NCERT Solutions for Class 9 Maths Chapter 14 - Statistics

NCERT Solutions for Class 9 Maths Chapter 14, "Statistics," deals with the concepts related to the collection, organization, analysis, interpretation, and presentation of data. The chapter consists of the following exercises:

Exercise 14.1: This exercise discusses the range of a given data set and the calculation of mean, median, and mode.

Exercise 14.2: This exercise covers cumulative frequency and frequency distribution tables.

Exercise 14.3: This exercise explains the graphical representation of data using histograms and frequency polygons.

Exercise 14.4: This exercise deals with the calculation of cumulative frequency, quartiles, and interquartile range.

Access NCERT Solutions for class 9 Mathematics Chapter 14 – Statistics

Exercise 14.1

1. Give five examples of data that you can collect from day to day life.

Ans: In our day-to-day life, we collect the following data.

i. Number of boys and girls in our classroom.

ii. Rainfall in Mumbai in the last \[{\text{10}}\] months.

iii. The number of hours we study daily.

iv. Number of runs scored by Virat Kohli.

v. Number of times we exercise in a week.

2. Classify the data in Q1 above as primary or secondary data.

Ans: Primary data is information gathered by the investigator himself with a specific goal in mind, whereas secondary data is information gathered from a source that already has the information stored.

The data in \[2\]and \[4\] can be classified as secondary data, while the data in \[1,3\]and \[5\] can be classified as primary data.

Exercise  14.2

1. The blood group of \[30\] students of class VIII are recorded as follows:

A, B, O, O, AB, O, A, O, B, A, O, B, A, O, O, 

A, AB, O, A, A, O, O, AB, B, A, O, B, A, B, O.

Represent this data in the form of a frequency distribution table. Which is the most common, and which is the rarest, blood group among these students?

Ans: It can be observed that \[9\]students have their blood group as A, \[6\]as B, \[3\]as AB and \[12\] as O.

Therefore, the blood group of \[30\] students of the class can be represented as follows.

Hence, from the above table it is clear that the most common blood group is O (maximum number of students) and the rarest blood group is AB (minimum number of students).

2. The distance (in km) of \[40\] engineers from their residence to their place of work were found as follows:

5 3 10 20 25 11 13 7 12 31 

19 10 12 17 18 11 32 17 16 2 

7 9 7 8 3 5 12 15 18 3 

12 14 2 9 6 15 15 7 6 12

Construct a grouped frequency distribution table with class size 5 for the data given above taking the first interval as \[0 - 5\] (\[5\] not included). What main feature do you observe from this tabular representation?

Ans:

The classes in the table are not overlapping. Also, \[36\] out of \[40\] Engineers have their houses below\[{\text{20km}}\]distance.

3. The relative humidity (in %) of a certain city for a month of \[30\] days was as follows:

i. Construct a grouped frequency distribution table with classes \[84 - 86,86 - 88\]

Ans: We need to create a grouped frequency distribution table with a class size of two. \[84 - 86,86 - 88\]and \[88 - 90\].

By observing the data given above, the required table can be constructed as follows:

ii. Which month or season do you think this data is about?

Ans: The relative humidity is high, as can be seen. As a result, the data is for a month during the rainy season.

iii. What is the range of this data?

Ans: Range of data = Maximum value − Minimum value

\[ = 99.2 - 84.9 = 14.3\]

4. The heights of \[50\] students, measured to the nearest centimeters, have been found to be as follows: 

161 150 154 165 168 161 154 162 150 151 

162 164 171 165 158 154 156 172 160 170 

153 159 161 170 162 165 166 168 165 164 

154 152 153 156 158 162 160 161 173 166 

161 159 162 167 168 159 158 153 154 159

i. Represent the data given above by a grouped frequency distribution table, taking the class intervals as \[160 - 165,165 - 170\]etc.

