NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Ex 14.3) Exercise 14.3

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What’s there in Chapter 14 Statistics Exercise 14.3 of Class 10 Maths NCERT Solutions?

Statistics is an important branch of maths that is associated with the collection, classification, and representation of any type of data in order to get a better understanding and comprehension. On solving exercise 14.3 of the Class 10 Maths NCERT textbook, the students will get a clear understanding of the statistics. They will learn how to find the mean, mode and median of a grouped data. They will also learn to represent Cumulative Frequency Distribution graphically. Read the article till the end to find out more about class 10 NCERT Solutions chapter 14 statistics exercise 14.3.


Vedantu’s NCERT Solutions of Class 10 Maths Chapter 14 will act as a good company for the Young aspirants to come over their exam with the best performance, because these notes have all the required information in a structured way and explained in a simple language so that the slow learners can also learn easily. Every concept has a set of solved and unsolved questions. It helps the students to get enough practice and command on the concept then and there itself. Students can download NCERT Solution Class 10 Maths and benefit from our study material students can also find Class 10 Science Solutions for free on Vedantu.

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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.

Monthly Consumption (in units)

Number of Consumers

\[65-85\]

\[4\]

\[85-105\]

\[5\]

\[105-125\]

\[13\]

\[125-145\]

\[20\]

\[145-165\]

\[14\]

$165-185$ 

\[8\]

$185-205$

\[4\]

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:

Monthly

consumption

(in units)

Number of Consumers

\[{{f}_{i}}\] 

Class Mark \[{{x}_{i}}\]

${{d}_{i}}={{x}_{i}}-135$ 

${{u}_{i}}=\frac{{{d}_{i}}}{20}$ 

\[{{f}_{i}}{{u}_{i}}\]

\[65-85\] 

\[4\] 

\[75\] 

\[-60\] 

\[-3\] 

\[-12\] 

\[85-105\] 

\[5\]  

\[95\]  

\[-40\] 

\[-2\] 

\[-10\] 

\[105-125\]

\[13\] 

\[115\] 

\[-20\] 

\[-1~\] 

\[-13\]

\[125-145\]

\[20\]

\[135\] 

\[0~\] 

\[0~\] 

\[0~\]

\[145-165\]  

\[14\] 

\[155\] 

\[20\] 

\[1\]  

\[14\]

\[165-185\]

\[8\]

\[175\]

\[40\]

\[2\]

\[16\]

\[185-205\]

\[4\]

\[195\]

\[60\]

\[3\]

\[12\]

Total 

\[68\]




\[7\]

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.

Monthly consumption (in units)

Number of Consumers 

Cumulative Frequency 

\[65-85\] 

\[4\] 

\[4\]

\[85-105\] 

\[5\]  

\[4+5=9\]  

\[105-125\]

\[13\] 

\[9+13=22\] 

\[125-145\]

\[20\]

\[22+20=42\] 

\[145-165\]  

\[14\] 

\[42+14=56\] 

\[165-185\]

\[8\]

\[56+8=64\]

\[185-205\]

\[4\]

\[64+4=68\]

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 \[\mathbf{y}\].

Class interval 

Frequency

\[0-10\]

\[5\]

\[10-20\]

\[X\]

\[20-30\]

\[20\]

\[30-40\]

\[15\]

\[40-50\]

\[Y\]

$50-60$ 

\[5\]

Total

\[60\]

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.

Class interval 

Frequency

Cumulative frequency

\[0-10\]

\[5\]

\[5\]

\[10-20\]

\[X\]

\[5+x\]

\[20-30\]

\[20\]

\[25+x\]

\[30-40\]

\[15\]

\[40+x\]

\[40-50\]

\[Y\]

\[40+x+y\]

$50-60$ 

\[5\]

\[45+x+y\]

Total $(n)$ 

\[60\]


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.

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 $60$

\[100\]

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.

Age (in years) 

Number of Policy Holders $({{f}_{i}})$ 

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 $(n)$ 



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:

Length (in mm)

Number of leaves ${{f}_{i}}$ 

$118-126$ 

\[3\]

$127-135$

\[5\]

$136-144$

\[9\]

$145-153$

\[12\]

$154-162$

\[5\]

$163-171$

\[4\]

$172-180$

\[2\]

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

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.

Length (in mm)

Number of leaves ${{f}_{i}}$ 

Cumulative frequency

$117.5-126.5$ 

\[3\]

\[3\]

$126.5-135.5$

\[5\]

\[3+5=8\]

$135.5-144.5$

\[9\]

\[8+9=17\]

$144.5-153.5$

\[12\]

\[17+12=29\]

$153.5-162.5$

\[5\]

\[29+5=34\]

$162.5-171.5$

\[4\]

\[34+4=38\]

$171.5-180.5$

\[2\]

\[38+2=40\]

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:

Lifetime (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.

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.

Lifetime (in

hours)

Number of

Lamps

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$(n)$ 

\[400\]


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:

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.

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.

