Ch 4 Organisation of data Statistics grade 11

Page No 4.54:

Question 1:

Following are the figures of marks obtained by 40 students. You ae required to arrange them in ascending and in descending order.

15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10

ANSWER:

Marks in Ascending Order:
3, 3, 4, 5, 6, 6, 7, 8, 8, 8, 9, 10, 10, 10, 10, 11, 11, 12, 13, 14, 14, 14, 14, 15, 15, 16, 16, 17, 17, 18, 18, 18, 18 ,19, 19, 21, 22, 22, 22, 25

Marks in Descending Order:
25, 22, 22, 22, 21, 19, 19, 18, 18, 18, 18, 17, 17, 16, 16, 15, 15, 14, 14, 14, 14, 13, 12, 11, 11, 10, 10, 10, 10, 9, 8, 8, 8, 7, 6, 6, 5, 4, 3, 3

Page No 4.54:

Question 2:

Heights (in inches) of 35 students of a class in given below. Classify the following data in a discrete frequency series

58	60	72	68	58	55	58	72	55	68	66	60
60	55	76	66	58	72	68	60	55	68	55	58
62	66	72	58	60	68	55	58	65	72	68

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Height (in inches)	Tally Marks	Frequency
55 58 60 62 65 66 68 72	$ΙΙΙΙ Ι$ $ΙΙΙΙ Ι Ι$ $ΙΙΙΙ$ $Ι$ $Ι$ $ΙΙΙ$ $ΙΙΙΙ Ι$ $ΙΙΙΙ Ι$	6 7 5 1 1 3 6 6
		$\sum_{} f = 35$

Page No 4.54:

Question 3:

The following are the marks of the 30 students in statistics. prepare a frequency distribution taking the class−intervals as 10−20, 20−30 and so on.

12	33	23	25	18	35	37	49	54	51	37	15	33	42	45
47	55	69	65	63	46	29	18	37	46	59	29	35	45	27

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Marks	Tally Marks	Number of students (Frequency)
10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70	$ΙΙΙΙ$ $ΙΙΙΙ$ $ΙΙΙΙ ΙΙ$ $ΙΙΙΙ ΙΙ$ $ΙΙΙΙ$ $ΙΙΙ$	4 5 7 7 4 3
		$\sum_{} f = 30$

Page No 4.54:

Question 4:

Following are the marks (out of 100) obtained by 50 students in statistics:

70	55	51	42	57	45	60	47	63	53
33	65	39	82	55	64	58	61	65	42
50	52	53	45	45	25	36	59	63	39
65	54	49	54	64	75	42	41	52	35
30	35	15	48	26	20	40	55	46	18

make a frequency distribution taking the first class interval as 0-10.

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Marks	Tally Marks	Number of students (Frequency)
0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90	- $ΙΙ$ $ΙΙΙ$ $ΙΙΙΙ ΙΙ$ $ΙΙΙΙ ΙΙΙΙ ΙΙ$ $ΙΙΙΙ ΙΙΙΙ ΙΙΙΙ$ $ΙΙΙΙ ΙΙΙΙ$ $ΙΙ$ $Ι$	0 2 3 7 12 14 9 2 1
		$\sum_{} f = 50$

Page No 4.54:

Question 5:

Prepare a frequency table taking class intervals 20−24, 25−29, 30−34 and so on, from the following data:

21	20	55	39	48	46	36	54	42	30
29	42	32	40	34	31	35	37	52	44
39	45	37	33	51	53	52	46	43	47
41	26	52	48	25	34	37	33	36	27
54	36	41	33	23	39	28	44	45	38

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Class Intervals	Tally Marks	Frequency
20 - 24 25 - 29 30 - 34 35 - 39 40 - 44 45 - 49 50 - 54 55 - 59	$ΙΙΙ$ $ΙΙΙΙ$ $ΙΙΙΙ ΙΙΙ$ $ΙΙΙΙ ΙΙΙΙ Ι$ $ΙΙΙΙ ΙΙΙ$ $ΙΙΙΙ ΙΙ$ $ΙΙΙΙ ΙΙ$ $Ι$	3 5 8 11 8 7 7 1
		$\sum_{} f = 50$

Page No 4.55:

Question 6:

From the following data, calculate the lower limit of the first class and upper limit of the last class.

Daily Wages	Less than 120	120−140	140−160	160−180	Above 180
No. of Workers	35	12	10	40	13

ANSWER:

In the given question, the size of each class is 20. Thus, maintaining the uniformity, we take the first class as 100-120 and the last class as 180-200. Therefore, the lower limit of the first class-interval is 120 − 20 = 100 and the upper limit of the last class is 180 + 20 = 200.

