NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.3

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Chapter 14 Ex.14.3 Question 1

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
\(65 – 85\) \(4\)
\(85 – 105\) \(5\)
\( 105 – 125\) \(13\)
\(125 – 145\) \(20\)
\(145 – 165\) \(14\)
\(165 – 185\) \(8\)
\(185 – 205\) \(4\)

Solution

Video 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.

Class mark

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

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

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

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

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

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

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

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 ,\(cf\)\(=22\)

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

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.

Chapter 14 Ex.14.3 Question 2

If the median of the distribution given below is \(28.5\), find the values of \(x\) and \(y.\)

Class interval Frequency
\(0 – 10\) \(5\)
\(10 – 20\) \(x\)
\(20 – 30\) \(20\)
\(30 – 40\) \(15\)
\(40 – 50\) \(y\)
\( 50 – 60\) \(5\)
Total  \(60\)

 

Solution

Video Solution

What is known?

The median of the distribution is \(28.5\)

What is the unknown?

The values of \(x\) and \(y.\)

Reasoning:

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:

The cumulative frequency for the given data is calculated as follows.

Class interval  Frequency Frequency
\(0 – 10\) \(5\) \(5\)
\(20 – 30\) \(x\) \(5 + x\)
\(30 – 40\) \(20\) \(25 + x\)
\(40 – 50\) \(15\) \(40 + x\)
\(40 – 50\) \(y\) \(40 + x + y\)
\( 50 – 60\) \(5\) \(45 +x+y\)
\(n=\) \(60\)

From the table, it can be observed that

\(n =60\)    \(\Rightarrow \frac{n}{2}=30\)

\[\begin{align} \\ {45+x+y}&={60} \\ {x+y} &={15}.............(i) \end{align}\]
Median of the data is given as \(28.5\) which lies in interval \(20 - 30.\)

Therefore, median class \(= 20 - 30\)

Class size (\(h\)) \(= 10\)

Lower limit of median class (\(l\)\(=20\)

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

Cumulative frequency of class preceding the median class, (\(cf\)\(=5 + x\)

\[\begin{align}{\text { Median }}&\!=\!{l\!+\!\left(\frac{\frac{n}{2}\!-\!c f}{f}\right)\!\times\!h} \\ {28.5}&\!=\!{20\!+\!\!\left(\frac{\frac{60}{2}-(5+x)}{20}\right) \!\!\times\!\!10} \\ {8.5}&={\left(\frac{25-x}{2}\right)} \\ {17}&={25-x} \\ {x}&={8}\end{align}\]

Putting \(x=8\) equation, (i)

\[\begin{align} {8+y}&={15} \\ {y}&={7}\end{align}\]

Hence, the values of \(x\) and \(y\) are \(8\) and \(7\) respectively.

Chapter 14 Ex.14.3 Question 3

A life insurance agent found the following data for distribution of ages of \(100\) policy holders. Calculate the median age, if policies are given only to persons having age \(18\)  years onwards but less than \(60\) year.

Age (in years) Number of policy holders
Below \(20\) \(2\)
Below \(25\) \(6\)
Below \(30\) \(24\)
Below \(35\) \(45\)
Below \(40\) \(78\)
Below \(45\) \(89\)
Below \(50\) \(92\)
Below \(55\) \(98\)
Below \(60\) \(100\)

 

Solution

Video Solution

What is known?

The data for distribution of ages of \(100\) policy holders. The policies are given only to persons having age \(18\) years onwards but less than \(60\) years.

What is unknown?

The median age

Reasoning:

Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. 

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:

Class intervals with their respective cumulative frequency can be defined as below.

