Proving a function using Central Limit Theorem?
920901
You can get it the following way:\n\nThe cdf of Xbar is \nF(y) = P((X1+X2+...+Xn)/n < y) = P(X1+X2+...+Xn < n*y) = \nP(sqrt(n)*(X1+X2+...+Xn-mu)/ sigma < sqrt(n)*(n*y-mu)/sigma) = Phi(sqrt(n)*(n*y-mu)/sigma), where Phi(.) is the cdf of the standard normal distribution.\n\nDifferentiating this with respect to y yields the pdf of Xbar:\nf(y) = phi(sqrt(n)*(n*y-mu)/sigma) *n*sqrt(n)/sigma, where phi(.) is the pdf of the standard normal distribution.\n\nAs a consequence of the CLT, for large values of n, this does not depend on the distribution of the original X1,X2,...,Xn, only on its mean and standard deviation. However, the CLT is only an approximation.\n\nIf you'd like to get the accurate pdf of Xbar, you need to use the convolution formula for the pdf of the sum of IID random variables. This is pretty nasty, so I'd rather suggest that you use moment generating functions or characteristic functions instead. (It is very simple to compute the mgf and the chf of the sum of IID random variables; in fact, the most well-known proof of the CLT is also based on this technique.)
695801
I forgot too much about this stuff to prove convergence.  However, if you can take care of that side of things...\n\nKnowing that the sequence converges, then consecutive terms are nearly equal as n grows larger.  This means\n\nlim(n to inf)  X_(n+1) = X_n\n\n0.5(a+X_n) = X_n\n\nX_n = a.\n\nEDIT: I got something better, you can show that\n\nX_n = a/2 + a/4 + a/8 + ... + a/2^n + X_0/2^n\n\nwhere X_0 is the initial term.  Of course X_0/2^n converges to zero so you're left with a geometric series.
254992
Well, you know that R=lim A_n /A_(n+1).\n\nNow do ratio tests on the problems:\n\nFor 2), convergence happens if lim |A_(n+1) z^3(n+1))/(A_n z^(3n)| <1. Thus, |z|^3 /R <1, so |z|<R^(1/3). This means the radius of convergence is R^(1/3).\n\nThe rest should be easy now.
810602
The central limit theorem states that given a distribution with a mean μ and variance σ², the sampling distribution of the mean approaches a normal distribution with a mean (μ) and a variance σ²/N as N, the sample size, increases. \n\nThe amazing and counter-intuitive thing about the central limit theorem is that no matter what the shape of the original distribution, the sampling distribution of the mean approaches a normal distribution. \n\nFurthermore, for most distributions, a normal distribution is approached very quickly as N increases. Keep in mind that N is the sample size for each mean and not the number of samples. Remember in a sampling distribution the number of samples is assumed to be infinite.\n\nDEGREE OF FREEDOM:\nThat measure of variability which merely expresses the number of options available within a variable or space. \n\nDegrees of freedom are the equivalent of currency in statistics - you earn a degree of freedom for every data point you collect, and you spend a degree of freedom for each parameter you estimate.
612882
You first need the probablity function of the basic variable... Lets call it P(x) where x is the variable... Then to get the epectation take the integral of x*P(x). The limits of integration should be the min and max values that x can be.
303094
OK.  From any given population, you can take many, many different samples of size n.  For each of those samples, you can find the mean.  So if you take 1000 samples of size 50,  Then you would have 1000 sample means.  Now each of those sample means are probably going to be different from the overall population mean, since samples are going to look different from the population.  Also, each of the 1000 sample means are going to look different from each other because the mean is coming from different samples.\n\nSo if I were to draw a histogram (hopefully you know what that is; it is a bar graph that you can use to look at the distribution of numerical data).  Anyway, if I were to draw a histogram of my 1000 sample means,  You should notice the following three things...\n\n1.  The center of your histogram should be the same as you population mean (so if the mean of the population is 20, then the histogram of my 1000 sample means should center around 20).\n\n2.  As the sample size gets larger, the spread of the histogram should get smaller (so the spread of the means of samples of size 50 would be smaller than for the means of samples of size 25).  This is because with larger samples, you would have more information about the population, and thus you should be able to get closer to the population mean.\n\n3.  As the sample size gets larger, the shape of the histogram will look more and more like a normal curve.\n\nThis is because of the Central Limit Theorem, that states that if you take a sample of size n from any population with a finite variance, the distribution of the mean of the sample would get closer and closer to a normal curve as n gets larger.  The mean of the normal would be the same as the population mean, and the standard deviation of the normal (usually called the standard error) would  be the standard deviation of the population divided by the square root of n.  Notice that as n gets larger, the standard error gets smaller, just as described before.  So typically you can approximate these histrgrams with a normal curve if n > 25.\n\nThis phenomenon is hard to describe in words alone.  Looking at pictures of these histograms that I described would help you understand it better.