Page No 4.55:

Question 7:

Calculate the missing class-intervals from the following distribution:

Class-Interval	Below 20	20−50	50−90	90−140	More than 140
Frequency	7	12	16	2	4

ANSWER:

In the given question, the class-intervals are of different width. Thus, the missing class-intervals cannot be determined.

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Question 8:

Convert the following 'more than' cumulative frequency distribution into a 'less than' cumulative frequency distribution

Class-Interval (More than)	10	20	30	40	50	60	70	80
Frequency	124	119	107	84	55	31	12	2

ANSWER:

The 'more than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.

Class-Interval	Frequency
10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90	124 − 119 = 5 119 − 107 = 12 107 − 84 = 23 84 − 55 = 29 55 − 31 = 24 31 − 12 = 19 12 − 2 = 10 2

From the above distribution, we can make the 'less than' cumulative frequency distribution as follows.

Class-Interval	Frequency
Less then 20 Less then 30 Less then 40 Less then 50 Less then 60 Less then 70 Less then 80 Less then 90	5 5 + 12 = 17 17 + 23 = 40 40 +29 = 69 69 + 24 = 93 93 + 19 = 112 112 + 10 = 122 122 + 2 = 124

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Question 9:

Convert the following cumulative frequency series into simple frequency series

Marks	No. of Students
Less than 20	10
Less than 40	18
Less than 60	25
Less than 80	45
Less than 100	55

ANSWER:

The 'less than' cumulative frequency distribution can be presented in the form of a 'simple frequency distribution' as follows.

Class-Interval	Frequency
0 - 20 20 - 40 40 - 60 60 - 80 80 - 100	10 18 − 10 = 8 25 − 18 = 7 45 − 25 = 20 55 − 45 = 10

Page No 4.55:

Question 10:

Prepare a frequency distribution from the following data:

Mid-Points	25	35	45	55	65	75
Frequency	6	10	9	12	8	5

ANSWER:

The class interval can be calculated from the mid-points using the following adjustment formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
$Value of adjustment = \frac{35 - 25}{2} = 5$

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 25 – 5 = 20

Upper limit of first class = 25 + 5 = 30

Thus, the first class interval is (20 − 30). Similarly, we can calculate the remaining class intervals.

Frequency Distribution

Mid Value	Class Interval	Frequency
25 35 45 55 65 75	20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	6 10 9 12 8 5
		$\sum_{}^{} f = 50$

Page No 4.55:

Question 11:

The ages of 20 husbands and wives are given below. Form a two-way frequency table showing the relationship between the ages of husbands and wives with the class-intervals 20−25, 25−30, etc.

S.No	Age of Husband	Age of Wife	S. No.	Age of Husband	Age of Wife
1	28	23	11	27	24
2	37	30	12	39	34
3	42	40	13	23	20
4	25	26	14	33	31
5	29	25	15	36	29
6	47	31	16	32	35
7	37	35	17	22	23
8	35	25	18	29	27
9	23	21	19	38	34
10	41	38	20	48	47

ANSWER:

The given data can be presented in the form of a bivariate frequency distribution as follows.

Age of wife (Y)	Age of husband (X)						Total
	20 − 25	25 − 30	30 − 35	35 − 40	40 − 45	45 − 50	Total

20 − 25							5
25 − 30							5
30 − 35							5
35 − 40							3
40 − 45							1
45 − 50							1
Total	3	5	2	6	3	1	20

Page No 4.56:

Question 12:

The following data give the points scored in a tennis match by two player X and Y at the end of twenty games:

(10,12)	(2,11)	(7,9)	(15,19)	(17,21)	(12,8)	(16,10)	(14,14)	(22,18)	(16,4)
(15,16)	(22,20)	(19,15)	(7,18)	(11,11)	(12,18)	(10,10)	(5,13)	(11,7)	(10,10)

Taking class intervals as: 5−9, 10−14, 15−19..., for both X and Y, construct a Bivariate Frequency Distribution.

ANSWER:

The given data can be presented in the form of a bivariate frequency distribution as follows.

Player Y	Player X					Total
	0 − 4	5 − 9	10 − 14	15 − 19	20 − 24	Total

0 − 4						1
5 − 9						3
10 − 14						8
15 − 19						6
20 − 24						2
Total	1	3	8	6	2	20

Page No 4.56:

Question 13:

In a survey, it was found that 64 families bought milk in the following quantities in a particular month. Prepare a frequency distribution with classes as 5−9, 10−14 etc.

