OP Malhotra Class 10 Maths Solutions Chapter 19 Histogram and Ogive Ex 19(d)

Parents can use ICSE Class 10 Maths Solutions S Chand Chapter 19 Histogram and Ogive Ex 19(d) to provide additional support to their children.

S Chand Class 10 ICSE Maths Solutions Chapter 19 Histogram and Ogive Ex 19(d)

Question 1.
Find out the median by plotting the following in the form of an ogive, where ‘m’ denotes the mid-mark of a class :

m	12.5	17.5	22.5	27.5	32.5	37.5	42.5	47.5	52.5
f	2	22	19	14	3	4	6	1	1

Solution:
∵ 12.5, 17.5, etc. are class marks
∴ The corresponding classes will be 10-15, 15-20, etc.

Class	m	f	c.f
10-15	12.5	2	2
15-20	17.5	22	24
20-25	22.5	19	43
25-30	27.5	14	57
30-35	32.5	3	60
35-40	37.5	4	64
40-45	42.5	6	70
45-50	47.5	1	71
50-55	52.5	1	72

Plot the points (15, 2), (20, 24), (25, 43), (30, 57), (35, 60), (40, 64), (45, 70), (50, 71), (55, 72) on the graph and join them with free hand to get on ogive as shown.

Here n = 72 which is even
∴ Median = \(\frac { n }{ 2 }\)
= \(\frac { 72 }{ 2 }\) = 36
From 36 on 7-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw a perpendicular, to x-axis which meet it at M which is median.
∴ Median = 23.5 (approx)

Question 2.
From the data given below draw an ogive and find the values of median and quartiles from the graph.

Years under	10	20	30	40	50	60
Number of persons	12	31	44	55	65	70

Solution:

Years under	10	20	30	40	50	60
Number of persons(c.f.)	12	31	44	55	65	70

Now plot the points (10, 12), (20, 31), (30, 44), (40, 55), (50, 65) and (60, 70) on the graph and join them with free hand to get an ogive as shown.

(i) Herein = 70
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 70 }{ 2 }\)th = 35th term.
From 35 (A) on y-axis, draw a line parallel to x-axis meeting the curve at P and from P, draw PL ⊥ x-axis
∴ L is the median which is = 23 years
∴Meadian = 23 years

(ii) First Quartile (Q₁) = \(\frac { n }{ 4 }\) = \(\frac { 7 0 }{ 4 }\) = 17.5
From 17.5 (B) on y-axis, draw a line parallel to x-axis meeting the curve at Q and from Q, draw QM ⊥ x-axis
∴ M is the Q₁ which is 13 years

(iii) Similarly for Q₃ = \(\frac { 3n }{ 4 }\) = \(\frac { 3×40 }{ 4 }\) = \(\frac { 210 }{ 4 }\) = 52.5
From 52.5 (C) on y-axis, draw a line parallel to x-axis meeting the curve at R and from R, RN ⊥ x-axis
Then N is the Q₃ which is = 37 years

Question 3.
From the following find out the median with the help of ogive curve :

Profit per shop less than	10	20	30	40	50	60
No. of shops	12	30	57	77	94	100

Solution:
Plot the points (10, 12), (20, 30), (30, 57), (40, 77), (50, 94) and (60, 100) on the graph and join them to get an ogive.
Here n = 100
∴ Median = \(\frac { n }{ 2 }\) =\(\frac { 100 }{ 2 }\)th = 50
From 50 (A) on y-axis, draw a line parallel to x-axis which meet the curve at P From P, draw PL ⊥ x-axis which meet it at L
L is the median which is 27.8 (approx)
∴ Median 27.8 or = 28 (approx)

Question 4.
The following are the marks obtained by 50 students in Statistics :

Marks less than	10	20	30	40	50	60
No. of students	4	10	30	40	47	50

Draw an ogive and find the median marks from it.
Solution:
Plot the points (10, 4), (20, 10), (30, 30), (40, 40), (50, 47) and (60, 50) on the graph and join them with free hand to get an ogive as shown on the graph,
Here n = 50
∴ Median = \(\frac { 50 }{ 2 }\) = 25th
From 25 on y-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw a line perpendicular to x-axis which meet it at L
L is the median which is 27.8 = 28
Hence median = 28

