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Q.Calculate the mean deviation about median for the following data. Class: 0-100\text{-}10, 10-2010\text{-}20, 20-3020\text{-}30, 30-4030\text{-}40, 40-5040\text{-}50, 50-6050\text{-}60; Frequency: 66, 77, 1515, 1616, 44, 22

Rajasthan RbseRajasthan Board Senior Secondary Part-I Examination 2026Subjective· 4mImportance★★★★★
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The mean deviation about the median is 10.1610.16.

ClassFrequency ffMidpoint xix_iCumulative frequency
00-1010665566
1010-20207715151313
2020-3030151525252828
3030-4040161635354444
4040-50504445454848
5050-60602255555050

N=50N = 50, so N/2=25N/2 = 25. The cumulative frequency first reaching 2525 is 2828, in the class 2020-3030 — this is the median class (L=20L=20, cumulative freq before it cf=13cf=13, class frequency f=15f=15, class width h=10h=10).

Median =L+N2−cff×h=20+25−1315×10=20+1215×10=20+8=28= L + \dfrac{\frac{N}{2}-cf}{f}\times h = 20+\dfrac{25-13}{15}\times10 = 20+\dfrac{12}{15}\times10 = 20+8 = 28

Absolute deviations ∣xi−28∣|x_i - 28| and f∣xi−28∣f|x_i-28|:

| xix_i | ∣xi−28∣|x_i-28| | ff | f∣xi−28∣f|x_i-28| |

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

| 55 | 2323 | 66 | 138138 | …

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