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Q.

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

Height (in cms)95-105105-115115-125125-135135-145145-155
No. of Boys91326301210
Andhra Pradesh BieapBIEAP Intermediate Board 2025Subjective· 7mImportance★★★★★
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Find the mean using class midpoints, then take the frequency-weighted average of the absolute deviations of each midpoint from the mean.

Class midpoints xix_i and frequencies fif_i:

Class95-105105-115115-125125-135135-145145-155
xix_i100110120130140150
fif_i91326301210

N=∑fi=9+13+26+30+12+10=100N=\sum f_i = 9+13+26+30+12+10=100.

Mean: ∑fixi=900+1430+3120+3900+1680+1500=12530\sum f_ix_i = 900+1430+3120+3900+1680+1500=12530.

xˉ=12530100=125.3.\bar x = \frac{12530}{100} = 125.3.

Absolute deviations ∣xi−xˉ∣|x_i-\bar x| and their weighted sum:

| xix_i | ∣xi−125.3∣|x_i-125.3| | fif_i | fi∣xi−125.3∣f_i|x_i-125.3| |

|---|---|---|---|

| 100 | 25.3 | 9 | 227.7 |

| 110 | 15.3 | 13 | 198.9 | …

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