NCERT Solutions Class 9 Maths Chapter 12 Statistics

Class 9 chapter 12 statistics deals with information about how data can be presented graphically in the form of bar graphs, histogram and frequency polygons. The NCERT class 9 statistics enables the students to know the collection, analysis, interpretation as well as presentation of data information.

By practicing class 9 maths statistics solutions, students not only strengthen the understanding of the concept being studied but also enhance their problem solving ability. Every ncert class 9 statistics question is described clearly, giving the students an understanding of the concepts used in statistics. Such an exhaustive practice makes sure that learners are prepared for their exams because statistics is an important subject in mathematics which is essential for higher classes.

1.0NCERT Solutions for Class 9 Science Chapter 12 PDF

In the below table, students can find chapter 12 statistics class 9 pdf which can be downloaded so students can access it anytime and anywhere.

2.0NCERT Solutions Class 9 Maths Chapter 12 Statistics: All Exercises

In addition, answering such questions enables a student to discover his/her areas of weakness and be in a position to attend to them long before their examinations. This way the students get the confidence they need to succeed in exams through enhancing their statistics competence.

3.0What Will Students Learn in Chapter 12: Statistics?

• A guide on how to gather and categorize raw information into valuable information.
• Arithmetic mean, median and mode for describing data.
• Methods including bar graphs, histograms and frequency polygon to display data.
• Frequency and relative frequency, making cumulative frequency tables and cumulative frequency graphs.
• Applying statistical methods to solve real problems that may involve data analysis and/or data interpretation.

4.0NCERT Questions with Solutions

Exercise : 12.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

2. Which method did you use for finding the mean, and why?

Sol.

We have, $\mathrm{N}=\Sigma {\mathrm{f}}_{\mathrm{i}}=20$ and $\Sigma {\mathrm{f}}_{\mathrm{i}}=162$. Then mean of the data is $\overline{\mathrm{x}}=\frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}=\frac{1}{20}\times 162=8.1$ Hence, the required mean of the data is 8.1 plants.

We find the mean of the data by direct method because the figures are small.

• Consider the following distribution of daily wages of 50 workers of a factory.

3. Find the mean daily wages of the workers of the factory by using an appropriate method.

Sol.

We have $\sum {\mathrm{f}}_{\mathrm{i}}=50$ and $\sum {\mathrm{f}}_{\mathrm{i}}=27260$ Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{27260}{50}$ $=545.2$

4. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency $f$.

Sol. We may prepare the table as given below :

It is given that mean $=18$. From the table, we have $\mathrm{a}=18,\mathrm{N}=44+\mathrm{f}$ and $\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}=2\mathrm{f}-40$ Now, mean $=\mathrm{a}+\frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{d}}_{\mathrm{i}}$ Then substituting the values as given above, we have $18=18+\frac{1}{44+\mathrm{f}}\times (2\mathrm{f}-40)$ $\Rightarrow 0=\frac{2\mathrm{f}-40}{44+\mathrm{f}}\Rightarrow \mathrm{f}=20$.

5. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{2277}{30}=75.9$.

6. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

7. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Sol.

$\mathrm{a}=57,\mathrm{h}=3,\mathrm{N}=400$ and $\Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=25$. By step deviation method, Mean $=\mathrm{a}+\mathrm{h}\times \frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=57+3\times \frac{1}{400}\times 25$ $=57.19$

• The table below shows the daily expenditure on food of 25 households in a locality.

8. Find the mean daily expenditure on food by a suitable method.

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{{\mathrm{i}}_{\mathrm{i}}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{5275}{25}=211$

9. To find out the concentration of ${\mathrm{SO}}_2$ in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below :

10. Find the mean concentration of ${\mathrm{SO}}_2$ in the air.

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{2.96}{30}=0.0986$

11. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{499}{40}=12.475$ 12. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{x}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{2430}{35}=69.43$

Exercise: 12.2

1. The following table shows the ages of the patients admitted in a hospital during a year :

2. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Sol. From the given data, we have the modal class 35-45. $\{\because$ It has largest frequency among the given classes of the data}

So, $\ell =35,{\mathrm{f}}_{\mathrm{m}}=23,{\mathrm{f}}_1=21,{\mathrm{f}}_2=14$ and $\mathrm{h}=10$

Mode $=\ell +\left\{\frac{{\mathrm{f}}_{\mathrm{m}}-{\mathrm{f}}_1}{2{\mathrm{f}}_{\mathrm{m}}-{\mathrm{f}}_1-{\mathrm{f}}_2}\right\}\times \mathrm{h}$ $=35+\left\{\frac{23-21}{46-21-14}\right\}\times 10=35+\frac{20}{11}$ $=36.8$ years Now, let us find the mean of the data :

$\mathrm{a}=30,\mathrm{h}=10,\mathrm{N}=80$ and $\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=43$ Mean $=\mathrm{a}+\mathrm{h}\times \frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=30+10\times \frac{1}{80}\times 43$ $=30+5.37=35.37$ years Thus, mode $=36.8$ years and mean $=35.37$ years. So, we conclude that the maximum number of patients admitted in the hospital are of the age 36.8 years (approx), whereas on an average the age of a patient admitted to the hospital is 35.37 years.

3. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components : Determine the modal lifetimes of the components. Sol. Modal class of the given data is 60-80. Here, $\ell =60,{\mathrm{f}}_1=61,{\mathrm{f}}_0=52,{\mathrm{f}}_2=38$ and $\mathrm{h}=$

• Mode $=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h}$ $=60+\left\{\frac{61-52}{122-52-38}\right\}\times 20$ $=60+\frac{9\times 20}{32}=60+\frac{45}{8}$ $=60+5.625$ $=65.625$ hours
• 4. 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:

Sol.

Mean $=\frac{\sum {\mathrm{f}}_{\mathrm{i}}{\mathrm{X}}_{\mathrm{i}}}{\sum {\mathrm{f}}_{\mathrm{i}}}=\frac{532500}{200}=2662.5$ Modal class $=1500-2000$ Mode $=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h}$ $=1500+\left\{\frac{40-24}{2\times 40-24-33}\right\}\times 500$ $=1500+\frac{16}{80-57}\times 500=1847.83$.

5. The following distribution gives the statewise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures.

Sol. Modal class is (30-35) and its frequency is 10. So, $\ell =30,{\mathrm{f}}_1=10,{\mathrm{f}}_0=9,{\mathrm{f}}_2=3,\mathrm{h}=5$. Mode $=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h}$ $=30+\left\{\frac{10-9}{20-9-3}\right\}\times 5=30+\frac{5}{8}=30.6$

$\mathrm{a}=32.5,\mathrm{h}=5,\mathrm{N}=35$ and $\Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=-23$. By step-deviation method, Mean $=\mathrm{a}+\mathrm{h}\times \frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}$ $=32.5+5\times \frac{1}{35}\times (-23)$ $=32.5-\frac{23}{7}=32.5-3.3=29.2$ Hence, Mode $=30.6$ and Mean $=29.2$ We conclude that most states/U.T. have a student teacher ratio of 30.6 and on an average, the ratio is 29.2 .

The given distribution shows the number of runs scored by some top batsmen of the world in one day international cricket matches :

6. Find the mode of the data. Sol. Modal class $=4000-5000$ Mode $=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h}$ $=4000+\left\{\frac{18-4}{2\times 18-4-9}\right\}\times 1000$ $=4000+\left\{\frac{14}{23}\right\}\times 1000$ $=4608.69$
7. 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 the mode of the data.

Sol. Modal class $=40-50$ Mode $=40+\left\{\frac{20-12}{2\times 20-12-11}\right\}\times 10$ $=40+\left\{\frac{8}{40-23}\right\}\times 10$ $=40+4.706$ $=44.706$

Exercise: 12.3

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.

Sol. (i)

$\mathrm{n}=68$ gives $\frac{\mathrm{N}}{2}=34$ So, we have the median class (125-145) $\ell =125,\mathrm{N}=68,\mathrm{f}=20,\mathrm{cf}=22,\mathrm{h}=20$ Median $=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h}$ $=125+\left\{\frac{34-22}{20}\right\}\times 20=137$ units. (ii) Modal class is (125 - 145) having maximum frequency ${\mathrm{f}}_1=20,{\mathrm{f}}_0=13,{\mathrm{f}}_2=14$, $\ell =125$ and $\mathrm{h}=20$ $$\begin{gathered} \text{ Mode }=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h} \\ =125+\left\{\frac{20-13}{40-13-14}\right\}\times 20=125+ \end{gathered}$$ $\frac{7\times 20}{13}$ $=125+\frac{140}{13}=125+10.76=135.76\text{ units }$ (iii)

$\mathrm{N}=68,\mathrm{a}=135,\mathrm{h}=20$ and $\Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=7$ By step-deviation method. Mean $=\mathrm{a}+\mathrm{h}\times \frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}$ $=135+20\times \frac{1}{68}\times 7$ $=135+\frac{35}{17}=135+2.05$ $=137.05$ units

• If the median of the distribution given below is 28.5 , find the values of $x$ and $y$.

Sol.

