NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.2
Exercise 14.2
Question 1
The following table shows the ages of the patients admitted in a hospital during a year:
Age ( in years) | \(5-15\) | \(15 - 25\) | \(25 - 35\) | \(35 - 45\) | \(45 - 55\) | \(55 - 65\) |
Number of Patients | \(6\) | \(11\) | \(21\) | \(23\) | \(14\) | \(5\) |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution
What is known?
The ages of the patients admitted in a hospital during a year.
What is unknown?
The mode and the mean of the data and their comparison and interpretation.
Reasoning:
We will find the mean by direct method.
Mean,\(\overline x = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\)
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, \(f_2\)
Steps:
To find Mean
We know that,
Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Age (in years) |
Number of patients \(f_i\) |
\(x_i\) |
\(f_i\, x_i\) |
5 – 15 |
6 |
10 |
6 |
15 – 25 |
11 |
20 |
220 |
25 – 35 |
21 |
30 |
630 |
35 – 45 |
23 |
40 |
920 |
45 – 55 |
14 |
50 |
700 |
55 – 65 |
5 |
60 |
300 |
\(\Sigma f_i = 80\) | \(\sum {{f_i}{x_i}} = 2830\) |
From the table it can be observed that,
\[\begin{array}{l}\sum {{f_i} = 80} \\\sum {{f_i}{x_i}} = 2830\end{array}\]
Mean,\(\overline x = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\)
\(\begin{array}{l} = \frac{{2830}}{{80}}\\ = 35.37\end{array}\)
To find mode
We know that,Modal Class is the class with highest frequency
Age (in years) |
Number of patients \(f_i\) |
5 – 15 |
6 |
15 – 25 |
11 |
25 – 35 |
21 |
35 – 45 |
23 |
45 – 55 |
14 |
55 – 65 |
5 |
From the table, it can be observed that the maximum class frequency is \(23,\) belonging to class interval \(35 − 45.\)
Therefore, Modal class \(=35 − 45\)
Class size,\(h=10\)
Lower limit of modal class,\(l=35\)
Frequency of modal class,\(f_1\)
Frequency of class preceding modal class,\(f_0=23\)
Frequency of class succeeding the modal class,\(f_2=14\)
Mode,\( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align}&= 35 + \left( {\frac{{23 - 21}}{{2 \times 23 - 21 - 14}}} \right) \times 10\\ &= 35 + \left( {\frac{2}{{46 - 35}}} \right) \times 10\\ &= 35 + \frac{2}{{11}} \times 10\\&= 35 + 1.8\\&= 36.8\end{align}\]
So the modal age is \(36.82\) years which means maximum patients admitted to the hospital are of age \(36.82\) years .
Mean age is \(35.37\) and average age of the patients admitted is \(35.37\) years.
Question 2
The following data gives information on the observed lifetimes (in hours) of \(225\) electric components
Lifetime (in hours) | \(0 - 20\) | \(20 - 40\) | \(40 - 60\) | \(60 - 80\) | \(80 -100\) | \(100 -120\) |
Frequency | \(10\) | \(35\) | \(52\) | \(61\) | \(38\) | \(29\) |
Determine the modal lifetimes of the components.
Solution
What is known?
The observed lifetimes (in hours) of \(225\) electric components.
The modal lifetimes of the components.
Reasoning:
Modal Class is the class with highest frequency.
\[\begin{align} {\bf{Mode}} = l + \left( \frac{{f_1} - {f_0}}{2{f_1} - {f_0} - {f_2}} \right) \times h\end{align}\]
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,\(f_2\)
Steps:
Lifetime (in hours) |
Frequency |
\(0 – 20\) \(20 – 40\) \(40 – 60\) \(60 – 80\) \(80 – 100\) \(100 – 120\) |
\(10\) \(35\) \(52\) \(61\) \(38\) \(29\) |
From the table, it can be observed that the maximum class frequency is \(61,\) belonging to class interval \(60 − 80\)
Therefore, Modal class\(=60 − 80\)
Class size,\(h=20\)
Lower limit of modal class,\(l=60\)
Frequency of modal class,\(f_1=61\)
Frequency of class preceding modal class,\(f_0=52\)
Frequency of class succeeding the modal class,\(f_2=38\)
Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align}&= 60 + \left( {\frac{{61 - 52}}{{2 \times 61 - 52 - 38}}} \right) \times 20\\&= 60 + \left( {\frac{9}{{122 - 90}}} \right) \times 20\\&= 60 + \frac{9}{{32}} \times 20\\&= 60 + 5.625\\&= 65.625\end{align}\]
Hence,the modal lifetimes of the components are \(65.62\) hours.
Question 3
The following data gives the distribution of total monthly household expenditure of \(200\) families of a village. Find the modal monthly expenditure of the families. Also find the mean monthly expenditure.
