\displaystyle \textbf{Question 1: }\text{Draw an ogive to represent the following frequency}
\displaystyle \text{distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Class interval} & 0-4 & 5-9 & 10-14 & 15-19 & 20-24 \\ \hline \text{No. of students} & 2 & 6 & 10 & 5 & 3 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The given class intervals are inclusive. Converting them into continuous}
\displaystyle \text{class intervals and finding cumulative frequencies, we get}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Class interval} & \text{Frequency} & \text{Less than} & \text{Cumulative frequency} \\ \hline -0.5-4.5 & 2 & 4.5 & 2 \\ \hline 4.5-9.5 & 6 & 9.5 & 8 \\ \hline 9.5-14.5 & 10 & 14.5 & 18 \\ \hline 14.5-19.5 & 5 & 19.5 & 23 \\ \hline 19.5-24.5 & 3 & 24.5 & 26 \\ \hline \end{array}
\displaystyle \text{For the less-than ogive, plot the points}
\displaystyle (-0.5,0),\ (4.5,2),\ (9.5,8),\ (14.5,18),\ (19.5,23),\ (24.5,26).
\displaystyle \text{Join these points by a smooth curve to obtain the required ogive.}

\displaystyle \textbf{Question 2: }\text{The monthly profits (in Rs.) of }100\text{ shops are distributed}
\displaystyle \text{as follows. Draw the frequency polygon for it.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Profits per shop} & 0-50 & 50-100 & 100-150 & 150-200 & 200-250 & 250-300 \\ \hline \text{No. of shops} & 12 & 18 & 27 & 20 & 17 & 6 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are calculated by }x_i=\frac{\text{lower limit}+\text{upper limit}}{2}.
\displaystyle \begin{array}{|c|c|c|} \hline \text{Profit interval} & \text{Class mark }(x_i) & \text{Frequency }(f_i) \\ \hline 0-50 & 25 & 12 \\ \hline 50-100 & 75 & 18 \\ \hline 100-150 & 125 & 27 \\ \hline 150-200 & 175 & 20 \\ \hline 200-250 & 225 & 17 \\ \hline 250-300 & 275 & 6 \\ \hline \end{array}
\displaystyle \text{To close the polygon, take the imaginary classes }-50-0\text{ and }300-350
\displaystyle \text{with zero frequency. Their class marks are }-25\text{ and }325,\text{ respectively.}
\displaystyle \text{Hence, plot the points}
\displaystyle (-25,0),\ (25,12),\ (75,18),\ (125,27),\ (175,20),\ (225,17),
\displaystyle (275,6),\ (325,0).
\displaystyle \text{Join these points successively by straight line segments to obtain the}
\displaystyle \text{required frequency polygon.}

\displaystyle \textbf{Question 3: }\text{The following distribution gives the daily income of }50
\displaystyle \text{workers of a factory. Convert the above distribution to a less than type}
\displaystyle \text{cumulative frequency distribution and draw its ogive.}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Daily income (in Rs.)} & 100-120 & 120-140 & 140-160 & 160-180 & 180-200 \\ \hline \text{Number of workers} & 12 & 14 & 8 & 6 & 10 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2018]}
\displaystyle \text{Answer:}
\displaystyle \text{The less than type cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Daily income} & \text{Frequency} & \text{Income less than} & \text{Cumulative frequency} \\ \hline 100-120 & 12 & 120 & 12 \\ \hline 120-140 & 14 & 140 & 26 \\ \hline 140-160 & 8 & 160 & 34 \\ \hline 160-180 & 6 & 180 & 40 \\ \hline 180-200 & 10 & 200 & 50 \\ \hline \end{array}
\displaystyle \text{Take the class }80-100\text{ before the first class with zero frequency.}
\displaystyle \text{Hence, the points to be plotted are}
\displaystyle (100,0),\ (120,12),\ (140,26),\ (160,34),\ (180,40),\ (200,50).
\displaystyle \text{Plot the upper class limits on the }X\text{-axis and cumulative frequencies}
\displaystyle \text{on the }Y\text{-axis. Join the points by a smooth curve to obtain the ogive.}