Median $=28.5$ lies in the class-interval (20-30). Then median class is (20-30). So, we have $\ell =20,\mathrm{f}=20,\mathrm{cf}=5+\mathrm{x},\mathrm{h}=10$, $\mathrm{N}=60$ Median $=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h}=28.5$ $28.5=20+\left\{\frac{30-(5+x)}{20}\right\}\times 10$ $\Rightarrow 8.5=\frac{25-\mathrm{x}}{2}\Rightarrow 17=25-\mathrm{x}\Rightarrow \mathrm{x}=8$ Find the given table, we have i.e., $x+y+45=60$ or $x+y=15$ $\Rightarrow y=15-x=15-8=7$, i.e., $y=7$

2, 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.

Sol.

Here, $\ell =35,\mathrm{N}=100,\mathrm{f}=33,\mathrm{cf}=45,\mathrm{h}=5$ Median $=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h}$ $=35+\left\{\frac{50-45}{33}\right\}\times 5$ $=35+\frac{25}{33}=35+0.76$ $=35.76$ years.

3. The length of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table.

4. Find the median length of the leaves.

Sol. The given series is in inclusive form. We may prepare the table in exclusive form and prepare the cumulative frequency table as given below :

Here, $\mathrm{N}=40$ $\Rightarrow \frac{\mathrm{N}}{2}=20$ The cumulative frequency just greater than 20 is 29 and the corresponding class is 144.5-153.5. So, the median class is 144.5-153.5. $\therefore \ell =144.5,\mathrm{N}=40,\mathrm{cf}=17,\mathrm{f}=12$ and $\mathrm{h}=9$ Therefore, median $=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h}$ $=144.5+\frac{(20-17)}{12}\times 9$ $=144.5+\frac{3\times 9}{12}$ $=144.5+2.25=146.75$ Hence, median length of leaves is 146.75 mm .

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

5. Find the median life time of a lamp.

Sol.

$\frac{\mathrm{N}}{2}=\frac{400}{2}=200$ Median class $=3000-3500$ Median $=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h}$ $=3000+\left\{\frac{200-130}{86}\right\}\times 500=3406.98$ hours Hence, median life time of a lamp 3406.98 hours.

• 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:

6. 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.

Sol.

(i) Here,

$\ell =7,\mathrm{N}=100,\mathrm{f}=40,\mathrm{cf}=36,\mathrm{h}=3$

Median $=\ell +\left\{\frac{\frac{N}{2}-cf}{f}\right\}\times h$ $=7+\left\{\frac{50-36}{40}\right\}\times 3$ $=7+\frac{21}{20}=8.05$ (ii) Modal class is (7-10).

$$\begin{gathered} \ell =7,{\mathrm{f}}_1=40,{\mathrm{f}}_0=30,{\mathrm{f}}_2=16,\mathrm{h}=3 \\ \text{ Mode }=\ell +\left\{\frac{{\mathrm{f}}_1-{\mathrm{f}}_0}{2{\mathrm{f}}_1-{\mathrm{f}}_0-{\mathrm{f}}_2}\right\}\times \mathrm{h} \\ =7+\left\{\frac{40-30}{80-30-16}\right\}\times 3 \\ =7+\frac{30}{34}=7.88 \end{gathered}$$ (iii) Here, $\mathrm{a}=8.5,\mathrm{h}=3,\mathrm{n}=100$ and $\Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}=-6$.

Mean $=\mathrm{a}+\mathrm{h}\times \frac{1}{\mathrm{N}}\times \Sigma {\mathrm{f}}_{\mathrm{i}}{\mathrm{u}}_{\mathrm{i}}$ $=8.5+3\times \frac{1}{100}\times (-6)$. $=8.5-\frac{18}{100}=8.5-0.18=8.32$

• The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Sol.

$\frac{\mathrm{N}}{2}=\frac{30}{2}=15$ Median class $=55-60$

$$\begin{gathered} \text{ Median }=\ell +\left\{\frac{\frac{\mathrm{N}}{2}-\mathrm{cf}}{\mathrm{f}}\right\}\times \mathrm{h} \\ =55+\left\{\frac{15-13}{6}\right\}\times 5 \\ =56.67\mathrm{kg} \end{gathered}$$

Exercise: 12.4

1. The following distribution gives the daily income of 50 workers of a factory.

Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.

Sol.

$N=50$ gives $\frac{N}{2}=25$ On the graph, we will plot the points $(120,12),(140,26),(160,34),(180,40)$, $(200,50)$.

• During the medial check up of 35 students of a class, their weights were recorded as follows:

2. Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

Sol.

To draw the 'less than' type ogive, we plot the points $(38,0),(40,3),(42,5),(44,9)$, $(46,14),(48,28),(50,32)$ and $(52,35)$ on the graph.