Ans: The class intervals and \[160 - 165,165 - 170\]so on must be taken into account when creating a grouped frequency distribution table. The needed table can be built as follows using the data provided above.

ii. What can you conclude about their heights from the table?

Ans: It is observed that more than \[50\% \] of the students are shorter than \[165{\text{cm}}\].

5. A study was conducted to find out the concentration of sulphur dioxide in the air in parts per million (ppm) of a certain city. The data obtained for \[30\] days is as follows: 

0.03 0.08 0.08 0.09 0.04 0.17 

0.16 0.05 0.02 0.06 0.18 0.20 

0.11 0.08 0.12 0.13 0.22 0.07 

0.08 0.01 0.10 0.06 0.09 0.18 

0.11 0.07 0.05 0.07 0.01 0.04

i. Make a grouped frequency distribution table for this data with class intervals as \[0.00 - 0.04,0.04 - 0.08\] and so on.

ii. For how many days, was the concentration of Sulphur dioxide more than \[0.11\] parts per million?

Ans: The number of days with a SO2 concentration greater than 0.11 is equal to the number of days with a concentration between \[0.12 - 0.16,0.16 - 0.20\] and \[0.20 - 0.24\].

Required number of days \[ = 2 + 4 + 2 = 8\]

Therefore, the concentration of SO2 is more than \[0.11\]ppm, for \[8\]days.

6. Three coins were tossed \[30\] times simultaneously. Each time the number of heads occurring was noted down as follows: 

0 1 2 2 1 2 3 1 3 0 

1 3 1 1 2 2 0 1 2 1 

3 0 0 1 1 2 3 2 2 0 

Prepare a frequency distribution table for the data given above.

Ans: The desired frequency distribution table can be produced as follows by examining the above data.

7.  The value of π up to 50 decimal places is given below: 

3.14159265358979323846264338327950288419716939937510

i. Make a frequency distribution of the digits from \[0\] to \[9\] after the decimal point.

Ans: The required table can be generated as follows by observing the digits following the decimal point:

ii. What are the most and the least frequently occurring digits?

Ans: The table above shows that the lowest frequency is 2 of digit 0, while the highest frequency is 8 of digit 3 and 9. As a result, 3 and 9 are the most often occurring digits, whereas 0 is the least frequently occurring digit.

8. Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows: 

1 6 2 3 5 12 5 8 4 8 

10 3 4 12 2 8 15 1 17 6 

3 2 8 5 9 6 8 7 14 12

i. Make a grouped frequency distribution table for this data, taking class width 5 and one of the class intervals as \[5 - 10.\]

Ans:  Our class interval will be \[0 - 5,5 - 10,10 - 15\]....

The grouped frequency distribution table can be constructed as follows:

ii. How many children watched television for \[15\] or more hours a week?

Ans: The number of children who watched TV for \[15\] hours or more per week is 2 (i.e., the number of children in the \[15 - 20\] class interval).

9. A company manufactures car batteries of a particular type. The lives (in years) of \[40\] such batteries were recorded as follows: 

2.6 3.0 3.7 3.2 2.2 4.1 3.5 4.5 

3.5 2.3 3.2 3.4 3.8 3.2 4.6 3.7 

2.5 4.4 3.4 3.3 2.9 3.0 4.3 2.8 

3.5 3.2 3.9 3.2 3.2 3.1 3.7 3.4 

4.6 3.8 3.2 2.6 3.5 4.2 2.9 3.6 

Construct a grouped frequency distribution table for this data, using class intervals of size 0.5 starting from the intervals \[2 - 2.5\].

Ans: Starting with the class interval 22.5, a grouped frequency table with a class size of 0.5 must be formed.

Therefore, class intervals will be \[2 - 2.5,2.5 - 3,3 - 3.5\]

By observing the data given above, the required grouped frequency distribution table can be constructed as follows:

Exercise  14.3

1. A survey conducted by an organisation for the cause of illness and death among the women between the ages \[15 - 44\] (in years) worldwide, found the following figures (in %)

i. Represent the information given above graphically.