Question 5.
100 pupils in a school have heights as tabulated below :

Height in cm	121-130	131-140	141-150	151-160	161-170	171-180
No. of pupils	12	16	30	20	14	8

Draw the ogive for the above data and from it determine the median. (Use graph paper).
Solution:
Writing the given class-interval in exclusive form,

Height (in cm) (class)	Frequency (f)	c.f.
120.5-130.5	12	12
130.5-140.5	16	28
140.5-150.5	30	58
150.5-160.5	20	78
160.5-170.5	14	92
170.5-180.5	8	100

Plot the points (130.5,12), (140.5,28), (150.5,58), (160.5,78), (170.5,92) and (180.5,100) on the graph and join them with free hand to get an ogive as shown.
Here n = 100
∴ Median =\(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) =50th
∴ From 50 on y-axis, draw a line parallel to x-axis which meets the curve at P From P, draw PL ⊥x-axis
Then L is the median which is =147.8
= 148 (approx)
∴ Median =148 (approx)

Question 6.
Draw an ogive curve from the following data and find out (a) the median wage and (b) number of workers earning less than Rs. 55 per week :

Weekly wages (Rs.)	0-20	20-40	40-60	60-80	80-100
No. of workers	40	51	60	38	7

Solution:

Weekly wages	No. of workers (f)	c.f.
0-20	40	40
20-40	51	91
40-60	60	151
60-80	38	189
80-100	7	196

We plot the points (20, 40), (40, 91), (60, 151), (80, 189) and (100, 196) on the graph and join them with free hand to get an ogive as shown Here n= 196
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 196 }{ 2 }\) =98

From 98 on y-axis, draw a line parallel to x-axis is which meets the curve at P. From P, draw PL ⊥ x-axis
Then L is the median which is
∴ Median =42.3

(ii) From 55 on x-axis, draw a perpendicular on x-axis which meets the curve at Q. From Q, draw a line parallel to x-axis which meets y-axis at B
Then B =136
∴ Number of workers getting less then Rs. 55 per week =136

Question 7.
Represent the following data by a cumulative frequency curve and locate the median and quartiles graphically :

Income (Rs.)	50-60	40-50	30-40	20-30	10-20	0-10
No. of persons	100	200	200	150	100	50

Solution:
Rewriting the data in order

Income (Rs.)	0-10	10-20	20-30	30-40	40-50	50-60
No. of persons	50	100	150	200	200	100

Now writing in c.f.

Income (Rs.)	No. of persons (f)	c.f.
0-10	50	50
10-20	100	150
20-30	150	300
30-40	200	500
40-50	200	700

Plot the points (10, 50), (20, 150), (3o, 300), (40, 500), (50, 700) and (60, 800) on the graph and join them with free hand to get an ogive as shown Here n = 800
∴ Median = \(\frac { n }{ 2 }\)th = \(\frac { 800 }{ 2 }\)th = 400th term
From 400 on y-axis, draw a line parallel to x-axis which meets the curve at P. From P, draw PL ⊥ x-axis
∴ L is the median which is 35
Hence median = Rs. 35

(i) Q₁ = \(\frac { n }{ 4 }\) = \(\frac { 800 }{ 4 }\) = 200
From 200 on y-axis, draw a line parallel to x-axis which meets the curve at Q.
From Q draw QM ⊥ x-axis
∴ M is the Q₁ which is 23

(ii) Q₃ = \(\frac { 3n }{ 4 }\) = \(\frac { 3×800 }{ 4 }\) = 600
From 600 on y-axis, draw a line parallel to income in Rs. x-axis which meets the curve at R. From R, draw RN ⊥ x-axis
Then N is Q₃ which is 45
Hence Q₁ = Rs. 23 and Q₃ = Rs. 45

Question 8.
Draw the ogive for the following frequency distribution. Estimate the median from your graph :

Class	1-10	11-20	21-30	31-40	41-50	51-60	61-70
Frequency	2	5	7	12	8	7	4

Solution:
Writing the class in exclusive form, and in c.f.

