Question

# The table shows the salaries of $$280$$ persons.Salary(In thousand)No. of Persons$$5-10$$$$49$$$$10-15$$$$133$$$$15-20$$$$63$$$$20-25$$$$15$$$$25-30$$$$6$$$$30-35$$$$7$$$$35-40$$$$4$$$$40-45$$$$2$$$$45-50$$$$1$$Calculate the median salary of the data.

Solution

## Salary(In thousand) Frequency$$(f)$$      Cumulative frequency$$(cf)$$       $$5-10$$                          $$49$$                                   $$49$$$$10-15$$   $$133$$                                $$182$$$$15-20$$   $$63$$                                  $$245$$$$20-25$$   $$15$$                                  $$260$$$$25-30$$   $$6$$                                    $$266$$$$30-35$$   $$7$$                                    $$273$$$$35-40$$   $$4$$                                    $$277$$$$40-45$$   $$2$$                                    $$279$$$$45-50$$   $$1$$                                    $$280$$Total no. of persons $$N=49+133+63+15+6+7+4+2+1=280$$Now, $$\dfrac{N}{2}=140$$$$cf\ge \dfrac{N}{2}$$ is $$182$$Median class is $$10-15$$$$f=133$$Lower limit of median class $$l=10$$$$c.f.$$ above the median class is $$49$$Width of the interval $$h=5$$Median $$=l+\left(\dfrac{\frac{N}{2}-c.f}{f}\right)h$$$$=10+\left(\dfrac{140-49}{133}\right)5=10+\dfrac{91}{133}\times 5=13.42$$Hence, median $$=13.42$$Mathematics

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