Question

# A life insurance agent found the following data for distribution of ages of $$100$$ policy holders. Calculate the median age, if policies are only given to person having age $$18$$ years onwards but less than $$60$$ years.Age (in years)No. of policy holdersBelow $$20$$$$2$$Below $$25$$$$6$$Below $$30$$$$24$$Below $$35$$$$45$$Below $$40$$$$78$$Below $$45$$$$89$$Below $$50$$$$92$$Below $$55$$$$98$$Below $$60$$$$100$$

A
62.12 years
B
53.53 years
C
35.76 years
D
None of these

Solution

## The correct option is D $$35.76$$ yearsAge (in years)       Number of policy holders     $$f_1$$Cumulative frequencyBelow $$20$$$$2=2$$$$2$$$$20-25$$$$(6-2)=4$$$$6$$$$25-30$$$$(24-6)=18$$$$24$$$$30-35$$$$(45-24)=21$$$$45$$$$35-40$$$$(78-45)=33$$$$78$$$$40-45$$$$(89-78)=11$$$$89$$$$45-50$$$$(92-89)=3$$$$92$$$$50-55$$$$(98-92)=6$$$$98$$$$55-60$$$$(100-98)=2$$$$100$$Total$$n=100$$Here, $$l=35, n=100, f=33, cf=45, h=5$$ Median$$=l+\left \{\cfrac {\frac {n}{2}-cf}{f}\right \}\times h$$$$=35+\left \{\cfrac {50-45}{33}\right \}\times 5$$$$=35+\cfrac {25}{33}$$$$=35+0.76$$$$=35.76$$ yearsMathematicsRS AgarwalStandard X

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