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Q.Find the mean deviation about the mean for the following data: xix_i: 2,5,7,8,10,352, 5, 7, 8, 10, 35 fif_i: 6,8,10,6,8,26, 8, 10, 6, 8, 2

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 7mImportance★★★★★
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xˉ=32040=8\bar x=\dfrac{320}{40}=8; then M.D.=∑fi∣xi−8∣40=14040=3.5\text{M.D.}=\dfrac{\sum f_i|x_i-8|}{40}=\dfrac{140}{40}=3.5.

Data: xi=2,5,7,8,10,35x_i=2,5,7,8,10,35 with fi=6,8,10,6,8,2f_i=6,8,10,6,8,2, and N=∑fi=40N=\sum f_i=40.

Mean:

∑fixi=2(6)+5(8)+7(10)+8(6)+10(8)+35(2)=12+40+70+48+80+70=320,\sum f_ix_i=2(6)+5(8)+7(10)+8(6)+10(8)+35(2)=12+40+70+48+80+70=320,

xˉ=32040=8.\bar x=\dfrac{320}{40}=8.

Absolute deviations ∣xi−8∣=6,3,1,0,2,27|x_i-8|=6,3,1,0,2,27, and

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