8 Math Data Handling

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Finish Line & Beyond

Data Handling

1. Data mostly available to us in an unorganised form is called raw data.

2. In order to draw meaningful inferences from any data, we need to organise thedata systematically.

3. Frequency gives the number of times that a particular entry occurs.4. Raw data can be ‘grouped’ and presented systematically through ‘grouped fre-

quency distribution’.5. Grouped data can be presented using histogram. Histogram is a type of bar dia-

gram, where the class intervals are shown on the horizontal axis and the heights of the bars show the frequency of the class interval. Also, there is no gap between the

bars as there is no gap between the class intervals.6. Data can also presented using circle graph or pie chart.A circle graph shows the

relationship between a whole and its part.7. There are certain experiments whose outcomes have an equal chance of occur-

ring.

8. Arandom experiment is one whose outcome cannot be predicted exactly in ad-vance.9. Outcomes of an experiment are equally likely if each has the same chance of oc-

curring.

10. Probability of an event =

experimenttheof outcomesof number Total

eventanmakethatoutcomesof Number , when the outcomes are equally likely.

11. One or more outcomes of an experiment make an event.12. Chances and probability are related to real life.

Exercise 1

1. For which of these would you use a histogram to show the data?

(a) The number of letters for different areas in a postman’s bag.(b) The height of competitors in an athletics meet.

(c) The number of cassettes produced by 5 companies.(d) The number of passengers boarding trains from 7:00 a.m. to 7:00 p.m. at a

station.Give reasons for each.

Answer: Data in (b) and (d) can be represented using a histogram. These data can

be arranged in class intervals. This is what is required to depict a histogram. There

should be no gap between two bars.

2. The shoppers who come to a departmental store are marked as: man (M),woman (W), boy (B) or girl (G). The following list gives the shoppers whocame during the first hour in the morning:

W W W G B W W M G G M M W W W W G B M W B G G M W W M M W W

W M W B W G M W W W W G W M M W W M W G W M G W M M B G G WMake a frequency distribution table using tally marks. Draw a bar graph to illustrate

it.

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Answer:

Shopper Tally Marks Number

M 15

WIII

28

B 5

GII

12

3. The weekly wages (in Rs) of 30 workers in a factory are.

830, 835, 890, 810, 835, 836, 869, 845, 898, 890, 820, 860, 832, 833, 855, 845,804, 808, 812, 840, 885, 835, 835, 836, 878, 840, 868, 890, 806, 840

Using tally marks make a frequency table with intervals as 800–810, 810–820 and

so on.

Answer:

Interval Tally Marks Total

800-810 III 3

810-820 II 2

820-830 I 1

830-840IIII

9

840-850 5

850-860 I 1

860-870 III 3

870-880 I 1

880-890 I 1

890-900 IIII 4

4. Draw a histogram for the frequency table made for the data in Question

3, and answer the following questions.(i) Which group has the maximum number of workers?

(ii) How many workers earn Rs 850 and more?(iii) How many workers earn less than Rs 850?

Answer:








800-810810-820820-830 830-840 840-850850-860 860-870870-880880-890 890-900

0

1

2

3

4

5

6

7

8

9

10

Total

(i) The group earning Rs. 830-840 per week is having the most number of workers.

(ii) 10 workers are earning Rs. 850 or more.(iii) 20 workers are earning less than Rs. 850.

5. The number of hours for which students of a particular class watchedtelevision during holidays is shown through the given graph.

Answer the following.(i) For how many hours did the maximum number of students watch TV?

(ii) How many students watched TV for less than 4 hours?(iii) How many students spent more than 5 hours in watching TV?

1—2 2—3 3—4 4—5 5—6 6—7

0

5

10

15

20

25

30

35

4

8

22

32

8

6

Students

Answer: (i) Most of the students watched TV for 4 to 6 hours.

(ii) 34 students watched TV for less than 4 hours.

(iii) 14 students watch TV for more than 5 hours

Exercise 2








1. A survey was made to find the type of music that a certain group of young peopleliked in a city. Adjoining pie chart shows the findings of this survey.

20.00%

10.00%

30.00%

40.00%

Music Lovers

Semi Classical

Classical

Folk

Light

From this pie chart answer the following:

(i) If 20 people liked classical music, how many young people were surveyed?

Answer: When 10 people liked classical music, number of people surveyed is 100

∴ When 1 person liked the classical music, number of people surveyed is10100

∴ When 20 people like the classical music, number of people surveyed =

2002010

100 =

(ii) Which type of music is liked by the maximum number of people?

Answer: Light music is liked by maximum number of people.

(iii) If a cassette company were to make 1000 CD’s, how many of each type wouldthey make?

