data: 10 female endurance athletes ages: 13.6, 13.9, 14.0, 14.2, 14.9, 15.0, 150, 15.1,15.4, 16.4 Sample mean: 13.9 ✓ + 13.6 median: 14.9 +15=29.9/2=14,95 14.0 19.2 14.9 15 = √ 15.1 15.4 16.4 ✓ mean 147.5/10 14.75 = X Σ(x-x²)² n-1 (13.6-14.75)+ (13.9-14.75) + (14-14.75)² + (14.2-14.75) (14.9-14.75)2 + (15-14.75)² + (15-14.75)² + (15.1-14.75) +(15.4-14.75)2 + (16.4-14.75) ² V 9 - (1.3225) + (0.72) + (0.56)+ (0.30) + (0.0225) + (0.0625) +(0,025) LO22)+(0,4225)+(2122) 9 5 -7/6.3165/9 = (0.837 Variance

Big Ideas Math A Bridge To Success Algebra 1: Student Edition 2015
1st Edition
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:HOUGHTON MIFFLIN HARCOURT
Chapter2: Solving Linear Inequalities
Section2.5: Solving Compound Inequalities
Problem 40E
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I am not sure how to get the variance for this problem. The answer is 0.703 yr^2. Can someone show me which formula and how they got the answer? Thx
data: 10 female endurance athletes ages:
13.6, 13.9, 14.0, 14.2, 14.9, 15.0, 15.0, 15.1,15.4, 16.4
Sample mean: 13.9
+
13.6
median: 14.9 +15=29.9/2-14,95
14.0
19.2
14.9
3=
Lis
15
15.1
15.4
16.4
≤ (x-x²) ²
n-1
mean
147.5/10=(14.75 = X
=
(13.6-14.75) + (13.9-14.75) + (14-14.75) ²³+ (14.2-14.75)³²
(14.9-14.75)2 + (15-14.75)² + (15-14.75)² + (15.1-14.75)²
+ (15.4-14.75)2 + (16.4-14.75) ²
(1.3225) + (0.72) + (0.56)+ (0.30) + (0.0225) + (0.0625)
+ (0.0625) + (0.122) + (0.4225) + (2.722)
9
5 -76.3165/9 = (0.837
Variance
Transcribed Image Text:data: 10 female endurance athletes ages: 13.6, 13.9, 14.0, 14.2, 14.9, 15.0, 15.0, 15.1,15.4, 16.4 Sample mean: 13.9 + 13.6 median: 14.9 +15=29.9/2-14,95 14.0 19.2 14.9 3= Lis 15 15.1 15.4 16.4 ≤ (x-x²) ² n-1 mean 147.5/10=(14.75 = X = (13.6-14.75) + (13.9-14.75) + (14-14.75) ²³+ (14.2-14.75)³² (14.9-14.75)2 + (15-14.75)² + (15-14.75)² + (15.1-14.75)² + (15.4-14.75)2 + (16.4-14.75) ² (1.3225) + (0.72) + (0.56)+ (0.30) + (0.0225) + (0.0625) + (0.0625) + (0.122) + (0.4225) + (2.722) 9 5 -76.3165/9 = (0.837 Variance
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