# Statistics

## Class 9 NCERT Maths

### NCERT

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

##### Solution :

In our day to day life, we can collect the following data.$\\$ 1. Number of females per $1000$ males in various states of our country$\\$ 2. Weights of students of our class$\\$ 3. Production of wheat in the last $5$ years in our country$\\$ 4. Number of plants in our locality$\\$ 5. Rainfall in our city in the last $12$ months.

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

##### Solution :

The given data is very large. So, we construct a group frequency of class size $5$.$\\$ Therefore the class interval will be $0-5,5-10,10-15$ and so on.$\\$ The data is represented in table as:$\\$ The classes in the table are not overlapping. Also, $36$ out of $40$ Engineers have their houses below $20km$ distance.

5   The relative humidity (in %) of a certain city for a month of 30 days was as follows:$\\$ $98.1 \ 98.6 \ 99.2 \ 90.3 \ 86.5 \ 95.3 \ 92.9 \ 96.3 \ 94.2 \ 95.1$$\\ 89.2 \ 92.3 \ 97.1 \ 93.5 \ 92.7 \ 95.1 \ 97.2 \ 93.3 \ 95.2 \ 97.3$$\\$ $96.2 \ 92.1 \ 84.9 \ 90.2 \ 95.7 \ 98.3 \ 97.3 \ 96.1 \ 92.1 \ 89$$\\ (i) Construct a grouped frequency distribution table with classes 84-86, 86-88$$\\$ (ii) Which month or season do you think this data is about?$\\$ (iii) What is the range of this data?

##### Solution :

(i) A grouped frequency distribution table of class size $2$ has to be constructed. The class intervals will be $84-86, 86-88,$ and $88-90, ...$ By observing the data given above, the required table can be constructed as follows:$\\$ (ii) It can be observed that the relative humidity is high. Therefore, the data is about a month of rainy season.$\\$ (iii) Range of data = Maximum value - Minimum value$\\$ $= 99.2 - 84.9 = 14.3$

6   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.$\\$ (ii) What can you conclude bout their heights from the table?

##### Solution :

(i) A grouped frequency distribution table has to be constructed taking class intervals $160-165, 165-170,$ etc. By observing the data given above, the required table can be constructed as follows.$\\$ $\begin{array}{|c|c|} \hline \text{Height (in cm)} & \text{Number of students(frequency)}\\ \hline 150-155 & 12 \\ \hline 155-160 &9 \\ \hline 160 -165 & 14 \\ \hline 165-170 & 10 \\ \hline 1700-175 & 5 \\ \hline \text{Total} & 50 \\ \hline \end{array}$ $\\$ (ii)It can be concluded that more than $50\%$ of the students are shorter than $165 cm.$

7   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?

##### Solution :

Taking class intervals as $0.00-0.04, 0.04-0.08,$ and so on, a grouped frequency table can be constructed as follows :$\\$ $\begin{array}{|c|c|}\hline \text{Concentration of SO_2(in ppm)} & \text{Number of days(frequency)}\\ \hline 0.00-0.04 & 4 \\ \hline 0.04-0.08 & 9 \\ \hline 0.08-0.12 & 9 \\ \hline 0.12 - 0.16 & 2\\ \hline 0.16-0.20 & 4 \\ \hline 0.20 -0.24 & 2 \\ \hline \text{Total} & 30 \\ \hline \end{array}$ $\\$ The number of days for which the concentration of $SO _2$ is more than $0.11$is the number of days for which the concentration is in between$0.12-0.16, 0.16-0.20, 0.20-0.24.$ $\\$ Required number of days =$2 + 4 + 2 = 8$ $\\$ Therefore, for $8$ days, the concentration of $SO_2$ is more than $0.11 ppm.$

8   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

##### Solution :

By observing the data given above, the required frequency distribution table can be constructed as follows.$\\$ $\begin{array}{|c|c|} \hline \text{Number of head} & \text{Number of times(frequency )} \\ \hline 0 & 6 \\ \hline 1 & 10 \\ \hline 2 & 9 \\ \hline 3 & 5 \\ \hline \text{Total} & 30 \\ \hline \end{array}$ $\\$

9   The value of $\pi$ 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.$\\$ (ii) What are the most and the least frequently occurring digits?

