statistics - testlabz class ix 4 question bank ans. frequency distribution to exclusive form of the...

Math Class IX 1 Question Bank

STATISTICS Q.1 Following data gives the number of children in 40 families :

1, 2, 6, 5, 1, 3, 2, 6, 2, 3, 4, 2, 0, 4, 4, 3, 2, 2, 0, 0, 1, 2, 2, 4, 4, 3, 2, 1, 0, 5, 1, 2, 4,

3, 4, 1, 1, 6, 2, 2.

Represent it in the form of a frequency distribution.

Ans. Below is given a frequency distribution of the given data.

No. of Children Tally-Marks Frequency

0 |||| 4

1 |||| || 7

2 |||| |||| || 12

3 |||| 5

4 |||| || 7

5 || 2

6 ||| 3

Q.2. The marks obtained by 40 students of a class in an examination are given

below. Present the data in the form of a frequency distribution using equal

class-size, one such class being 10–15 (15 not included)

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

18, 2, 5, 7, 10, 16, 8, 16, 17, 8, 23, 21, 6, 23, 15.

Ans. Frequency distribution of the given data can be represented as follows :

Class-Interval Tally-Marks Frequency

1 - 5 |||| | 6

5 - 10 |||| |||| 10

10 - 15 |||| ||| 8

15 - 20 |||| ||| 8

20 - 25 |||| ||| 8

Q.3. Construct a frequency table with equal class intervals from the following

data on the monthly wages (in rupees) of 28 labourers working in a factory,

taking one of the class intervals as 210 - 230 (230 not included).

220, 268, 258, 242, 210, 268, 272, 242, 311, 290, 300, 320, 319, 304, 302, 318,

306, 292, 254, 278, 210, 240, 280, 316, 306, 215, 256, 236.


Ans. Frequency distribution of the given data is as follows :

Class-Interval Tally-Marks Frequency

210 - 230 |||| 4

230 - 250 |||| 4

250 - 270 |||| 5

270 - 290 ||| 3

290 - 310 |||| || 7

310 - 330 |||| 5

Q.4. The weekly wages (in rupees) of 30 workers in a factory are given below :

630, 635, 690, 610, 635, 636, 639, 645, 698, 690, 620, 660, 632, 633, 655, 645,

604, 608, 612, 640, 685, 635, 636, 678, 640, 668, 690, 606, 640, 690.

Represent the data in the form of a frequency distribution with class size 10.

Ans. From the given data, Lowest data = 604 and Largest data = 698

∴ Range of data 698 604 94= − =

Frequency distribution of the given data is as follows :

Class Interval Tally-Marks Frequency

600 - 610 ||| 3

610 - 620 || 2

620 - 630 | 1

630 - 640 |||| |||| 9

640 - 650 |||| 5

650 - 660 | 1

660 - 670 || 2

670 - 680 | 1

680 - 690 | 1

690 - 700 |||| 5

Total 30

Q.5. Construct the cumulative frequency table from the frequency table given

below :

Class Interval Frequency

0 - 6 7

6 - 12 11

12 - 18 8

18 - 24 14

24 - 30 12

Ans. Cumulative frequency table from the given frequency table is as follows :


Class Interval Frequency Cumulative

Frequency

0 - 6 7 7

6 - 12 11 18

12 - 18 8 26

18 - 24 14 40

24 - 30 12 52

Q.6 Construct a frequncy table from the following data :

Age (in years) Number of Students

Less than 10 6

Less than 20 14

Less than 30 30

Less than 40 52

Less than 50 65

Less than 60 70

Ans. Frequency table of the given data is as follows :

Class Interval Cumulative

Frequency

Frequency

0 - 10 6 6

10 - 20 14 8

20 - 30 30 16

30 - 40 52 22

40 - 50 65 13

50 - 60 70 5

Q.7 Convert the following frequency distribution to exclusive form :


30 - 34 7

35 - 39 9

40 - 44 13

45 - 49 6

50 - 54 3

55 - 59 10

Use this table to find :

(i) The true class limits of the fourth class interval.

(ii) The class boundaries of the fifth class interval.

(iii) The class mark of the third class interval.

(iv) The class size of the sixth class interval.


Ans. Frequency distribution to exclusive form of the given frequency distribution is

as follows :


29.5 - 34.5 7

34.5 - 39.5 9

39.5 - 44.5 13

44.5 - 49.5 6

49.5 - 54.5 3

54.5 - 59.5 10

(i) The true class limits of the fourth class interval is 44.5-49.5.