Ans: The graph of the information presented above can be produced as follows by depicting causes on the x-axis and family fatality rate on the y-axis, and selecting an acceptable scale (1 unit = 5% for the y axis).

All the rectangle bars are of the same width and have equal spacing between them.

ii. Which condition is the major cause of women’s ill health and death worldwide?

Ans: Reproductive health issues are the leading cause of women's illness and mortality globally, affecting 31.8% of women.

iii. Try to find out, with the help of your teacher, any two factors which play a major role in the cause in (ii) above being the major cause

Ans: The factors are as follows:

a. Lack of medical facilities

b. Lack of correct knowledge of treatment

2. The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:

i. Represent the information above by a bar graph.

Ans: The graph of the information presented above may be built by choosing an appropriate scale (1 unit = 100 girls for the y-axis) and representing section (variable) on the x-axis and number of girls per thousand boys on the y-axis.

Here, all the rectangle bars are of the same length and have equal spacing in between them.

ii. In the classroom discuss what conclusions can be arrived at from the graph.

Ans: The largest number of females per thousand boys (i.e., 970) is found in ST, while the lowest number of girls per thousand boys (i.e., 910) is found in urban areas.

In addition, the number of females per thousand boys is higher in rural regions than in cities, in backward districts than in non-backward districts, and in SC and ST districts than in non-SC/ST districts.

3. Given below are the seats won by different political parties in the polling outcome of a state assembly elections:

i. Draw a bar graph to represent the polling results.

Ans:

Here, all the rectangle bars are of the same length and have equal spacing in between them.

ii. Which political party won the maximum number of seats?

Ans: From the above graph it is clear that Political party ‘A’ won the maximum number of seats.

4. The length of\[40\] leaves of a plant are measured correct to one millimeter, and the obtained data is represented in the following table:

i. Draw a histogram to represent the given data.

Ans: The length of leaves is represented in a discontinuous class interval with a difference of \[1\] between them, as can be seen. To make the class intervals continuous, \[\dfrac{1}{2} = 0.5\] must be added to each upper class limit and \[0.5\] must be subtracted from the lower class limits.

The above histogram may be built using the length of leaves on the x-axis and the number of leaves on the y-axis.

On the y-axis, one unit symbolises two leaves.

ii. Is there any other suitable graphical representation for the same data?

Ans: Frequency polygon is another good graphical representation of this data.

iii. Is it correct to conclude that the maximum number of leaves are 153 mm long? Why?

Ans: No, because the maximum number of leaves (i.e.\[12\]) has a length of \[144.5{\text{mm}}\] to \[153.5{\text{mm}}\] It is not necessary for all of them to be \[153{\text{mm}}\]long.

5. The following table gives the life times of neon lamps: 

\[\boxed{\begin{array}{*{20}{c}}  {{\text{ Length (in hours) }}}&{{\text{ Number of lamps}}} \\   {300 - 400}&{14} \\   {400 - 500}&{56} \\   {500 - 600}&{60} \\   {600 - 700}&{86} \\   {700 - 800}&{74} \\   {800 - 900}&{62} \\   {900 - 1000}&{48} \\   {}&{} \end{array}}\]

i. Represent the given information with the help of a histogram.

Ans: The histogram of the given data may be produced by plotting the life duration (in hours) of neon lamps on the x-axis and the number of lamps on the y-axis. Here,1

Here, 1 unit on the y-axis represents 10 lamps.

ii. How many lamps have a lifetime of more than \[700\] hours?

Ans: It may be deduced that the number of neon lamps with a lifetime more than \[700\]is equal to the sum of the numbers of neon lamps with lifetimes of \[700,800\]and \[900\]. As a result, there are \[184\] neon bulbs with a lifetime of more than \[700\] hours \[(74 + 62 + 48 = 184)\].

