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Exercise 13.1 · Q5
Q.

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

xix_ifif_i
57
104
156
203
255
Bihar BsebTextbookSubjective· 3mImportance★★★★★est
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Mean deviation about the mean measures the average absolute distance of each data point from the central mean. For this data, the mean is 1414, and the mean deviation is 6.326.32.

The mean deviation about the mean tells you, on average, how far each observation lies from the arithmetic mean — but without caring whether it’s above or below. That’s why we take absolute differences. It’s a simple, intuitive measure of spread, especially useful when you want to avoid squaring the deviations (as variance does).

Here’s the step-by-step:

  1. Find the mean (xˉ\bar{x}). The mean of a frequency distribution is

xˉ=∑fixi∑fi.\bar{x} = \frac{\sum f_i x_i}{\sum f_i}.

Compute ∑fixi\sum f_i x_i:

5×7=35,10×4=40,15×6=90,20×3=60,25×5=125.5 \times 7 = 35,\quad 10 \times 4 = 40,\quad 15 \times 6 = 90,\quad 20 \times 3 = 60,\quad 25 \times 5 = 125.

Sum: 35+40+90+60+125=35035 + 40 + 90 + 60 + 125 = 350.

Total frequency: 7+4+6+3+5=257 + 4 + 6 + 3 + 5 = 25.

So

xˉ=35025=14.\bar{x} = \frac{350}{25} = 14.

  1. Find the absolute deviations ∣xi−xˉ∣|x_i - \bar{x}|. For each xix_i:

∣5−14∣=9,∣10−14∣=4,∣15−14∣=1,∣20−14∣=6,∣25−14∣=11.|5 - 14| = 9,\quad |10 - 14| = 4,\quad |15 - 14| = 1,\quad |20 - 14| = 6,\quad |25 - 14| = 11.

  1. Multiply each absolute deviation by its frequency fif_i.

7×9=63,4×4=16,6×1=6,3×6=18,5×11=55.7 \times 9 = 63,\quad 4 \times 4 = 16,\quad 6 \times 1 = 6,\quad 3 \times 6 = 18,\quad 5 \times 11 = 55.

Sum these: 63+16+6+18+55=15863 + 16 + 6 + 18 + 55 = 158. …

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