(ii) The class boundaries of the fifth class interval is 49.5-54.5.

(iii) The class mark of the third class interval 39.5 44.5 84

42.2 2

+= = =

(iv) The class size of th sixth class interval 59.5 54.5 5= − =

Q.8. Find the actual lower class limits and upper class limits of the classes :

10 - 19, 20 - 29, 30 - 39 and 40 - 49. Ans. Classes are 10 - 19, 20 - 29, 30 - 39 and 40 - 49.

Difference between upper limit of one class and lower limit of next class 1=

∴ Adjustment of factor 1

0.52

= =

Subtracting the adjustment factor from the lower limits and adding it to all the

upper limits.

Thus, classes become,

9.5 - 19.5, 19.5 - 29.5, 29.5 - 39.5 and 39.5 - 49.5.

Hence, actual lower limits are 9.5, 19.5, 29.5 and 39.5 and actual upper limits are

19.5, 29.5, 29.5 and 49.5.

Q.9. Find the actual lower and upper class limits and also the class marks of the

classes : 1.1 - 2.0, 2.1 - 3.0 and 3.1 - 4.0. Ans. Classes are 1.1 - 2.0, 2.1 - 3.0 and 3.1 - 4.0.

Difference between upper limit of one class and lower limit of next class.

2.1 2.0 0.1= − =

∴ Adjustment factor of class 0.1

0.05.2

= =

Subtracting the adjustment factor from lower limits and adding it to the upper

limits. 1.1 0.05 1.05− = and 2.0 0.5 2.05.+ =

Thus, classes become 1.05 - 2.05, 2.05 - 3.05 and 3.05 - 4.05.

Hence, actual lower limits are 1.05, 2.05 and 3.05 and actual upper limits are

2.05, 3.05 and 4.05.

In this way class marks are :


1.05 2.05 3.10

1.552 2

+= = ,

2.05 3.05 5.102.55

2 2

+= =

and 3.05 4.05 7.10

3.55.2 2

+= =

Hence, class marks are 1.55, 2.55 and 3.55.

Q.10. Use the table given below to find :

(a) The actual class limits of the fourth class.

(b) The class boundaries of the sixth class.

(c) The class-mark of the third class.

(d) The upper and lower limits of the fifth class.

(e) The size of the third class.


30 - 34 7

35 - 39 10

40 - 44 12

45 - 49 13

50 - 54 8

55 - 59 4

Ans. The fourth class if 45-49 and next class is 50 - 54

Adjustment factor 50 49 1

0.52 2

−= = =

Thus, actual limits of class 45 - 49 is 45 0.5 44.5− = and 49 0.5 49.5.+ =

Hence, class is 44.5-49.5.

(b) Sixth class is 55 - 59.

Adjustment factor 0.5=

∴ Class boundaries are 55 0.5 54.5− = and 59. 0.5 59.5+ =

i.e. 54.5-59.5.

(c) Third class is 40 - 44

Actual class limits are 39.5 - 44.5

∴ Class mark 39.5 44.5 84.0

42.0 422 2

+= = = =

(d) Fifth class is 50 - 54

∴ Lower and upper limits are 50 and 54.

(e) Third class is 50 - 54

Actual class becomes 49.5 - 54.5

∴ Size of class 54.5 49.5 5= − =


Q.11. Construct a cumulative frequency distribution table from the frequency

table given below : (i)


0 - 8 9

8 - 16 13

16 - 24 12

24 - 32 7

32 - 40 15

40 - 48 6

(ii)


1 - 10 12

11 - 20 18

21 - 30 23

31 - 40 15

41 - 50 10

Ans. (i)

Class Interval Frequency Cumulative Frequency

0 - 8 9 9

8 - 16 13 22

16 - 24 12 34

24 - 32 7 41

32 - 40 15 56

40 - 48 6 62

Total 62

(ii)


1 - 10 12 12

11 - 20 18 30

21 - 30 23 53

31 - 40 15 68

41 - 50 10 78

Total 78


Q.12. Construct a frequency distribution table from the following cumulative

frequency distribution : (i)

Class Interval Cumulative Frequency

10-19 8

20-29 19

30-39 23

40-49 30

(ii)

