NCERT Solutions for Class 10 Maths Chapter 14: Statistics - Exercise 14.3

Exercise: 14.3

1. The following frequency distribution gives the monthly consumption of electricity of \[\mathbf{68}\]  consumers of a locality. Find the median, mean and mode of the data and compare them.

Ans: The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is \[32.5\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

To find the class mark for each interval, the following relation is used.

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

Class size (\[h\]) of this data is:

\[h=85-65\]

\[h=20\]

[\[{{d}_{i}},\]\[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] can be calculated according to step deviation method as follows:

It can be observed from the above table

\[\sum{{{f}_{i}}=68}\] 

$\sum{{{f}_{i}}{{u}_{i}}}=7$ 

Class size \[\left( h \right)\text{ }=\text{ }20\] 

Substituting \[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean 

The required mean:

\[\overline{X}=a+\left(\frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}}\right)\times h\]

\[\Rightarrow\overline{X}=135+\left(\frac{7}{68} \right)\times (20)\]

\[\Rightarrow\overline{X}=135+\frac{140}{68}\] 

\[\Rightarrow\overline{X}=137.058\]

Hence, the mean of given data is $137.058$.

For Mode

Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

From the table, it can be noticed that the maximum class frequency is $20$ ,

Belongs to class interval \[125-145\].

Modal class = \[125-145\]

 \[l=125\] 

Class size \[\text{h}=20\] 

\[{{f}_{1}}=20\] 

\[{{f}_{0}}=13\] 

\[{{f}_{2}}=14\] 

Substituting these values in the formula of mode we get:

\[M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h\]

\[\Rightarrow M=125+\left( \frac{20-13}{2(20)-13-14} \right)\times 20\]

\[\Rightarrow  M=125+\frac{7}{13}\times 20\]

\[\Rightarrow  M=135.76\]

Hence, the value of mode is $135.76$

For Median

We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

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

It can be observed from the given table

\[n\text{ }=\text{ }68\] 

$\frac{\text{n}}{2}=34$

Cumulative frequency just greater than $\frac{\text{n}}{2}$ is  $42$ , belonging to

interval \[125-145\].

Therefore, median class = \[125-145\].

\[l=\text{ }125\] 

\[\text{h }=\text{ }20\] 

\[f=20\] 

\[cf=22\] 

Substituting these values in the formula of median we get:

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

\[\Rightarrow  m=125+\left( \frac{34-22}{20} \right)\times 20\]

\[\Rightarrow  m=125+12\]

\[\Rightarrow  m=137\]

Hence, median, mode, mean of the given data is \[137,\text{ }135.76,\] and \[137.05\] respectively.

Mean, mode and median are almost equal in this case.

2. If the median of the distribution is given below is\[\mathbf{28}.\mathbf{5}\], find the values of \[\mathbf{x}\] and \[\mathbf{y}\].

Ans: We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

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

It is given that the value of $\text{n}$ is $60$

From the table, it can be noticed that the cumulative frequency of last entry is \[45+x+y\]

Equating \[45+x+y\] and $\text{n}$, we get:

\[45\text{ }+\text{ }x\text{ }+\text{ }y\text{ }=\text{ }60\] 

\[x\text{ }+\text{ }y\text{ }=\text{ }15\text{ }\]……(1)

It is given that.

Median of the data is given \[28.5\] which lies in interval \[20-30\].

Therefore, median class = \[20-30\]

\[l=\text{ 20}\] 

\[cf=5+x\] 

\[f=\text{ }20\] 

\[h=10\] 

Substituting these values in the formula of median we get:

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

$\Rightarrow 28.5=20+\left( \frac{\frac{60}{2}-(5+x)}{20} \right)\times 10$ 

$\Rightarrow 8.5=\left( \frac{25-x}{2} \right)$ 

$\Rightarrow 17=25-x$ 

$\Rightarrow x=8$ 

Substituting $x=8$ in equation (1), we get:

\[8\text{ }+\text{ }y\text{ }=\text{ }15\] 

\[y\text{ }=\text{ }7\] 

Hence, the values of \[x\] and \[y\] are \[8\]  and \[7\]  respectively.