Class	Frequency	c.f.
0.5-10.5	2	2
10.5-20.5	5	7
20.5-30.5	7	14
30.5-40.5	12	26
40.5-50.5	8	34
50.5-60.5	7	41
60.5-70.5	4	45

Plot the points (10.5, 2), (20.5, 7), (30.5, 14), (40.5, 26), (50.5, 34), (60.5, 41) and (70.5, 45) on the graph and join them with free hand to get an ogive as shown Here n = 45 which is odd
∴ Median = \(\frac { n + 1 }{ 2 }\)th = \(\frac { 45 + 1 }{ 2 }\) = 23th

From 23 on y-axis, draw a line parallel to x-axis which meets the curve at P.
From P, draw PL ⊥ x-axis, then L is the median which is (38.1)
Hence median = 38.1

Question 9.
Attempt this question on graph paper.

Age (yrs)	5-15	15-25	25-35	35-45	45-55	55-65	65-75
No. of casualties	6	10	15	13	24	8	7

due to accidents
(i) Construct the ‘less than’ cumulative frequency curve for the above data, using 2 cm = 10 years, on one axis and 2 cm = 10 casualties on the other.
(ii) From your graph determine.
(a) The median and
(b) the upper quartile
Solution:

Age (in years)	No. of casulaties (f)	c.f.
5-15	6	6
15-25	10	16
25-35	15	31
35-45	13	44
45-55	24	68
55-65	8	76
65-75	7	83

Plot the points (15, 6), (25, 16), (35, 31), (45, 44), (55, 68), (65, 76) and (75, 83) on the graph and join them with free hand to form an ogive (less than)

(i) Median, Here n = 83
∴ \(\frac { n }{ 2 }\) = \(\frac { 83 }{ 2 }\) = 41.5
From 41.5 on y-axis, draw a line parallel to x-axis meeting the curve at P.
From P draw a perpendicular PM on x-axis
M is the median which is 43 years.

(ii) Upper quartile (Q₃) = \(\frac { 3 }{ 4 }\) × n = \(\frac { 3 }{ 4 }\) × 83
= \(\frac { 249 }{ 4 }\) = 62.25

From 62.25 ony-axis, draw a line parallel to x-axis meeting the curve at Q. From Q, draw a perpendicular on x-axis meeting it at N.
N is the upper quardant which is 52 years
∴ Q₃ = 52 years

Question 10.
Marks scored by 400 students in an examination are as follows :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	10	20	22	40	55	75	80	58	28	12

Draw an ogive and from it determine:
(i) the median mark,
(ii) pass marks if 80% of the students pass examination.
Solution:

Marks	No. of casulaties	c.f.
0-10	10	10
10-20	20	30
20-30	22	52
30-40	40	92
40-50	55	147
50-60	75	222
60-70	80	302
70-80	58	360
80-90	28	388
90-100	12	400

Plot the points (10, 10), (20, 30), (30, 52), (40, 92), (50,. 147), (60, 222), (70, 302), (80, 360), (90, 388), (100,400) on the graph and join them with free hand to form an ogive as shown below:

Here n = 400

(i) Median = \(\frac { n }{ 2 }\) = \(\frac { 400 }{ 2 }\) = 200
From 200 (A) on y-axis draw a line parallel to x-axis is meeting the curve at P. From P, draw a perpendicular on x-axis meeting it at M.
M is median which is 57.5
∴ Median = 257.5
∵ 80% students passed

(ii) No. of students passed = 80% of 400 = \(\frac { 80 }{ 100 }\) × 400 = 320
∴ No. of students who did not pass = 400 – 320 = 80
From a point 80 on y-axis, draw a line parallel to x-axis meeting the curve at Q.
From Q₁ draw a line perpendicular to x-axis meeting at N. Then N is the required pass marks.
Which is 36
∴ Pass marks = 36%

Question 11.
Draw an ogive for the following frequency distribution. Use your ogive to estimate
(i) the median
(ii) the number of students who obtained more than 75% marks (use square paper to solve this question).

