Find the mean age of 100 residents of a town from the following data.
Age equal and above (in years)010203040506070Number of persons100907550251550
Find the mean age of 100 residents of a town from the following data.
Age equal and above (in years)010203040506070Number of persons100907550251550
Here, we observe that all 100 residents of a town have age equal and above 0. Since 90 residents of a town have age equal and above 10.
So, 100−90=10 residents lie in the interval 0−10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table.
Class intervalNumber of persons(fi)Class marks(xi)ui=xi−ahfiui0−10100−90=105−3−3010−1090−75=1515−2−3020−3075−50=2525−1−2530−4050−25=2535=a0040−5025−15=104511050−6015−5=105522060−705−0=565315N=∑fi=100∑fiui=−40
Class mark of any class interval =Upper limit + Lower limit2
Here, (assumed mean) a=35
and (class width) h=10
By step deviation method, we have
Mean (¯x)=a+∑fiui∑fi×h=35+(−40)100×10=35−4=31
Hence, the required mean age is 31 years.
Here, we observe that all 100 residents of a town have age equal and above 0. Since 90 residents of a town have age equal and above 10.
So, 100−90=10 residents lie in the interval 0−10 and so on. Continue in this manner, we get frequency of all class intervals. Now, we construct the frequency distribution table.
Class intervalNumber of persons(fi)Class marks(xi)ui=xi−ahfiui0−10100−90=105−3−3010−1090−75=1515−2−3020−3075−50=2525−1−2530−4050−25=2535=a0040−5025−15=104511050−6015−5=105522060−705−0=565315N=∑fi=100∑fiui=−40
Class mark of any class interval =Upper limit + Lower limit2
Here, (assumed mean) a=35
and (class width) h=10
By step deviation method, we have
Mean (¯x)=a+∑fiui∑fi×h=35+(−40)100×10=35−4=31
Hence, the required mean age is 31 years.
