Question:-

The following table represents the number of illiterate females in the age group (10 - 34) in a town:

[tex]\boxed{\begin{array}{c|c|c|c|c|c|} \bf{Age\:Group} & \sf{10 - 14} & \sf{15 - 19} & \sf{20-24} & \sf{25 - 29} & \sf{30 - 34} \\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} \\ \bf{Number\:of\:females} & \sf{300} & \sf{380} & \sf{800} & \sf{580} & \sf{290} \end{array}}[/tex]

Find the mean.

Solution:-

Here the class interval is not regular.

Hence,

Difference between two class intervals = Lower class of second interval - upper class of first interval

Difference = 15 - 14 = 1

Now,

Divide difference by 2 = 1/2 = 0.5

Now, Subtract 0.5 from the lower class of each interval and add 0.5 to the upper class of each interval.

New table is as follows:-

[tex]\boxed{\begin{array}{c|c|c|c|c|c|} \bf{Age\:Group} & \sf{9.5 - 14.5} & \sf{14.5 - 19.5} & \sf{19.5 - 24.5} & \sf{24.5 - 29.5} & \sf{29.5 - 34.5} \\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} \\ \bf{Number\:of\:females} & \sf{300} & \sf{380} & \sf{800} & \sf{580} & \sf{290} \end{array}}[/tex]

Now the class intervals are in regular form.

Now, lets creat a complete table of the given data:

[tex]\boxed{\begin{array}{c|c|c|c} \bf{\underline{Age\:group}} & \bf{\underline{No.\:of\:females\:(f_i)}} & \bf{\underline{Class\:mark\:(x_i)}} & \bf{\underline{f_i x_i}} \\\\ \sf{9.5 - 14.5} & \sf{300} & \sf{\dfrac{9.5 + 14.5}{2} = 12} & \sf{3600} \\ \\ \sf{14.5 - 19.5} & \sf{380} & \sf{\dfrac{14.5 + 19.5}{2} = 17} & \sf{6460} \\ \\ \\ \sf{19.5 - 24.5} & \sf{800} & \sf{\dfrac{19.5 + 24.5}{2} = 22} & \sf{17600} \\\\ \sf{24.5 - 29.5} & \sf{580} & \sf{\dfrac{24.5 + 29.5}{2} = 27} & \sf{15660} \\\\ \sf{29.5 - 34.5} & \sf{290} & \sf{\dfrac{29.5 + 34.5}{2} = 32} & \sf{9280} \\\\ \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} & \dfrac{\qquad\qquad}{} \\\\ & \sf{\sum f_i = 2350} & & \sf{\sum f_i x_i = 52600}\end{array}}[/tex]

Now, we have:

• ∑fᵢ = 2350
• ∑fᵢxᵢ = 52600

We already know:

• [tex]\dag\boxed{\rm{\red{Mean = \dfrac{\sum f_i x_i}{\sum f_i}}}}[/tex]

Therefore:

[tex]\sf{Mean = \dfrac{52600}{2350}}[/tex]

[tex] = \sf{Mean = 22.4}[/tex]

∴ The mean of the given data is 22.4.

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