Textbook - Statistics and Probability Questions in English

(N/A) 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 in our class.
$3.$ Production of wheat in the last $10$ years in our country.
$4.$ Number of plants in our locality.
$5.$ Rainfall in our city in the last $10$ years.

(N/A) The information which is collected by the investigator himself with a definite objective in his mind is called primary data. Conversely,when the information is gathered from a source that already has the information stored,it is called secondary data.
$1.$ Number of females per $1000$ males: Secondary data (collected from government records).
$2.$ Weights of students of our class: Primary data (collected by direct measurement).
$3.$ Production of wheat in the last $10$ years: Secondary data (collected from agricultural reports).
$4.$ Number of plants in our locality: Primary data (collected by direct observation).
$5.$ Rainfall in our city in the last $10$ years: Secondary data (collected from meteorological records).
Thus,$1, 3,$ and $5$ are secondary data,while $2$ and $4$ are primary data.

(N/A) The data in this form is called raw data.
By looking at it in this form,can you find the highest and the lowest marks?
Did it take you some time to search for the maximum and minimum scores? It would be less time-consuming if these scores were arranged in ascending or descending order. So,let us arrange the marks in ascending order as:
$25, 36, 42, 55, 60, 62, 73, 75, 78, 95$
Now,we can clearly see that the lowest marks are $25$ and the highest marks are $95$.
The difference between the highest and the lowest values in the data is called the range of the data. So,the range in this case is $95 - 25 = 70$.
Presentation of data in ascending or descending order can be quite time-consuming,particularly when the number of observations in an experiment is large.

(N/A) Recall that the number of students who have obtained a certain number of marks is called the frequency of those marks. For instance,$4$ students got $70$ marks. So the frequency of $70$ marks is $4$. To make the data more easily understandable,we write it in a table,as given below:

Marks	Number of students (i.e.,the frequency)
$10$	$1$
$20$	$1$
$36$	$3$
$40$	$4$
$50$	$3$
$56$	$2$
$60$	$4$
$70$	$4$
$72$	$1$
$80$	$1$
$88$	$2$
$92$	$3$
$95$	$1$
< strong>Total	$30$

This table is called an ungrouped frequency distribution table,or simply a frequency distribution table. Note that you can also use tally marks in preparing these tables.

(N/A) To present such a large amount of data so that a reader can make sense of it easily,we condense it into groups like $20-29, 30-39, . . ., 90-99$ (since our data ranges from $23$ to $98$). These groupings are called 'classes' or 'class-intervals',and their size is called the class-size or class width,which is $10$ in this case. In each of these classes,the least number is called the 'lower class limit' and the greatest number is called the 'upper class limit'. For example,in $20-29$,$20$ is the lower class limit and $29$ is the upper class limit.
Using tally marks,the data above can be condensed into a grouped frequency distribution table as follows:

Number of plants survived	Number of schools (frequency)
$20-29$	$3$
$30-39$	$14$
$40-49$	$12$
$50-59$	$8$
$60-69$	$18$
$70-79$	$10$
$80-89$	$23$
$90-99$	$12$
Total	$100$

Presenting data in this form simplifies and condenses it,enabling us to observe important features at a glance. We can observe that $50\%$ or more plants survived in $8 + 18 + 10 + 23 + 12 = 71$ schools.

(N/A) To include weights like $35.5\, kg$ and $40.5\, kg$,we must convert the discontinuous class intervals into continuous ones.
$1$. Calculate the difference between the upper limit of a class and the lower limit of the next class (e.g.,$36 - 35 = 1$).
$2$. Divide this difference by $2$ to get the adjustment factor $(1 / 2 = 0.5)$.
$3$. Subtract $0.5$ from each lower limit and add $0.5$ to each upper limit to make the classes continuous.
The new continuous intervals are: $30.5-35.5, 35.5-40.5, 40.5-45.5, 45.5-50.5, 50.5-55.5, 55.5-60.5, 60.5-65.5, 65.5-70.5, 70.5-75.5$.
By convention,the upper limit value is included in the next class interval. Therefore,$35.5$ is included in $35.5-40.5$ and $40.5$ is included in $40.5-45.5$.
The updated table is:

Weights (in $kg$)	Number of students
$30.5-35.5$	$9$
$35.5-40.5$	$6$
$40.5-45.5$	$15$
$45.5-50.5$	$3$
$50.5-55.5$	$1$
$55.5-60.5$	$2$
$60.5-65.5$	$2$
$65.5-70.5$	$1$
$70.5-75.5$	$1$
Total	$40$

(N/A) By counting the occurrences of each blood group in the given data,we find:
- Blood group $A$: $9$ students
- Blood group $B$: $6$ students
- Blood group $AB$: $3$ students
- Blood group $O$: $12$ students
Total number of students = $9 + 6 + 3 + 12 = 30$.
The frequency distribution table is as follows:

Blood group	Number of students
$A$	$9$
$B$	$6$
$AB$	$3$
$O$	$12$
Total	$30$

From the table,it is clear that the most common blood group is $O$ (with $12$ students) and the rarest blood group is $AB$ (with $3$ students).

(N/A) To construct a grouped frequency distribution table with a class size of $5$,we use intervals: $0-5, 5-10, 10-15, 15-20, 20-25, 25-30, 30-35$.

Distance (in $km$)	Number of engineers
$0-5$	$5$
$5-10$	$11$
$10-15$	$11$
$15-20$	$9$
$20-25$	$1$
$25-30$	$1$
$30-35$	$2$
Total	$40$

Observations:
$1$. Most of the engineers (i.e.,$5+11+11+9 = 36$ engineers) live within a distance of $20\, km$ from their workplace.
$2$. Very few engineers (i.e.,$1+1+2 = 4$ engineers) live at a distance of $20\, km$ or more from their workplace.

(N/A) $(i)$ $A$ grouped frequency distribution table of class size $2$ is constructed. The class intervals are $84-86, 86-88, 88-90, \dots$
By observing the data,the frequency distribution table is as follows:

Relative humidity (in $\%$)	Number of days (frequency)
$84-86$	$1$
$86-88$	$1$
$88-90$	$2$
$90-92$	$2$
$92-94$	$7$
$94-96$	$6$
$96-98$	$7$
$98-100$	$4$
Total	$30$

$(ii)$ It can be observed that the relative humidity is high. Therefore,the data is about a month of the rainy season.
$(iii)$ Range of data = Maximum value $-$ Minimum value = $99.2 - 84.9 = 14.3$.

(N/A) $(i)$ $A$ grouped frequency distribution table is constructed by taking class intervals of size $5$ starting from $150$. By tallying the data provided,we obtain the following frequency distribution table:

Height (in $cm$)	Number of students (frequency)
$150-155$	$12$
$155-160$	$9$
$160-165$	$14$
$165-170$	$10$
$170-175$	$5$
Total	$50$

$(ii)$ From the table,it can be concluded that $35$ students (i.e.,$12+9+14 = 35$) have a height less than $165\, cm$. Since $35/50 = 70\%$,we can conclude that $70\%$ of the students are shorter than $165\, cm$.

(8) To construct the grouped frequency distribution table,we count the number of observations falling into each class interval. Note that the upper limit of each class is excluded from that class.

Concentration of $SO_2$ (in $ppm$)	Number of days (frequency)
$0.00 - 0.04$	$4$
$0.04 - 0.08$	$9$
$0.08 - 0.12$	$9$
$0.12 - 0.16$	$2$
$0.16 - 0.20$	$4$
$0.20 - 0.24$	$2$
Total	$30$

$(ii)$ The number of days for which the concentration of $SO_2$ is more than $0.11$ $ppm$ corresponds to the sum of frequencies for the intervals $0.12 - 0.16, 0.16 - 0.20$,and $0.20 - 0.24$.
Required number of days $= 2 + 4 + 2 = 8$.
Therefore,for $8$ days,the concentration of $SO_2$ was more than $0.11$ $ppm$.

(N/A) To prepare the frequency distribution table,we count the occurrences of each outcome (number of heads) in the given data set.
$1$. Count of $0$ heads: $6$ times.
$2$. Count of $1$ head: $10$ times.
$3$. Count of $2$ heads: $9$ times.
$4$. Count of $3$ heads: $5$ times.
Sum of frequencies: $6 + 10 + 9 + 5 = 30$.

