To draw a bar graph of given data, first we make a frequency table, which is as follows

Clearly, from the bar graph, grapes is liked by 4 students. So, the least popular fruit among the students is grapes.

Saturday

Micromax

10:11

Required 'double bar graph' is shown below

It is inferred that more people prefer cricket and less athletics

If 40 is the pass mark, then students who getting less than 40 marks have failed. Hence, from the bargraph we can see students getting marks between 30-39 have failed i.e. 4 students have failed

Given, the enrolment of a school during six consecutive years
1555,1670,1750,2013,2540,2820
Now, sum of enrolments
= 1555 + 1670 + 1750 + 2013 + 2540 + 2820 = 12348
Number of years = 6
\(\therefore\)  Mean enrolment of the school for this period
 = \(\frac { sum\quad of\quad enrolments }{ number\quad of\quad years } =\frac { 12348 }{ 6 } =2058\)
Hence, the mean enrolment of the school for this period is 2058.

To find the mean number of points per game for C, we divide the total points by 3, because C did not play 3rd game.

The frequency table is shown below.

It is clear from the table that, the highest number = 9

Total number of plates = 6
i.e. The total number of outcomes = 6
.. Probability of the plate \(\boxed { 3 } \) is taken out. \({ P }_{ \boxed { 3 } }=\frac { 1 }{ 6 } \)

He obtained less than pass marks in English and Social Science

Let us start the scale at 0. The greatest value in the data is 55, therefore we can end the scale at the value greater than 55 (such as at 60). Since all the bars would lie between °and 60, therefore we choose the scale such that the distance between °and 60 is neither too long nor too small. Here, the most appropriate scale is '1 unit for 10 marks'. We draw the bar graph as follows:

Arranging in ascending order:
6,8,9,12,13,15,18,20,21,25
Since number of observations is even, the median is given by finding the average or mean of the two middle most observations:
So, median \(=\frac{13+15}{2}=\frac{28}{2}=14\)

Frequency distribution of ages of 25 students:

Age	Tally marks	Frequency
14	IIII	4
15		8
16		10
17	III	3
Total		25

Rahul score runs per inning are as follows: 34,37,47,49,54,61
\(Mean=\frac{Sum\ of\ the\ data}{Number\ of\ data}\)
\(=\frac{34+37+47+49+54+61}{6}\)
\(=\frac{282}{6}=47\)

First ten even natural numbers = 2, 4, 6,8,10, 12, 14, 16, 18,20
\(mean=\frac { sum\ of\ all\ obsevations }{ number\ of\ observations } \)
Sum of all observations = 2 + 4 + 6 + 8 + 10+ 12 +14+16+18+ 20=110
\(\therefore\) Mean = \(\frac { 110 }{ 10 } =11\)

\(\frac { 1 }{ 4 } \)

Total number of marbles = 8, Total number of possible outcomes = 8
\(\therefore\)   P(drawing marble with number 2) = \(\frac { Number\ of\ favourble\ outcomes }{ Number\ of\ possible\ outcomes } =\frac { 1 }{ 8 } \)

Given, marks obtained by a group of students in a Science test
85,76,90,85,39,48,56,95,81,75
On rearranging the marks in ascending order, we get
39,48,56,75,76,81,85,85,90,95
It is clear from the ascending order of marks that, highest marks = 95 and lowest marks = 39

(a) There are 11 outcomes namely M, M, A, A, T, T, H, E,I, C, S.
(b) (i) Probability of getting \('M'=\frac{2}{11}\)
(ii) Probability of getting a vowel \(=\frac{4}{11}\)
(iii) Probability of getting a consonant \(=\frac{7}{11}\)
(iv) Probability of getting \(X=0=\frac{0}{11}\)

We go through the following steps to construct the bar graphs:
(a) We draw two lines perpendicular to each other on a graph paper and call them horizontal and vertical axes as shown in Fig. below :

(b) Along the horizontal axis, we mark the 'years' and along the vertical axis, we mark the 'pass percentage.
(c) We choose a suitable scale to determine the heights of bars. Here, we choose the scale as 1 big division to represent 10.
(d) First we draw the bars for Class X results and then bars for Class XII results for different years.
Bars for X and XII class results are shaded separately and the shading is shown in the top right comer of the graph paper.