Class Interval Cumulative Frequency

5-10 18

10-15 30

15-20 46

20-25 73

25-30 90

Ans. (i) Frequency Distribution Table is as follows :

Class Interval Cumulative Frequency Frequency f

10-19 8 8

20-29 19 19 8 11− =

30-39 23 23 19 4− =

40-49 30 30 23 7− =

Total 30

(ii) Frequency Distribution Table is as follows :

Class Interval Cumulative Frequency Frequency f

5-10 18 18

10-15 30 30 18 12− =

15-20 46 46 30 16− =

20-25 73 73 46 27− =

25-30 90 90 73 17− =

Total 90

Q.13. Construct a frequency table from the following data

Marks Number of Students

Less than 10 6

Less than 20 15

Less than 30 30

Less than 40 39

Less than 50 53

Less than 60 70

Ans. In the given data number of students represents the cumulative frequency.


Thus, Classes are 0 - 10, 10 - 20, 20 - 30, 30 - 40, 40 - 50 and 50 - 60.

Hence, frequency table is as follows :

Marks No. of Students Frequency f

0 - 10 6 6

10 - 20 15 15 6 9− =

20 - 30 30 30 15 15− =

30 - 40 39 39 30 9− =

40 - 50 53 53 39 14− =

50 - 60 70 70 53 17− =

Total 70

Q.14. Construct the frequency distribution table from the following cumulative

frequency table :

Ages Number of Students

Below 4 0

Below 7 85

Below 10 140

Below 13 243

Below 16 300

(i) State the number of students in the age group 10-13.

(ii) State the age group which has the least number of students. Ans. Since, there is no students below age of 4 years, hence starting the classes from

lower limits as 4.

Ages No. of Students Frequency f

4 - 7 85 85

7 - 10 140 140 85 55− =

10 - 13 243 243 140 103− =

13 - 16 300 300 243 57− =

Total 300

(i) Number of students in the age group 10-13 = 103.

(ii) Age group which has the least number of students is 7-10.


Q.15. Fill in the blanks in the following table.


25 - 34 ... 15

35 - 44 ... 28

45 - 54 21 ...

55 - 64 16 ...

65 - 74 ... 73

75 - 84 12 ...

Ans.


25 - 34 15 15

35 - 44 28 15 13− = 28

45 - 54 21 28 21 49+ =

55 - 64 16 49 16 65+ =

65 - 74 73 65 8− = 73

75 - 84 12 73 12 85+ =

Q.16. Draw a bar graph (chart) to represent the following data :

Name of

the Animal

Goat Horse Cow Camel Elephant

Age (in

years)

12 15 15 26 60

Ans. Taking name of the animals along x-axis and age (in years) along y-axis, the

required bar graph (chart) is as shown.


Q.17. Construct a histogram for the following frequency distribution :

Class Interval 0 - 5 5 - 10 10 - 15 15 - 20

Frequency 5 3 6 2

Ans. Taking class intervals along x-axis and

frequency along y-axis. Below is given

histogram of the given data. Scale 1 cm

= 1 frequency.

Q.18. Construct a histogram for the following frequency distribution :

Daily earning

(in Rs.)

80-120 120-160 160-200 200-240

Number of

Workers

4 8 10 2

Ans. Taking daily earnings (in Rs.) along x-

axis and number of workers along y-

axis. Below is given a histogram of the

given data. scale 1 cm = 2 workers.


Q.19. Draw a histogram for the following frequency distribution :

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

No. of students 4 9 5 12 3

Ans. Taking marks along x-axis and

number of students along y-axis. Below

is given the histogram of the given

data. Scale 1 cm = 2 students.

Q.20. Construct a frequency polygon from the following data :

Class

Interval

1-5 6-10 11-15 16-20 21-25

Frequency 5 8 12 7 4

Ans. Taking class intervals along x-axis and

frequency along y-axis. Then, we will

draw a histogram and take mid-points

of each class interval and by joining

them, we get the frequency polygon as

given. Scale 1 cm = 2.

By converting the class intervals in

exclusive form, we get

0.5 - 5.5, 5.5 - 10.5, 10.5 - 15.5, 15.5 -

20.5 and 20.5 - 25.5.


Q.21. The following table shows the marks obtained by the students of a class in

an examination.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50

No. of

Students

15 32 55 35 13

Draw a frequency polygon.