19	16	22	9	22	12	39	19	6	24	16	18	7	17	20	25
10	24	20	21	10	7	18	28	14	23	25	34	22	14	33	23
13	36	11	26	11	37	30	13	22	21	32	21	31	17	16	23
15	27	17	21	23	14	24	8	9	15	29	20	18	28	26	12

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Quantity	Tally Marks	Frequency
5 − 9		6
10 − 14		11
15 − 19		13
20 − 24		18
25 − 29		8
30 − 34		5
35 − 39		3
40 − 44
		$\sum_{}^{} f = 64$

Page No 4.56:

Question 14:

You are given below a mid value series, convert it into a continuous series.

Mid-Value	15	25	35	45	55
Frequency	8	12	15	10	5

ANSWER:

The class interval can be calculated from the mid-points using the following adjustment formula.

The value obtained is then added to the mid point to obtain the upper limit and subtracted from the mid-point to obtain the lower limit.

For the given data, the class interval is calculated by the following value of adjustment.
$Value of adjustment = \frac{25 - 15}{2} = 5$

Thus, we add and subtract 5 to each mid-point to obtain the class interval.

For instance:

The lower limit of first class = 15 – 5 = 10

Upper limit of first class = 15 + 5 = 20

Thus, the first class interval is (10 − 20). Similarly, we can calculate the remaining class intervals.

Mid value	Class Interval	Frequency
15 25 35 45 55	10 − 20 20 − 30 30 − 40 40 − 50 50 − 60	8 12 15 10 5
		$\sum_{}^{} f = 50$

Page No 4.56:

Question 15:

For the data given below, prepare a frequency distribution table with classes 100−110, 110-120, etc.

125	108	112	126	110	132	136	130	140	155
120	130	136	138	125	111	112	125	140	148
147	137	145	150	142	135	137	132	165	154

ANSWER:

In the form of frequency distribution, the given data can be arranged as follows.

Class Interval	Telly Marks	Frequency
100 − 110		1
110 − 120		4
120 − 130		5
130 − 140		10
140 − 150		6
150 − 160		3
160 − 170		1
		$\sum_{}^{} f = 30$

Page No 4.56:

Question 16:

Prepare a bivariate frequency distribution for the following data for 20 students:

Marks in Maths	10	11	10	11	11	14	12	12	13	10
Marks in Statistics	20	21	22	21	23	23	22	21	24	23
Marks in Maths	13	12	11	12	10	14	14	12	13	10
Marks in Statistics	24	23	22	23	22	22	24	20	24	23

ANSWER:

The data can be presented in the form of a bivariate distribution as follows.

Marks in Statistics	Marks in Maths
	10	11	12	13	14	Total

20						2
21						3
22						5
23						6
24						4
Total	5	4	5	3	3	20

Page No 4.56:

Question 17:

In a school, no student has scored less than 25 marks or more than 60 marks in an examination. Their cumulative frequencies are as follows:

Less than	60	55	50	45	40	35	30
Total frequency	120	115	90	75	65	45	32

Find the frequencies in the class intervals 25−30, 30−35,....55−60.

ANSWER:

Marks	Cumulative Frequency
Less than 60 Less than 55 Less than 50 Less than 45 Less than 40 Less than 35 Less than 30	120 115 90 75 65 45 32

This can be presented in the form of a simple frequency distribution as follows.

Marks	Number of Students
25 − 30 30 − 35 35 − 40 40 − 45 45 − 50 50 − 55 55 − 60	32 45 − 32 = 13 65 − 45 = 20 75 − 65 = 10 90 − 75 = 15 115 − 90 = 25 120 − 115 = 5

Page No 4.56:

Question 18:

Marks scored by 50 students are given below:

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53

(a) Arrange the marks in ascending order.
(b) Represent the marks in the form of discrete frequency distribution.
(c) Construct an inclusive frequency distribution with first class as 10−19. Also construct class boundaries.
(d) Construct a frequency distribution with exclusive class-intervals, taking the lowest class as 10−20.
(e) Convert the exclusive series constructed in (d) into 'less than' and 'more than' cumulative frequency distribution.

ANSWER:

(a)

Marks arranged in ascending order
16 18 20 24 27 30 32 35 36 37 38 38 40 42 43 43 43 45 45 45 46 46 46 48 48	50 51 52 52 53 53 54 54 55 55 55 58 58 60 62 62 64 64 65 65 66 66 72 78 82

(b)

Discrete Frequency Distribution

Marks	Frequency (f)
16 18 20 24 27 30 32 35 36 37 38 40 42 43 45 46 48 50 51 52 53 54 55 58 60 62 64 65 66 72 78 82	1 1 1 1 1 1 1 1 1 1 2 1 1 3 3 3 2 1 1 2 2 2 3 2 1 2 2 2 2 1 1 1
	$\sum_{}^{} f = 50$

Class Interval (Marks)	No. of Students
10 − 19 20 − 29 30 − 39 40 − 49 50 − 59 60 − 69 70 − 79 80 − 89	2 3 7 13 13 9 2 1
	$\sum_{}^{} f = 50$