Given numbers are \(\frac{1}{2}\) and \(\frac{1}{3}\)
1st number=\(\frac { Sum\ of\ given\ numbers }{ Number\ of\ digits } =\frac { \frac { 1 }{ 2 } +\frac { 1 }{ 3 } }{ 2 } =\frac { \frac { 3+2 }{ 6 } }{ 2 } \)
=\(\frac { 5 }{ 6\times 2 } =\frac { 5 }{ 12 } \)
=\(\frac { 5 }{ 6\times 2 } =\frac { 5 }{ 12 } \)
=\(\frac { 1 }{ 2 } >\frac { 5 }{ 12 } >\frac { 1 }{ 3 } \)
2nd number=\(\frac { Given\ number+1st\ number }{ 2 } \)
=\(\frac { \frac { 1 }{ 2 } +\frac { 5 }{ 12 } }{ 2 } =\frac { \frac { 6+5 }{ 12 } }{ 2 } =\frac { 11 }{ 12\times 2 } =\frac { 11 }{ 24 } \)
\(\frac { 1 }{ 2 } >\frac { 11 }{ 24 } >\frac { 5 }{ 12 } >\frac { 1 }{ 3 } \)
Similarly, 3rd number=\(\frac { \frac { 5 }{ 12 } +\frac { 1 }{ 3 } }{ 2 } =\frac { \frac { 5+4 }{ 12 } }{ 2 } =\frac { 9 }{ 12\times 2 } =\frac { 9 }{ 24 } \)
\(\frac { 1 }{ 2 } >\frac { 11 }{ 24 } >\frac { 5 }{ 12 } >\frac { 9 }{ 24 } >\frac { 1 }{ 3 } \)
4th number=\(\frac { \frac { 5 }{ 12 } +\frac { 11 }{ 24 } }{ 2 } =\frac { \frac { 10+11 }{ 24 } }{ 2 } =\frac { 21 }{ 24\times 2 } =\frac { 21 }{ 48 } \)
\(\frac { 1 }{ 2 } >\frac { 11 }{ 24 } >\frac { 21 }{ 48 } >\frac { 5 }{ 12 } >\frac { 9 }{ 24 } >\frac { 1 }{ 3 } \)
5th number=\(\frac { \frac { 5 }{ 12 } +\frac { 9 }{ 24 } }{ 2 } =\frac { \frac { 10+9 }{ 24 } }{ 2 } =\frac { 10+9 }{ 24\times 2 } =\frac { 19 }{ 48 } \)
\(\frac { 1 }{ 2 } >\frac { 11 }{ 24 } >\frac { 21 }{ 48 } >\frac { 5 }{ 12 } >\frac { 19 }{ 48 } >\frac { 9 }{ 24 } >\frac { 1 }{ 3 } \)
Hence 5 numbers between \(\frac { 1 }{ 2 } \) and \(\frac { 1 }{ 3 } \)are \(\frac { 11 }{ 24 } ,\frac { 21 }{ 48 } ,\frac { 5 }{ 12 } ,\frac { 19 }{ 48 } \) and \(\frac { 9 }{ 24 } \)

From the given table, age of children in group A
=7yr,7yr,9yr,8yr, 10yr, 10yr
Age of children in group 8 
= 7yr , 9yr, 11 yr, 12yr, 12yr, 12yr
(a) Mode in group A = 7 yr and 10 yr
( \(\because\)  7 y rand 10 yr occurs 2 times)
Range in group A = Maximum value - Minimum value
=10-7=3
Mode in group 8= 12 yr
Range in group 8=12 -7 = 5
(b) If both groups are combined together 7, 7, 7, 9, 9,11,8, 12, 10, 12, 10, 12
Mode = 7 and 12 [\(\because\)  7 and 12 occurs 3 times]
Range = Maximum value - Minimum value
=12-7=5

On the basis of given data, we can draw the following double bar graph:

The demand of Mathematics subject books rose from 1999 to 2002
= 540-450=90
The demand of Hindi subject books rose from 1999 to 2002
= 610 - 490 =120 books
\(\because\) 120 > 90
So, the demand of Hindi books rose faster than Mathematics books.

In each class, number of students < 500

2008 ⟶500 - 500 = 0

200, 300, 400, 500⟶ 4.

600 -100 = 500

A coin has two faces

Friday⟶60.