Number of heads	Number of times (frequency)
$0$	$6$
$1$	$10$
$2$	$9$
$3$	$5$
Total	$30$

(N/A) $(i)$ By observing the digits after the decimal point,the frequency distribution table is constructed as follows:

Digit	Frequency
$0$	$2$
$1$	$5$
$2$	$5$
$3$	$8$
$4$	$4$
$5$	$5$
$6$	$4$
$7$	$4$
$8$	$5$
$9$	$8$
Total	$50$

$(ii)$ From the table,the least frequency is $2$ (for digit $0$),and the maximum frequency is $8$ (for digits $3$ and $9$). Thus,the most frequently occurring digits are $3$ and $9$,and the least frequently occurring digit is $0$.

(N/A) $(i)$ The class intervals with a width of $5$ are $0-5, 5-10, 10-15, 15-20$.
The grouped frequency distribution table is as follows:

Hours	Number of children
$0-5$	$10$
$5-10$	$13$
$10-15$	$5$
$15-20$	$2$
Total	$30$

$(ii)$ The number of children who watched $TV$ for $15$ or more hours a week corresponds to the class interval $15-20$. From the table,this value is $2$.

(N/A) To construct a grouped frequency distribution table with a class size of $0.5$,we start from the interval $2 - 2.5$.
The class intervals will be $2 - 2.5, 2.5 - 3.0, 3.0 - 3.5, 3.5 - 4.0, 4.0 - 4.5, 4.5 - 5.0$.
By tallying the data provided,we get the following frequency distribution table:

Lives of batteries (in years)	Number of batteries
$2.0 - 2.5$	$2$
$2.5 - 3.0$	$6$
$3.0 - 3.5$	$14$
$3.5 - 4.0$	$11$
$4.0 - 4.5$	$4$
$4.5 - 5.0$	$3$
Total	$40$

(N/A) Note that the variable here is the 'month of birth',and the value of the variable is the 'Number of students born'.
$(i)$ By observing the bar graph,the height of the bar corresponding to the month of November is $4$. Thus,$4$ students were born in the month of November.
$(ii)$ By observing the bar graph,the tallest bar corresponds to the month of August,which has a height of $6$. Thus,the maximum number of students were born in the month of August.

(N/A) We draw the bar graph for this data in the following steps. Note that the unit in the second column is thousand rupees. So,'$4$' against 'Grocery' means $4000$ rupees.
$1.$ We represent the Heads (variable) on the horizontal axis,choosing any scale,since the width of the bar is not important. For clarity,we take equal widths for all bars and maintain equal gaps in between.
$2.$ We represent the expenditure (value) on the vertical axis. Since the maximum expenditure is $5000$,we can choose the scale as $1$ unit = $1000$ rupees.
$3.$ To represent our first Head,i.e.,Grocery,we draw a rectangular bar with width $1$ unit and height $4$ units.
$4.$ Similarly,other Heads are represented,leaving a gap of $1$ unit in between two consecutive bars.
Here,you can easily visualize the relative characteristics of the data at a glance,e.g.,the expenditure on education is more than double that of medical expenses. Therefore,in some ways,it serves as a better representation of data than the tabular form.

(N/A) No,the graph is giving us a misleading picture. In a histogram,the areas of the rectangles are proportional to the frequencies. This condition is satisfied when the widths of all class intervals are equal. However,in this case,the widths of the rectangles are varying,so the histogram does not give a correct representation. For example,it shows a greater frequency in the interval $70-100$ than in $60-70$,which is incorrect.
To correct this,we must modify the lengths of the rectangles so that their areas become proportional to the frequencies. The steps are:
$1$. Select the minimum class size,which is $10$ in this case.
$2$. Modify the lengths of the rectangles to be proportionate to the class size of $10$. For a class size of $20$,the length is $7$. For a class size of $10$,the length becomes $\frac{7}{20} \times 10 = 3.5$.
Following this method,we get the modified table:

Marks	Frequency	Width	Length of rectangle
$0-20$	$7$	$20$	$\frac{7}{20} \times 10 = 3.5$
$20-30$	$10$	$10$	$\frac{10}{10} \times 10 = 10$
$30-40$	$10$	$10$	$\frac{10}{10} \times 10 = 10$
$40-50$	$20$	$10$	$\frac{20}{10} \times 10 = 20$
$50-60$	$20$	$10$	$\frac{20}{10} \times 10 = 20$
$60-70$	$15$	$10$	$\frac{15}{10} \times 10 = 15$
$70-100$	$8$	$30$	$\frac{8}{30} \times 10 = 2.67$

These lengths represent the "proportion of students per $10$ marks interval". Plotting these values gives the correct histogram.