Marks	0-9	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
No. of students	5	9	16	22	26	18	11	6	4	3

Solution:
Writing the classes in exclusive form,

Marks	No. of casulaties (f)	c.f.
0.5-9.5	5	5
9.5-19.5	9	14
19.5-29.5	16	30
29.5-39.5	22	52
39.5-49.5	26	78
49.5-59.5	18	96
59.5-69.5	11	107
69.5-79.5	6	113
79.5-89.5	4	117
89.5-99.5	3	120

Now plot the points (9.5, 5), (19.5, 14), (29.5, 30), (39.5, 52), (49.5, 78), (59.5, 96), (69.5, 107), (79.5, 113), (89.5, 117), and (99.5, 120) on the graph and join them with free hand to get an ogive as shown below :
Here n = 120

(i) Median: \(\frac { n }{ 2 }\) = \(\frac { 120 }{ 2 }\) = 60
From 60 on the y-axis draw a line parallel to x-axis meeting the curves at P. From P, draw a perpendicular on x-axis meeting it at M.
M is the median which is 43
∴ Median = 43

(ii) From 75 on the x-axis draw a line parallel toy-axis meeting the curve at Q. From Q, draw a line perpendicular to y-axis meeting at N.
Now Q = 110
Total no. of students getting more them 75% marks = 120 – 110= 10

Question 12.
In a public collection towards the erection of a memorial, 1000 people contributed sums of money varying from Rs. 1 to Rs. 100 (in units of Re. 1). The following table gives the frequency distribution of contribution :

Contribution in Rs.)	No. of people	Contribution (in Rs.)	No. of people
1-10	30	51-60	180
11-20	60	61-70	140
21-30	80	71-80	70
31-40	170	81-90	40
41-50	200	91-100	30

Using a suitable scale, draw on graph paper an ogive (cumulative frequency graph) and use it to answer the following :
(i) Estimate the median.
(ii) If it is agreed to allow only those who contributed Rs. 45 or more to attend the unveilling ceremony, what percentage would attend ?
Solution:
Writing the classes in exclusive forms

Contribution (in Rs.)	No. of people (f)	c.f.
0.5-10.5	30	30
10.5-20.5	60	90
20.5-30.5	80	170
30.5-40.5	170	340
40.5-50.5	200	540
50.5-60.5	180	720
60.5-70.5	140	860
70.5-80.5	70	930
80.5-90.5	40	970
90.5-100.5	30	1000

Plot the points (10.5, 30), (20.5, 90), (30.5, 170), (40.5, 340), (50.5, 540), (60.5, 720), (70.5, 860), (80.5, 930), (90.5, 970) and (100.5, 1000) on the graph and join them with free hand to get an ogive as shown.
Here n = 1000
∴ Median = \(\frac { n }{ 2 }\) = \(\frac { 1000 }{ 2 }\) = 500th term
From 500 on y-axis; draw a line parallel to x-axis which meets the curve at P. From P, draw PL ⊥ x-axis then L is the median which is 49
∴ Median = 49 rupees

(ii) From Rs. 45 on x-axis, draw a perpendicular which meets the curve at Q and from Q, draw a parallel line to x-axis which meets y-axis at B Then B is 440
∴ Persons who could attend the ceremony = 1000 – 440 = 560
Percentage = \(\frac { 560 × 100 }{ 1000 }\) = 56%

Self-Evaluation and Revision (LATEST ICSE QUESTIONS)

Question 1.
Draw a histogram to represent the following data :

Pocket money (in ₹)	No. of students
150-200	10
200-250	5
250-300	7
300-350	4
350-400	3

Solution:

Pocket money (in ₹)	No. of students
150-200	10
200-250	5
250-300	7
300-350	4
350-400	3

Represent the pocket money along x-axis
and number of students along y-axis and
form the histogram as shown

Question 2.
Using a graph paper, draw an Ogive for the following distribution which shows a record of the weight in kilogram of 200 students.