Age (in years)   Number of policy holders \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) Cumulative frequency (cf)
\(18 – 20\) \(2\) \(2\)
\( 20 – 25\) \(6 - 2 = 4\) \(6\)
\(25 – 30\) \(24 -6 =18\) \(24\)
\(30 – 35\) \(45 - 24 = 21\) \(45\)
\(35 – 40\) \(78 - 45 = 33\) \(78\)
\(40 – 45\) \(89 - 78 =11\) \(89\)
\(45 – 50\) \(92 - 89 = 3\) \(92\)
\( 50 – 55\) \(98 - 92 = 6\) \(98\)
\(55 – 60\) \(100  - 98 = 2\) \(100\)
Total (n)\(= 100\) \(100\)  

From the table, it can be observed that

 \(n =100\)   \(\Rightarrow \frac n{2}=50\)

Cumulative frequency \((cf)\) just greater than \(50\) is \(78, \)belonging to class-interval \(35-40.\)

Therefore, median class \(=35 - 40\)

Class size (\(h\)) \(= 5\)

Lower limit of median class (\(l\)\(=35\)

Frequency of median class (\(f\)\(=33\)

Cumulative frequency of class preceding median class  (\(cf\)\(=45\)

\[\begin{align} \text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=35+\left(\frac{50-45}{33}\right) \times 5 \\ &=35+\frac{25}{33} \\ &=35.76\end{align}\]

Therefore, median age is \(35.76\) years.

Chapter 14 Ex.14.3 Question 4

The lengths of \(40\) leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm) Number or leaves \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\)
\(118 – 126\) \(3\)
\(127 – 135\) \(5\)
\(136 – 144\) \(9\)
\(145 – 153\) \(12\)
\(154 – 162\) \(5\)
\(163 – 171\) \(4\)
\(172 – 180\) \(2\)

Find the median length of the leaves.

Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to \(117.5\!-\!126.5,\!126.5\!-\!135.5\dots\;\!171.5\!-\!180.5\))

Solution

Video Solution

What is known?

The lengths of \(40\) leaves of a plant measured in millimetre.

What is unknown?

The median length of the leaves.

Reasoning:

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:

Length (in mm)   Number or leaves \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\)

Cumulative frequency

\(cf\)

\(117.5 – 126.5\) \(3\) \(3\)
\(126.5 – 135.5\) \(5\) \( 3+5=8\)
\(135.5 - 144.5\) \(9\) \(8+9=17\)
\(144.5 – 153.5\) \(12\) \(17 + 12 = 29\)
\(153.5 - 162.5\) \(5\) \(29 + 5 = 34\)
\(162.5 - 171.5\) \(4\) \(34 + 4 = 38\)
\(171.5 – 180.5\) \(2\) \(38 + 2 = 40\)
\(n=40\)

From the table, it can be observed that

\(n = 40{\rm{    }} \Rightarrow \frac{n}{2} = 20\)

Cumulative frequency \((cf) \) just greater than \(20\) is \(29,\) belonging to class \(144.5 – 153.5\)

Therefore, median class \(=144.5 – 153.5\)

Class size\(, h = 9\)

Lower limit of median class\(, l = 144.5\)

Frequency of median class\(, f = 12\)

Cumulative frequency of class preceding median class\(, cf = 17\)

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

\[\begin{align} &= 144.5 + \left( {\frac{{20 - 17}}{{12}}} \right) \times 9\\ &= 144.5 + \frac{3}{{12}} \times 5\\ &= 144.5 + \frac{5}{4}\\ &= 144.5 + 1.25\\ &= 145.75\end{align}\]

Therefore, median length of leaves is \(146.75 \rm mm.\)

Chapter 14 Ex.14.3 Question 5

The following table gives the distribution of the life time of \(400\) neon lamps:

Life time (in hours) Number of lamps
\(1500 - 2000\) \(14\)
\(2000 - 2500\) \(56\)
\(2500 - 3000\) \(60\)
\(3000 - 3500\) \(86\)
\(3500 - 4000\) \(74\)
\(4000 - 4500\) \(62\)
\(4500 - 5000\) \(48\)

Find the median life time of a lamp.

Solution

Video Solution

What is known?

The life time of \(400\) neon lamps.

What is unknown?

The median life time of a lamp.

