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Q.Find the mean deviation from the mean for a continuous frequency distribution. Sales (in Rs. thousand): 40-50, 50-60, 60-70, 70-80, 80-90, 90-100
Number of companies: 5, 15, 25, 30, 20, 5

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 7mImportance★★★★★
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Use class mid-points to find the mean, then compute ∑f∣x−xˉ∣N\frac{\sum f|x-\bar x|}{N}.

ClassMid-value xxFrequency fffxfx
40-50455225
50-605515825
60-7065251625
70-8075302250
80-9085201700
90-100955475

N=∑f=100N=\sum f = 100, ∑fx=225+825+1625+2250+1700+475=7100\sum fx = 225+825+1625+2250+1700+475 = 7100

Mean: xˉ=7100100=71\bar x = \dfrac{7100}{100}=71

Absolute deviations ∣x−xˉ∣|x-\bar x| and f∣x−xˉ∣f|x-\bar x|:

xx∣x−71∣\vert x-71\vertfff∣x−71∣f\vert x-71\vert
45265130
551615240
65625150

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