##### Solution :

(i)By observation of the digits after decimal point, the required table can be constructed as follows:$\\$ $\begin{array}{|c|c|}\hline \text{Digit } & \text{Frequency}\\ \hline 0 & 2 \\ \hline 1 & 5 \\ \hline 2 & 5 \\ \hline 3 & 8 \\ \hline 4 & 4 \\ \hline 5 & 5 \\ \hline 6 & 4 \\ \hline 7 & 4 \\ \hline 8 & 5 \\ \hline 9 & 8 \\ \hline \text{Total} & 50 \\ \hline \end{array}$ $\\$ (ii) It can be observed from the above table that the least frequency is $2$ of digit $0$, and the maximum frequency is $8$ of digit $3$ and $9$. Therefore, the most frequently occurring digits are $3$ and $9$ and the least frequently occurring digit is $0.$

10   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 $5-10.$ $\\$ (ii) How many children watched television for $15$ or more hours a week?

##### Solution :

(i) Our class intervals will be $0-5, 5-10, 10-15, ...$ $\\$ The grouped frequency distribution table can be constructed as follows:$\\$ $\begin{array}{|c|c|} \hline \text{Hours } & \text{Number of children}\\ \hline 0-5 & 10 \\ \hline 5-10 & 13 \\ \hline 10 -15 & 5 \\ \hline 15 -20 & 2 \\ \hline \text{Total} & 30 \\ \hline \end{array}$ $\\$ (ii) The number of children who watched TV for $15$ or more hours a week is $2$ (i.e., the number of children in class interval $15-20)$.

11   A company manufactures car batteries of a particular type. The lives (in years) of $40$ such batteries were recorded as follows: $\\$ $2.6 \quad 3.0 \quad 3.7 \quad 3.2\quad 2.2\quad 4.1\quad 3.5\quad 4.5$ $\\$ $3.5 \quad 2.3\quad 3.2\quad 3.4\quad 3.8\quad 3.2\quad 4.6\quad 3.7\quad$ $\\$ $2.5\quad 4.4\quad 3.4\quad 3.3\quad 2.9 \quad 3.0\quad 4.3\quad 2.8\quad$ $\\$ $3.5\quad 3.2\quad 3.9\quad 3.2\quad 3.2\quad 3.1\quad 3.7\quad 3.4$ $\\$ $4.6\quad 3.8\quad 3.2\quad 2.6\quad 3.5\quad 4.2\quad 2.9\quad 3.6$ $\\$ Construct a grouped frequency distribution table for this data, using class intervals of size $0.5$ starting from the intervals $2-2.5.$

##### Solution :

A grouped frequency table of class size 0.5 has to be constructed, starting from class interval $2-2.5.$ $\\$ Therefore, the 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 :$\\$ $\begin{array}{|c|c|} \hline \text{Lives of batteries(in hours)}& \text{Number of batteries }\\ \hline 2-2.5 & 2 \\ \hline 2.5-3.0 & 6 \\ \hline 3.0-3.5 & 14 \\ \hline 3.5 - 4.0 & 11 \\ \hline 4.0-4.5 & 4 \\ \hline 4.5 -5.0 & 3 \\ \hline \text{Total}& 40 \\ \hline \end{array}$

12   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 $\%$):$\\$ $\begin{array}{|c|c|} \hline \text{S.No.} & \text{Causes} & \text{Female fatality rate (\%)} \\ \hline 1.& \text{Reproductive health conditions} & 31.8 \\ \hline 2.& \text{Neuropsychiatric conditions }& 25.4 \\ \hline 3.& \text{Injuries} & 12.4 \\ \hline 4.& \text{Cardiovascular conditions }& 4.3 \\ \hline 5.& \text{Respiratory conditions} & 4.1 \\ \hline 6. & \text{Other causes} & 22.0 \\ \hline \end{array}$ $\\$ (i) Represent the information given above graphically.$\\$ (ii) Which condition is the major cause of women’s ill health and death worldwide?$\\$ (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.

##### Solution :

(i) By representing causes on x-axis and family fatality rate on y-axis and choosing an appropriate scale ($1$ unit = $5\%$ for y axis), the graph of the information given above can be constructed as follows.

