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ssrs export to pdf barcode font Standard deviation in Software
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EAN 13 Recognizer In .NET Using Barcode reader for .NET Control to read, scan read, scan image in VS .NET applications. Generate Bar Code In Java Using Barcode printer for BIRT reports Control to generate, create barcode image in Eclipse BIRT applications. (c) The theoretical mean of the sampling distribution of means, given by mX, equals the population mean , which is 67.45 inches (see Problem 5.33), in close agreement with the value 67.58 inches of part (b). The theoretical standard deviation (standard error) of the sampling distribution of means, given by sX, equals s> !n, where the population standard deviation s 2.92 inches (see Problem 5.40) and the sample size n 4. Since s> !n 2.92> !4 1.46 inches, we have close agreement with the value 1.41 inches of part (b). Discrepancies are due to the fact that only 30 samples were selected and the sample size was small. 5.44. The standard deviation of the weights of a very large population of students is 10.0 lb. Samples of 200 students each are drawn from this population, and the standard deviations of the weights in each sample are computed. Find (a) the mean, (b) the standard deviation of the sampling distribution of standard deviations. We can consider that sampling is either from an infinite population or with replacement from a finite population. From Table 51, page 160, we have: (a) (b) sS mS s 22n s 10.0 lb 10 2400 0.50 lb 5.45. What percentage of the samples in Problem 5.44 would have standard deviations (a) greater than 11.0 lb, (b) less than 8.8 lb The sampling distribution of standard deviations is approximately normal with mean 10.0 lb and standard deviation 0.50 lb. (a) 11.0 lb in standard units (11.0 10.0)>0.50 2.0. Area under normal curve to right of z (0.5 0.4772) 0.0228; hence, the required percentage is 2.3%. (b) 8.8 lb in standard units (8.8 10.0)>0.50 2.4. Area under normal curve to left of z (0.5 0.4918) 0.0082; hence, the required percentage is 0.8%. 2.0 is 2.4 is 5.46. A sample of 6 observations is drawn at random from a continuous population. What is the probability that the last 2 observations are less than the first 4 Assume that the population has density function f (x). The probability that 3 of the first 4 observations are greater than u while the 4th observation lies between u and u du is given by 4C3 B
3u f (x) dx R f (u) du
CHAPTER 5 Sampling Theory
B 3 f (x) dx R
u ` 3
The probability that the last 2 observations are less than u (and thus less than the first 4) is given by Then the probability that the first 4 are greater than u and the last 2 are less than u is the product of (1) and (2), i.e., (3) 4C3 B
Since u can take on values between ` and `, the total probability of the last 2 observations being less than the first 4 is the integral of (3) from ` to `, i.e., 3u f (x) dx R f (u) du B 3 `f (x) dx R
` u ` 3 u 2
4C3 3 To evaluate this, let (5) Then (6) When u `, v 1, and when u dv
B 3 f (x) dx R B 3 f (x) dx R f (u) du ` u ` v
3 `f(x) dx
f (u) du
3u f (x) dx
`, v
1 4C3 3 0 0. Therefore, (4) becomes v)3dv (3) (4) (7) 1 15 v2(1 which is the required probability. It is of interest to note that the probability does not depend on the probability distribution f(x). This is an example of nonparametric statistics since no population parameters have to be known. Another method Denote the observations by x1, x2, . . . , x6. Since the population is continuous, we may assume that the xi s are distinct. There are 6! ways of arranging the subscripts 1, 2, . . . , 6, and any one of these is as likely as any other one to result in arranging the corresponding xi s in increasing order. Out of the 6!, exactly 4! 2! arrangements would have x1, x2, x3, x4 as the smallest 4 observations and x5, x6 as the largest 2 observations. The required probability is, therefore, 4! 6! 2! 1 15 5.47. Let {X1, X2, . . . , Xn} be a random sample of size n drawn without replacement from a finite population of size N. Prove that if the population mean and variance are m and s2, then (a) E(Xj) m, s2 >(N 1). (b) Cov(Xj, Xk) Assume that the population consists of the set of numbers (a1, a2, c, aN), where the a s are not necessarily distinct. A random sampling procedure is one under which each selection of n out of N a s has the same probability (i.e., 1> N Cn). This means that the Xj are identically distributed: d a1 a2 ( aN prob. 1>N prob. 1>N (j prob. 1>N 1, 2, c, n)