Number

of Letters

Number of

Surnames

Cumulative Frequency

$1-4$ 

\[6\]

\[6\]

\[4-7\]

\[30\]

\[30+6=36\]

\[7-10\]

\[40\]

\[36+40=76\]

\[10-13\]

\[16\]

\[76+16=92\]

\[13-16\]

\[4\]

\[92+4=96\]

\[16-19\]

\[4\]

\[96+4=100\]

Total$(n)$ 

\[100\]


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:

Number of letters

Number of surnames

\[{{f}_{i}}\] 


  \[{{x}_{i}}\]

${{d}_{i}}={{x}_{i}}-11.5$ 

${{u}_{i}}=\frac{{{d}_{i}}}{20}$ 

\[{{f}_{i}}{{u}_{i}}\]

$1-4$ 

\[6\]

\[2.5\] 

\[-9\] 

\[-3\] 

\[-18\] 

\[4-7\]

\[30\]

\[5.5\]  

\[-6\] 

\[-2\] 

\[-60\] 

\[7-10\]

\[40\]

\[8.5\] 

\[-3\] 

\[-1~\] 

\[-40\]

\[10-13\]

\[16\]

\[11.5\] 

\[0~\] 

\[0~\] 

\[0~\]

\[13-16\]

\[4\]

\[14.5\] 

\[3\] 

\[1\]  

\[4\]

\[16-19\]

\[4\]

\[17.5\]

\[6\]

\[2\]

\[8\]

Total 

\[100\]




\[-106\]

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

Number

of Letters

Frequency $({{f}_{i}})$ 

$1-4$ 

\[6\]

\[4-7\]

\[30\]

\[7-10\]

\[40\]

\[10-13\]

\[16\]

\[13-16\]

\[4\]

\[16-19\]

\[4\]

Total$(n)$ 

\[100\]

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.

Weight

(in kg)

\[40-45\] 

\[45-50\]  

\[50-55\] 

\[55-60\] 

\[60-65\]

\[65-70\] 

\[70-75\]

Number

of

students

\[2\]

\[3\]

\[8\]

\[6\]

\[6\]

\[3\]

\[2\]

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 :

Weight (in kg)

Number

of Students

Cumulative Frequency

\[40-45\] 

\[2\]

\[2\]

\[45-50\]

\[3\]

\[2+3=5\]

\[50-55\]

\[8\]

\[5+8=13\]

\[55-60\]

\[6\]

\[13+6=19\]

\[60-65\]

\[6\]

\[19+6=25\]

\[65-70\]

\[3\]

\[25+3=28\]

\[70-75\]

\[2\]

\[28+2=30\]

Total$(n)$ 

\[30\]


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

Chapter 14 - Statistics Exercises in PDF Format

Exercise 14.1

9 Questions & Solutions (9 Long Answers)

Exercise 14.2

6 Questions & Solutions (6 Long Answers)

Exercise 14.4

3 Questions & Solutions (3 Long Answers)

 

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

FAQs on NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Ex 14.3) Exercise 14.3

1. What do you Mean by Median, Mean, and Mode?

Mean:-  It refers to average. Mean is the average of a given set of data. We can calculate by dividing the sum of all values by several values.

Mean = (n1 + n2 +........n10)/10.


Median:- Simply, the middle term of the set is known as the set’s median. If it has an even number of terms in the set, the average of two middle terms is the median.


Mode:- The highest repetitive term of the set is called the mode of the set.

2. Explain the Relationship Between Mean, Median, and Mode?

As these three are statistical measures, all are interrelated but can't say dependent all the time. The relation can be understood easily by plotting graphs. Apart from that, after several observations, it is clear that the difference between mean and mode is equal to the thrice of the difference between mean and median. It can be represented as,

Mean – Mode = 3 (Mean – Median).


We will also get a symmetrical curve when we plot the graph while showing the relation between them. It also considers frequency distribution on the other side.

3. Can you please brief me about the Class 10 Maths Exercise 14.3?

Exercise 14.3 of Class 10 Maths basically covers the concepts of mean, median and mode. There are a total of seven questions in Exercise 14.3. All these questions ask for the application of mean, median, and mode. Practising these questions will help you to understand the concept and how to apply them practically. Also, these types of questions are most likely to appear in Board exams as well.

4. How can I score excellent grades in Maths in Class 10 Boards?

Scoring a good set of marks is something every 10th grader dreams off. To achieve the same, it is important to devise a study plan. First, list down the important chapters which hold the maximum weightage of marks. Choose the study materials wisely. Practice the question given in NCERT Solutions. Along with them also revise the concept by practising previous years papers and mock tests. You can refer to the official website of Vedantu for the study materials. They are also available on Vedantu App, and all the resources are available free of cost.

5. How many questions are there in Class 10 Maths Chapter 14 Exercise 14.3?

Exercise 14.3 of Class 10 Maths Chapter 14 'Statistics' consists of seven questions. These questions revolve around the concept of mean, median and mode. The questions require the application of the formulas of mean, median and mode. If you have any doubts or queries regarding the exercise, you can visit Vedantu’s website or download its app to get the solutions to the NCERT Questions. The NCERT Solutions Class 10 Chapter 14 can help you solve the exercises easily.

6. What is Statistics Class 10 Maths?

Statistics Class 10 Maths is a branch of mathematics concerned with the collection, classification, and representation of any type of data to facilitate analysis and comprehension. There are various data representations in statistics which include bar graphs, pie charts, histograms, and frequency polygons. Class 10 Statistics comprises calculating the central tendency of grouped data with the help of three measures, i.e. mean, median and mode.

7. Which topics should I focus on most for the Class 10 Chapter 14 CBSE board exam?

CBSE Class 10 Maths Chapter 14 'Statistics' is an important chapter that can assure you with good grades. This chapter is easy to understand as well as fun to study. The main topics that are covered in this chapter are data analysis and calculation of central tendency of grouped data through the measurement of mean, median and mode. You can expect questions that ask the application of these concepts.

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