Weight	40-45	45-50	50-55	55-60	60-65	65-70	70-75	75-80
Frequency	5	17	22	45	51	31	20	9

Use the ogive to estimate the following :
(i) The percentage of students weighing 55 kg or more;
(ii) The weight above which the heaviest 30% of the students fall;
(iii) The number of students who are
(1) under weight
(2) over weight, if 55.70 kg is considered as the standard weight.
Solution:

Weight	Frequency	c.f.
40-45	5	5
45-50	17	22
50-55	22	44
55-60	45	89
60-65	51	140
65-70	31	171
70-75	20	191

Plot the points (45, 5), (50, 22), (55, 44), (60, 89), (65, 140), (70, 171), (75, 191) and (80, 200) on the graph and join them with free hand to get an ogive as shown
From 55 on x-axis, draw a perpendicular which meets the curve at A. From A, draw a line parallel to x-axis which meets j’-axis at D which is 44.

(i) Percentage number of students weighing 55 kg or more = 200 – 44 = 156 (From ogive)
∴ Percentage = \(\frac { 156 × 100 }{ 200 }\) = 78%

(ii) The weight above which the heaviest 30% of the students fall = \(\frac { 200 × 30 }{ 100 }\) = 60
∴ Heighest students from the graph will be 65 kg and more upto 80 kg = 31 + 20 + 9 = 60

(iii) Number of students who are under weight i.e. below 55.70 kg = 47
Over weight i.e. above 55.70 kg = 200 – 47 = 153

Question 3.
If the mean of the following distribution is 7.5 find the missing frequency f:

Variable	5	6	7	8	9	10	11	12
Frequency	20	17	f	10	8	6	7	6

Solution:

Variable (x)	Frequency (f)	f x
5 6 7 8 9 10 11 12	20 17 f 10 8 6 7 6	100 102 7 f 80 72 60 77 72
Total	74+f	563+7 f

Here mean = 7.5
But mean = \(\frac{\sum f x}{\sum f}\)
∴ 7.5 = \(\frac{563+7 f}{74+f} \)
⇒ 7.5 (74 + f) = 563 + 7f ⇒ 555 + 7.5f = 563 + 7f ⇒ 7.5f – 7f = 563 – 555
⇒ 0.5f = 8 ⇒ f = \(\frac { 8 }{ 0.5 }\) = \(\frac { 8 × 10 }{ 5 }\) = \(\frac { 8 × 2 }{ 1 }\) ⇒ f = 16

Question 4.
The median of the following observation 11, 12, 14, 18, (x + 4), 30, 32, 35, 41 arranged in ascending order is 24. Find x.
Solution:
Here n = 9 which is odd
∴ Median = \(\frac { n+ 1 }{ 2 }\)th term = \(\frac { 9 + 1 }{ 2 }\) = 5th term, which is x + 4
∴ x + 4 = 24 ⇒ x = 24 – 4 = 20
∴ x = 20

Question 5.
Find the mean of the following distribution :

Class interval	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	10	6	8	12	5	9

Solution:

Class interval	Mid value (x)	Frequency (f)	f.x.
20-30 30-40 40-50 50-60 60-70 70-80	25 35 45 55 65 75	10 6 8 12 5 9	250 210 360 660 325 675
Total		50	2480

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{2480}{50}\) = 49.6

Question 6.
The daily wages of 160 workers in a building project are given below :

Wages in Rs.	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of workers	12	20	30	38	24	16	12	8

Using a graph, draw an ogive for the above distribution. Use your ogive to estimate :
(i) the median wage of the workers;
(ii) the upper quartile wage of the workers;
(iii) the lower quartile wage of the workers;
(iv) the percentage of workers who earn more than Rs. 45 a day.
Solution:

Wages (in Rs.)	No. of workers (f)	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80	12 20 30 38 24 16 12 8	12 32 62 100 124 140 152 160

Plot the points (10, 12), (20, 32), (30, 62), (40, 100), (50, 124), (60, 140), (70, 152) and (80, 160) on the graph and join them with free hand to get an ogive as shown (i) Here n = 160
∴ Median = \(\frac{ n }{ 2 }\) = \(\frac{ 160 }{ 2 }\) = 80th
From 80 on y-axis, draw a line parallel to x-axis meeting the curve at P. From P, draw PL perpendicular on x-axis which meets it at L.
L is median which is 35
∴ Median = Rs. 35