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

Ans: We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

In this case, class width is not the constant. We are not required to adjust the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age \[18\]  years onwards but less than \[60\]  years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

From the table, it can be observed that \[n\text{ }=\text{ }100\] .

Thus, 

$\frac{\text{n}}{2}=50$

Cumulative frequency (\[cf\]) just greater than $\frac{n}{2}$ is\[78\] ,

belongs interval \[35\text{ }-\text{ }40\] .

Therefore, median class = \[35\text{ }-\text{ }40\]

\[l=35\] 

\[\text{h}=5\] 

\[f=33\] 

\[cf=45\] 

Substituting these values in the formula of median we get:

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

\[\Rightarrow  m=35+\left( \frac{50-45}{33} \right)\times 5\]

\[\Rightarrow m=35+\left( \frac{25}{33} \right)\]

\[\Rightarrow  m=35.76\]

Hence, median age of people who get the policies is \[35.76\] years.

4. The lengths of \[\mathbf{40}\]  leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

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\[\mathbf{117}.\mathbf{5}\text{ }-\text{ }\mathbf{126}.\mathbf{5},\text{ }\mathbf{126}.\mathbf{5}\text{ }-\text{ }\mathbf{135}.\mathbf{5}...\text{ }\mathbf{171}.\mathbf{5}\text{ }-\text{ }\mathbf{180}.\mathbf{5}\])

Ans: We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

The given data does not have continuous class intervals. It can be noticed that the difference between two class intervals is\[1\] . Therefore, we will add $0.5$ in the upper class and subtract $0.5$ in the lower class.

Continuous class intervals with respective cumulative frequencies can be represented as follows.

It can be observed from the given table

\[n=\text{40}\] 

$\frac{\text{n}}{2}=20$

From the table, it can be noticed that the cumulative frequency just greater than

$\frac{n}{2}$ is \[29\] , Belongs to interval $144.5-153.5$ .

median class = $144.5-153.5$

\[l=144.5\] 

\[\text{h}=9\] 

\[f=12\] 

\[cf=17\] 

Substituting these values in the formula of median we get:

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

\[m=144.5+\left( \frac{20-17}{12} \right)\times 9\]

\[m=144.5+\left( \frac{9}{4} \right)\]

\[m=146.75\]

Hence, median length of leaves is \[146.75\] mm.

5. The following table gives the distribution of the life time of \[\mathbf{400}\]  neon lamps:

Find the median life time of a lamp.

Ans: We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows.

It can be observed from the given table

\[n\text{ }=400\] 

$\frac{\text{n}}{2}=200$

It can be observed that the cumulative frequency just greater than

$\frac{n}{2}$ is \[290\] ,Belongs to interval $3000-3500$ .

Median class = $3000-3500$

\[l=3000\] 

\[f=86\] 

\[cf=130\] 

\[h=500\] 

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

\[m=3000+\left( \frac{200-130}{86} \right)\times 500\] 

\[m=3000+\left( \frac{70\times 500}{86} \right)\]

\[m=3406.976\]

Hence, median life time of lamps is \[3406.98\] hours.

6.\[\mathbf{100}\] 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:

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.

Ans: For median

We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows.

It can be observed from the given table

\[n\text{ }=100\] 

$\frac{\text{n}}{2}=50$

It can be noticed that the cumulative frequency just greater than

 $\frac{n}{2}$ is \[76\] , Belongs to interval \[7-10\] .

Median class = \[7-10\]

\[l=7\] 

\[cf=36\] 

\[f=40\] 

\[\text{h}=3\] 

Substituting these values in the formula of median we get:

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

\[m=7+\left( \frac{50-36}{40} \right)\times 3\]

\[m=7+\left( \frac{14\times 3}{40} \right)\]

\[m=8.05\]

Hence, the median number of letters in the surnames is $8.05$.