6. The following table gives the distribution of students of two sections according to the mark obtained by them:

Represent the marks of the students of both the sections on the same graph by two frequency polygons. From the two polygons compare the performance of the two sections.

Ans: We can find the class marks of the given class intervals by using the following formula.

\[{\text{Class mark  = }}\dfrac{{{\text{Upper class limit  +  Lower class limit}}}}{2}\]

The frequency polygon can be constructed as follows, with class markings on the x-axis and frequency on the y-axis, and an appropriate scale \[(1{\text{ unit  =  3 for the y - axis}})\].

It can be observed that the performance of students of section ‘A’ is better than the students of section ‘B’ in terms of good marks.

7. The runs scored by two teams A and B on the first 60 balls in a cricket match are given below:

\[\boxed{\begin{array}{*{20}{c}} {{\text{ }}\underline {{\text{Number of balls}}} {\text{ }}}&{{\text{ }}\underline {{\text{Class mark}}} {\text{ }}}&{{\text{ }}\underline {{\text{Team A}}} {\text{ }}}&{{\text{ }}\underline {{\text{Team B}}} {\text{ }}} \\   {0.5 - 6.5}&{3.5}&2&5 \\   {6.5 - 12.5}&{9.5}&1&6 \\   {12.5 - 18.5}&{15.5}&8&2 \\   {18.5 - 24.5}&{21.5}&9&{10} \\   {24.5 - 30.5}&{27.5}&4&5 \\  {30.5 - 36.5}&{33.5}&5&6 \\   {36.5 - 42.5}&{39.5}&6&3 \\  {42.5 - 48.5}&{45.5}&{10}&4 \\   {48.5 - 54.5}&{51.5}&6&8 \\  {54.5 - 60.5}&{57.5}&2&{10} \\   {}&{}&{}&{} \end{array}}\]

Represent the data of both the teams on the same graph by frequency polygons. 

(Hint: First make the class intervals continuous.)

Ans: A frequency polygon can be created by plotting class grades on the x-axis and running times on the y-axis.

8. A random survey of the number of children of various age groups playing in park was found as follows:

\[\boxed{\begin{array}{*{0}{c}}  {\underline {{\text{Age (in years)}}} }&{\underline {{\text{Number of children }}} }      \\   {{\text{1 - 2}}}&{\text{5}} \\   {{\text{2 - 3}}}&{\text{3}} \\   {{\text{3 - 5}}}&{\text{6}} \\ {{\text{5 - 7}}}&{{\text{12}}} \\  {{\text{7 - 10}}}&{\text{9}} \\   {{\text{10 - 15}}}&{{\text{10}}} \\   {{\text{15 - 17}}}&{\text{4}}    \end{array}}\]

Draw a histogram to represent the data above.

Ans:

9. \[100\] surnames were randomly picked up from a local telephone directory and a frequency distribution of the number of letters in the English alphabet in the surnames was found as follows:

i. Draw a histogram to depict the given information.

Ans:

The histogram can be generated using the number of letters on the x-axis and the fraction of the number of surnames per 2 letters interval on the y-axis, as well as an acceptable scale (1 unit = 4 students for the y axis).

ii. Write the class interval in which the maximum number of surnames lie.

Ans: The maximum number of surnames in the class interval is 6-8 since it contains 44 surnames, which is the maximum for this data.

Exercise 14.4

1. The following number of goals was scored by a team in a series of \[10\] matches: 

\[2,3,4,5,0,1,3,3,4,3\] Find the mean, median and mode of these scores.

