Question

# Find the mean, median and mode of the following data:Classes:$$0-50$$$$50-100$$$$100-150$$$$150-200$$$$200-250$$$$250-300$$$$300-350$$Frequency:$$2$$$$3$$$$5$$$$6$$$$5$$$$3$$$$1$$

Solution

## Class intervalMid value  $$x_i$$ Frequency   $$f_i$$$$f_ix_i$$ $$cf$$  $$0-50$$$$35$$ $$2$$ $$50$$ $$2$$  $$50-100$$$$75$$ $$3$$ $$225$$ $$5$$  $$100-150$$$$125$$ $$5$$ $$625$$ $$10$$  $$150-200$$$$175$$ $$6$$ $$1050$$ $$16$$  $$200-250$$$$225$$ $$5$$ $$1127$$ $$21$$  $$250-300$$$$275$$ $$3$$ $$825$$ $$24$$  $$300-350$$$$325$$ $$1$$ $$325$$ $$25$$    $$\sum f_i=25$$ $$\sum f_ix_i=4225$$ $$\Rightarrow$$  $$Mean=\dfrac{\sum f_ix_i}{\sum f_i}=\dfrac{4225}{25}=169$$$$\Rightarrow$$  We have $$N=25$$\$. Then, $$\dfrac{N}{2}=12.5$$$$\Rightarrow$$ So, median class is $$150-200$$.$$l =$$ lower limit of the modal class$$h =$$ size of the class intervals$$f =$$ frequency of the modal class$$f_1 =$$ frequency of the class preceding the modal class$$f_2 =$$ frequency of the class succeed in the modal class.$$\therefore$$  $$l=150,\,h=200-150=50,\,f=6,\,cf=10$$$$\Rightarrow$$  $$Median=l+\dfrac{\dfrac{N}{2}-cf}{f}\times h$$$$\Rightarrow$$  $$Median=150+\dfrac{12.5-10}{6}\times 50$$$$\Rightarrow$$  $$Median=150+\dfrac{125}{6}$$$$\therefore$$   $$Median=150+20.83=170.83$$. $$\Rightarrow$$  Here maximum frequency is $$6$$, then the corresponding class $$150-200$$ is the modal class.$$\Rightarrow$$  $$l=150,\,h=50,\,f=6,\,f_1=5\,f_2=5$$$$\Rightarrow$$  $$Mode=l+\dfrac{f-f_1}{2f-f_1-f_2}\times h$$$$\Rightarrow$$  $$Mode=150+\dfrac{6-5}{2\times 6-5-5}\times 50$$$$\Rightarrow$$  $$Mode=150+\dfrac{50}{2}$$$$\therefore$$   $$Mode=150+25=175$$MathematicsRS AgarwalStandard X

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