Answer: Number of CD’s for different musicSemi Classical = 200%201000 =Classical = 100%101000 =Folk = 300%301000 =Light = 400%401000 =

2. A group of 360 people were asked to vote for their favourite season from the

three seasons rainy, winter and summer.








Season No. of Votes

Summer 90

Rainy 120Winter 150

(i) Which season got the most votes?

(ii) Find the central angle of each sector.(iii) Draw a pie chart to show this information.

Answer: (i) Winter got the most number of votes.

(ii) Central Angle can be calculated as follows:

°360VotesTotal

Votesof Number Seasons

Summer’s Central Angle = °90360360

90

Rainy’s Central Angle = °120360360

120

Winter’s Central Angle = °150360360

150

(iii) Pie Chart is as follows:

25%

33%

42%

Favourite Season

Summer

Rainy

Winter








3. Draw a pie chart showing the following information. The table shows the colourspreferred by a group of people.

Colours Number of People

Blue 18

Green 9

Red 6

Yellow 3

Total 36

Calculate respective proportion of each colour and use that proportion to find thecentral angle.

Answer: Following is the required pie chart:

50%

17%

8%

25%

Favourite Colour

Blue

Red

Yellow

Green

Proportion of Blue =2

1

36

18 =

So, Central Angle for Blue = °1802

1360

Proportion of Red =6

1

36

6 =

So, Central Angle for Red = °6061360

Proportion of Yellow =12

1

36

3 =

So, Central Angle for Yellow = °3012

1360








Proportion of Green =4

1

36

9 =

So, Central Angle for Green = °904

1360

4. The adjoining pie chart gives the marks scored in an examination by a

student in Hindi, English, Mathematics, Social Science and Science. If thetotal marks obtained by the students were 540, answer the following ques-

tions. All values are shown in degrees.

90

55

70

80

65

Marks Scored

Hindi

English

Math

Social

Science

Science

(i) In which subject did the student score 105 marks?

Central Angle = °70360540

105

In the pie chart 70° is the angle for Math, so the student scored 105 marks in math.

(ii) How many more marks were obtained by the student in Mathematics

than in Hindi?

The difference in central angle between Math and Hindi = 90°-70°=20°

So, Difference in Scores in Hindi And Math = 30540360

20 =°°

(iii) Examine whether the sum of the marks obtained in Social Science and

Mathematics is more than that in Science and Hindi.

Answer: Total of Angles of Social Science and Math = 80°+70°=150°Total of Angles of Science and Hindi = 65°+90°=155°

It is clear that sum of marks obtained in Social Science and Math is less than that inScience and Hindi.








5. The number of students in a hostel, speaking different languages is givenbelow. Display the data in a pie chart.

Language Hindi English Marathi Tamil Bengali Total

Number of Students

40 12 9 7 4 72

Answer:

40

12

9

7

4

Language Spoken

Hindi

English

Marathi

Tamil

Bengali








So, probability of favourable outcome =5

1

(b) Probability of getting an ace from a well shuffled deck of 52 playing cards?

Answer: Total Number of events = 52Number of Favourable outcomes = 4 (because there are 4 aces in a pack of cards)

Probability of favourable outcome:13

1

52

4=

(c) Probability of getting a red candies. (See figure below)

Answer: Total number of events= 7Number of Favourable outcomes = 4

Probability of favourable outcomes =

7

4

4. Numbers 1 to 10 are written on ten separate slips (one number on oneslip), kept in a box and mixed well. One slip is chosen from the box without

looking into it. What is the probability of .(i) getting a number 6?

(ii) getting a number less than 6?(iii) getting a number greater than 6?

(iv) getting a 1-digit number?

Answer: Total number of events = 10Number of favourable events for getting number 6 = 1

Probability of getting number 6 =

10

1

Number of favourable outcomes for getting a number less than 6 = 5

Probability of getting a number less than 6 =2

1

10

5 =

Probability of getting a number greater than 6 =5

2

10

4 =

Probability of getting a 1-digit number =10

9








5. If you have a spinning wheel with 3 green sectors, 1 blue sector and 1

red sector, what is the probability of getting a green sector? What is theprobability of getting a non blue sector?

Answer: Total number of events = 5

Probability of getting green =5

3

Probability of Getting Blue =5

1

So, probability of getting non-Blue =5

4

5

15

5

11 =−=

6. Find the probabilities of the events given in Question

(i) Probability of getting prime numbers=2

1

6

3 =

(ii) Probability of getting non-prime numbers=2

1

6

3 =

(iii) Probability of getting a number greater than 5 =6

1

(iv) Probability of getting a number not greater than 5 =6

5





8 Math Data Handling

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