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Exercise 13.1 · Q10
Q.

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

Height in cmsNumber of boys
95-1059
105-11513
115-12526
125-13530
135-14512
145-15510
Uttar Pradesh UpmspTextbookSubjective· 5mImportance★★★★★est
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Mean =125.3= 125.3 cm, and the mean deviation about the mean is 1128.8100=11.288≈11.29\dfrac{1128.8}{100} = 11.288 \approx 11.29 cm.

Mean Deviation About the Mean (Grouped Data)

M.D.(xˉ)=1N∑fi ∣xi−xˉ∣,xˉ=∑fixiN\text{M.D.}(\bar{x}) = \frac{1}{N}\sum f_i\,|x_i - \bar{x}|,\qquad \bar{x} = \frac{\sum f_i x_i}{N}

Each class is represented by its mid-point xix_i.

Step-by-Step Solution

1. Mid-points, fixif_i x_i and the mean.

Classxix_ifif_ifixif_i x_i
95–1051009900
105–115110131430
115–125120263120
125–135130303900
135–145140121680
145–155150101500
TotalN=100N=1001253012530

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

2. Weighted absolute deviations fi ∣xi−125.3∣f_i\,|x_i - 125.3|.

xix_ifif_i∣xi−125.3∣\lvert x_i-125.3\rvertfi∣xi−125.3∣f_i\lvert x_i-125.3\rvert
100925.3227.7

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