(N/A) To draw a frequency polygon,we first represent the data using a histogram. We mark the mid-points of the tops of the rectangles as $B, C, D, E, F, G, H, I, J, K$ respectively.
Since the first class is $0-10$,to close the polygon,we extend the horizontal axis in the negative direction to include an imaginary class-interval $(-10) - 0$ with frequency $0$. We mark the mid-point of this interval as the starting point. The line segment from this point joins to the first mid-point $B$. Similarly,we take an imaginary class-interval $100-110$ with frequency $0$ and mark its mid-point $L$. The last mid-point $K$ is joined to $L$.
The resulting figure $OABCDEFGHIJKL$ is the required frequency polygon.
Alternatively,frequency polygons can be drawn using class-marks. The class-mark is calculated as:
$\text{Class-mark} = \frac{\text{Upper limit} + \text{Lower limit}}{2}$

(N/A) Since we want to draw a frequency polygon without a histogram,let us find the class-marks of the classes given above,that is of $140-150, 150-160, \dots$
For $140-150$,the upper limit $= 150$ and the lower limit $= 140$.
So,the class-mark $= \frac{150+140}{2} = \frac{290}{2} = 145$.
Continuing in the same manner,we find the class-marks of the other classes as well.
The new table obtained is as follows:

Classes	Class-marks	Frequency
$140-150$	$145$	$5$
$150-160$	$155$	$10$
$160-170$	$165$	$20$
$170-180$	$175$	$9$
$180-190$	$185$	$6$
$190-200$	$195$	$2$

We can now draw a frequency polygon by plotting the class-marks along the horizontal axis,the frequencies along the vertical axis,and then plotting and joining the points $B(145, 5), C(155, 10), D(165, 20), E(175, 9), F(185, 6)$ and $G(195, 2)$ by line segments.
We should not forget to plot the point corresponding to the class-mark of the class $130-140$ (just before the lowest class $140-150$) with zero frequency,that is,$A(135, 0)$,and the point $H(205, 0)$ which occurs immediately after $G(195, 2)$.
So,the resultant frequency polygon will be $ABCDEFGH$.

(A) $(i)$ By representing causes on the $x-$axis and the female fatality rate on the $y-$axis and choosing an appropriate scale ($1 \text{ unit} = 5\%$ for the $y-$axis),a bar graph can be constructed.
$(ii)$ Reproductive health conditions are the major cause of women's ill health and death worldwide,as $31.8\%$ of women are affected by them.
$(iii)$ Two factors that play a major role in this are:
$1.$ Lack of access to quality medical facilities.
$2.$ Lack of awareness and correct knowledge regarding treatment and reproductive health.

(N/A) $(i)$ By representing the section (variable) on the $x$-axis and the number of girls per thousand boys on the $y$-axis,the bar graph can be constructed by choosing an appropriate scale (e.g.,$1$ unit $= 100$ girls on the $y$-axis).
$(ii)$ From the graph,it can be observed that:
$1$. The maximum number of girls per thousand boys $(970)$ is in the $ST$ section.
$2$. The minimum number of girls per thousand boys $(910)$ is in urban areas.
$3$. The number of girls per thousand boys is higher in rural areas compared to urban areas.
$4$. The number of girls per thousand boys is higher in backward districts compared to non-backward districts.
$5$. The number of girls per thousand boys is higher in $SC$ and $ST$ sections compared to the non-$SC/ST$ section.

(A) $(i)$ By taking political parties on the $x$-axis and seats won on the $y$-axis and choosing an appropriate scale ($1$ unit = $10$ seats for the $y$-axis),the required bar graph can be constructed as shown below.
Here,the rectangular bars are of equal width and have equal spacing between them.
$(ii)$ Political party '$A$' won the maximum number of seats ($75$ seats).

(N/A) $(i)$ It can be observed that the length of leaves is represented in a discontinuous class interval having a difference of $1$ between them.
Therefore,$1/2 = 0.5$ has to be added to each upper class limit and subtracted from each lower class limit to make the class intervals continuous.