Ans. Taking marks along x-axis and

number of students, along y-axis. First

we draw a histogram and then by

joining the mid-points of consecutive

rectangle, we will get a frequency

polygon. Scale: 1 cm = 10 students.

Q.22. The heights of boys in a school are given below :

Height (in

cm)

140 - 145 145 - 150 150 - 155 155 - 160 160 - 165

No. of boys 24 32 16 20 44

Draw a histogram and a frequency

polygon to represent the above data. Ans. We will take heights (in cm) on x-axis

and number of boys on the y-axis. First

we will draw a histogram and then by


rectangle, we get a frequency polygon.

Scale : 1 cm = 8 boys.


Q.23. Draw a histogram and frequency polygon to represent the following data :

Weight (in

kg)

35 - 40 40 - 45 45 - 50 50 - 55 55 - 60

No. of

Workers

6 17 30 8 3

Ans. Taking weights (in kg) along x-axis

and no. of workers along y-axis. First

we draw a histogram and then by


rectangle, we get a frequency polygon.

Scale : 1 cm = 5 workers.

Q.24. Each of the 25 students in a class was given a home assignment comprising

10 questions in mathematics. The data given below, show the number of

questions solved and submitted by individual students on the next day.

1, 4, 5, 6, 0, 9, 3, 2, 3, 4, 6, 4, 5, 2, 7, 5, 2, 2, 3, 5, 1, 0, 7, 6, 3.

(a) Taking classes as 0-2, 2-4, 4-6, ... etc., make a frequency table for the

above distribution.

(b) Draw frequency polygon to represent the given data.

Ans. (a) Frequency distribution table for the above data is as follows :

Class Interval Tally Marks Frequency

0-2 |||| 4

2-4 |||| ||| 8

4-6 |||| || 7

6-8 |||| 5

8-10 | 1

Total 25

(b) To draw frequency polygon :


(i) Draw the histogram for the given data by taking class intervals along x-axis

and frequency along y-axis.

(ii) Mark the mid-point of top of each

rectangle of histogram.

(iii) Mark the mid-point of immediately

higher class interval (i.e. 10-12)

with frequency zero.

(iv) Join the consecutive mid-points

marked by straight lines. Also, join

the mid-point of immediately lower

class with frequency zero to mid-

point of rectangle for 0-2. Required

frequency polygon is as follow.

Q.25. (a) Draw up a frequency distribution table, for the numbers given below,

using the intervals 10-19, 20-29, ... etc.

33, 48, 50, 67, 84, 41, 57, 96, 69, 59, 19, 52, 53, 54, 66, 28, 56, 63, 77, 49, 30, 54,

45, 87, 65, 15, 34, 39, 45, 49, 60, 32, 47, 29, 69, 39, 46, 47, 58, 27.

(b) Use this table to draw a combined histogram and frequency polygon.

Ans. (a) Frequency distribution table

Class Interval Tally Marks Frequency

10-19 || 2

20-29 ||| 3

30-39 |||| | 6

40-49 |||| |||| 9

50-59 |||| |||| 9

60-69 |||| || 7

70-79 | 1

80-89 || 2

90-99 | 1

Total 40

Now, we convert this table is exclusive form by removing the gap in between

the class intervals.

Here, adjustment factor 1

[20 19] 0.52

= − =


The exclusive form of the above frequency distribution is as follows :


9.5-19.5 2

19.5-29.5 3

29.5-39.5 6

39.5-49.5 9

49.5-59.5 9

59.5-69.5 7

69.5-79.5 1

79.5-89.5 2

89.5-99.5 1

Total 40

Now, to draw the histogram and frequency

polygon we proceed as :

(i) Draw the histogram for the given data by

taking class intervals along x-axis and

frequency along y-axis.

(ii) Mark the mid-point of top of each

rectangle of histogram.

(iii) Mark the mid-point of immediately lower

class interval (i.e. 0.5-9.5) and mid-point

of immediately higher class interval (i.e.

99.5-109.5).

(iv) Join the consecutive mid-points by

straight lines to obtain the frequency

polygon.

Q.26. The following figures represent the amount of money spent in various

years by a company on research and development. Represent the data with a

suitable diagram.

Years 1986-87 1987-88 1988-89 1989-90

Amount (in

crores)

45 70 95 90

Ans. Taking years along x-axis and amount (in crores) along y-axis, the following

Bar-diagram is drawn for the given data. Or

the given data can also be represented by Pie Diagram as follow :


Years Amount in

Crores

Central Angle

1986-87 45 45360 54

300× = °

1987-88 70 70360 84

300× = °

1988-89 95 95360 114

300× = °

1989-90 90 90360 108

300× = °

Total 300 360°

Q.27. Find the mean of first six natural numbers.

