Question

please answer this question if you know the answer then only answer else reported answer all of them​

Answer 1

Question : -

The marks obtained by 40 students of class VIII in an examination are given below :

18,8,12,6,8,16,12,5,23,2,16,23,2,10,20,12,9,7,6,5,3,5,13,21,13,15,20,24,1,7,21,16,13,18,23,7,3,18,17,16

Represent the data in form of an frequency distribution table using the same class size, one such class being 15 - 20 . ( where 20 is not included )?

ANSWER

Given : -

The marks obtained by 40 students of class VIII in an examination are given below :

18,8,12,6,8,16,12,5,23,2,16,23,2,10,20,12,9,7,6,5,3,5,13,21,13,15,20,24,1,7,21,16,13,18,23,7,3,18,17,16

Required to find : -

• Frequency distribution table ?

Solution : -

Here,

We need to construct/prepare a frequency distribution table for the above data .

The data is , Marks scored by the students ;

18,8,12,6,8,16,12,5,23,2,16,23,2,10,20,12,9,7,6,5,3,5,13,21,13,15,20,24,1,7,21,16,13,18,23,7,3,18,17,16

It is mentioned that , the size of class interval should be like the class 15 - 20 ( where 20 isn't included )

The meaning of 20 isn't included is we should consider 20 in 15 - 20 class interval instead we should consider it in the next consecutive interval .

Same in the case of all numbers which end at the each class interval .

Now,

Let's form a frequency distribution table with a class interval size of 5 .

$\huge{\begin{tabular}{|c|c|c|} \cline{1-3}\multicolumn{3}{|c|}{ Frequency Distribution table } \\ \cline{1-3} 0 - 5 & \mid \mid \mid \mid & 4 \\\cline{1-3} 5 - 10 & \not{ \mid \mid \mid \mid } \: \not{||||} \: | & 11 \\ \cline{1-3} 10 - 15 & \not{ \mid \mid \mid \mid} \: \: \mid \mid & 7 \\ \cline{1-3} 15 - 20 & \not{ \mid \mid \mid \mid} \: \: \not{ \mid \mid \mid \mid }& 10 \\ \cline{1-3} 20 - 25 & \not{ \mid \mid \mid \mid} \: \: \mid \mid \mid & 8 \\ \cline{1 - 3} \end{tabular}}$

Question - 2

ANSWER

a ) How many pedestrians passed the street before 9 a.m. ?

Answer : 120 + 160 + 180 + 170 = 630

b ) How many pedestrians passed the street after 9 a.m. ?

Answer : 100 + 140 + 120 + 130 = 490

c ) what is the ratio of no. of pedestrians passing between 8 - 8:15 a.m. to no. of pedestrians passing between 9 - 9:15 a.m. ?

Answer :

120 : 100

12 : 10

6 : 5

d ) What is the % increase in the number of pedestrians from 8 - 8:15 a.m. to 8:15 - 8:30 a.m. ?

Answer :

No. of pedestrians from 8 - 8:15 a.m. = 120

No. of pedestrians from 8:15 - 8:30 a.m. = 160

This implies ;

160 - 120

40

=> ( 40/120 ) x 100

=> 4000/120

=> 400/12

=> 33.33 . . .

=> 33.3 % ( approx. )

3rd Question

Answer

Total sum of all frequencies =

90 + 75 + 140 + 190 + 105

600

Divide 360° by 600

360°/600 =

36°/60

0.6°

This implies ;

Percentage of Birds = 90 x 0.6 = 54°

Percentage of Reptiles = 75 x 0.6 = 45°

Percentage of water animals = 140 x 0.6 = 84°

Percentage of Beast animals = 190 x 0.6 = 114°

Percentage of land animals = 105 x 0.6 = 63°

Note : -

To verify your calculations let's perform a verification step .

Sum of all the percentages should be equal to 360°

54° + 45° + 84° + 114° + 63° = 360°

360° = 360°

For pie chart refer attachment !

4th Question

ANSWER

Here, in this question we are asked to plot a histogram .

So,

In histogram ,

We should need to take the class intervals on the x - axis .

The frequency should be taken on y - axis .

The scale of the graph should be need to mentioned accuracy .

However,

The scale of the graph which I had plotted is ;

Scale

on x - axis , 1cm = 500 units

on y - axis , 1cm = 2 units

If the the class interval begins with 0 then the bars will be coming attached to y - axis . if not they won't come attached with y - axis .

Graphs are widely used in markets to check out the profits and losses .

For histogram , refer to the attachment .