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The following series relates to the daily income of workers employed in a firm. Compute the highest income of lowest 50% of workers, ie; median.Daily income(x) :10 - 1415 - 1920 - 2425 - 2930 - 3435 - 39No. of cities (f) :51015201015 |
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Answer»
Since the series of data given is of inclusive type, first we shall convert the class intervals into exclusive type using a correction factor. C.F – LL of a CI – UL of the previous CI = \(\frac{15 - 14}{2}\) = 0.5 Where LL = Lower Limit UL = Upper Limit CI = Class Interval CF = Cumulative frequency
Median class = Size of \([\frac{N}{2}]\) th item = \([\frac{65}{2}]\)th item = 32.5 th item so, it lies between 24.5 = 29.5 Medain = L + \([\frac{\frac{N}{2}-Cf}{f}]\times i\) Where L – Lower limit C.f = Cumulative frequency of class preceding the median class L – Length of the median class N = Total of frequency f = frequency of the median class Median class = 24.5 – 29.5 I = 24.5; i = 5; c.f = 30; N = 65; f = 20 Medain = 24.5 +\([\frac{\frac{65}{2}-30}{20}]\times 5\) Medain = Rs. 25.125 Highest income of lowest 50% of workers |
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