# NCERT Solutions For Class 10 Maths Chapter 14 Exercise 14.2

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## Chapter 14 Ex.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.

## Chapter 14 Ex.14.2 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.

What is unknown?

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.

## Chapter 14 Ex.14.2 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 $$₹ 2662.50$$

## Chapter 14 Ex.14.2 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 $$29.26$$

## Chapter 14 Ex.14.2 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.

What is unknown?

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 $$4608.7$$

## Chapter 14 Ex.14.2 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

What is unknown?

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 $$40.7$$

Ncert Class 10 Exercise 14.2
Ncert Solutions For Class 10 Maths Chapter 14 Exercise 14.2

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