All the rectangle bars are of the same width and have equal spacing between them.$\\$ (ii) Reproductive health condition is the major cause of women’s ill health and death worldwide as $31.8\%$ of women are affected by it.$\\$ (iii) The factors are as follows.$\\$ 1. Lack of medical facilities$\\$ 2. Lack of correct knowledge of treatment$\\$

13   The following data on the number of girls (to the nearest ten) per thousand boys in different sections of Indian society is given below:$\\$ $\begin{array}{|c|c|}\hline \text{Section}& \text{Number of girls per thousand boys} \\ \hline \text{Scheduled Caste (SC)} & 940 \\ \hline \text{Scheduled Tribe(ST)}& 970 \\ \hline \text{Non SC/ST}& 920 \\ \hline \text{Backward districts}& 950 \\ \hline \text{Non-Backward districts }& 920 \\ \hline \text{Rular} & 930 \\ \hline \text{Urban}& 910 \\ \hline \end{array}$ $\\$ (i) Represent the information above by a bar graph.$\\$ (ii) In the classroom discuss what conclusions can be arrived at from the graph.

##### Solution :

(i) By representing section (variable) on x-axis and number of girls per thousand boys on y-axis, the graph of the information given above can be constructed by choosing an appropriate scale ($1$ unit = $100$ girls for y-axis)

Here, all the rectangle bars are of the same length and have equal spacing in between them.$\\$ (ii) It can be observed that maximum number of girls per thousand boys (i.e., $970$) is for $ST$ and minimum number of girls per thousand boys (i.e., $910$) is for urban.$\\$ Also, the number of girls per thousand boys is greater in rural areas than that in urban areas, backward districts than that in non-backward districts, $SC$ and $ST$ than that in non-$SC/ST.$

14   Given below are the seats won by different political parties in the polling outcome of a state assembly elections: $\\$ $\begin{array}{|c|c|}\hline \text{Political Party}& A & B & C & D & E & F \\ \hline \text{Seat Won} & 75 & 55 & 37 & 29 7 10 & 37 \\ \hline \end{array}$ $\\$ (i) Draw a bar graph to represent the polling results.$\\$ (ii) Which political party won the maximum number of seats?

##### Solution :

(i) By taking polling results on x-axis and seats won as y-axis and choosing an appropriate scale ($1$ unit = $10$ seats for y-axis), the required graph of the above information can be constructed as follows.

Here, the rectangle bars are of the same length and have equal spacing in between them. $\\$ (ii) Political party $‘A’$ won maximum number of seats.

15   The length of $40$ leaves of a plant are measured correct to one millimetre, and the obtained data is represented in the following table:$\\$ $\begin{array}{|c|c|}\hline \text{Length (in mm)} & \text{Number of leaves}\\ \hline 118-126& 3 \\ \hline 127-135 & 5 \\ \hline 136-144 & 9 \\ \hline 145-153 & 12 \\ \hline 154-162 & 5 \\ \hline 163 -171 & 4 \\ \hline 172-180 & 2 \\ \hline \end{array}$ $\\$(i) Draw a histogram to represent the given data.$\\$ (ii) Is there any other suitable graphical representation for the same data?$\\$ (iii) Is it correct to conclude that the maximum number of leaves are $153 mm$ long? Why?

##### Solution :

(i) It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of $1$ in between them. Therefore, $\dfrac{1}{2}=0.5$ has to be added to each upper class limit and also have to subtract $0.5$ from the lower class limits so as to make the class intervals continuous. $\\$ $\begin{array}{|c|c|}\hline \text{Length (in mm)}&\text{Number of leaves} \\ \hline 117.5-126.5 & 3 \\ \hline 126.5-135.5 & 5 \\ \hline 135.5-144.5 & 9 \\ \hline 144.5-153.5 & 12 \\ \hline 153.5 -162.5 & 5 \\ \hline 162.5-171.5 & 4 \\ \hline 171.5-180.5 & 2 \\ \hline \end{array}$ $\\$

Taking the length of leaves on x-axis and the number of leaves on y-axis, the histogram of this information can be drawn as above.$\\$ Here,$1$ unit on y-axis represents $2$ leaves.$\\$ (ii) Other suitable graphical representation of this data is frequency polygon.$\\$ (iii) No, as maximum number of leaves (i.e., 12) has their length in between $144.5 mm$ and $153.5 mm.$ It is not necessary that all have their lengths as $153 mm.$

16   The following table gives the life times of neon lamps:$\\$ $\begin{array}{|c|c|}\hline \text{Length (in hours)} &\text{Number of lamps}\\ \hline 300-400 & 14 \\ \hline 400-500 & 56 \\ \hline 500-600 & 60 \\ \hline 600-700 & 86 \\ \hline 700-800 & 74 \\ \hline 800 - 900 & 62 \\ \hline 900-1000 & 48 \\ \hline \end{array}$ $\\$ (i) Represent the given information with the help of a histogram.$\\$ (ii) How many lamps have a lifetime of more than $700$ hours?