For Mean

The mean can be found as given below:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)h$

Suppose the assured mean $\left( a \right)$ of the data is \[11.5\].

Class mark \[\left( {{x}_{i}} \right)\] for each interval is calculated as follows:

Class mark \[\left( {{x}_{i}} \right)=\frac{\text{Upper class limit+Lower class limit}}{2}\] 

Class size (\[h\]) of this data is:

\[h=4-1\]

\[h=3\]

\[{{d}_{i}},\]\[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] can be calculated according to step deviation method as follows:

It can be observed from the above table

$\sum{{{f}_{i}}{{u}_{i}}}=-106$

\[\sum{{{f}_{i}}=100}\] 

Substituting \[{{u}_{i}},\]and \[{{f}_{i}}{{u}_{i}}\] in the formula of mean 

The required mean:

$\overline{X}=a+\left( \frac{\sum{{{f}_{i}}{{u}_{i}}}}{\sum{{{f}_{i}}}} \right)\times h$ 

$\overline{X}=11.5+\left( \frac{-106}{100} \right)\times 3$ 

\[\Rightarrow \overline{X}=11.5-3.18\]

\[\Rightarrow \overline{X}=8.32\]

Hence the mean of number of letters in the surnames is $8.32$.

For Mode

Mode can be calculated as

$M=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$

Where

\[l=\] Lower limit of modal class

${{\text{f}}_{1}}=$ Frequency of modal class

\[{{f}_{0}}=\] Frequency of class preceding the modal class

\[{{f}_{2}}=\] Frequency of class succeeding the modal class

\[h=\]Class size

The data in the given table can be written as

From the table, it can be observed that the maximum class frequency is \[40\]

Belongs to \[7-10\] class intervals.

Therefore, modal class = \[7-10\]

\[l=7\] 

\[h=3\] 

\[{{f}_{1}}=40\] 

\[{{f}_{0}}=30\] 

\[{{f}_{2}}=16\] 

Substituting these values in the formula of mode we get:

$m=l+\left( \frac{{{f}_{1}}-{{f}_{0}}}{2{{f}_{1}}-{{f}_{0}}-{{f}_{2}}} \right)\times h$ 

\[m=7+\left( \frac{40-30}{2(40)-30-16} \right)\times 3\] 

\[m=7+\left( \frac{10}{34} \right)\times 3\] 

\[m=7+\frac{30}{34}\]

\[m=7.88\]

Hence, modal size of surname is \[7.88\].

7.The distribution below gives the weights of \[\mathbf{30}\]  students of a class. Find the median weight of the students.

Ans: We can calculate the median as given below:

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

Where

\[l=\] Lower limit of median class

\[h=\] Class size

\[f=\] Frequency of median class

\[cf=\] cumulative frequency of class preceding median class

The cumulative frequencies with their respective class intervals are as follows :

It can be observed from the given table

\[n\text{ }=30\] 

$\frac{\text{n}}{2}=15$

Cumulative frequency just greater than $\frac{n}{2}$ is \[19\], Belongs to class interval \[55-60\].

Median class = \[55-60\]

\[l=55\] 

\[f=6\] 

\[cf=13\] 

\[h=5\] 

Substituting these values in the formula of median we get:

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

$m=55+\left( \frac{15-13}{6} \right)\times 5$ 

$m=55+\left( \frac{10}{6} \right)$ 

$m=56.67$ 

Hence, median weight is \[56.67\] kg.

NCERT Solutions for Class 10 Maths Chapter 14 Exercise 14.3 Free PDF

The notes are available in your PDF format on the official website of Vedantu for free of cost. Students can avail either a physical copy or digital copy without being bothered about the internet connection. The material is accomplished by well-experienced faculty after thorough verification.

Maths NCERT Solutions Class 10 Chapter 14 Exercise 14.3 has explained each question of the exercise in a simple manner so that the students can practice these and solve the problems easily and quickly. Time being also counts for the 10th-grade students during the time of final examinations. Keeping this in mind, students need to practice the number of times thoroughly by understanding the logic and applying techniques.

