5 Marks Questions

Question

Draw a frequency polygon for the following frequency distribution: Class interval 1-10 11-20 21-30 31-40 41-50 51-60 Frequency 8 3 6 12 2 7

Accepted Answer

The given frequency distribution table is as below:

Class intervals	$1-10$	$11-20$	$21-30$	$31-40$	$41-50$	$51-60$
Frequency	$8$	$3$	$6$	$12$	$2$	$7$

This table has inclusive class intervals and so these are to be converted into exclusive class intervals (i.e true class limits).
These are$ (0.5-10.5), (10.5-20.5), (20.5-30.5), (30.5-40.5), (40.5-50.5)$ and $(50.5-60.5)$ In order to draw a frequency polygon, we need to determine to determine the class marks.
Class marks of a class interval $=\frac{\text{Lower limit}+\text{upper limit}}{2}$ Take imaginary class interval $(-9.5-0.5)$ at the beginning and $(60.5-70.5)$ at the end, each with frequency zero. So, we have the following table

Class intervals	True class intervals	class marks	Frequency
$(-9)-0$	$(-9.5)-0.5$	$-4.5$	$0$
$1-10$	$0.5-10.5$	$5.5$	$8$
$11-20$	$10.5-20.5$	$15.5$	$3$
$21-30$	$20.5-30.5$	$25.5$	$6$
$31-40$	$30.5-40.5$	$35.5$	$12$
$41-50$	$40.5-50.5$	$45.5$	$2$
$51-60$	$50.5-60.5$	$55.5$	$7$
$61-70$	$60.5-70.5$	$65.5$	$0$

Now, take class marks along $x$-axis and their corresponding frequency along $y$-axis. Mark the points ans join them. Thus, we obtain a complete frequency polygon as shown below:

Political party	$A$	$B$	$C$	$D$	$E$	$F$
Seats won	$75$	$55$	$37$	$29$	$10$	$37$

Age $($in years$)$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Number of patients	$90$	$50$	$60$	$80$	$50$	$30$

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Class interval	$1-10$	$11-20$	$21-30$	$31-40$	$41-50$	$51-60$
Frequency	$8$	$3$	$6$	$12$	$2$	$7$

Bar Graph, Histogram and Frequency Polygon — Maths STD 9 — Question

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