Question

# A survey regarding the height (in cm) of $$51$$ girls of class X of a school was conducted and the following data was obtained:Height in cmNumber of GirlsLess than 140$$4$$Less than 145$$11$$Less than 150$$29$$Less than 155$$40$$Less than 155$$46$$Less than 165$$51$$Find the median height.

Solution

## Height (in cm) f C.F. below 140 4 4 140-145 711 145-150 18 29 150-155 11 40 155-160 6 46 160-165 5 51 $$N = 51 \Rightarrow \cfrac{N}{2} = \cfrac{51}{2} = 25.5$$As 29 is just greater than 25.5, therefore median class is 145-150.$$Median = l + \cfrac{\cfrac{N}{2} + C}{f} \times h$$Here, $$l =$$ lower limit of median class $$= 145$$$$C = C.F.$$ of the class preceding the median class $$= 11$$$$h =$$ higher limit - lower limit $$= 150 - 145 = 5$$$$f =$$ frequency of median class $$= 18$$$$\therefore \; median = 145 + \cfrac{25.5 - 11}{18} \times 5 = 145 + 4.03 = 149.03$$Mathematics

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