(d)
Exclusive Frequency Distribution

Class Interval	No. of Students
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80 80 −90	2 3 7 13 13 9 2 1
	$\sum_{}^{} f = 50$

(e)
Less than frequency distribution

Marks	Cumulative Frequency
Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90	2 2 + 3 = 5 5 + 7 = 12 12 + 13 = 25 25 + 13 = 38 38 + 9 = 47 47 + 2 = 49 49 + 1 = 50

More than frequency distribution

Marks	Cumulative Frequency
More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70 More than 80	50 50 − 2 = 48 48 − 3 = 45 45 − 7 = 38 38 − 13 = 25 25 − 13 = 12 12 − 9 = 3 3 − 2 =1

Page No 4.57:

Question 19:

From the following data of the ages of different persons, prepare less than and more than cumulative frequency distribution.

Age (in years)	10−20	20−30	30−40	40−50	50−60	60−70	70−80
No. of Persons	5	12	10	6	4	11	2

ANSWER:

Age (in years)	No. of Persons
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	5 12 10 6 4 11 2
	$\sum_{} f = 50$

Less than frequency distribution

Age	Cumulative frequency
Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80	5 5 + 12 = 17 17 + 10 = 27 27 + 6 = 33 33 + 4 = 37 37 + 11 = 48 48 + 2 = 50

More than frequency distribution

Age	Cumulative frequency
More than 10 More than 20 More than 30 More than 40 More than 50 More than 60 More than 70	50 50 − 5 = 45 45 − 12 = 33 33 − 10 = 23 23 − 6 = 17 17 − 4 = 13 13 − 11 = 2

15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10

58	60	72	68	58	55	58	72	55	68	66	60
60	55	76	66	58	72	68	60	55	68	55	58
62	66	72	58	60	68	55	58	65	72	68

12	33	23	25	18	35	37	49	54	51	37	15	33	42	45
47	55	69	65	63	46	29	18	37	46	59	29	35	45	27

70	55	51	42	57	45	60	47	63	53
33	65	39	82	55	64	58	61	65	42
50	52	53	45	45	25	36	59	63	39
65	54	49	54	64	75	42	41	52	35
30	35	15	48	26	20	40	55	46	18

21	20	55	39	48	46	36	54	42	30
29	42	32	40	34	31	35	37	52	44
39	45	37	33	51	53	52	46	43	47
41	26	52	48	25	34	37	33	36	27
54	36	41	33	23	39	28	44	45	38

19	16	22	9	22	12	39	19	6	24	16	18	7	17	20	25
10	24	20	21	10	7	18	28	14	23	25	34	22	14	33	23
13	36	11	26	11	37	30	13	22	21	32	21	31	17	16	23
15	27	17	21	23	14	24	8	9	15	29	20	18	28	26	12

125	108	112	126	110	132	136	130	140	155
120	130	136	138	125	111	112	125	140	148
147	137	145	150	142	135	137	132	165	154

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53

15	18	16	14	10	6	5	3	8	7
22	18	14	19	17	8	6	4	10	3
12	16	15	13	11	10	18	22	14	19
11	18	22	14	25	21	17	8	9	10

58	60	72	68	58	55	58	72	55	68	66	60
60	55	76	66	58	72	68	60	55	68	55	58
62	66	72	58	60	68	55	58	65	72	68

12	33	23	25	18	35	37	49	54	51	37	15	33	42	45
47	55	69	65	63	46	29	18	37	46	59	29	35	45	27

70	55	51	42	57	45	60	47	63	53
33	65	39	82	55	64	58	61	65	42
50	52	53	45	45	25	36	59	63	39
65	54	49	54	64	75	42	41	52	35
30	35	15	48	26	20	40	55	46	18

21	20	55	39	48	46	36	54	42	30
29	42	32	40	34	31	35	37	52	44
39	45	37	33	51	53	52	46	43	47
41	26	52	48	25	34	37	33	36	27
54	36	41	33	23	39	28	44	45	38

19	16	22	9	22	12	39	19	6	24	16	18	7	17	20	25
10	24	20	21	10	7	18	28	14	23	25	34	22	14	33	23
13	36	11	26	11	37	30	13	22	21	32	21	31	17	16	23
15	27	17	21	23	14	24	8	9	15	29	20	18	28	26	12

125	108	112	126	110	132	136	130	140	155
120	130	136	138	125	111	112	125	140	148
147	137	145	150	142	135	137	132	165	154

40	45	38	24	46	42	45	18	53	64
45	32	52	54	78	65	52	64	66	43
48	55	50	43	48	20	27	65	37	55
51	55	62	66	38	16	60	58	46	35
72	62	54	58	30	36	43	82	46	53