Given the set of schools shown below, find the following:
a. Find the unweighted mean center.
b. Find the weighted mean center.
c. Find the weighted standard distance for the schools.
d. Find the relative standard distance, assuming that the study area ranges from coor-
dinates (0,0) in the southwest to (6,6) in the northeast. Repeat assuming the north-
east is defined by the point (10,10).
A. xmean = (5.4 + 1.2 + 0.5 + 2.1) / 4 = 2.3
ymean = (5.3 + 3.4 + 3.5 + 2.2) / 4 = 3.5 (2.3, 3.5)
B. xmean = (5.4 * 400 + 1.2 * 1000 + 0.5 * 3000 + 2.1 * 1500)/(400 + 1000 + 3000 + 1500) =1.4
ymean = (5.3 * 400 + 3.4 * 1000 + 3.5 * 3000 + 2.2 * 1500)/(400 + 1000 + 3000 + 1500) =3.3 (1.4, 3.3)
C. xdist = 400 * (5.4 - 1.4)2 + 1000 * (1.2 - 1.4)2 + 3000 * (0.5 - 1.4)2 + 1500 *
(2.1 - 1.4)2 =9594.4
ydist = 400 *(5.3 - 3.3)2 + 1000 *(3.4 - 3.3)2 + 3000 *(3.5 - 3.3)2 + 1500 *
(2.2 - 3.3)2 =3541.2
WSD = ?(xdist + ydist )/(400 + 1000 + 3000 + 1500) = 1.5
D. sd = ?(distance((5.4, 5.3), (2.3, 3.5))...+ distance((2.1, 2.2), (2.3, 3.5))) /4 = ?7.8/4 = 1.4
For the (6,6) area: sd,rel = sd /(?A/?) = 1.4 /(?6 * 6/?) = 0.4
For the (10,10) area: sd,rel = sd /(?A/?) = 1.4 /(?10 * 10/?) = 0.2
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Answer:
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