##### Solution :

(i) By taking life time (in hours) of neon lamps on x-axis and the number of lamps on y-axis, the histogram of the given information can be drawn as follows.$\\$

(ii) It can be concluded that the number of neon lamps having their lifetime more than $700$ is the sum of the number of neon lamps having their lifetime as $700-800, 800-900,$ and $900-1000$. Therefore, the number of neon lamps having their lifetime more than $700$ hours is $184 (74 + 62 + 48 = 184)$

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

17   The following table gives the distribution of students of two sections according to the mark obtained by them:$\\$ $\begin{array}{|c|}\hline &\qquad\text{Section A} \qquad && \text{Section B}\qquad & \end{array}$ $\\$ $\begin{array}{|c|c|}\hline \text{Marks} & \text{Frequency} & \text{Marks} & \text{Frequency}\\ \hline 0-10 & 3 & 0-10& 5 \\ \hline 10-20 & 9 & 10-20 & 19 \\ \hline 20-30 & 17 & 20-30 & 15 \\ \hline 30-40 & 12 & 30-40 & 10 \\ \hline 40-50 & 9 & 40-50 & 1 \\ \hline \end{array}$ $\\$ 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.

##### Solution :

We can find the class marks of the given class intervals by using the following formula.$\\$ Classmark=$\dfrac{\text{ Upperclasslimit+Lowerclasslimit}}{ 2}$ $\\$ $\begin{array}{|c|}\hline &\qquad \qquad\text{Section A} \qquad \qquad \qquad\qquad && \qquad \quad \text{Section B}\qquad \qquad& \end{array}$ $\\$ $\begin{array}{|c|c|}\hline \text{Marks} &\text{Class marks}& \text{Frequency} & \text{Marks}&\text{Class marks } & \text{Frequency}\\ \hline 0-10&5 & 3 & 0-10&5& 5 \\ \hline 10-20 &15& 9 & 10-20 &15& 19 \\ \hline 20-30 &25& 17 & 20-30 &25& 15 \\ \hline 30-40 &35& 12 & 30-40 &35& 10 \\ \hline 40-50 &45& 9 & 40-50 &45& 1 \\ \hline \end{array}$ $\\$ Taking class marks on x-axis and frequency on y-axis and choosing an appropriate scale ($1$ unit = $3$ for y-axis), the frequency polygon can be drawn as follows.

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.

18   The runs scored by two teams A and B on the first $60$ balls in a cricket match are given below: $\\$ $\begin{array}{|c|c|}\hline \text{Number of balls} & \text{Team A}& \text{Team B} \\ \hline 1-6 & 2 & 5 \\ \hline 7-12 & 1 & 6 \\ \hline 13-18 & 8 & 2 \\ \hline 19-24 & 9 & 10 \\ \hline 25-30 & 4 & 5 \\ \hline 31-36 & 5 & 6 \\ \hline 37 -42 & 6 & 3 \\ \hline 43 -48 & 10 & 4 \\ \hline 49-54 & 6 & 8 \\ \hline 55-60 & 2 & 10 \\ \hline \end{array}$ $\\$ Represent the data of both the teams on the same graph by frequency polygons. [Hint: First make the class intervals continuous.]

##### Solution :

It can be observed that the class intervals of the given data are not continuous. There is a gap of $1$ in between them. Therefore, $\dfrac{1}{2}=0.5$ has to be added to the upper class limits and $0.5$ has to be subtracted from the lower class limits.$\\$ Also, class mark of each interval can be found by using the following formula.$\\$ Class mark= $\dfrac{\text{Upper class limit + Lower class limit}}{2}$ $\\$ Continuous data with class mark of each class interval can be represented as follows:

$\begin{array}{|c|c|}\hline \text{Number of balls} & \text{Class Marks }&\text{Team A}& \text{Team B} \\ \hline 1-6 &3.5& 2 & 5 \\ \hline 7-12&9.5 & 1 & 6 \\ \hline 13-18&15.5 & 8 & 2 \\ \hline 19-24 &21.5& 9 & 10 \\ \hline 25-30&27.5 & 4 & 5 \\ \hline 31-36 &33.5& 5 & 6 \\ \hline 37 -42 &39.5& 6 & 3 \\ \hline 43 -48 & 45.5& 10 & 4 \\ \hline 49-54 &51.5 & 6 & 8 \\ \hline 55-60 &57.5 & 2 & 10 \\ \hline \end{array}$ $\\$ By taking class marks on x-axis and runs scored on y-axis, a frequency polygon can be constructed as follows.

