In a city,the weekly observations made in a study on the cost of living index are given in the following table:

Cost of living index	Number of weeks
$140-150$	$5$
$150-160$	$10$
$160-170$	$20$
$170-180$	$9$
$180-190$	$6$
$190-200$	$2$
Total	$52$

Draw a frequency polygon for the data above (without constructing a histogram).

(N/A) Since we want to draw a frequency polygon without a histogram,let us find the class-marks of the classes given above,that is of $140-150, 150-160, \dots$
For $140-150$,the upper limit $= 150$ and the lower limit $= 140$.
So,the class-mark $= \frac{150+140}{2} = \frac{290}{2} = 145$.
Continuing in the same manner,we find the class-marks of the other classes as well.
The new table obtained is as follows:

Classes	Class-marks	Frequency
$140-150$	$145$	$5$
$150-160$	$155$	$10$
$160-170$	$165$	$20$
$170-180$	$175$	$9$
$180-190$	$185$	$6$
$190-200$	$195$	$2$

We can now draw a frequency polygon by plotting the class-marks along the horizontal axis,the frequencies along the vertical axis,and then plotting and joining the points $B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6)$ and $G(195, 2)$ by line segments.
We should not forget to plot the point corresponding to the class-mark of the class $130-140$ (just before the lowest class $140-150$) with zero frequency,that is,$A(135, 0)$,and the point $H(205, 0)$ which occurs immediately after $G(195, 2)$.
So,the resultant frequency polygon will be $ABCDEFGH$.