199925
Probability function of a poission distribution\n==> P(X=x) = e^(-m)*m^(x)/x! \n[x= 0,1,2,.... ]  countably infinite\n\nMean = 4 ==> m = 4\n\nProbability that x = 2\n==> P(X=2)= e^(-4)*4^(2)/2!\n          = \n\nProbability that x <= 2\n==> P(X<=2)= P(X=0) + P(X=1) + P(X=2)\n==> P(X<=2)\n= e^(-4)*4^(0)/0! + e^(-4)*4^(1)/1! + e^(-4)*4^(2)/2!\n= e^(-4) {1/1 + 4/1 + 16/2}\n= e^(-4) {1 + 4 + 8}\n= e^(-4) {13}\n\nProbability that x > 2\n==> P(X > 2)\n= P(X=3) + P(X=4) + ...... \n(Or) = 1 - {P(X<=2)}\n\n\nhttp://www.futureaccountant.com/probability/study-notes/\n\nhttp://www.schoolingkids.com/probability/
567661
the Chebyshev's inequality goes Pr(|X-mean| > or = k.sd)< or = 1/k^2\n\ni m assuming u meant standard deviation when u said 4.3818, which is 2 sd.\n\nThe prob of points outside 2sd at both sides is not more than 1/k^2.. or 1/2^2=1/4.\n\nSo, the probability of points within 2sd will be at least 1-1/4=3/4.\n\nhow many data points? if u have 100 data points in total, there will be at least 100x3/4=75 data points within the mean+/-4.3818 range.\n\nNote: The exactly probability depends on the type of distribution the data follows, for example, for Normal/Gaussian, 95% lies within +/- 1.96sd. Chebyshev's inequality states the minimum possible for any type of distribution.
551595
Gaussian distribution
747397
Checking by method of differences\n\n. x ............. f(x) ...... Δ1 ...... Δ2\n. 0 .......... 1104\n. 1 .......... 1104 ........ 0\n. 2 .......... 1100 ........ 4 ....... -4\n. 3 .......... 1092 ........ 8 ....... -4\n. 4 .......... 1080 ...... 12 ....... -4\n. 5 .......... 1064 ...... 16 ....... -4\n. 6 .......... 1044 ...... 20 ....... -4\n. 7 .......... 1020 ...... 24 ....... -4\n. 8 ............ 992 ...... 28 ....... -4\n. 9 ............ 960 ...... 32 ....... -4\n10 ............ 924 ...... 36 ....... -4\n11 ............ 884 ...... 40 ....... -4\n12 ............ 840 ...... 44 ....... -4\n13 ............ 792 ...... 48 ....... -4\n14 ............ 740 ...... 52 ....... -4\n15 ............ 684 ...... 56 ....... -4\n16 ............ 624 ...... 60 ....... -4\n17 ............ 560 ...... 64 ....... -4\n18 ............ 492 ...... 68 ....... -4\n19 ............ 420 ...... 72 ....... -4\n20 ............ 344 ...... 76 ....... -4\n21 ............ 264 ...... 80 ....... -4\n22 ............ 180 ...... 84 ....... -4\n23 .............. 92 ...... 88 ....... -4\n24 ................ 0 ...... 92 ....... -4\n\nSo the relation is quadratic (as Δ2 is constant which imples the 2nd derivative is constant)\n\nSo the relation is of the form\n\nf(x) = ax^2 + bx + c\n\nNow f(0) = c = 1104\nf(1) = a + b + c = 1104\nSo a + b + 1104 = 1104\nie a + b = 0\n\nf(2) = 4a + 2b + c = 1100\n4a + 2b + 1104 = 1100\nSo 4a + 2b = -4\n\nie 2a + 2a + 2b = 2a + 2(a + b) = -4\nSo 2a + 0 = -4\nie a = -2 and b = 2\n\nThus f(x) = -2x^2 + 2x + 1104\n= -2(x^2 - x - 552)\n= -2(x - 24)(x + 23)\n\nTest f(10) = -2*(10 - 24)(10 + 23)\n= -2 * (-14) * (33)\n= 28 * 33\n= 924 YESSSSSSSSSSS!!
945372
Get some help here:\nhttp://en.wikipedia.org/wiki/Brownian_movement
907438
I could tell you the answer or I could give you an idea of what to consider when you do it yourself.\n\na) Binomial numbers and p = 0.7, q= 0.3 might help here.  You do not know the size of the population of the mailing list so you might need to make an assumption that it is large enough to assume a Normal distribution.  I think the central limit theorem might help here.\n\nb) Binomial theory, and conditional probabilities.  Look it up in your text book!  Again if the size of the population is large enough you might be able to simplify the problem.
245944
The integral of a function over a definite interval.
1047841
By the def\n\nlim f(x + h ) - f(x)\n              /h\nPlug in (x+h) into f\nlim (x+h)^2 + 5(x +h) - (x^2 + 5x)\n                        /h\nFoil\nlim x^2 + 2xh + h^2 + 5x + 5h - x^2 - 5x\n                          /h\n\nCancel like terms\nlim 2xh + h^2 + 5h\n           /h\n\n\nReduce everything by h\nlim 2x + h + 5\n\nTake the limit and\nPlug in 0 for h\n2x + 5
566883
First, since 1/x is not defined for x = 0, before you can show  anything about your functions, you have to define values for g(0) and f(0). I'm going to assume that either the original statement of the problem had that or that you are supposed to find the value that makes the problem solvable.\n\nTo be true you have to define the problems with f(x) = 0 and g(x) = 0\n\n\nFor the continutiy of f(x) given epsilon > 0 find delta > 0 such that |f(x) - f(0)| < epsilon for all |x-0| < delta.\n\n|f(x) - f(0)| = |xsin(1/x) - 0| < |x| = |x - 0|\n\nGiven epsilon, pick delta = epsilon.\nIf |x-0| < delta, then |f(x) - f(0)| < epsilon.\n\n\nAs for the differentiability of g(x),\n\nf(0+h) - f(0)) = (0+h)^2sin(1/(0+h)) - 0\n\nlim(x->0)(f(0+h) - f(0))/h = \nlim(x->0)((0+h)^2sin(1/(0+h)) - 0)/h = \nlim(x->0) h^2(sin(1/h)/h = lim hsin(1/h) = 0\n\nSo the limit exists and is 0.
952174
That is the central limit theorem.  Each distribution can be uniform, but the distributions themselves are normally distributed in their range, their value, their frequency.\nRead the attached link for more details.
477523
The no of degrees of freedom for a chi-square distribution ia n-1