Length (in $mm$)	Number of leaves
$117.5-126.5$	$3$
$126.5-135.5$	$5$
$135.5-144.5$	$9$
$144.5-153.5$	$12$
$153.5-162.5$	$5$
$162.5-171.5$	$4$
$171.5-180.5$	$2$

Taking the length of leaves on the $x$-axis and the number of leaves on the $y$-axis,the histogram is drawn as shown in the figure.
$(ii)$ Another suitable graphical representation for this data is a frequency polygon.
$(iii)$ No,it is not correct. The maximum number of leaves $(12)$ have their lengths in the range of $144.5 \, mm$ to $153.5 \, mm$. It does not mean that all these leaves are $153 \, mm$ long.

(N/A) $(i)$ By taking the life time (in hours) of neon lamps on the $x$-axis and the number of lamps on the $y$-axis,the histogram of the given information can be drawn as shown in the figure.
Here,$1$ unit on the $y$-axis represents $10$ lamps.
$(ii)$ The number of neon lamps having a lifetime of more than $700$ hours is the sum of the number of lamps in the intervals $700-800$,$800-900$,and $900-1000$.
Therefore,the number of neon lamps having a lifetime of more than $700$ hours is $74 + 62 + 48 = 184$.

(N/A) We can find the class marks of the given class intervals by using the following formula:
$\text{Class mark} = \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$

Marks	Class mark	Frequency (Section $A$)	Frequency (Section $B$)
$0-10$	$5$	$3$	$5$
$10-20$	$15$	$9$	$19$
$20-30$	$25$	$17$	$15$
$30-40$	$35$	$12$	$10$
$40-50$	$45$	$9$	$1$

By plotting the class marks on the $x$-axis and frequency on the $y$-axis,we draw the frequency polygons for both sections.
From the graph,it can be observed that the frequency polygon for section $A$ is shifted more towards the right compared to section $B$. This indicates that the students of section $A$ have performed better than the students of section $B$ in terms of obtaining higher marks.

(N/A) 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,$1/2 = 0.5$ has to be added to the upper class limits and $0.5$ has to be subtracted from the lower class limits.
Also,the class mark of each interval can be found by using the following formula:
Class mark $= \frac{\text{Upper class limit} + \text{Lower class limit}}{2}$
Continuous data with the class mark of each class interval can be represented as follows:

Number of balls	Class mark	Team $A$	Team $B$
$0.5-6.5$	$3.5$	$2$	$5$
$6.5-12.5$	$9.5$	$1$	$6$
$12.5-18.5$	$15.5$	$8$	$2$
$18.5-24.5$	$21.5$	$9$	$10$
$24.5-30.5$	$27.5$	$4$	$5$
$30.5-36.5$	$33.5$	$5$	$6$
$36.5-42.5$	$39.5$	$6$	$3$
$42.5-48.5$	$45.5$	$10$	$4$
$48.5-54.5$	$51.5$	$6$	$8$
$54.5-60.5$	$57.5$	$2$	$10$

By taking class marks on the $x$-axis and runs scored on the $y$-axis,a frequency polygon can be constructed as shown in the graph.

(N/A) 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:

Age (in years)	Number of children	Width of class	Height of rectangle
$1-2$	$5$	$1$	$(5 \times 1) / 1 = 5$
$2-3$	$3$	$1$	$(3 \times 1) / 1 = 3$
$3-5$	$6$	$2$	$(6 \times 1) / 2 = 3$
$5-7$	$12$	$2$	$(12 \times 1) / 2 = 6$
$7-10$	$9$	$3$	$(9 \times 1) / 3 = 3$
$10-15$	$10$	$5$	$(10 \times 1) / 5 = 2$
$15-17$	$4$	$2$	$(4 \times 1) / 2 = 2$

Taking the age of children on the $x$-axis and the proportion of children per $1$ year interval on the $y$-axis,the histogram is constructed based on the calculated heights.

(N/A) $(i)$ 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:

Number of letters	Frequency (Number of surnames)	Width of class	Length of rectangle
$1-4$	$6$	$3$	$\frac{6 \times 2}{3} = 4$
$4-6$	$30$	$2$	$\frac{30 \times 2}{2} = 30$
$6-8$	$44$	$2$	$\frac{44 \times 2}{2} = 44$
$8-12$	$16$	$4$	$\frac{16 \times 2}{4} = 8$
$12-20$	$4$	$8$	$\frac{4 \times 2}{8} = 1$

By taking the number of letters on the $x$-axis and the proportion of the number of surnames per $2$ letters interval on the $y$-axis,the histogram is constructed as shown in the provided image.
$(ii)$ The class interval in which the maximum number of surnames lies is $6-8$,as it contains $44$ surnames,which is the maximum frequency for this data.