Reasoning:

Median Class is the class having Cumulativefrequency \((cf)\) 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:

Life time (in hours) Number of lamps \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) Cumulative frequency
\(1500 - 2000\) \(14\) \(14\)
\(2000 - 2500\) \(56\) \(14 +56 =70\)
\(2500 - 3000\) \(60\) \(70 +60 =130\)
\(3000 - 3500\) \(86\) \(130 +86 =216\)
\(3500 - 4000\) \(74\) \(216 + 74 = 290\)
\(4000 - 4500\) \(62\) \(290 + 62 = 352\)
\(4500 - 5000\) \(48\) \(352 + 48 = 400\)
Total (n) = 400

From the table, it can be observed that

\(n = 400{\rm{    }} \Rightarrow \frac{n}{2} = 200\)

Cumulative frequency \((cf)\) just greater than \(200 \) is \(216,\)belonging to class \(3000 – 3500.\)

Therefore, median class \(=3000 – 3500\)

Class size (\(h\)) \(=500\).

Lower limit (\(l\)) of median class \(=3000\).

Frequency (\(f\)) of median class \(=86\).

Cumulative frequency (\(cf\)) of class preceding median class \(=130\).

\[\begin{align}\begin{aligned} \text {Median} &=l+\left(\frac{\frac{n}{2}-c f}{f}\right) \times h \\ &=\!3000\!+\!\!\left(\!\frac{200-130}{86}\!\right)\!\!\times\!\!500 \\ &=3000+\frac{70 \times 500}{86} \\ &=3406.976 \end{aligned}\end{align}\]

Therefore, median life time of lamps is \(3406.98\) hours.

Chapter 14 Ex.14.3 Question 6

\(100\) surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters \(1 - 4\) \(4 - 7\) \(7 - 10\) \(10 - 13\) \(13 - 16\) \(16 - 19\)
Number of surnames \(6\) \(30\) \(40\) \(16\) \(4\) \(4\)

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

Solution

Video Solution

What is known?

The frequency distribution of the number of letters in the English alphabets for \(100\)  surnames.

What is unknown?

The median and mean number of letters in the surnames and the modal size of the surnames

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 median.

Number of letters    Frequency \(\begin{align}\mathbf{f}_{\mathbf{i}}\end{align}\) Cumulative frequency (\(cf\))
\(1 - 4\) \(6\) \(6\)
\(4 - 7\) \(30\) \(30 +6= 36\)
\(7 - 10\) \(40\) \(36 + 40 = 76\)
\(10 - 13\) \(16\) \(76 + 16 = 92\)
\(13 - 16\) \(4\) \(92 + 4 = 96\)
\(16 - 19\) \(4\) \(96 + 4 = 100\)
Total (n)\(= 100\)

From the table, it can be observed that

\(n = 100{\rm{    }} \Rightarrow \frac{n}{2} = 50\)

Cumulative frequency \((cf)\) just greater than \(50\) is \(76,\) belonging to class \(7 – 10.\)

Therefore, median class \(=7 – 10\)

Class size\(, h = 3\)

Lower limit of median class\(, l = 7\)

Frequency of median class\(, f = 40\)

Cumulative frequency of class preceding median class\(, cf = 36\)

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

\[\begin{array}{l}
 = 7 + \left( {\frac{{50 - 36}}{{40}}} \right) \times 3\\
 = 7 + \frac{{14}}{{40}} \times 3\\
 = 7 + \frac{{21}}{{20}}\\
 = 7 + 1.05\\
 = 8.05
\end{array}\]

To find the mean

Class mark, \({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)

Taking assumed mean\(, a = 11.5\)

Number of letters Number of Surnames \(f_i\) Class mark \(\mathbf{x}_{\mathbf{i}}\) \(\mathbf{d}_{\mathbf{i}}=\mathbf{x}_{\mathbf{i}}-\mathbf{a}\) \(\mathbf{u}_{i}=\frac{\mathbf{d}_{i}}{h}\) \(\mathbf{f}_{\mathbf{i}} \mathbf{u}_{\mathbf{i}}\)
\(1 - 4\) \(6\) \(2.5\) \(– 9\) \(– 3\) \(– 18\)
\(4 - 7\) \(30\) \(5.5\) \(– 6\) \(– 2\) \(– 60\)
\(7 - 10\) \(40\) \(8.5\) \(– 3\) \(– 1\) \(– 40\)
\(10 - 13\) \(16\) \(11.5\) \(0\) \(0\) \(0\)
\(13 - 16\) \(4\) \(14.5\) \(3\) \(1\) \(4\)
\(16 - 19\) \(4\) \(17.5\) \(6\) \(2\) \(8\)
Total \(100\)       \(-106\)