19   A random survey of the number of children of various age groups playing in park was found as follows:$\\$ $\begin{array}{|c|c|}\hline \text{Age(in years)} & \text{Number of children} \\ \hline 1-2 & 5\\ 2-3 & 3\\ 3-5 & 6\\ 5-7 & 12\\ 7-10 & 9 \\ 10-15 & 10\\ 15-17 & 4\\ \hline \end{array}$ $\\$ Draw a histogram to represent the data above. $\\$

##### Solution :

Here, it can be observed that the data has class intervals of varying width. The proportion of children per $1$ year interval can be calculated as follows.$\\$ $\begin{array}{|c|c|}\hline \text{Age(in years)} & \text{Frequency (Number of children)} & \text{Width of class} & \text{Length of rectangle } \\ \hline 1-2 & 5 & 1 & \dfrac{5*1}{1}=5 \\ \hline 2-3 & 3 & 1 & \dfrac{3*1}{1}=3 \\ \hline 3-5 & 6 & 2 & \dfrac{6 * 1}{2}=3 \\ \hline 5-7 & 12 & 1 & \dfrac{12* 1}{2}=6 \\ \hline 7-10 & 9 & 3 & \dfrac{9*1}{3}=3 \\ \hline 10-15 & 10 & 5 & \dfrac{10* 1}{5}=2 \\ \hline 15-17 & 4 & 2 & \dfrac{4*1}{2}=2 \\ \hline \end{array}$ $\\$ Taking the age of children on x-axis and proportion of children per $1$ year interval on y-axis, the histogram can be drawn as follows.

20   $100$ 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: $\\$ $\begin{array}{|c|c|}\hline \text{Number of letters} & \text{Number of surnames} \\ \hline 1-4 & 6 \\ \hline 4-6 & 30 \\ \hline 6-8 & 44 \\ \hline 8-12 & 16 \\ \hline 12-20 & 4 \\ \hline \end{array}$ $\\$ (i) Draw a histogram to depict the given information. $\\$ (ii) Write the class interval in which the maximum number of surname lie.

##### Solution :

(i) Here, it can be observed that the data has class intervals of varying width. The proportion of the number of surnames per $2$ letters interval can be calculated as follows.$\\$ $\begin{array}{|c|c|}\hline \text{Number of letters} & \text{Frequeny (Number of surnames)} & \text{Width of class } & \text{Length of Rectangle }\\ \hline 1-4 & 6 & 3 & \dfrac{6*2}{3}=4 \\ \hline 4-6 & 30 &2 & \dfrac{30* 2}{2}=30\\ \hline 6-8 & 44 & 2 & \dfrac{44*2}{2}=44 \\ \hline 8-12 & 16& 4 & \dfrac{16*2}{4}=8 \\ \hline 12-20 & 4 & 8 & \dfrac{4*2}{8}=1 \\ \hline \end{array}$ $\\$ By taking the number of letters on x-axis and the proportion of the number of surnames per $2$ letters interval on y-axis and choosing an appropriate scale ($1$ unit = $4$ students for y axis), the histogram can be constructed as follows. $\\$ (ii) The class interval in which the maximum number of surnames lies is $6-8$ as it has $44$ surnames in it i.e., the maximum for this data.

21   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.

##### Solution :

The number of goals scored by the team is $2, 3, 4, 5, 0, 1, 3, 3, 4, 3$ Mean of data $=\dfrac{\text{Sum of all observations }}{\text{Total number of observations}}$ $\\$ Mean score $=\dfrac{ 2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3}{ 10}$ $=\dfrac{28}{10}=2.8$ $\\$ $= 2.8$ goals $\\$ Arranging the number of goals in ascending order,$0, 1, 2, 3, 3, 3, 3, 4, 4, 5$ $\\$ The number of observations is $10,$ which is an even number. Therefore, median score will be the mean of $10$ i.e., $5^{th}$ and $\dfrac{10}{2} +1$ i.e., $6^{th}$ observation while arranged in ascending or descending order . $\\$ Median score =$\dfrac{ 5^{th} \text{ observation} + 6^{th}\text{ observation}}{ 2}$ $\\$ $=\dfrac{3+3}{2}=\dfrac{6}{2}=3$ $\\$ Mode of data is the observation with the maximum frequency in data. $\\$ Therefore, the mode score of data is $3$ as it has the maximum frequency as $4$ in the data.

