Ex.14.3 Q1 Statistics Solution - NCERT Maths Class 10

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Question

The following frequency distribution gives the monthly consumption of electricity of \(68\) consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units) Number of consumers
\(65 – 85\) \(4\)
\(85 – 105\) \(5\)
\( 105 – 125\) \(13\)
\(125 – 145\) \(20\)
\(145 – 165\) \(14\)
\(165 – 185\) \(8\)
\(185 – 205\) \(4\)

  

Text Solution

  

What is known?

The frequency distribution of the monthly consumption of electricity of \(68\) consumers of a locality.

What is unknown?

The median, mean and mode of the data and the comparison between them.

Reasoning:

We will find the mean by step-deviation method.

Mean,\(\overline x  = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)

Modal Class is the class with highest frequency

Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)

Where,

Class size,\(h\)

Lower limit of modal class,\(l\)

Frequency of modal class,\(f_1\)

Frequency of class preceding modal class,\(f_0\)

Frequency of class succeeding the modal class,\(f_2\)

Median Class is the class having Cumulative frequency \((cf)\) just greater than \(\frac{n}{2}\)

Median \( = l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\)

Class size,\(h\)

Number of observations,\(n\)

Lower limit of median class,\(l\)

Frequency of median class,\(f\)

Cumulative frequency of class preceding median class,\(cf\)

Steps:

To find the class marks, the following relation is used.

\[\begin{align}\text{Class mark}\,\,{x}_{{i}}=\frac{\text { Upper class limit }+\text { Lower class limit }}{2}\end{align}\] 

Class size , \(h=20\)

 Taking assumed mean,\(a=\) \(135\)

 \( d_i\;u_i\) and \( f_i\;u_i\) are calculated according to step deviation method as follows:

Monthly consumption (in units)

Number of consumers

\[{f_i}\] 

Class mark

\[{x_i}\]

\[{d_i} = {x_i} - a\] \[{u_i} = \frac{{{d_i}}}{h}\] \[{f_i}{u_i}\]
\(65 – 85\) \(4\) \(75\) \(-60\) \(-3\) \(-12\)
\(85 – 105\) \(5\) \(95\) \(-40\) \(-3\) \(-10\)
\( 105 – 125\) \(13\) \(115\) \(-20\) \(-1\) \(-13\)
\(125 – 145\) \(20\) \(135\) \(0\) \(0\) \(0\)
\(145 – 165\) \(14\) \(155\) \(20\) \(1\) \(14\)
\(165 – 185\) \(8\) \(175\) \(40\) \(2\) \(16\)
\(185 – 205\) \(4\) \(195\) \(60\) \(3\) \(12\)
Total \(68\)       \(7\)

From the table, we obtain

\[\begin{array}{l}
\sum {{f_i} = 68} \\
\sum {{f_i}{u_i}}  = 7
\end{array}\]

Mean, \(\overline {\rm{x}}  = a + \left( {\frac{{\sum {{f_i}{u_i}} }}{{\sum {{f_i}} }}} \right) \times h\)

\[\begin{array}{c}
 = 135 + \left( {\frac{7}{{68}}} \right) \times 20\\
 = 135 + \frac{{140}}{{68}}\\
 = 135 + 2.05\\
 = 137.05
\end{array}\]

To find mode

Monthly consumption (in units)

Number of consumers

\(65 – 85\)

\(85 – 105\)

\(105 – 125\)

\(125 – 145\)

\(145 – 165\)

\(165 – 185\)

\(185 – 205\)

\(4\)

\(5\)

\(13\)

\(20\)

\(14\)

\(8\)

\(4\)

From the table, it can be observed that the maximum class frequency is \(20\), belonging to class interval \(125 - 145.\)

Class size (\(h\)) \(=20\)

Modal class =\(125 - 145\)

Lower limit of modal class \(l\)=\(125\)

Frequency  of modal class \((f_1)\)=\(20\)
Frequency  of class preceding modal class \((f_0)\)=\(13\)
Frequency  of class succeeding the modal class \((f_2)\)=\(14\)

Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)

\[\begin{array}{l}
 = 125 + \left( {\frac{{20 - 13}}{{2 \times 20 - 13 - 14}}} \right) \times 20\\
 = 125 + \left( {\frac{7}{{40 - 27}}} \right) \times 20\\
 = 125 + \frac{7}{{13}} \times 20\\
 = 125 + \frac{{140}}{{13}}\\
 = 125 + 10.76\\
 = 135.76
\end{array}\]

To find the median of the given data, cumulative frequency is calculated as follows.

Monthly consumption (in units)

Number of consumers

\(f\)

Cumulative Frequency

\(cf\)

\(65 – 85\) \(4\) \(4\)
\(85 – 105\) \(5\) \(4 + 5 = 9\)
\( 105 – 125\) \(13\) \(9 + 13 = 22\)
\(125 – 145\) \(20\) \(22 + 20 = 42\)
\(145 – 165\) \(14\) \(42 + 14 = 56\)
\(165 – 185\) \(8\) \(56 + 8 = 64\)
\(185 – 205\) \(4\) \( 64 + 4 = 68\)
\(n=68\)

From the table, we obtain

\(n =68\) \(\Rightarrow \frac{{n}}{2}=34\)

Cumulative frequency (\(cf\) )just greater than \(\frac{{n}}{2}\) is \(42,\) belonding to class-interval \(125-145\)

Therefore,median class = \(125-145\)

Class size, \(h\) \(= 20\)

Lower limit  of median class ,\(l\) \(=125\)

Frequency  of median class ,\(f\) \(=20\)

Cumulative frequency  of class preceding median class ,\(cf\)\(=22\)

\[\begin{align}\begin{aligned} \text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=125+\left(\frac{34-22}{20}\right) \times 20 \\ &=125+12 \\ &=137 \end{aligned}\end{align}\]

Therefore, median, mode, mean of the given data is \(137, 135.76, \)and \(137.05\) respectively.The three measures are approximately the same in this case.

  
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