Ans. First six natural numbers are 1, 2, 3, 4, 5, 6.

Here, 6n =

∴ Mean 1 2 3 4 5 6

6

ixx

n

Σ + + + + += =

21 73.5

6 2= = =

Q.28. Find the mean of first ten odd natural numbers. Ans. First ten odd natural numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19.

Here, 10n =

∴ Mean 1 3 5 7 9 11 13 15 17 19

10

ixx

n

Σ + + + + + + + + += =

10010

10= =


Q.29. Find the mean of all factors of 10.

Ans. Factors of 10 = 1, 2, 5, 10.

Here, 4n =

∴ Mean 1 2 5 10

4

ixx

n

Σ + + += =

184.5

4= =

Q.30. Find the mean of x x x x+ 3, + 5, + 7, + 9 and x + 11.

Ans. Mean of 3, 5, 7, 9, 11x x x x x+ + + + +

Here, 5n =

∴ Mean 3 5 7 9 11

5

ix x x x x xx

n

Σ + + + + + + + + += =

5 357

5

xx

+= = +

Q.31. If different values of variable ix are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5

and 11.1; find :

(i) the mean x (ii) the value of n

i

i

x x∑=1

( - )

Ans. Different values of variable ix are 9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5 and

11.1.

Here, 10n =

(i) ∴ Mean ixx

n

Σ= =

9.8 5.4 3.7 1.7 1.8 2.6 2.8 8.6 10.5 11.1

10

+ + + + + + + + +=

58.05.8

10= =

(ii) 1 2 3 4 5

1

( ) ( ) ( ) ( ) ( ) ( )n

i

i

x x x x x x x x x x x x

=

− = − + − + − + − + −∑

6 7 8 9 10( ) ( ) ( ) ( ) ( )x x x x x x x x x x+ − + − + − + − + −

1 2 3 4 5 6 7 8 9 10( ) 10x x x x x x x x x x x= + + + + + + + + + −

58.0 10 5.8 58.0 58.0 0= − × = − =

Q.32. The mean of 15 observations is 32. Find the resulting mean, if each

observation is : (i) increased by 3 (ii) decreased by 7

(iii) multiplied by 2 (iv) divided by 0.5

(v) increased by 60% (vi) decreased by 20% Ans. Mean of 15 observations = 32

Total sum 32 15 480x= Σ = × =

(i) Resulting mean when each observation is increased by 3 32 3 35= + =

(ii) Resulting mean when each observation is decreased by 7 32 7 25= − =

(iii) Resulting mean when each observation is multiplied by 2 32 2 64= × =


(iv) Resulting mean when each observation is divided by 0.5 32 0.5 64= ÷ =

(v) Resulting mean when each observation is increased by 60%

32 60% of 32= +60 192

32 32 32100 10

= + × = + 32 19.2 51.2= + =

(vi) Resulting mean when each observation is decreased by 20%

20 32

32 32 32100 5

= − × = − 32 6.4 25.6= − =

Q.33. The mean of 5 numbers is 18. If one number is excluded the mean of the

remaining numbers becomes 16. Find the excluded number.

Ans. Mean of 5 numbers 18=

∴ Total of 5 numbers 18 5 90= × = . After excluding one number

The mean of remaining 4 numbers 16= , Total of 4 numbers 16 4 64= × =

Excluding number 90 64 26= − =

Q.34. If the mean of observation x x x x, + 2, + 4, + 6 and x + 8 is 11, find :

(i) the value of x (ii) the mean of first three observation.

Ans. Mean of , 2, 4, 6, 8x x x x x+ + + + is 11

Here, 5n =

(i) ∴ 2 4 6 8 11 5x x x x x+ + + + + + + + = ×

⇒ 5 20 55 5 55 20 35x x+ = ⇒ = − =

⇒ 35

7.5

x = = Hence, 7x =

(ii) Mean of first 3 numbers2 4 3 6

3 3

x x x x+ + + + += = 2 7 2 9x= + = + = ( 7)x =∵

Q.35. The mean of 100 observations is 40. It is found that an observation 53 was

misread as 83. Find the correct mean.

