Question

The height of trees in a forest is given below. Draw a histogram to represent
the data.
Height: 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55
No.of trees: 10, 15, 25, 30, 45, 50, 35, 20

Answer 1

Answer:

$\large\begin{tabular}{|c|c|}\cline{1-2}\sf Height &\sf No. of Trees\\\cline{1-2}16 - 20 &10\\21 - 25&15 \\26 - 30&25 \\31 - 35&30 \\36 - 40&45 \\41 - 45&50 \\46 - 50&35 \\51 - 55&20 \\\cline{1-2}\end{tabular}$

Whenever we draw a histogram Gaps between the Class Interval should be equal.

But in this Question Class Interval are discontinuous, so we will make it continuous first.

⠀

⇒ Adjustment Factor = ½[Lower limit of a class Interval - Upper limit of the preceding class interval]

⇒ Adjustment Factor = ½[21 - 20]

⇒ Adjustment Factor = ½ × 1

⇒ Adjustment Factor = 0.5

⠀

Now we will Subtract 0.5 from each lower limit and Add 0.5 in each upper limit, Hence New Data :

$\rule{200}{1}$

$\large\begin{tabular}{|c|c|}\cline{1-2}\sf Height &\sf No. of Trees\\\cline{1-2}15.5 - 20.5 &10\\20.5 - 25.5&15 \\25.5 - 30.5&25 \\30.5 - 35.5&30 \\35.5 - 40.5&45 \\40.5 - 45.5&50 \\45.5 - 50.5&35 \\50.5 - 55.5&20 \\\cline{1-2}\end{tabular}$

• X axis = Height ; 5 cm each
• Y axis = No. of tress ; 5 cm each

$\setlength{\unitlength}{9mm}\begin{picture}\thicklines\put(8,1){\vector(1,0){8.2}}\put(8,.9){\vector(0,1){7.5}}\multiput(7.9,1)(0,.6){12}{\line(1,0){.2}}\multiput(9.2,.9)(.8,0){9}{\line(0,1){.2}}\put(9.2,1){\dashbox{0.01}(.8,1.2)}\put(10,1){\dashbox{0.01}(.8,1.8)}\put(10.8,1){\dashbox{0.01}(.8,2.9)}\put(11.6,1){\dashbox{0.01}(.8,3.5)}\put(12.4,1){\dashbox{0.01}(.8,5.3)}\put(13.2,1){\dashbox{0.01}(.8,5.9)}\put(14,1){\dashbox{0.01}(.8,4.2)}\put(14.8,1){\dashbox{0.01}(.8,2.4)}}\put(7.6,0.8){\sf0}\put(7.6,1.4){\sf5}\put(7.4,2.1){\sf10}\put(7.4,2.7){\sf15}\put(7.4,3.3){\sf20}\put(7.4,3.9){\sf25}\put(7.4,4.5){\sf30}\put(7.4,5.1){\sf35}\put(7.4,5.7){\sf40}\put(7.4,6.3){\sf45}\put(7.4,6.9){\sf50}\put(7.4,7.5){\sf55}\put(9,.6){\small\sf15.5}\put(9.8,.6){\small\sf20.5}\put(10.6,.6){\small\sf25.5}\put(11.4,.6){\small\sf30.5}\put(12.2,.6){\small\sf35.5}\put(13,.6){\small\sf40.5}\put(13.8,.6){\small\sf45.5}\put(14.6,.6){\small\sf50.5}\put(15.4,.6){\small\sf55.5}\put(10,0){\vector(1,0){4}}\put(11.5,-.5){\sf Height}\put(7,2){\vector(0,1){4}}\put(6.5,4.8){\sf T}\put(6.55,4.5){\sf r}\put(6.52,4.2){\sf e}\put(6.52,3.9){\sf e}\put(6.52,3.6){\sf s}\end{picture}$

Answer 2

${\colorbox{cyan}{✿QuesTion}}$

The height of trees in a forest is given below. Draw a histogram to represent

the data.

Height: 16-20, 21-25, 26-30, 31-35, 36-40, 41-45, 46-50, 51-55

No.of trees: 10, 15, 25, 30, 45, 50, 35, 20

${\colorbox{pink}{✿ AnsWer}}$

${\boxed{\underline{\tt{ \orange{Required \: answer \: :-}}}}}$

In this type of question, The given class interval is discontinuous (Inclusive form). So, we must convert this class interval in continuous form.

In Histogram, the bars should be continuous placed without any space.

Therefore, we should make the class interval continuous.

FOR THIS,WE NEED :-

$: \longrightarrow \: { \boxed{ \boxed{ \underline{ \red{adjustment \: factor}}}}}$

${ \boxed{ adjustment \: factor = \frac{1}{2}(lower \: limit \: of \: class \: interval) - (upper \: limit \: of \: the \: preceding \: class \: interval)}}$

$: \longmapsto(21 - 20)$

$: \longmapsto{ \blue {\tt{0.5}}}$

IN THE ABOVE CLASS INTERVAL, WE SUBTRACT 0.5 FROM LOWER LIMIT AND ADD 0.5 IN UPPER LIMIT.

REWRITING THE FOLLOWING CLASS TABLE:-

⑅ HEIGHTS IN METRES:-

• 15.5 - 20.5

• 20.5 - 25.5

• 25.5 - 30.5

• 30.5 - 35.5

• 35.5 - 40.5

• 40.5 - 45.5

• 45.5 - 50.5

• 50.5 - 55.5

⑅ NUMBER OF TREES:-

• 10

• 15

• 25

• 30

• 45

• 50

• 35

• 20

NOW,

THE ABOVE TABLE HAS BECAME CONTINUOUS FREQUENCY DISTRIBUTION TABLE.

THE HISTOGRAM FOR THIS QUESTION IS IN THE

[ATTACHMENT].