\displaystyle \textbf{Question 4: }\text{The following table gives production yield per hectare of}
\displaystyle \text{wheat of }100\text{ farms of a village. Draw less than ogive and more than ogive.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Production yield in kg per hectare} & 50-55 & 55-60 & 60-65 & 65-70 & 70-75 & 75-80 \\ \hline \text{Number of farms} & 2 & 8 & 12 & 24 & 38 & 16 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2020]}
\displaystyle \text{Answer:}
\displaystyle \text{For the less than ogive, the cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Production yield less than} & \text{Frequency} & \text{Cumulative frequency} \\ \hline 55 & 2 & 2 \\ \hline 60 & 8 & 10 \\ \hline 65 & 12 & 22 \\ \hline 70 & 24 & 46 \\ \hline 75 & 38 & 84 \\ \hline 80 & 16 & 100 \\ \hline \end{array}
\displaystyle \text{Hence, plot the points for the less than ogive:}
\displaystyle (50,0),\ (55,2),\ (60,10),\ (65,22),\ (70,46),\ (75,84),\ (80,100).
\displaystyle \text{For the more than ogive, the cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Production yield more than} & \text{Frequency} & \text{Cumulative frequency} \\ \hline 50 & 2 & 100 \\ \hline 55 & 8 & 98 \\ \hline 60 & 12 & 90 \\ \hline 65 & 24 & 78 \\ \hline 70 & 38 & 54 \\ \hline 75 & 16 & 16 \\ \hline 80 & 0 & 0 \\ \hline \end{array}
\displaystyle \text{Hence, plot the points for the more than ogive:}
\displaystyle (50,100),\ (55,98),\ (60,90),\ (65,78),\ (70,54),\ (75,16),\ (80,0).
\displaystyle \text{Plot both sets of points on the same axes and join each set by a smooth}
\displaystyle \text{curve to obtain the less than ogive and the more than ogive.}

\displaystyle \textbf{Question 5: }\text{During the medical check-up of }35\text{ students of a class,}
\displaystyle \text{their weights were recorded as follows. Draw a less than type ogive for the}
\displaystyle \text{given data. Hence, obtain the median weight from the graph and verify the}
\displaystyle \text{result by using the formula.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Weight (in kg) less than} & 38 & 40 & 42 & 44 & 46 & 48 & 50 & 52 \\ \hline \text{Number of students} & 0 & 3 & 5 & 9 & 14 & 28 & 32 & 35 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{The given data is already a less than type cumulative frequency distribution.}
\displaystyle \text{Hence, the points to be plotted for the less than ogive are}
\displaystyle (38,0),\ (40,3),\ (42,5),\ (44,9),\ (46,14),\ (48,28),
\displaystyle (50,32),\ (52,35).
\displaystyle \text{Plot these points and join them by a smooth curve to obtain the ogive.}
\displaystyle N=35\quad\Rightarrow\quad\frac{N}{2}=\frac{35}{2}=17.5
\displaystyle \text{Mark }17.5\text{ on the }Y\text{-axis and draw a line parallel to the }X\text{-axis}
\displaystyle \text{to meet the ogive. From this point, draw a perpendicular to the }X\text{-axis.}
\displaystyle \text{The corresponding value on the }X\text{-axis is approximately }46.5\text{ kg.}
\displaystyle \therefore\ \text{Median weight from the graph}\approx46.5\text{ kg.}
\displaystyle \text{To verify by formula, first obtain the frequencies by successive differences.}
\displaystyle \begin{array}{|c|c|c|} \hline \text{Weight (in kg)} & \text{Frequency }(f) & \text{Cumulative frequency }(cf) \\ \hline 38-40 & 3 & 3 \\ \hline 40-42 & 2 & 5 \\ \hline 42-44 & 4 & 9 \\ \hline 44-46 & 5 & 14 \\ \hline 46-48 & 14 & 28 \\ \hline 48-50 & 4 & 32 \\ \hline 50-52 & 3 & 35 \\ \hline \end{array}
\displaystyle \frac{N}{2}=17.5,\text{ and the cumulative frequency just greater than }17.5\text{ is }28.
\displaystyle \therefore\ 46-48\text{ is the median class.}
\displaystyle l=46,\quad f=14,\quad F=14,\quad h=2
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =46+\frac{17.5-14}{14}\times2
\displaystyle =46+\frac{3.5}{14}\times2=46+0.5=46.5
\displaystyle \therefore\ \text{Median weight}=46.5\text{ kg.}
\displaystyle \text{Hence, the median obtained from the graph is verified by the formula.}

