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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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