Question

# The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data: Class intervalFrequencyClass intervalFrequency0−1002500−60020100−2005600−700f2200−300f1700−8009300−40012800−9007400−50017900−10004

Solution

## Class interval Frequency Cumulative frequency 0 – 100 2 2 100 – 200 5 7 200 – 300 f1 7 + f1 300 – 400 12 19 + f1 400 – 500 17 36 + f­1 (F) 500 – 600 20 (f) 56 + f1 600 – 700 f2 56 + f1 + f2 700 – 800 9 65 + f1 + f2 800 – 900 7 72 + f1 + f2 900 – 1000 4 76 + f1 + f2   N = 100   Given Median = 525 Then, median class = 500 – 600 L = 500, f = 20, F = 36 + f1, h = 600 – 500 = 100 5 = 14 – f1 f1 = 14 – 5 = 9 Given Sum of frequencies = 100 2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100 2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100 85 + f2 = 100 f2 = 100 – 85 = 15 f1 = 9 and f2 = 15 MathematicsRD SharmaStandard X

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