(ii) Upper quartile (Q₃) = \(\frac{3n}{4}\)
= \(\frac{3}{4}\) × 160 = 120
From 120 on y-axis, draw a parallel line to x-axis which meets the curve at Q. From Q, draw perpendicular on x-axis meeting it at M
M is Q₃ which is 47.50
∴ Upper quartile (Q₃) = Rs. 47.50

(iii) Lower quartile (Q₁) = \(\frac{ 1 }{ 4 }\)n = \(\frac{ 1 }{ 4 }\) × 160 = 40
From 40 on j-axis, draw a line parallel to x-axis meeting the curve at R. From R, draw a perpendicular on x-axis meeting it at N.
N is the lower quartile (Q₁) which is 22.5
∴ Lower quartile (Q₁) = Rs. 22.5

(iv) Number of workers earning more than Rs. 45 per day
From 45 on x-axis, draw a perpendicular which i.-.eets the curve at B. From B, draw a line parallel to x-axis meeting 7-axis at C which is 112
∴ Number of workers getting more than Rs. 45 = 160 – 112 = 48
Percentage = \(\frac{ 48 × 100 }{ 160 }\) = 30%

Question 7.
Find the mean of the following distribution :

Class inverval	0-10	10-20	20-30	30-40	40-50
Frequency	10	6	8	12	5

Solution:

Class interval	Mid value {x}	Frequency (f)	f.x
0-10 10-20 20-30 30-40 40-50	5 15 25 35 45	10 6 8 12 5	50 90 200 420 225
Total		41	985

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{985}{41}\) = 24.02

Question 8.
The table below shows the distributions of the scores obtained by 120 shooters in a shooting competition. Using a graph sheet, draw in ogive for the distributions.

Scores obtained	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of shooters	5	9	16	22	26	18	11	6	4	3

Use your ogive to estimate : (i) The median (ii) The inter quartile range (iii) The number of shooters who obtained more than 75% scores.
Solution:

Score obtained (class interval)	Number of students (f)	c.F.
0-10	5	5
10-20	9	14
20-30	16	30
30-40	22	52
40-50	26	78
50-60	18	96
60-70	11	107
70-80	6	113
80-90	4	117
90-100	3	120

Plot the points (10, 5), (20, 14). (30, 30). (40, 52), (50, 78), (60, 96), (70, 107), (80, 113), (90, 117), and (100, 120) on the graph and join them with free hand to get an ogive as shown below Here n = 120

(i) Median = \(\frac{n}{2}\)th = \(\frac{120}{2}\) = 60th term
From 60 on y-axis draw a line parallel to.x-axis meeting the curve at P. From P, draw a perpendi-cular to x-axis meeting it at L. L, is median which is 43
∴ Median = 43

(ii) Inner quartile (Q₁) = \(\frac{n}{4}\)th = \(\frac{120}{4}\) = 30 th term
From 30 on y-axis draw a line parallel to x-axis meeting the curve at Q. From Q, draw perpendi-cular on x-axis meeting it at M.
M is Q₁ which is 30
Upper quartile (Q₃) = \(\frac{3n}{4}\)th = \(\frac{3×120}{4}\)th = 90th term
From 90 on y-axis, draw a line parallel to x-axis meeting the curve at R. From R draw perpendi-cular on x-axis meeting it at N. Then N is Q₃ which is 57
∴ Q₃ = 57
Now interquartile range = Q₃ – Q₁ = 57 – 30 = 27

(iii) No. of shooter getting more then 75% score
From 75 on x-axis, draw a perpendicular meeting the curve at B and from B, draw a line parallel to x-axis meeting v-axis at C.
C is 110
∴ Number of shooter getting more them 75% score = 120 – 110= 10

Question 9.
Using a graph paper, draw an Ogive for the following distribution which shows the marks obtained in the General Knowledge Paper by 100 students.

