Presentation of Data Test

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Presentation of Data Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Data was collected on a student's typing rate and graph was drawn as shown below. Approximately how many words had this student typed in $$30$$ seconds?

A
$$20$$

B
$$24$$

C
$$28$$

D
$$34$$

Solution

The correct answer is option (c). approximately $$28$$ words were typed in $$30$$ seconds.
Question 2
1 / -0

Data represented using circle is known as

A
Bar Graph

B
Histogram

C
Pictograph

D
Pie chart

Solution

A pie chart represents the data in circular form.
Hence, option $$D$$ is correct.
Question 3
1 / -0

Look at the histogram and answer the following .
How many workers earn more than Rs. 850?

A
7

B
8

C
9

D
10

Solution

From the graph, we get the following table.

$$\text{Wages (Rs.)}$$ $$\text{Number of workers}$$
800-810 3
810-820 2
820-830 1
830-840 9
840-850 5
850-860 1
860-870 3
870-880 1
880-890 1
890-900 4
The workers who earn more than $$850$$ are $$n(850-860)+n(860-870)+n(870-880)+n(880-890)+n(890-900)=1+3+1+1+4=10$$
Question 4
1 / -0

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are the:

A
upper limits of the classes

B
lower limits of the classes

C
class marks of the classes

D
upper limits of perceeding classes

Solution

For drawing a frequency polygon of a continuous frequency distribution,
we draw the class marks of the classes on the abscissae and the frequency of the classes on the ordinates.

We know, class marks are the mean of lower and upper limit of the class intervals.

Hence, option $$C$$ is correct.
Question 5
1 / -0

Directions For Questions

A group of $$360$$ people were asked to vote for their favourite season from the three seasons rainy, winter and summer.

...view full instructions

Find the central angle of pie chart for winter.

A
$$90$$

B
$$120$$

C
$$150$$

D
None of these

Solution

Central Angle for Summer $$= \dfrac { 90 }{ 360 } \times 360 = 90$$ degree

Central Angle for Rainy $$= \dfrac { 120 }{ 360 } \times 360 = 120$$ degree

Central Angle for Winter $$= \dfrac { 150 }{ 360 } \times 360 = 150$$ degree
Question 6
1 / -0

Class mark of a particular class is 10.5 and class size is 7, then class interval is :

A
10.5 -17.5

B
3.5 -10.5

C
7 -17.5

D
7 -14

Solution

Class mark is the mid point of the class interval.
Class size is the upper limit minus lower limit.
Given that class mark of a class is $$10.5$$ and
class size is $$7$$
By option verification,
Opition A class mark is $$\dfrac{10.5+17.5}{2}=\dfrac{28}{2}=14$$ False

Option B class mark is $$\dfrac{10.5+3.5}{2}=\dfrac{14}{2}=7$$ False

Option C class mark is $$\dfrac{17.5+7}{2}=\dfrac{24.5}{2}=12.25$$ False

Option D class size is $$14-7=7$$ and
Class mark is $$\dfrac{14+7}{2}=\dfrac{21}{2}=10.5$$ True

Therefore the class interval is $$7-14$$
Question 7
1 / -0

Class mark of class interval 60 - 70 is

A
60

B
70

C
65

D
75

Solution

Class mark is the mid point of the class interval.
Therefore, the class mark for the interval $$60-70$$ is $$\dfrac{60+70}{2}=\dfrac{130}{2}=65$$
Question 8
1 / -0

If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5, then the class corresponding to the class mark 33.5 is :

A
16 - 23

B
23-30

C
30-37

D
37-41

Solution

Class mark is the mid point of the class interval.
Class size is the upper limit minus lower limit.
Given that class mark of a class is $$6.5$$ and
the class size is $$26.5-19.5=33.5-26.5=40.5-33.5=7$$
By option verification,
Opition A class mark is $$\dfrac{16+23}{2}=\dfrac{39}{2}=19.5$$ False

Option B class mark is $$\dfrac{30+23}{2}=\dfrac{53}{2}=26.5$$ False

Option C class mark is $$\dfrac{30+37}{2}=\dfrac{67}{2}=33.5$$ and
class size is $$37-30=7$$ True

Option D class mark is $$\dfrac{37+41}{2}=\dfrac{78}{2}=39$$ False

Therefore the class interval is $$30-37$$
Question 9
1 / -0

The class mark of the class 29.5 $$-$$ 30.5 is :

A
30

B
30.5

C
31.5

D
31

Solution

Class mark is the mid point of the class interval.

Therefore, the class mark for the interval $$29.5-30.5$$ is $$\dfrac{29.5+30.5}{2}=\dfrac{60}{2}=30$$
Question 10
1 / -0

The class mark of the class 100 $$-$$ 150:

A
$$100$$

B
$$125$$

C
$$130$$

D
$$150$$

Solution

Class mark is the mid point of the class interval.
Therefore the class mark of the class $$100-150$$ is $$\dfrac{100+150}{2} = \dfrac{250}{2}=125$$

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Re-Attempt Weekly Quiz Competition

$$\text{Wages (Rs.)}$$	$$\text{Number of workers}$$
800-810	3
810-820	2
820-830	1
830-840	9
840-850	5
850-860	1
860-870	3
870-880	1
880-890	1
890-900	4

Presentation of Data Test - 21

Result

Presentation of Data Test - 21

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SHARING IS CARING

Question 1

$$20$$

$$24$$

$$28$$

$$34$$

Solution

Question 2

Bar Graph

Histogram

Pictograph

Pie chart

Solution

Question 3

7

8

9

10

Solution

Question 4

upper limits of the classes

lower limits of the classes

class marks of the classes

upper limits of perceeding classes

Solution

Question 5

$$90$$

$$120$$

$$150$$

None of these

Solution

Question 6

10.5 -17.5

3.5 -10.5

7 -17.5

7 -14

Solution

Question 7

60

70

65

75

Solution

Question 8

16 - 23

23-30

30-37

37-41

Solution

Question 9

30

30.5

31.5

31

Solution

Question 10

$$100$$

$$125$$

$$130$$

$$150$$

Solution

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