Expenditure (in ₹) |
Number of families |
\(1000 – 1500\) \(1500 – 2000\) \(2000 – 2500\) \(2500 – 3000\) \(3000 – 3500\) \(3500 – 4000\) \(4000 – 4500\) \(4500 – 5000\) |
\(24\) \(40\) \(33\) \(28\) \(30\) \(22\) \(16\) \(7\) |
Solution
What is known?
The total monthly household expenditure of \(200\) families of a village.
What is unknown?
The Modal and mean monthly expenditure of the families.
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,\(f_2\)
Steps:
To find mean
We know that,
Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Class size,\(h=500\)
Taking assumed mean, \(a=2750\)
Expenditure (in Rs) | No of families (\(f_i\)) | \(X_i\) | \(u_i = \frac {(x_i - a)} {h}\) | \(f_iu_i\) |
\(1000 - 1500\) | \(24\) | \(1250\) | \(-3\) | \(-72\) |
\(1500 - 2000\) | \(40\) | \(1750\) | \(-2\) | \(-80\) |
\(2000 - 2500\) | \(33\) | \(2250\) | \(-1\) | \(-33\) |
\(2500 -3000\) | \(28\) | \(2750(a)\) | \(0\) | \(0\) |
\(3000 - 3500\) | \(30\) | \(3250\) | \(1\) | \(30\) |
\(3500 - 4000\) | \(22\) | \(3750\) | \(2\) | \(44\) |
\(4000 - 4500\) | \(16\) | \(4250\) | \(3\) | \(48\) |
\(4500 - 5000\) | \(7\) | \(4750\) | \(4\) | \(28\) |
\(\Sigma f_i= 200\) | \(\Sigma f_iu_i= -35\) |
From the table,we obtain
\[\begin{array}{l}\sum {{f_i} = 200} \\\sum {{f_i}{u_i}} = - 35\end{array}\]
\[\begin{align} {\text{ Mean}\,(\overline x )} &= a + \left( \frac{\Sigma f_iu_i}{\Sigma f_i} \right)h\\ & = 2750 + \left( {\frac{{ - 35}}{{200}}} \right) \times 500\\& = \frac{{2750 + }}{{2750 + }}( - 0.175) \times 500\\ &= 2750 + ( - 87.5)\\ &= 2662.50 \end{align}\]
To find mode
Expenditure (in ₹) |
Number of families |
\(1000 – 1500\) \(1500 – 2000\) \(2000 – 2500\) \(2500 – 3000\) \(3000 – 3500\) \(3500 – 4000\) \(4000 – 4500\) \(4500 – 5000\) |
\(24\) \(40\) \(33\) \(28\) \(30\) \(22\) \(16\) \(7\) |
From the table, it can be observed that the maximum class frequency is \(40,\) belonging to class interval \(1500 − 2000\)
Therefore, Modal class\(=1500 − 2000\)
Class size,\(h=500\)
Lower limit of modal class,\(l=1500\)
Frequency of modal class,\(f_1=40\)
Frequency of class preceding modal class,\(f_0=24\)
Frequency of class succeeding the modal class,\(f_2=33\)
\[\begin{align} {\text {Mode}} = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\end{align}\]
\[\begin{align}& = 1500\!+\!\left( {\frac{{40 - 24}}{{2 \times 40\!-\!24\!-33}}} \right)\!\times\!500\\ &= 1500 + \left( {\frac{{16}}{{80 - 57}}} \right) \times 500\\ &= 1500 + \frac{{16}}{{23}} \times 500\\ &= 1500 + 347.83\\ &= 1847.83\end{align}\]
The modal monthly expenditure of the families is \(₹ 1847.83\)
and the mean monthly expenditure of the families is \(₹ 2662.50\)
Question 4
The following data gives the state- wise teacher- student ratio in higher secondary schools of India. Find the mode and mean of the data and interpret the two.
Number of students per teacher |
Number of states / U.T. |
\(15 – 20\) \(20 – 25\) \(25 – 30\) \(30 – 35\) \(35 – 40\) \(40 – 45\) \(45 – 50\) \(50 – 55\) |
\(3\) \(8\) \(9\) \(10\) \(3\) \(0\) \(0\) \(2\) |
Solution
What is known?
The state- wise teacher- student ratio in higher secondary schools of India.
What is unknown?
The mode and mean of the data and their interpretation.
Reasoning:
We will find the mean by direct method.