From the table, we obtain

\(\begin{align} \Sigma f_{i} u_{i} &=-106 \\ \Sigma f_{i} &=100\\ \text{Class size,} h &=3 \end{align}\)

\[\begin{align} \operatorname{Mean}, \overline{x} &=a+\left(\frac{\sum f_{i} u_{i}}{\sum f_{i}}\right) h \\ &=11.5+\left(\frac{-106}{100}\right) \times 3 \\ &=11.5-3.18 \\ &=8.32 \end{align}\] 

To find mode

Number of letters Number of Surnames
\(1 - 4\) \(6\)
\(4 - 7\) \(30\)
\(7 - 10\) \(40\)
\(10 - 13\) \(16\)
\(13 - 16\) \(4\)
\(16 - 19\) \(4\)
 \(n\)\(= 100\)

From the table, it can be observed that the maximum class frequency is \(40,\) belonging to class interval \(7 - 10.\)

Class size (\(h\)) \(=3.\)

Modal class \(=7 - 10.\)

Lower limit of modal class (\(l\)\( =7.\)

Frequency of modal class \(f_1\)\(\) \(=40.\)

Frequency of class preceding the modal class \(f_0\) \(= 30.\)

Frequency of class succeeding the modal class, \(f_2\) \(=16.\)

\[\begin{align} \text { Mode } &=l+\left(\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right) \times h \\ &=\!7\!+\!\left[\!\frac{40-30}{2(40)-30-16}\right]\!\!\times\!\!3 \\ &=7+\frac{10}{34}\!\!\times\!3 \\ &=7+\frac{30}{34} \\ &=7.88\end{align}\]

Therefore, median and mean number of letters in surnames is \(8.05\) and \(8.32\) respectively while modal size of surnames is \(7.88.\)

Chapter 14 Ex.14.3 Question 7

The distribution below gives the weights of \(30\) students of a class. Find the median weight of the students.

Weight (in kg) \(40-45\) \(45-50 \) \(50-55\) \(55-60\) \(60-65\) \(65-70\) \(70-75\)
Number of students \(2\) \(3\) \(8\) \(6\) \(6\) \(3\) \(2\)

Solution

Video Solution

What is known?

The weights of \(30\) students of a class.

What is unknown?

The median weight of the students.

Reasoning:

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:

Weight (in kg)

Number of students

\(f\)

Cumulative frequency

\(cf\)

\(40-45\) \(2\) \(2\)
\(45-50 \) \(3\) \(2 + 3 = 5\)
\(50-55\) \(8\) \(5 + 8 = 13\)
\(55-60\) \(6\) \(13 + 6 = 19\)
\(60-65\) \(6\) \(19 + 6 = 25\)
\(65-70\) \(3\) \(25 + 3 = 2\)
\(70-75\) \(2\) \(28 + 2 = 30\)
\(n=30 \)

From the table, it can be observed that

\(n = 30{\rm{    }} \Rightarrow \frac{n}{2} = 15\)

Cumulative frequency \((cf)\) just greater than \(15\) is \(19,\) belonging to class \(55 – 60.\)

Therefore, median class \(=55 – 60\)

Class size\(, h = 5\)

Lower limit of median class\(, l = 55\)

Frequency of median class\(, f = 6\)

Cumulative frequency of class preceding median class, \(cf = 13\)

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

\[\begin{align}
& = 55 + \left( {\frac{{15 - 13}}{6}} \right) \times 5\\
& = 55 + \frac{2}{6} \times 5\\
 &= 55 + \frac{5}{3}\\
 &= 55 + 1.67\\
 &= 56.67
\end{align}\]

Therefore, median weight is \(56.67 \rm\,kg.\)

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