Ans. Mean of 100 observations 40=

∴ Total of 100 observations

Difference by misreading 83 53 30= − =

Hence, actual total 4000 30 3970= − = and actual mean 3970

39.7100

= =

Q.36. The mean of 200 items was 50. Later on, it was discovered that two items

were misread as 92 and 8 instead of 192 and 88. Find the correct mean.

Ans. Mean of 200 items 50=

∴ Total 50 200 10000= × =

Difference in two items by misreading (192 88) (92 8)= + − +

280 100 180= − =

∴ Actual total 10000 180 10180= + =


Actual mean 10180 5090

50.9200 100

= = =

Q.37. Find the median of :

(i) 25, 16, 26, 16, 32, 31, 19, 28 and 35.

(ii) 241, 243, 147, 350, 327, 299, 261, 292, 271, 258 and 257.

(iii) 63, 17, 50, 9, 25, 43, 21, 50, 14 and 34.

(iv) 233, 173, 189, 208, 194, 204, 194, 185, 200 and 220.

Ans. (i) 25, 16, 26, 16, 32, 31, 19, 28 and 35.

Arranging in ascending order

16, 16, 19, 25, 26, 28, 31, 32, 35

Here, N is 9 which is odd.

∴ Median N 1

2

+= th term

9 1

2

+= th term

10

2= th term 5= th term

Which is 26. Hence, median 26=

(ii) 241, 243, 147, 350, 327, 299, 261, 292, 271, 258 and 257.

Arranging in ascending order,

147, 241, 243, 257, 258, 261, 271, 292, 299, 327, 350.

Here, N 11= which is odd,

∴ Median N 1

2

+= th term

11 1

2

+= th

12

2= th 6= th term.

Which is 261.

Hence, median 261.=

(iii) 63, 17, 50, 9, 25, 43, 21, 50, 14 and 34


9, 14, 17, 21, 25, 34, 43, 50, 50, 63

Here N 10= which is even

∴ Median 1 10 10

th 1 th term2 2 2

= + +

1(5th term 6th term)

2= +

1 1

(25 34) 592 2

= + = ×

29.5=

Hence, median 29.5=

(iv) 233, 173, 189, 208, 194, 204, 194, 185, 200 and 220.


173, 185, 189, 194, 194, 200, 204, 208, 220, 233.


Here, N 10= which is even

∴ Median 1 10 10

th term 1 th term2 2 2

= + +


2= +

1

(194 200)2

= +1

394 1972

= × =

Hence, median 197=

Q.38. The following data has been arranged in ascending order. If their median

is 63, find the value of x.

34, 37, 53, 55, x, x + 2, 77, 83, 89 and 100

Ans. Data in ascending order is

34, 37, 53, 55, x, x + 2, 77, 83, 89, 100

Here, N 10=

∴ Median 1 10 10

th term 1 th term2 2 2

= + +


2= +

1 1

( 2) (2 2)2 2

x x x= + + = + 1x= +

But median is given 63=

∴ 1 63x + = ⇒ 63 1 62x = − =

Hence, 62.x =

Q.39. Find the mean and the median of the following set of numbers :

8, 0, 5, 3, 2, 9, 1, 5, 4, 7, 2, 5

Ans. Arranging in descending order

0, 1, 2, 2, 3, 4, 5, 5, 5, 7, 8, 9

Here, N 12= which is even

∴ Mean ixx

n

Σ=

0 1 2 2 3 4 5 5 5 7 8 9

12

+ + + + + + + + + + +=

51 17

4.2512 4

= = =

and Median 1 12 12

th 1 th term2 2 2

= + +

1

(6th term 7th term)2

= +


1 9

(4 5) 4.52 2

= + = =

Q.40. The mean weight of 60 students of a class is 52.75 kg. If the mean weight of

35 of them is 54 kg, find the mean weight of the remaining students.

Ans. Mean weight of 60 students 52.75= kg

∴ Total weight of 60 students 52.75 60 kg 3165 kg= × =

Mean weight of 35 students among them 54 kg=

∴ Total weight of 35 students 54 35 kg 1890 kg= × =

∴ Remaining students 60 35 25= − =

and total weight of 25 students 3165 1890 1275 kg= − =

∴ Mean weight of 25 students 1275 25 51 kg= ÷ =

statistics - testlabz class ix 4 question bank ans. frequency distribution to exclusive form of the...

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