CIV171-SPRING-2024-ASSIGNMENT-2-GONZALO-LAVALLE

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Northern Virginia Community College *

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171

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Statistics

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Apr 3, 2024

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1 CIV -171 SURVEYING I — ASSIGNMENT 2 • 5 points total NAME: DATE: Express 5,377,700 ft 2 in: (a) acres (b) hectares (c) square Gunter’s chains 1 2 Give answers to the following problems in the correct number of significant figures: (a) sum 23.15, 0.984, 124, and 12.5 (b) sum of 2.115, 23.04, 13.8, and 199.66 (c) product of 127.08, and 13.1 (q) quotient of 4466.83 divided by 35.61 GONZALO 1/29/2024 123.455 49.96 1234.545 160.598= 5 238.615= 6 1,664.748= 7 125.4375175512496= 16

2 CIV -171 SURVEYING I — ASSIGNMENT 2 • 5 points total NAME: DATE: 3 The observations 124.53, 124.55, 142.51, and 124.52 are obtained when taping the length of a line. What should the observer consider doing before a mean length is determined from the set of observations? 4 The difference in elevation between A and B was observed four times as: 32.05, 32.03, 32.08, and 32.01 ft. The observations were given weights of 2, 1, 3 and 2, respectively, by the observer. (a) Calculate the weighted mean for distance AB. (b) What difference results if later judgment revises the weights to 2, 3, 1, and 1, respectively? GONZALO 1/29/2024 A) Total Weight =2+1+3+2 = 8 Weight for AB = 32.05x(2/8) + 32.03x(1/8) + 32.08(3/8) + 32.01x(2/8) = 32.05ft Weight distance AB= 32.05 ft B) Total Weight = 2+3+1+1 = 7 Weight = 32.05x(2/7) + 32.03x(3/7) + 32.08x(1/7) + 32.01x(1/7) = 32.04ft Difference = 32.05 - 32.04 =0.01 ft

3 5 A distance XY is observed repeatedly using the same equipment and procedures, and the results in meters, are listed as: 65.401, 65.400, 65.402, 65.396, 65.406, 65.401, 65.401, 65.405, and 65.404. (a) Calculate the line’s most probable length (b) Calculate the standard deviation (c) Calculate the standard deviation of the mean CIV -171 SURVEYING I — ASSIGNMENT 2 • 5 points total NAME: DATE: GONZALO 1/29/2024 A) =65.401 + 65.400 + 65.402 + 65.396 + 65.406 + 65.401 + 65.401 + 65.405 + 65.404 / 9 =588.616 / 9 = 65.4018 Most porbable lenght = 65.4018 B) X= 65.4018 S= ½Swfi"ukipkhkec"gn"ÁA A) =65.401 + 65.400 + 65.402 + 65.396 + 65.406 + 65.401 + 65.401 + 65.405 + 65.404 / 9 =588.616 / 9 = 65.4018 Most porbable lenght = 65.4018 B) X= 65.4018 S= V(65.401-65.4018)2 + (65.400-65.4018)2 + (65.402-65.4018)2 + (65.396-65.4018)2 + (65.406-65.4018)2 + (65.401-65.4018)2 + (65.401-65.4018)2 + (65.405-65.4018)2 + (65.404-65.4018)2 /9-1 S= V0.000008945 S= 0.00317 C) Standard deviation= 0.022317 / V10 = 0.001

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