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What is the variance and standard deviation for: 13, 15, 12, 10, 4, 16, 17, 22, 9
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What is the variance and standard deviation for: 13, 15, 12, 10, 4, 16, 17, 22, 9

User Ernest S Kirubakaran
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\begin{gathered} 13,15,12,10,4,16,17,22,9 \\ S\tan dard\text{ deviation}=\sigma=? \\ Variance\text{ }\sigma^2=? \\ \sigma^2=\frac{\sum ^N_{i\mathop{=}1}(x_i-\bar{x})^2)}{N} \\ N=\text{ total of data=9} \\ \bar{x}=(13+15+12+10+4+16+17+22+9)/(9)=(118)/(9) \\ For\text{ }x_i=13 \\ (x_i-\bar{x})^2=(13-(118)/(9))^2=((-1)/(9))^2=(1)/(81) \\ For\text{ }x_i=15 \\ (x_i-\bar{x})^2=(15-(118)/(9))^2=((17)/(9))^2=(289)/(81) \\ For\text{ }x_i=12 \\ (x_i-\bar{x})^2=(12-(118)/(9))^2=(-(10)/(9))^2=(100)/(81) \\ For\text{ }x_i=10 \\ (x_i-\bar{x})^2=(10-(118)/(9))^2=(-(28)/(9))^2=(784)/(81) \\ For\text{ }x_i=4 \\ (x_i-\bar{x})^2=(4-(118)/(9))^2=(-(82)/(9))^2=(6724)/(81) \\ For\text{ }x_i=16 \\ (x_i-\bar{x})^2=(16-(118)/(9))^2=((26)/(9))^2=(676)/(81) \\ For\text{ }x_i=17 \\ (x_i-\bar{x})^2=(17-(118)/(9))^2=((35)/(9))^2=(1225)/(81) \\ For\text{ }x_i=22 \\ (x_i-\bar{x})^2=(22-(118)/(9))^2=((80)/(9))^2=(6400)/(81) \\ For\text{ }x_i=9 \\ (x_i-\bar{x})^2=(9-(118)/(9))^2=(-(37)/(9))^2=(1369)/(81) \\ \sum ^N_{i\mathop{=}1}(x_i-\bar{x})^2)=(17568)/(81)=(1952)/(9) \\ \sigma^2=((1952)/(9))/(9)=(1952)/(81)\approx24.1 \\ \text{The variance is }24.1 \\ For\text{ standard deviation } \\ \sigma=\sqrt{\frac{\sum^N_{i\mathop{=}1}(x_i-\bar{x})^2)}{N}} \\ \sigma=√(24.1) \\ \sigma\approx4.9 \\ \text{The standard deviation is 4.9} \end{gathered}

User Maulik Savaliya
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