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Q.

Find the mean deviation about the mean for the following distribution.

xix_i1030507090
fif_i42428168
Telangana TsbieTelangana Board of Intermediate Education 2026Subjective· 7mImportance★★★★★
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Mean =50=50; ∑fi∣xi−50∣=1280\sum f_i|x_i-50|=1280, so MD =1280/80=16=1280/80=16.

Data: xi=10,30,50,70,90x_i = 10, 30, 50, 70, 90 with fi=4,24,28,16,8f_i = 4, 24, 28, 16, 8.

∑fi=4+24+28+16+8=80\sum f_i = 4 + 24 + 28 + 16 + 8 = 80.

∑fixi=10(4)+30(24)+50(28)+70(16)+90(8)=40+720+1400+1120+720=4000\sum f_i x_i = 10(4) + 30(24) + 50(28) + 70(16) + 90(8) = 40 + 720 + 1400 + 1120 + 720 = 4000.

Mean xˉ=400080=50\bar{x} = \dfrac{4000}{80} = 50.

Absolute deviations ∣xi−50∣=40,20,0,20,40|x_i - 50| = 40, 20, 0, 20, 40.

fi∣xi−50∣=4(40),24(20),28(0),16(20),8(40)=160,480,0,320,320f_i|x_i - 50| = 4(40), 24(20), 28(0), 16(20), 8(40) = 160, 480, 0, 320, 320.

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