Mean,\(\overline x = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\)
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,\(f_2\)
Steps:
To find mean
We know that,
Class mark,\({x_i} = \frac{{{\text{Upper class limit }} + {\text{ Lower class limit}}}}{2}\)
Class size,\(h=500\)
Taking assumed mean,\(a=2750\)
Number of students per teacher |
Number of states / U.T. \(f_i\) |
\[{x_i}\] | \[{f_i}{x_i}\] |
15 – 20 |
3 |
17.5 |
52.5 |
20 – 25 |
8 |
22.5 |
180 |
25 – 30 |
9 |
27.5 |
247.5 |
30 – 35 |
10 |
32.5 |
325 |
35 – 40 |
3 |
37.5 |
112.5 |
40 – 45 |
0 |
42.5 |
0 |
45 – 50 |
0 |
47.5 |
0 |
50 – 55 |
2 |
52.5 |
105 |
\[\sum {{f_i}} = 35\] | \[\sum {{f_i}{x_i}} = 1024\] |
Mean, \(\overline x = \frac{{\sum {{f_i}{x_i}} }}{{\sum {{f_i}} }}\)
\[\begin{array}{l}
= \frac{{1024}}{{35}}\\
= 29.26
\end{array}\]
To find mode
Number of students per teacher |
Number of states / U.T. |
\(15 – 20\) \(20 – 25\) \(25 – 30\) \(30 – 35\) \(35 – 40\) \(40 – 45\) \(45 – 50\) \(50 – 55\) |
\(3\) \(8\) \(9\) \(10\) \(3\) \(0\) \(0\) \(2\) |
From the table, it can be observed that the maximum class frequency is \(10,\) belonging to class interval \(30 − 35\)
Therefore, Modal class\(=30 − 35\)
Class size,\(h=5\)
Lower limit of modal class,\(l=30\)
Frequency of modal class,\(f_1=10\)
Frequency of class preceding modal class,\(f_0=9\)
Frequency of class succeeding the modal class,\(f_2=3\)
Mode\( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align} &= 30 + \left( {\frac{{10 - 9}}{{2 \times 10 - 9 - 3}}} \right) \times 5\\&= 30 + \left( {\frac{1}{{20 - 12}}} \right) \times 5\\&= 30 + \frac{5}{8}\\&= 30 + 0.625\\&= 30.625\\&= 30.6\end{align}\]
The modal teacher- student ratio is \(30.6\) and mean teacher- student ratio is \(29.26.\)
Most states/U.T. have a teacher- student ratio of \(30.6\) and on an average the ratio is \(29.26\)
Question 5
The given distribution shows the number of runs scored by some top batsman of the world in one- day international cricket matches.
Runs scored |
Number of batsmen |
\(3000 – 4000\) \(4000 – 5000\) \(5000 – 6000\) \(6000 – 7000\) \(7000 – 8000\) \(8000 – 9000\) \(9000 – 10000\) \(10000 –11000\) |
\(4\) \(18\) \(9\) \(7\) \(6\) \(3\) \(1\) \(1\) |
Find the mode of the data.
Solution
What is known?
The number of runs scored by some top batsman of the world in one- day international cricket matches.
The mode of the data.
Reasoning:
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,\(f_2\)
Steps:
From the table, it can be observed that the maximum class frequency is \(18,\) belonging to class interval \(4000 − 5000\)
Therefore, Modal class\(=4000 − 5000\)
Class size,\(h=1000\)
Lower limit of modal class,\(l=4000\)
Frequency of modal class,\(f_1=18\)
Frequency of class preceding modal class,\(f_0=4\)
Frequency of class succeeding the modal class,\(f_2=9\)
Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align} &= 4000\!+\!\!\left( {\frac{{18 - 4}}{{2 \times 18 - 4 - 9}}} \right)\!\!\times\!\!1000\\ &= 4000 + \left( {\frac{{14}}{{36 - 13}}} \right) \times 1000\\ &= 4000 + \frac{{14}}{{23}} \times 1000\\& = 4000 + 608.695\\ &= 4608.695\\& = 4608.7\end{align}\]
Hence the mode is \(4608.7\)
Question 6
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find mode of the data.
Number of cars | \(0 -10\) | \(10 - 20\) | \(20 - 30\) | \(30 - 40\) | \(40 - 50\) | \(50 - 60\) | \(60 - 70\) | \(70 - 80\) |
Frequency | \(7\) | \(14\) | \(13\) | \(12\) | \(20\) | \(11\) | \(15\) | \(8\) |
Solution
What is known?
The number of cars passing through a spot on a road for \(100\) periods each of \(3\) minutes
The Mode of the data.
Reasoning:
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,\(f_2\)
Steps:
From the table, it can be observed that the maximum class frequency is \(20,\) belonging to class interval \(40 − 50\)
Therefore, Modal class\(=40 − 50\)
Class size,\(h=10\)
Lower limit of modal class,\(l=40\)
Frequency of modal class,\(f_1=20\)
Frequency of class preceding modal class,\(f_0=12\)
Frequency of class succeeding the modal class,\(f_2=11\)
Mode \( = l + \left( {\frac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right) \times h\)
\[\begin{align} &= 40 + \left( {\frac{{20 - 12}}{{2 \times 20 - 12 - 11}}} \right) \times 10\\ &= 40 + \left( {\frac{8}{{40 - 23}}} \right) \times 10\\ &= 40 + \frac{8}{{17}} \times 10\\ &= 40 + 4.705\\ &= 40.705\\ &= 40.7\end{align}\]
Hence the mode is \(40.7\)
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