The table shows the age distributions of the members of a club. Age (years) 10-14 15-19 20-24 25-29 30-34 35-39 Frequency 7 18 25 17 9 4 (a) Calculate, corr...

Answer Details

(a) Mean age

Take the class mid-value (class-mark) \(x\) of each interval and form \(fx\).

\[ \bar{x}=\frac{\sum fx}{\sum f}=\frac{1835}{80}=22.9375 \]

Mean age \( \approx \mathbf{22.9\ years}\) (to 1 d.p.).

(b)(i) Histogram

First convert each class to its continuous class boundaries, then draw bars whose heights equal the frequencies (the bars touch, since the boundaries are continuous).

(b)(ii) Modal age from the histogram

The tallest bar is the modal class \(20\text{-}24\) (boundaries \(19.5\text{-}24.5\)). To read the mode, join the top-left corner of the modal bar to the top-left corner of the bar after it, and the top-right corner of the modal bar to the top-right corner of the bar before it. From the point where these two lines cross, drop a vertical line to the age axis.

The vertical line meets the axis at approximately \(\mathbf{21.8\ years}\).

This agrees with the calculation \[ \text{Mode}=L+\frac{\Delta_1}{\Delta_1+\Delta_2}\times c =19.5+\frac{25-18}{(25-18)+(25-17)}\times 5 =19.5+\frac{7}{15}\times 5 =19.5+2.3=21.8\ \text{years}. \]

(c) Probability of being in the modal class

The modal class \(20\text{-}24\) contains \(25\) members out of a total of \(80\):

\[ P(\text{modal class})=\frac{25}{80}=\frac{5}{16}=0.3125 \]