(13) The mean (or average) of a set of observations is calculated as the sum of all observations divided by the total number of observations.
Let $x_i$ denote the $i$-th observation. Here,we have $5$ observations: $x_1 = 10, x_2 = 7, x_3 = 13, x_4 = 20, x_5 = 15$.
The mean $\bar{x}$ is given by the formula:
$\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$
Substituting the values:
$\bar{x} = \frac{10 + 7 + 13 + 20 + 15}{5}$
$\bar{x} = \frac{65}{5} = 13$
Thus,the mean time spent by these $5$ people in doing social work is $13$ hours in a week.

(C) The mean (average) of a set of observations is calculated by dividing the sum of all observations by the total number of observations.
Formula: $\bar{x} = \frac{\sum_{i=1}^{n} x_i}{n}$
Given,the number of students $n = 30$.
Sum of all marks = $10 + 20 + 36 + 92 + 95 + 40 + 50 + 56 + 60 + 70 + 92 + 88 + 80 + 70 + 72 + 70 + 36 + 40 + 36 + 40 + 92 + 40 + 50 + 50 + 56 + 60 + 70 + 60 + 60 + 88 = 1779$.
Mean $\bar{x} = \frac{1779}{30} = 59.3$.

(A) First,we arrange the given data in ascending order:
$144, 145, 147, 148, 149, 150, 152, 155, 160$
Since the number of observations $(n)$ is $9$,which is an odd number,the median is the value of the $\left(\frac{n+1}{2}\right)^{th}$ observation.
Median $= \left(\frac{9+1}{2}\right)^{th} = \left(\frac{10}{2}\right)^{th} = 5^{th}$ observation.
The $5^{th}$ observation in the ordered data is $149$.
Therefore,the median height is $149 \text{ cm}$.

(B) Arranging the points scored by the team in ascending order,we get:
$2, 5, 7, 7, 8, 8, 10, 10, 14, 15, 17, 18, 24, 27, 28, 48$.
There are $n = 16$ terms,which is an even number.
For an even number of observations,the median is the mean of the $(\frac{n}{2})$th and $(\frac{n}{2} + 1)$th terms.
Here,the $8$th term is $10$ and the $9$th term is $14$.
Median $= \frac{10 + 14}{2} = \frac{24}{2} = 12$.
Thus,the median of the points scored by the Kabaddi team is $12$.

(C) To find the mode,we first arrange the given data in ascending order:
$2, 3, 3, 4, 4, 4, 5, 5, 6, 6, 6, 7, 7, 7, 9, 9, 9, 9, 10, 10$
Next,we count the frequency of each observation:
$2$ appears $1$ time.
$3$ appears $2$ times.
$4$ appears $3$ times.
$5$ appears $2$ times.
$6$ appears $3$ times.
$7$ appears $3$ times.
$9$ appears $4$ times.
$10$ appears $2$ times.
The mode is the observation that occurs most frequently. Since $9$ appears $4$ times,which is the highest frequency,the mode is $9$.

(A) Mean = $\frac{5000+5000+5000+5000+15000}{5} = \frac{35000}{5} = 7000$.
So,the mean salary is $Rs. 7000$ per month.
To obtain the median,we arrange the salaries in ascending order:
$5000, 5000, 5000, 5000, 15000$.
Since the number of employees in the factory is $5$,the median is given by the $\left(\frac{5+1}{2}\right)^{th} = 3^{rd}$ observation.
Therefore,the median is $Rs. 5000$ per month.
To find the mode of the salaries,we observe that $5000$ occurs the maximum number of times in the data. So,the modal salary is $Rs. 5000$ per month.
Comparing the three measures of central tendency,we see that the mean salary of $Rs. 7000$ does not give an accurate estimate of the wages of the majority,while the median and mode of $Rs. 5000$ represent the data more effectively. Extreme values in the data affect the mean,which is a weakness of this measure.

