The arithmetic mean of the following frequency distribution is 50.


Class



0 - 10



10 - 20



20 - 30



30 - 40



40 - 50



Frequency



16



p



30



32



14



Find the value of p.

For equal class intervals, we will solve by finding mid points of these classes using direct method.


CLASS



MID - POINT(xi)



FREQUENCY(fi)



fixi



0 - 10



5



16



80



10 - 20



15



p



15p



20 - 30



25



30



750



30 - 40



35



32



1120



40 - 50



45



14



630



TOTAL




92 + p



2580 + 15p




We have got


Σfi = 92 + p and Σfixi = 2580 + 15p


∵ mean is given by



⇒ (∵ given: arithmetic mean is 50)


⇒ 4600 + 50p = 2580 + 15p


⇒ 50p – 15p = 2580 – 4600


⇒ 35p = -2020


⇒ p = -57.71


Thus, p is -57.71


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