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NCERT Exemplar · Q16
Q.

Calculate the mean deviation about the mean for the following frequency distribution:

Class interval0 - 44 - 88 - 1212 - 1616 - 20
Frequency46852
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Using class midpoints, the mean is xˉ=9.2\bar x = 9.2 and the mean deviation about the mean is 3.843.84.

Step-by-step solution

1. Midpoints and frequencies

Class intervalMidpoint xix_iFrequency fif_i
0 - 424
4 - 866
8 - 12108
12 - 16145
16 - 20182

Total frequency N=∑fi=25N = \sum f_i = 25.

2. Mean

xˉ=∑fixiN=2(4)+6(6)+10(8)+14(5)+18(2)25=8+36+80+70+3625=23025=9.2.\bar x = \frac{\sum f_i x_i}{N} = \frac{2(4)+6(6)+10(8)+14(5)+18(2)}{25} = \frac{8+36+80+70+36}{25} = \frac{230}{25} = 9.2.

3. Absolute deviations and weighted sum

xix_ifif_i∣xi−xˉ∣\lvert x_i-\bar x\rvertfi∣xi−xˉ∣f_i\lvert x_i-\bar x\rvert
247.228.8
663.219.2
1080.86.4
1454.824.0

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