  # For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find themean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.The mean of the population is(Round to two decimal places as needed.)The standard deviation of the population is(Round to two decimal places as needed.)Identify all samples of size 2 with the correct accompanying means below.A.580, 555, x = 567.5B. 504, 555, x 529.5C.555, 629, x = 592D.555, 580, x = 567.5E.555, 607, x = 581580, 623, x = 601.5F.G.607, 555, x 581H. 580, 629, x = 604.5stI.580, 580, x 580J.629, 580, x = 604.5K.504, 607, x = 555.5L. 607, 504, x= 555.5ilestioM. 580, 607, x - 593.5N. 629, 555, x = 592O. 607, 607, x = 607P.g Qu629, 629, x= 629Q. 504, 629, x = 566.5R. 504, 580, x = 542e-S. 607, 580, x = 593.5629, 607, x= 618T.U. 555, 504, x = 529.5ered V. 607, 629, x 618W. 555, 555, x 555Z. 629, 504, x 566.5X. 504, 504, x= 504580, 504, x = 542Y.607, 629, x 551ted QThe mean of the sampling distribution is(Round to two decimal nlaces as needed)Click to select your answer(s).eaway between two entries, use the z-score nairway between the corresponding z-scores.NOVTwX P3tv This Question: 1 pt7 of 14 (8 complete)This Quiz: 14 pts possibleFor the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find themean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population.The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2.M. 580, 607, X = 593.5N.629, 555, x = 592O. 607, 607, x= 607P.629, 629, x = 629Q. 504, 629, x = 566.5R. 504, 580, x = 542S. 607, 580, x = 593.5T.629, 607, x = 618U. 555, 504, x = 529.5tio607, 629, x = 618W. 555, 555, x = 555V.X. 504, 504, x 504Y. 580, 504, x 542607, 629, x 551Z.629, 504, x = 566.5eft fonThe mean of the sampling distribution is(Round to two decimal places as needed.)tThe standard deviation of the sampling distribution isile(Round to two decimal places as needed.)tioChoose the correct comparison of the population and sampling distribution below.QO A. The means are not equal and the standard deviation of the sampling distribution is larger.O B. The means are equal but the standard deviation of the sampling distribution is smaller.edO C. The means are equal but the standard deviation of the sampling distribution is larger.O D. The means are not equal and the standard deviation of the sampling distribution is smaller.ed QO E. The means and standard deviations are equal.?eaClick to select your answer(s).Priveway between two entries, use the z-score nalrway between the corresponding z-scores.WXPNOVtv

Question help_outlineImage TranscriptioncloseFor the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2. The mean of the population is (Round to two decimal places as needed.) The standard deviation of the population is (Round to two decimal places as needed.) Identify all samples of size 2 with the correct accompanying means below. A. 580, 555, x = 567.5 B. 504, 555, x 529.5 C. 555, 629, x = 592 D. 555, 580, x = 567.5 E. 555, 607, x = 581 580, 623, x = 601.5 F. G. 607, 555, x 581 H. 580, 629, x = 604.5 st I. 580, 580, x 580 J. 629, 580, x = 604.5 K. 504, 607, x = 555.5 L. 607, 504, x= 555.5 ile stio M. 580, 607, x - 593.5 N. 629, 555, x = 592 O. 607, 607, x = 607 P. g Qu 629, 629, x= 629 Q. 504, 629, x = 566.5 R. 504, 580, x = 542 e- S. 607, 580, x = 593.5 629, 607, x= 618 T. U. 555, 504, x = 529.5 ered V. 607, 629, x 618 W. 555, 555, x 555 Z. 629, 504, x 566.5 X. 504, 504, x= 504 580, 504, x = 542 Y. 607, 629, x 551 ted Q The mean of the sampling distribution is (Round to two decimal nlaces as needed) Click to select your answer(s). ea way between two entries, use the z-score nairway between the corresponding z-scores. NOV T wX P 3 tv fullscreen help_outlineImage TranscriptioncloseThis Question: 1 pt 7 of 14 (8 complete) This Quiz: 14 pts possible For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The word counts of five essays are 504, 629, 555, 607, and 580. Use a sample size of 2. M. 580, 607, X = 593.5 N. 629, 555, x = 592 O. 607, 607, x= 607 P. 629, 629, x = 629 Q. 504, 629, x = 566.5 R. 504, 580, x = 542 S. 607, 580, x = 593.5 T. 629, 607, x = 618 U. 555, 504, x = 529.5 tio 607, 629, x = 618 W. 555, 555, x = 555 V. X. 504, 504, x 504 Y. 580, 504, x 542 607, 629, x 551 Z. 629, 504, x = 566.5 eft f on The mean of the sampling distribution is (Round to two decimal places as needed.) tThe standard deviation of the sampling distribution is ile (Round to two decimal places as needed.) tio Choose the correct comparison of the population and sampling distribution below. Q O A. The means are not equal and the standard deviation of the sampling distribution is larger. O B. The means are equal but the standard deviation of the sampling distribution is smaller. ed O C. The means are equal but the standard deviation of the sampling distribution is larger. O D. The means are not equal and the standard deviation of the sampling distribution is smaller. ed Q O E. The means and standard deviations are equal. ? ea Click to select your answer(s). Prive way between two entries, use the z-score nalrway between the corresponding z-scores. WXP NOV tv fullscreen
check_circleExpert Solution
Step 1

Calculation of population mean and population standard deviation

Enter the given data in excel as shown below help_outlineImage TranscriptioncloseA 1 504 2 629 3 555 607 5 580 Population mean (u) 575 (From Excel -AVERAGE (A1:A5)) = Population standard Deviation (o) = 43.3728 (From Excel -STDEV.P(A1:A5)) Ln fullscreen
Step 2

Taking a samples of two and finding the mean of that

Enter the data in excel as shown below and the following for... help_outlineImage TranscriptioncloseA В A В C 1 x1 x2 1 х1 X-bar x2 504 2 504 2 504 -AVERAGE(A2:B2) AVERAGE(A3:B3) 504 629 3 504 3 504 629 555 4 504 4 504 AVERAGE(A4:B4) AVERAGE(A5:B5) 555 607 5 504 504 607 580 6 504 504 6 AVERAGE(A6:B6) AVERAGE(A7:B7) 580 629 7 629 7 629 629 555 8 629 AVERAGE(A8:B8) 8 629 555 607 9 629 629 -AVERAGE(A9:B9) 607 580 10 629 10 629 AVERAGE(A10:B10) 580 555 11 555 555 11 AVERAGE(A11:B11) 555 607 12 555 12 555 AVERAGE(A12:B12) 607 580 13 555 -AVERAGE(A13:B13)| 13 555 580 607 14 607 14 607 AVERAGE(A14:B14) 607 580 15 607 15 607 AVERAGE(A15:B15) -AVERAGE(A16:B16) 580 580 16 580 580 16 580 fullscreen

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