Class 10 Maths Chapter 14 Exercise 14.3

Class 10 Maths 14.3 Solutions Question 1

The first question of exercise 14.3 Class 10th maths has given data of electricity consumption of different months. Students need to find all the statistical measures like mean median and mode. Also need to correlate them. It is a straightforward question for the students to solve directly as the given intervals are continuous.

Class 10 Maths 14.3 Solutions Question 2

The next question of exercise 14.3 Class 10 ncert solutions consists of certain data with equal intervals. Also, the median value is given. Students need to find the values of X and y, which are missing terms in the table. This has to be done in a step by step procedure. It is also a short answer type of question that can be solved using algebraic expressions.

Class 10 Chapter 14 Exercise 14.3 Question 3

The third question is about a life insurance agent and his policies. The students are required to find the median of the given data. For this question, students need to check the intervals, whether they are continuous or not. As it is a straightforward question, students can use the median and find out the value of a policyholder’s median age.

Class 10th Exercise 14.3 Question Number 4

As the whole exercise is about to practice the median, the fourth sum is also needed to find the median length of leaves. The data is given about different lengths of different leaves. While finding the median of the data, it is necessary to find the value. 

Ex 14.3 Class 10 Maths NCERT Solutions Question 5

The fifth question of Maths Class 10 Statistics ex 14.3 is a simple question that is a short answer type. It is a direct question to find the median of the given data. The data is about the life of neon lamps. The frequency is relatively continuous and cumulative. Hence it is a simple task for the students who had practiced thoroughly with NCERT Solutions of Class 10 Maths Chapter 14.

Class 10 Maths Chapter 14 Exercise 14.3 Solutions Question 6

The sixth question of exercise 14.3 is a long answer type question that needs to find all statistical tools like median, mean, and mode of the given data. It is about surnames available in the directory. Even though the sum is easy and direct, it is lengthy and time-consuming. Students should be attentive while copying the data from the question also.

NCERT Solutions of Class 10 Maths Chapter 14 Question 7

It is the last question for this exercise. It is another short answer type of question. The data has been given, and students are asked to find the median. So students can directly apply the formula and find the median.

NCERT Solutions for Class 10 Maths Chapter 14 Exercises

Key Takeaways of NCERT Solutions Class 10 Maths Chapter 14 Exercise 14.3 free PDF

Students can learn many things and get advantages from PDFs. Students can gain good scores and knowledge, along with other benefits. Few of them are:

• It helps to learn every topic easily and slowly.

• Always strive for students' all-round development.

• Hard copy avoids unnecessary tensions like power fluctuations and internet disconnections.

• Well-experienced mathematicians prepared these PDFs with an updated syllabus.

• Several previous years' question papers were available.

Tips to Ace the Chapter 14 Statistics

Here are a few tips which may help you in understanding the concepts of the chapter in a better way:

• Start studying with a fresh and blank mind. This will help you in learning and understanding the concepts clearly.

• Take proper breaks between your study. This will help you in concentrating better.

• Drink lots of water and keep taking short meals (nuts, fruits and dry fruits).

• Try to avoid distractions.

• Before starting your studies, meditate for at least 15 minutes. This will help you in concentrating better.

• Finish one task at a time. Finish one topic and then move to another.

NCERT Solutions for Class 10 Maths

• Chapter 1 - Real Numbers

• Chapter 2 - Polynomials

• Chapter 3 - Pair of Linear Equations in Two Variables

• Chapter 4 - Quadratic Equations

• Chapter 5 - Arithmetic Progressions

• Chapter 6 - Triangles

• Chapter 7 - Coordinate Geometry

• Chapter 8 - Introduction to Trigonometry

• Chapter 9 - Some Applications of Trigonometry

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Areas Related to Circles

• Chapter 13 - Surface Areas and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability