Question statement
Compare the data sets using measures of centre and variation.
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Solution

Step 1:
$Data of set A = 0.8, 1.0, 1.1, 1.2, 1.5 Mean of set A = (0.8 + 1.0 + 1.1 + 1.2 + 1.5) \div 5 = 5.6 \div 5 = 1.12 Variation of set A = [{(0.8 - 1.12)}^{2} + {(1.0 - 1.12)}^{2} + {(1.1 - 1.12)}^{2} + {(1.2 - 1.12)}^{2} + {(1.5 - 1.15)}^{2}] \div 5 = [{(- 0.32)}^{2} + {(- 0.12)}^{2} + {(- 0.02)}^{2} + {(0.08)}^{2} + {(0.38)}^{2}] \div 5 = [0.10.24 + 0.0144 + 0.004 + 0.00.1444] \div 5 = 0.2716 \div 5 = 0.5432$
Step 2:
$Data of set B = 0.3, 0.4, 0.6, 0.9, 1.4 Mean of set B = 0.3 + 0.4 + 0.6 + . 9 + 1.4 \div 5 = 0.72 Variation of set B = [{(0.3 - 0.72)}^{2} + {(0.4 - 0.7)}^{2} + {(0.6 - 0.72)}^{2} + (0.9 - 0.72)^{2} + {(1.4 - 0.72)}^{2}] \div 5 {= [(- 0.42)}^{2} + {(- 0.32)}^{2} + {(- 0.12)}^{2} + {(0.18)}^{2} + {(0.68)}^{2}] \div 5 = [0.1764 + 0.1024 + 0.0144 + 0.0324 + 0.4624] \div 5 = 0.788 \div 5 = 0.1576$
Variation of set A and Variation of set B
$\Rightarrow 0.5432$ is greater than $0.1576$
Therefore, Variation of set A and Variation of set B
$\Rightarrow 0.5432$ is greater than $0.1576$ .

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