mean\n\nStandard Deviation = (250/15)^(1/2) \n                   = 5 * (2/3)^(1/2)\n                   = 5 * 0.816\n                   = 4.082
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.
731363
If v^2 is proportional to Fc, then v^2 = k*Fc, where k is a constant.  For each of your values of v and Fc, calculate k = v^2/Fc;\n\nYou will get the following values for k:\n\n39.51, 37.97, 33.61, 32\n\nThese average to 35.77 and the standard deviation is about 10%, which would be considered a good fit and that the formula\n\nV^2 = 35.77*Fc fits those values pretty well.
873308
OK - I will try and explain it, but forgive me if I get myself confused...  hehehe\n\nFirst step - make a bell curve.  Put a mark in the center and label that "7 hours" - that's your mean.  Next, put three marks on either side of that middle point - those will be your standard deviation points.  For the points on the right, ADD 5.3 to the mean for the first one, then add 5.3 again for the 2nd, etc...  Points on the left side of the mean have the 5.3 SUBTRACTED.  (Am I still making sense?  lol)\n\nNow - the 68-95-99.7 rule says that 68% of the freshmen will fall into the area between the first line on EITHER side of the mean, 95% between the second lines and 99.7% between the third lines on either side of the mean.\n\nI think you should be able to figure it from there - I hope so...  Right now I am so confused I think I gave myself a headache!  lol\n\nGood luck!
1107053
1.0 is a D average\n2.0 is a C average\nSo, since you have a little better than a D+ average, I would say no.
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.
360360
The average value of a function is given by\n\n(1/(b-a)) S f(x) dx, evaluated from a to b.\n\nSo for this question, the average will be given by\n\n(1/(4-0)) S 8+4e^-2t+te^-2t dt = (1/4)[8t - 2e^(-2t) + S te^-2t dt]\n\nLet u = e^(-t)\ndu =  -e^(-t) dt\n\nSo, S te^-2t dt = - S u^2 du\n= (-u^3)/3\n= -e^(-3t)/3\n\nSo your answer then becomes\n\n(1/4)[8t - 2e^(-2t) - e^(-3t)/3] = 2t - (1/2)e^(-2t) - e^(-3t)/12\n\nEvaluating from 0 to 4,\n= 2(4) - (1/2)e^(-2(4)) - e^(-3(4))/12 - [2(0) - (1/2)e^(-2(0)) - e^(-3(0))/12]\n= 8 - (1/2)e^(-8) - e^(-12)/12 + 1/2 + 1/12\n= 8 5/12 - (1/2)e^(-8) - e^(-12)/12\n~ $8.42
860146
The first value represents the (average) Minimun length recorded, the second the Average and the last the Maximum measured values. \nLength............141,5 mm..........161,5 mm..........181,1 mm. \nCircumference..117,5 mm.........127, 3 mm.........137,8 mm.\nKensey data are quite old. These statistic is the last one.