\displaystyle \textbf{Question 6: }\text{The annual rainfall record of a city for }66\text{ days is given}
\displaystyle \text{in the following table. Calculate the median rainfall using ogives of more than}
\displaystyle \text{type and less than type.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Rainfall (in cm)} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 \\ \hline \text{Number of days} & 22 & 10 & 8 & 15 & 5 & 6 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{For the less than ogive, the cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Rainfall less than (in cm)} & \text{Cumulative frequency} \\ \hline 10 & 22 \\ \hline 20 & 32 \\ \hline 30 & 40 \\ \hline 40 & 55 \\ \hline 50 & 60 \\ \hline 60 & 66 \\ \hline \end{array}
\displaystyle \text{Hence, plot the points for the less than ogive:}
\displaystyle (0,0),\ (10,22),\ (20,32),\ (30,40),\ (40,55),\ (50,60),\ (60,66).
\displaystyle \text{For the more than ogive, the cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Rainfall more than (in cm)} & \text{Cumulative frequency} \\ \hline 0 & 66 \\ \hline 10 & 44 \\ \hline 20 & 34 \\ \hline 30 & 26 \\ \hline 40 & 11 \\ \hline 50 & 6 \\ \hline 60 & 0 \\ \hline \end{array}
\displaystyle \text{Hence, plot the points for the more than ogive:}
\displaystyle (0,66),\ (10,44),\ (20,34),\ (30,26),\ (40,11),\ (50,6),\ (60,0).
\displaystyle \text{Plot both ogives on the same graph. Let them intersect at }P.
\displaystyle N=66\quad\Rightarrow\quad\frac{N}{2}=\frac{66}{2}=33
\displaystyle \text{The }x\text{-coordinate of the point of intersection }P\text{ is approximately }21.25.
\displaystyle \therefore\ \text{Median rainfall}\approx21.25\text{ cm.}
\displaystyle \text{Verification by formula:}
\displaystyle \text{The cumulative frequency just greater than }33\text{ is }40.
\displaystyle \therefore\ 20-30\text{ is the median class.}
\displaystyle l=20,\quad f=8,\quad F=32,\quad h=10
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =20+\frac{33-32}{8}\times10
\displaystyle =20+\frac{10}{8}=21.25
\displaystyle \therefore\ \text{Median rainfall}=21.25\text{ cm.}

\displaystyle \textbf{Question 7: }\text{Change the following distribution to a more than type}
\displaystyle \text{distribution. Hence, draw the more than type ogive for this distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Class interval} & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 & 80-90 \\ \hline \text{Frequency} & 10 & 8 & 12 & 24 & 6 & 25 & 15 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle N=10+8+12+24+6+25+15=100
\displaystyle \text{The more than type cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Class interval} & \text{Cumulative frequency} \\ \hline \text{More than }20 & 100 \\ \hline \text{More than }30 & 100-10=90 \\ \hline \text{More than }40 & 90-8=82 \\ \hline \text{More than }50 & 82-12=70 \\ \hline \text{More than }60 & 70-24=46 \\ \hline \text{More than }70 & 46-6=40 \\ \hline \text{More than }80 & 40-25=15 \\ \hline \text{More than }90 & 15-15=0 \\ \hline \end{array}
\displaystyle \text{Hence, the points to be plotted for the more than ogive are}
\displaystyle (20,100),\ (30,90),\ (40,82),\ (50,70),\ (60,46),\ (70,40),
\displaystyle (80,15),\ (90,0).
\displaystyle \text{Plot the lower class limits on the }X\text{-axis and cumulative frequencies}
\displaystyle \text{on the }Y\text{-axis. Join the points by a smooth curve to obtain the ogive.}

\displaystyle \textbf{Question 8: }\text{The following distribution gives daily income of }50\text{ workers}
\displaystyle \text{of a factory. Convert the distribution above to a less than type cumulative}
\displaystyle \text{frequency distribution and draw its ogive.}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Daily income (in Rs.)} & 200-220 & 220-240 & 240-260 & 260-280 & 280-300 \\ \hline \text{Number of workers} & 12 & 14 & 8 & 6 & 10 \\ \hline \end{array}
\displaystyle \hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle N=12+14+8+6+10=50
\displaystyle \text{The less than type cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Daily income less than (in Rs.)} & \text{Cumulative frequency} \\ \hline 220 & 12 \\ \hline 240 & 12+14=26 \\ \hline 260 & 26+8=34 \\ \hline 280 & 34+6=40 \\ \hline 300 & 40+10=50 \\ \hline \end{array}
\displaystyle \text{Hence, the points to be plotted for the less than ogive are}
\displaystyle (200,0),\ (220,12),\ (240,26),\ (260,34),\ (280,40),\ (300,50).
\displaystyle \text{Plot the upper class limits on the }X\text{-axis and cumulative frequencies}
\displaystyle \text{on the }Y\text{-axis. Join the points by a smooth curve to obtain the ogive.}

