Question

1

(1 point) Archeologists have discovered a sample of ancient Egyptian skulls from 150 A.C. A sample of 10 Egyptian skull breadths (in mm) from 150 A.C. is show below: Skull breadths Sample: 135, 127, 138, 127, 128, 141, 130, 128, 132, 134 Part 1: Calculate the mean and standard deviation for for the data set. Round the mean to 2 decimal places, and the standard deviation to 4 decimal places. 150 A.C. skull breadths mean: 150 A.C. skull breadths standard deviation:

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Answer 1

Solution(a)
Sample mean can be calculated as
Sample mean (Xbar) = $\sum$ Xi/n = (135+127+138+127+128+141+130+128+132+134)/10 = 1320/10 = 132
Sample standard deviation can be calculated as
Sample standard deviation S = sqrt($\sum$(Xi-mean)^2/(n-1)) = sqrt((135-132)^2 + (127-132)^2 + (138-132)^2 + (127-132)^2 + (128-132)^2 + (141-132)^2 + (130-132)^2 + (128-132)^2 + (132-132)^2 + (134-132)^2)/9)

X	(Xi-mean)	(Xi-mean)^2
135	3	9
127	-5	25
138	6	36
127	-5	25
128	-4	16
141	9	81
130	-2	4
128	-4	16
132	0	0
134	2	4

Sample standard deviation (S) = sqrt((9+25+36+25+16+81+4+16+0+4)/9) = sqrt(216/9) = 4.8990
Solution(b)

Sample mean can be calculated as
Sample mean (Xbar) = $\sum$ Xi/n = (136+124+137+122+120+119)/6 = 758/6 = 126.33
Sample standard deviation can be calculated as
Sample standard deviation S = sqrt($\sum$(Xi-mean)^2/(n-1)) = sqrt((136-126.33)^2 + (124-126.33)^2 + (137-126.33)^2 + (122-126.33)^2 + (120-126.33)^2 + (119-126.33)^2)/5) = sqrt(325.3334/5) = 8.0664

X	(Xi-mean)	(Xi-mean)^2
136	9.67	93.5089
124	-2.33	5.4289
137	10.67	113.8489
122	-4.33	18.7489
120	-6.33	40.0689
119	-7.33	53.7289