(D) The number of goals scored by the team is:
$2, 3, 4, 5, 0, 1, 3, 3, 4, 3$
Mean of data = $\frac{\text{Sum of all observations}}{\text{Total number of observations}}$
Mean score = $\frac{2+3+4+5+0+1+3+3+4+3}{10} = \frac{28}{10} = 2.8$ goals.
Arranging the number of goals in ascending order:
$0, 1, 2, 3, 3, 3, 3, 4, 4, 5$
The number of observations is $n = 10$,which is an even number.
Therefore,the median is the mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.
Median = $\frac{5^{th} \text{ observation} + 6^{th} \text{ observation}}{2} = \frac{3 + 3}{2} = \frac{6}{2} = 3$.
Mode is the observation with the maximum frequency.
In the given data,the value $3$ appears $4$ times,which is the maximum frequency.
Therefore,the mode is $3$.

(N/A) The marks of $15$ students in the mathematics test are:
$41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60$
$1.$ Mean of data = (Sum of all observations) / (Total number of observations)
$= (41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60) / 15$
$= 822 / 15 = 54.8$
$2.$ Arranging the scores in ascending order:
$39, 40, 40, 41, 42, 46, 48, 52, 52, 52, 54, 60, 62, 96, 98$
Since the number of observations $n = 15$ (which is odd),the median is the value of the $((n+1)/2)^{\text{th}}$ observation.
Median = $((15+1)/2)^{\text{th}} = 8^{\text{th}}$ observation.
The $8^{\text{th}}$ observation is $52$.
$3.$ Mode is the observation with the maximum frequency.
Here,$52$ appears $3$ times,which is the highest frequency.
Therefore,the mode is $52$.

(A) The total number of observations $(n)$ is $10$,which is an even number.
For an even number of observations,the median is the mean of the $(\frac{n}{2})^{th}$ and $(\frac{n}{2} + 1)^{th}$ observations.
Here,$n = 10$,so the median is the mean of the $5^{th}$ and $6^{th}$ observations.
$5^{th}$ observation = $x$
$6^{th}$ observation = $x + 2$
Median = $\frac{x + (x + 2)}{2} = 63$
$\frac{2x + 2}{2} = 63$
$x + 1 = 63$
$x = 62$

(B) Arranging the data in ascending order:
$14, 14, 14, 14, 17, 18, 18, 18, 22, 23, 25, 28$
It can be observed that the number $14$ appears $4$ times,which is the highest frequency in the given data set.
Therefore,the mode of the given data is $14$.

(C) The mean salary can be calculated using the formula:
Mean $= \frac{\sum f_{i} x_{i}}{\sum f_{i}}$
We calculate the product of salary $(x_i)$ and number of workers $(f_i)$ for each group:

Salary $(x_i)$	$f_i x_i$
$3000$	$3000 \times 16 = 48000$
$4000$	$4000 \times 12 = 48000$
$5000$	$5000 \times 10 = 50000$
$6000$	$6000 \times 8 = 48000$
$7000$	$7000 \times 6 = 42000$
$8000$	$8000 \times 4 = 32000$
$9000$	$9000 \times 3 = 27000$
$10000$	$10000 \times 1 = 10000$
Total	$\sum f_{i} x_{i} = 305000$

Given $\sum f_{i} = 60$.
Mean salary $= \frac{305000}{60} = 5083.33$.
Therefore,the mean salary of $60$ workers is Rs $5083.33$.

(N/A) The mean is an appropriate measure of central tendency when the data set does not contain extreme outliers and the values are relatively close to each other.
$(i)$ Consider the following example: The heights of the members of a family are $154.9 \text{ cm}, 162.8 \text{ cm}, 170.6 \text{ cm}, 158.8 \text{ cm}, 163.3 \text{ cm}, 166.8 \text{ cm}, 160.2 \text{ cm}$.
In this case,it can be observed that the observations in the given data are close to each other and there are no extreme values. Therefore,the mean is an appropriate measure of central tendency.

(N/A) When a data set contains a few observations that are significantly distant from the rest of the data (outliers),the mean is heavily influenced by these extreme values,making it an inappropriate measure of central tendency. In such cases,the median is a more robust and appropriate measure.
Consider the following data representing the marks obtained by $12$ students in a test:
$48, 59, 46, 52, 54, 46, 97, 42, 49, 58, 60, 99$
In this data set,the values $97$ and $99$ are significantly higher than the other marks. Because of these extreme values,the mean would be skewed upwards,failing to represent the typical performance of the students. Therefore,the median is a more appropriate measure of central tendency for this data.