22   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.

##### Solution :

The marks of $15$ students in mathematics test are $41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60$ Mean of data $=\dfrac{\text{ Sum of all observation}}{\text{ Total number of observation}}$ $\\$ $=\dfrac{41+39+48+52+46+62+54+40+96+52+98+40+42+52+60}{15}$ $\\$ $=\dfrac{822}{15}=54.8$ $\\$ Arranging the scores obtained by $15$ students in an ascending order, $39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98$ $\\$ As the number of observations is $15$ which is odd, therefore, the median of data will be $\dfrac{15+1}{2}=8^{th}$ observation whether the data is arranged in an ascending or descending order.$\\$ Therefore, median score of data = $52$ $\\$ Mode of data is the observation with the maximum frequency in data. Therefore, mode of this data is $52$ having the highest frequency in data as $3.$

23   The following observations have been arranged in ascending order. If the median of the data is $63$, find the value of $x.$ $29, 32, 48, 50, x, x + 2, 72, 78, 84, 95$

##### Solution :

It can be observed that the total number of observations in the given data is $10$ (even number). Therefore, the median of this data will be the mean of $\dfrac{10}{2}$ i.e., $5^{th}$ and $\dfrac{10 }{2}+1$ i.e., $6^{th}$ observation. $\\$ Therefore, median of data =$\dfrac{5^{th} \text{observation}+6^{th} \text{observation}}{2}$ $\\$ $\implies 63=\dfrac{x+x+2}{2}$ $\\$ $\implies 63=\dfrac{2x+2}{2}$ $\\$ $\implies 63 =x+1$ $\\$ $\implies x=62$ $\\$

24   Find the mode of $14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.$

##### Solution :

Arranging the data in an ascending order, $\\$ $14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28$ It can be observed that $14$ has the highest frequency, i.e. $4$, in the given data. Therefore, mode of the given data is $14$.

25   Find the mean salary of $60$ workers of a factory from the following table: $\\$ $\begin{array}{|c|c|}\hline \text{Salary (in Rs.)} & \text{Number of workers } \\ \hline 3000 & 16 \\ \hline 4000 & 12 \\ \hline 5000 & 10 \\ \hline 6000 & 8 \\ \hline 7000 & 6 \\ \hline 8000 & 4 \\ \hline 9000 & 3 \\ \hline 10000 & 1 \\ \hline \text{Total} & 60 \\ \hline \end{array}$ $\\$

##### Solution :

We know that $\\$ Mean=$\dfrac{\Sigma f_ix_i}{\Sigma f_i}$ $\\$ The value of $\Sigma f_ix_i$ and $\Sigma f_i$ can be calculated as follows.$\\$ $\begin{array}{|c|c|}\hline \text{Salary(in Rs.)(x_i)}& \text{Number of workers(f_i)} & f_ix_i \\ \hline 3000 & 16 & 3000*16=48000 \\ \hline 4000 & 12 & 4000*12 =48000 \\ \hline 5000 & 10 & 5000* 10=50000 \\ \hline 6000 & 8 & 6000*8=48000 \\ \hline 7000 & 6 & 7000*6 =42000 \\ \hline 8000 & 4 & 8000*4=32000 \\ \hline 9000 & 3 & 9000* 3 =27000 \\ \hline 10000 & 1 & 10000*1=10000 \\ \hline \text{Total} & \Sigma f_i=60 & \Sigma f_ix_i=305000 \\ \hline \end{array}$ $\\$ Mean salary =$\dfrac{305000}{60}$ $\\$ Therefore,mean salary of $60$ workers is Rs.$5083.33$

26   Give one example of a situation in which$\\$ (i) The mean is 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.

##### Solution :

When any data has a few observations such that these are very far from the other observations in it, it is better to calculate the median than the mean of the data as median gives a better estimate of average in this case.$\\$ (i) Consider the following example - the following data represents the heights of the members of a family.$\\$ $154.9 cm, 162.8 cm, 170.6 cm, 158.8 cm, 163.3 cm, 166.8 cm, 160.2 cm$ $\\$ In this case, it can be observed that the observations in the given data are close to each other. Therefore, mean will be calculated as an appropriate measure of central tendency.$\\$ (ii) The following data represents the marks obtained by $12$ students in a test. $48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, 99$ $\\$ In this case, it can be observed that there are some observations which are very far from other observations. Therefore, here, median will be calculated as an appropriate measure of central tendency.