\displaystyle \textbf{Question 9: }\text{The following table gives the height of trees. Draw less than}
\displaystyle \text{ogive and more than ogive.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline \text{Height less than} & 7 & 14 & 21 & 28 & 35 & 42 & 49 & 56 \\ \hline \text{Number of trees} & 26 & 57 & 92 & 134 & 216 & 287 & 341 & 360 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The given data itself gives the less than cumulative frequencies.}
\displaystyle \text{Hence, the points for the less than ogive are}
\displaystyle (0,0),\ (7,26),\ (14,57),\ (21,92),\ (28,134),\ (35,216),
\displaystyle (42,287),\ (49,341),\ (56,360).
\displaystyle \text{The class frequencies are obtained by successive differences:}
\displaystyle \begin{array}{|c|c|} \hline \text{Height interval} & \text{Frequency} \\ \hline 0-7 & 26 \\ \hline 7-14 & 57-26=31 \\ \hline 14-21 & 92-57=35 \\ \hline 21-28 & 134-92=42 \\ \hline 28-35 & 216-134=82 \\ \hline 35-42 & 287-216=71 \\ \hline 42-49 & 341-287=54 \\ \hline 49-56 & 360-341=19 \\ \hline \end{array}
\displaystyle \text{Thus, the more than cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Height more than} & \text{Cumulative frequency} \\ \hline 0 & 360 \\ \hline 7 & 334 \\ \hline 14 & 303 \\ \hline 21 & 268 \\ \hline 28 & 226 \\ \hline 35 & 144 \\ \hline 42 & 73 \\ \hline 49 & 19 \\ \hline 56 & 0 \\ \hline \end{array}
\displaystyle \text{Hence, the points for the more than ogive are}
\displaystyle (0,360),\ (7,334),\ (14,303),\ (21,268),\ (28,226),\ (35,144),
\displaystyle (42,73),\ (49,19),\ (56,0).
\displaystyle \text{Plot both sets of points on the same axes and join each set by a smooth curve.}

\displaystyle \textbf{Question 10: }\text{The annual profits earned by }30\text{ shops of a shopping}
\displaystyle \text{complex in a locality give rise to the following distribution. Draw both ogives}
\displaystyle \text{for the above data and hence obtain the median.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Profit (in lakhs of Rs.) more than or equal to} & 5 & 10 & 15 & 20 & 25 & 30 & 35 \\ \hline \text{Number of shops} & 30 & 28 & 16 & 14 & 10 & 7 & 3 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle N=30\quad\Rightarrow\quad\frac{N}{2}=\frac{30}{2}=15
\displaystyle \text{The given data is a more than type cumulative frequency distribution.}
\displaystyle \text{The ordinary frequencies are obtained by successive differences.}
\displaystyle \begin{array}{|c|c|} \hline \text{Profit (in lakhs of Rs.)} & \text{Frequency} \\ \hline 5-10 & 30-28=2 \\ \hline 10-15 & 28-16=12 \\ \hline 15-20 & 16-14=2 \\ \hline 20-25 & 14-10=4 \\ \hline 25-30 & 10-7=3 \\ \hline 30-35 & 7-3=4 \\ \hline 35-40 & 3-0=3 \\ \hline \end{array}
\displaystyle \text{Hence, the less than cumulative frequency distribution is:}
\displaystyle \begin{array}{|c|c|} \hline \text{Profit less than (in lakhs of Rs.)} & \text{Cumulative frequency} \\ \hline 5 & 0 \\ \hline 10 & 2 \\ \hline 15 & 14 \\ \hline 20 & 16 \\ \hline 25 & 20 \\ \hline 30 & 23 \\ \hline 35 & 27 \\ \hline 40 & 30 \\ \hline \end{array}
\displaystyle \text{The points for the less than ogive are}
\displaystyle (5,0),\ (10,2),\ (15,14),\ (20,16),\ (25,20),\ (30,23),
\displaystyle (35,27),\ (40,30).
\displaystyle \text{The points for the more than ogive are}
\displaystyle (5,30),\ (10,28),\ (15,16),\ (20,14),\ (25,10),\ (30,7),
\displaystyle (35,3),\ (40,0).
\displaystyle \text{Plot both ogives on the same graph. Let them intersect at }P.
\displaystyle \text{The point of intersection is approximately }P(17.5,15).
\displaystyle \therefore\ \text{Median annual profit}=Rs.\ 17.5\text{ lakhs.}

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