15. VEIGHTED KEABURENENTS. A dietanee AB is meaured ttvetimes an 610.03, 610.01, 610:05, 41e.04, and 610.02 metera.The measurenenta were given veights ot 3,2, 1, 2, and 3,tespectively, by the head tapenan. Determine the weightednean tor diotance AB. Alse, determine what ditterence re-sults it later judgaent revises the veights to i, 3, 1, 3,and 2.

Answered: 15. VEIGHTED KEABURENENTS. A dietanee…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Question

15. VEIGHTED KEABURENENTS. A dietanee AB is meaured ttve
times an 610.03, 610.01, 610:05, 41e.04, and 610.02 metera.
The measurenenta were given veights ot 3,2, 1, 2, and 3,
tespectively, by the head tapenan. Determine the weighted
nean tor diotance AB. Alse, determine what ditterence re-
sults it later judgaent revises the veights to i, 3, 1, 3,
and 2.

Expert Solution

Step 1

- Given data:

Five readings (x_i ) of distance AB are 610.03, 610.01, 610.05, 610.04 and 610.02

Respective weight (w_i ) of readings are 3, 2, 1, 2, and 3

To find:

Weighted mean for distance AB,

Solution:

Weighted mean for distance AB = $\frac{\sum_{}^{} x_{i} w_{i}}{\sum_{}^{} w_{i}}$

$W e i g h t e d m e a n f o r d i s \tan c e A B = \frac{(610.03 * 3) + (610.01 * 2) + (610.05 * 1) + (610.04 * 2) + (610.02 * 3)}{3 + 2 + 1 + 2 + 3} W e i g h t e d m e a n f o r d i s \tan c e A B = \frac{1830.09 + 1220.02 + 610.05 + 1220.08 + 1830.06}{11} W e i g h t e d m e a n f o r d i s \tan c e A B = \frac{6710.3}{11} W e i g h t e d m e a n f o r d i s \tan c e A B = 610.02727$