Ans: The number of goals scored by the team is 

\[2,3,4,5,0,1,3,3,4,3\]

\[{\text{Mean of data  =  }}\dfrac{{{\text{sum of all observations}}}}{{{\text{total number of observations}}}}\]

\[{\text{Mean score  =  }}\dfrac{{2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3}}{{10}}\]

\[{\text{Mean score  =  }}\dfrac{{28}}{{10}} = 2.8\]

\[ = 2.8{\text{ goals}}\]

Arranging the number of goals in ascending order,

\[0,{\text{ }}1,\,{\text{ }}2,{\text{ }}3,{\text{ }}3,{\text{ }}4,{\text{ }}4,\;5\]

The total number of observations is \[10\], an even number. As a result, the median score will be the mean of the \[\dfrac{{10}}{2}\] i.e., \[{5^{{\text{th}}}}\]and \[\dfrac{{10}}{2} + 1\]i.e.,\[{6^{{\text{th}}}}\] observations in ascending or descending order.

\[{\text{Median score  =  }}\dfrac{{{5^{{\text{th}}}}{\text{ observation  +  }}{{\text{6}}^{{\text{th}}}}{\text{ observation}}}}{2}\]

\[ = \dfrac{{3 + 3}}{2} = \dfrac{6}{2} = 3\]

The data mode is the observation having the highest frequency in the data.

As a result, the data's mode score is \[3\] because the data's maximum frequency is \[4\].

2. In a mathematics test given to \[15\] students, the following marks (out of \[100\]) are recorded:  \[41,39,48,52,46,62,54,40,96,52,98,40,42,52,60\] Find the mean, median and mode of this data.

Ans: In a mathematics test, \[15\] students received the following grades:

\[{\text{41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60}}\]

\[{\text{Mean of data  =  }}\dfrac{{{\text{sum of all observations}}}}{{{\text{total number of observations}}}}\]

            \[ = \dfrac{{{\text{41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60}}}}{{15}}\]

            \[ = \dfrac{{{\text{822}}}}{{15}} = 54.8\]

Arranging the scores obtained by \[15\] students in an ascending order,

\[{\text{39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98}}\]

The total number of observations is \[{\text{15}}\], an odd number. As a result, the median score will be \[\dfrac{{{\text{15 + 1}}}}{2} = \] \[{8^{{\text{th}}}}\] observations in ascending or descending order.

Therefore, median score of data \[ = 52\]

The data mode is the observation having the highest frequency in the data.

As a result, the data's mode score is \[52\] because the data's maximum frequency is \[3\].

3. The following observations have been arranged in ascending order. If the median of the data is \[63\], find the value of x. \[{\text{29, 32, 48, 50, x, x  +  2, 72, 78, 84, 95}}\]

Ans: The total number of observations is \[{\text{10}}\], an even number. As a result, the median score will be the mean of \[\dfrac{{10}}{2} + 1\] i.e., \[5\]and \[\dfrac{{10}}{2} + 1\]i.e., \[{6^{{\text{th}}}}\]  observations in ascending or descending order.

Therefore, \[{\text{Median score  =  }}\dfrac{{{5^{{\text{th}}}}{\text{ observation  +  }}{{\text{6}}^{{\text{th}}}}{\text{ observation}}}}{2}\]

= \[63 = \dfrac{{{\text{x + x + 2}}}}{2}\]

= \[63 = \dfrac{{{\text{2x + 2}}}}{2}\]

= \[63 = {\text{x + 1}}\]

\[ \Rightarrow {\text{x  =  62}}\]

4. Find the mode of \[{\text{14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18}}{\text{.}}\]

Ans: Arranging the data in an ascending order,

\[{\text{14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28}}\]

In the presented data, the number \[{\text{14}}\] has the highest frequency, i.e. \[{\text{14}}\], as noticed.

As a result, the given data's mode is \[{\text{14}}\].

5. Find the mean salary of \[{\text{60}}\] workers of a factory from the following table:

Ans: We know that,

\[{\text{Mean  =  }}\dfrac{{\sum\limits_{}^{} {{f_i}{x_i}} }}{{\sum\limits_{}^{} {{f_i}} }}\]

And the value of \[\sum\limits_{}^{} {{f_i}{x_i}} \]and \[\sum\limits_{}^{} {{f_i}} \]can be calculated as follows:

\[{\text{Mean salary  =  }}\dfrac{{305000}}{{60}}\]

Therefore, mean salary of \[60\]workers is Rs. \[5083.33\]

6. Give one example of a situation in which

i. The mean is an appropriate measure of central tendency.

