Frequency distribution of scores:

p = 12

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
We have, total marks 
= 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 +95
= 730
Number of students = 10
\(\therefore\) mean = \(\frac { total\ marks\ obtained\ by\ group }{ number\ of\ students } \)
= \(\frac { 730 }{ 10 } =73\)
Hence, mean marks obtained by the group is 73.

Let the weight (in kg) of 20 children of your classare as follows:
35, 35, 40, 33, 36, 40, 50, 38, 45, 34,
40, 40, 37, 34, 32, 48, 42, 35, 36, 40
Now, arranging the weight of 20 children of your class in descending order, we have
50, 48, 45, 42, 40, 40, 40, 40, 40, 38
37, 36, 36, 35, 35, 35, 34, 34, 33, 32
The most common weight is 40 kg, because five children have 40 kg weight.

(a) First four counting numbers are 1,2,3,4
\(\Rightarrow Mean=\frac{1+2+3+4}{4}\)
\(=\frac{10}{4}=2.5\)
(b) Arranging the numbers with same values together, we get
1,1,2,3,3,3,3,3,3,4,5,6,9
In this data maximum frequency is of 3 which is 6.
\(\therefore\) mode is 3
(c) \(Mean=\frac{98+99+100+0+1+2}{6}\)
\(=\frac{300}{6}=50.\)

Average number of points scared by A = 12.5 [as calculated in part (i)]
Average number of points scared by B = 4.5 [as calculated in part (iii)]
\(=\frac { sum\ of\ points\ in\ games }{ number\ of\ games } \)
= \(\frac { 8+11+13 }{ 3 } =\frac { 32 }{ 3 } \) =10.67
\(\because\)  12.5 > 10.67 > 4.5
Hence, A is the best performer.

The frequency table is shwn below.

From the above frequency table,
Highest number = 9
Lowest number = 1
\(\therefore\)  A range of data = Highest number - Lowest number
     =9-1=8
Hence, the range of the data is 8.

The frequency table is shown below.

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

We go through the following steps to construct the bar graph:
(a)Take a graph paper and draw two lines OX and OY perpendicular to each other. Call the horizontal line as OX and the vertical line as OY.
(b) Along the horizontal axis OX, mark "sections of Class VII" and along the vertical axis OYmark "No. of students".
(c) Along the horizontal axis OX, choose the uniform (equal) width of the bars and the uniform gap between them.
(d) Choose a suitable scale to determine the heights of the bars, according to the space available for the graph. Here, we choose 1 small division to represent 1 student.
(e) Calculate the heights of the various bars as follows:
Height of the bar for section A = 40 x 1
= 40 small divisions = 4 big divisions
Height of the bar for Section B = 48 x 1
= 48 small divisions
= 4 big divisions and 8 small divisions
Height of the bar for Section C = 52 x 1
= 52 small divisions
= 5 big divisions and 2 small divisions
Height of the bar for Section D = 45 small divisions
= 4 big divisions and 5 small divisions.
Height of the bar for Section E = 30 small divisions
= 3 big divisions.
(f) We draw the bars as shown in Fig. below and on the top of each bar, we write the number of students represented by it.

(a) For company A Number of tested watches=40
Number of leaked watches = 20
Now, required fraction = \(\frac { Leaked\quad watches }{ Tested\quad watches } =\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \)  [dividing numerator and denominator by 10]
For company B Number of tested watches = 40
Number of leaked watches = 10
Required fraction  = \(\frac { 10 }{ 40 } =\frac { 1 }{ 4 } \) [dividing numerator and denominator by 10]                     
For company C Number of tested watches = 40
Number of leaked watches = 15
Required fraction  = \(\frac { 15 }{ 40 } =\frac { 3 }{ 8 } \) [dividing numerator and denominator by 5]For company D Number of tested watches = 40
Number of leaked watches = 25
Required fraction = \(\quad \frac { 25 }{ 40 } =\frac { 5 }{ 8 } \) [dividing numerator and denominator by 5]
(b) From above discussion, it is clear that company B has better watches, because company B has least fraction of the number of watches that leaked to the number of watches that were tested.

A double bar graph can be used to compare two collections of data.

We know that on a die, there are six faces with numbers from 1 to 6. Clearly, all the numbers are less than 7. Hence, whatever number appears on throwing a die, it will be less than 7. In other words, we can say that it is certain to get a number less than 7 on throwing a die. Hence, its probability is 1

\(Mean= \frac { sum\ of\ observations }{ number\ of\ observatins } \)
\(6=\frac { sum\ of\ observations }{ 12 } \)
Sum of observations = 12 x 6 = 72

Mode = maximum occuring observation = 67  (occurs 2 times)

Total number of students = 8
Sum of the marks obtained =85 + 62 + 36 + 48 + x + 75 + 39 + 72=417+x
\(\therefore \)Mean marks =\(\frac { (417\quad +x) }{ 8 } \)
\(\Rightarrow \)\(\frac { (417\quad +x) }{ 8 } \)=60
\(\Rightarrow \)417 + x = 60 x 8 = 480
\(\Rightarrow \)x = 480 - 417 = 63
Thus, the required value of x = 63