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Exercise 13.1 · Q9
Q.

Find the mean deviation about the mean for the following data:

Income per day (in ₹)Number of persons
0-1004
100-2008
200-3009
300-40010
400-5007
500-6005
600-7004
700-8003
Tripura TbseTextbookSubjective· 5mImportance★★★★★est
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Using class midpoints, the mean income is ₹358, and the mean deviation about the mean is ₹157.92.

The mean deviation about the mean tells us, on average, how far each person's daily income lies from the average income — measured in absolute terms (we ignore the direction of the gap). Because the data is grouped into class intervals, we represent each class by its midpoint.

Step 1 — Class midpoints (xix_i).

The midpoint of a class is the average of its lower and upper limits, e.g. 0+1002=50\dfrac{0+100}{2}=50.

Midpoints: 50, 150, 250, 350, 450, 550, 650, 750.

Step 2 — Mean xˉ=∑fixi∑fi\bar{x}=\dfrac{\sum f_i x_i}{\sum f_i}.

Income (₹)fif_ixix_ifixif_i x_i
0–100450200
100–20081501200
200–30092502250
300–400103503500
400–50074503150
500–60055502750
600–70046502600
700–80037502250
Total5017900

xˉ=1790050=358.\bar{x}=\frac{17900}{50}=358.

Step 3 — Absolute deviations and their weighted sum ∑fi∣xi−xˉ∣\sum f_i\lvert x_i-\bar{x}\rvert. …

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