Ans: When a set of data has a few observations that are significantly different from the rest of the data, it is preferable to calculate the median rather than the mean, as the median provides a better approximation of the average in this case.

Consider the following example: the heights of a family's members are represented by the following data.

\[6{\text{154}}{\text{.9 cm, 162}}{\text{.8 cm, 170}}{\text{.6 cm, 158}}{\text{.8 cm, 163}}{\text{.3 cm, 166}}{\text{.8 cm, 160}}{\text{.2 cm}}\]

In this case, it can be observed that the observations in the given data are close to each other. Therefore, the mean will be calculated as an appropriate measure of central tendency.

ii. The mean is not an appropriate measure of central tendency, but the median is an appropriate measure of central tendency.

Ans: The following data represents the marks obtained by \[12\]students in a test.

\[{\text{48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, 99}}\]

It can be seen in this situation that some observations are considerably different from other observations. As a result, the median will be used as a suitable measure of central tendency in this case.

A Brief Overview of Class 9 Maths Chapter 14- Statistics

Statistics refers to the study of data interpretation, analysis, collection, and organisation. In various fields like industrial, scientific, or societal problems, Statistics is conventional and is used to measure statistical population and other model processes. Statistics deal with all the aspects of data, including the planning of the data and its collection to design various experiments and surveys.

In Class 9 Chapter 14, Maths, Mean, Median, and Mode are the three important measures of central tendency. Any data's mean can be calculated using the direct, assumed mean, and step deviation methods.

What are Grouped and Ungrpued Data?

Grouped Data

When raw data are grouped into different classes or categories, it is termed Grouped data.

Ungrouped Data

Ungrouped data is data that is in its raw or original form. These observations are hence not classified into any groups.

Maths Class 9 Chapter 14: Important Definitions

Frequency: 

The total number of times a particular data or observation occurs in the data.

Class Interval: 

All the data in a list/series can be divided and placed into groups or class intervals so that all the observations present in that particular range belong to only that group.

Class Width = Upper-Class limit – Lower Class limit

Mean: 

The average of “n” numbers can be calculated by dividing the sum of all the

numbers by n.

Mode

 It is the most frequently appearing observation. In the case of a class interval, the class

with the highest frequency is the modal class.

Median

The value of observation in the middle is called the median.

If the number of observations is odd (n observations), then median=[(n+1)/2] th observation.

If n is even the median = average of (n/2) th  and [(n+1)/2] th  observation.

Bar Graphs

Bar graphs are the pictorial representation of data. In it, uniform width bars are drawn on an axis (say, X-axis) by keeping equal spacing among all, depicting the variable. On another axis, the second variable is represented, and the heights of the bars depend on this variable. 

Histograms

Histograms are special Bar graphs. They are used for continuous class intervals on an axis (usually X-axis).

Frequency Polygon

Frequency polygon is a very special way of representing data. It is usually shown when a trend is shown. It requires a large set of data. 

A frequency polygon is nothing but data representation using Histograms, connecting the mid-points of upper-side bars by a line segment. The obtained figure is in the shape of a polygon, which is why it is called a Frequency Polygon.

We Cover All The Exercises in The Chapter Statistics Given Below

Kindly Note Solved Example Are Not Copied As Questions Are Good To Go

Extra Questions For Practice

1. Find the mean of the following marks obtained out of  50 by 5 students of Class 9th of a given school.

30, 35, 40, 47, 24

2. Consider the following marks obtained by the 10 student's in the English test as

30,  35,  40,  47,   24, 37, 24, 22, 43, and 33.

Find the highest and lowest marks from the above data.

3. The height of 9 students of a class in cm is given. Find the median of the following data.

155, 160, 145, 149, 150, 147, 152, 144, and 148.

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