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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 Girls
Less 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) C.F. 
below 140 
140-145 711 
145-150 18 29 
150-155 11 40 
155-160 46 
160-165 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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