Question

using a graph sheet draw a frequency polygon data
class: 25-35 , 35-45,45-55,55-65,65-75,75-85
frequency 5,8,12,9,6,4
​

Answer 1

Given :

$\boxed{\begin{array}{c|c|c}\bf \: Class \: Interval&\bf \: Mid-value \: of \: class(x) \: &\bf \: Frequency(f)\\\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad}{}&\dfrac{\qquad\qquad\qquad}{}\\\sf25 - 35&\sf 30&\sf5\\\sf35 - 45&\sf40&\sf8\\\sf45 - 55&\sf50&\sf12\\\sf55 - 65&\sf60&\sf9\\\sf65 - 75 &\sf70&\sf6\\\sf75 - 85 &\sf80&\sf4\end{array}}$

To Find :

Frequency Polygon.

Solution :

For drawing a frequency polygon we need to find the

$\\ :\boxed{\bf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 25-35:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 35
• Lower Limit = 25

$\\ :\implies\sf Mid\ Value\ of\ 25-35=\dfrac{35+25}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 25-35=\dfrac{60}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 25-35=\cancel{\dfrac{60}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 25-35=30.}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 35-45:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 45
• Lower Limit = 35

$\\ :\implies\sf Mid\ Value\ of\ 35-45=\dfrac{45+35}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 35-45=\dfrac{80}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 35-45=\cancel{\dfrac{80}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 35-45=40.}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 45-55:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 55
• Lower Limit = 45

$\\ :\implies\sf Mid\ Value\ of\ 45-55=\dfrac{45+55}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 45-55=\dfrac{100}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 45-55=\cancel{\dfrac{100}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 45-55=50.}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 55-65:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 65
• Lower Limit = 55

$\\ :\implies\sf Mid\ Value\ of\ 55-65=\dfrac{65+55}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 55-65=\dfrac{120}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 55-65=\cancel{\dfrac{120}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 55-65=60.}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 65-75:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 75
• Lower Limit = 65

$\\ :\implies\sf Mid\ Value\ of\ 65-75=\dfrac{75+65}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 65-75=\dfrac{140}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 65-75=\cancel{\dfrac{140}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 65-75=70.}$

$\bf\bullet{\underline{\underline{Mid\ Value\ of\ 75-85:-}}}$

$\\ :\implies\sf Mid\ Value\ of\ Class=\dfrac{Upper\ limit+Lower\ limit}{2}$

where,

• Upper Limit = 85
• Lower Limit = 75

$\\ :\implies\sf Mid\ Value\ of\ 75-85=\dfrac{85+75}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 75-85=\dfrac{160}{2}$

$\\ :\implies\sf Mid\ Value\ of\ 75-85=\cancel{\dfrac{160}{2}}$

$\\ \therefore\boxed{\bf Mid\ Value\ of\ 75-85=80.}$

We got all the mid values.

Now, plotting the points (30, 5), (40, 8), (50, 8), (60, 9), (70, 6), (80, 4). Join the plotted points by line segments. The end points (30, 5) and (80, 4) are joined to the mid points (15 – 25) and (85 – 95) respectively with frequency zero.

At the very beginning of the x-axis we have to add a kink because the frequency starts from (25 – 35).

Hence,

The frequency polygon for the above data is given in the attachment.

Answer 2

Answer:

using a graph sheet draw a frequency polygon data

class: 25-35 , 35-45,45-55,55-65,65-75,75-85

frequency 5,8,12,9,6,4