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80
No. of students	5	10	20	25	15	12	4	9

(i) the median,
(ii) the number of students who score marks above 65.
Solution:

Marks	Number of students (f)	c.f.
0-10	5	5
10-20	10	15
20-30	20	35
30-40	25	60
40-50	15	75
50-60	12	87
60-70	4	91
70-80	9	100

Plot the points (10, 5), (20, 15), (30, 35), (40, 60), (50, 75), (60, 87), (70, 91), (80, 100) on the graph and join them with free hand to get an ogive as shown on the graph Here n = 100

(i) Median = \(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50
From 50 on y-axis, draw a line parallel to x-axis meeting the curve at P.
From P, draw a perpendi-cular on x-axis meeting it at L.
L is median which is 35
∴ Median = 35 marks

(ii) Number of students getting more than 65 marks
From 65 on x-axis, draw a perpendicular meeting the curve at Q. From Q, draw a line parallel to x-axis meeting y-axis at M. M is 90
∴ Number of students getting above 65% marks = 100 – 90 = 10

Question 10.
The weights of 50 apples were recorded as given below. Calculate the mean weight, to the nearest gram, by step deivation method.

Weight in grams	Number of apples
80-85	5
85-90	8
90-95	10
95-100	12
100-105	8
105-110	4
110-115	3

Solution:

Weight (in gm)	Mid values (x)	No. of apples (f)	A = 97.5 d = x – A	f.d.
80-85 85-90 90-95 95-100 100-105 105-110 110-115	82.5 87.5 92.5 A=97.5 102.5 107.5 112.5	5 8 10 12 8 4 3	-15 -10 -5 0 5 10 15	-75 -80 -50 0 40 40 45
Total		50		-80

Mean = A + \(\frac{\sum f d}{\sum f}\) = 97.5 + \(\frac{-80}{50}\) = 97.5 – 1.6 = 95.9

Question 11.
Find the mean, median and mode of the following distribution :
8, 10, 7, 6, 10, 11, 6, 13, 10
Solution:
Mean = \(\frac{8+10+7+6+10+11+6+13+10}{9}\) = \(\frac{81}{9}\) = 9
arranging given nos. in ascending order.
6, 6, 7, 8. 10. 10, 10, 11, 13
Median = \(\frac{n+1}{2}\) th term = \(\frac{9+1}{2}\) = 5th term which is 10
∴ Median = 10
Mode = 10 (having highest frequency 3 times)

Question 12.
The following table gives the wages of workers in a factory.

Wages in Rs.	45-50	50-55	55-60	60-65	65-70	70-75	75-80
No. of workers	5	8	30	25	14	12	6

Calculate the mean by the short-cut method.
Solution:

Wages in Rs.	No. of Workers (f)	Mid. mark x	d = (x – A)	f × d
45-50 50-55 55-60 60-65 65-70 70-75 75-80	5 8 30 25 14 12 6	47.5 52.5 57.5 62.5 = A 67.5 72.5 77.5	-15 -10 -5 0 5 10 15	-75 -80 -150 0 70 120 90
Total	Σf =100			Σ fd = -25

Mean = A + \(\frac{\sum f d}{\sum f}\) = 62.5 + \(\frac{-25}{100}\) = 62.50 – 25 = 62.25

Question 13.
Attempt this question on graph paper. Marks obtained by 200 students in an examination are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	5	10	14	21	25	34	36	27	16	12

Draw an Ogive for the given distribution taking 2 cm = 10 marks. On one axis and 2 cm = 20 students on the other axis. From the graph, find :
(i) The Median
(ii) The upper Quartile
(iii) Number ofstudents scoring above 65 marks;
(iv) If 10 students qualify for merit scholarship, find the minimum marks required to qualify.
Solution:

Marks	No. of Students (fi)	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 10 14 21 25 34 36 27 16 12	5 15 29 50 75 109 145 172 188 200

No. of students (n) = 200
Plots the points (10. 5), (20, 15), (30, 29), (40, 50), (50, 75), (60, 109), (70, 145), (80, 172), (90, 188), (100, 200) on the graph and join them with free hand to get an ogive.

Mark a point P on y-axis corresponding to \(\frac{N}{2}\)
i.e., = \(\frac{200}{2}\) = 100
Draw PQ || x-axis, meeting the curve at Q.
Draw QR ⊥ x-axis, then the abscissa of R is 57

(i) ∴ Median = 57 marks
(ii) Mark a point A on y-axis corresponding to \(\frac{3N}{4}\) i.e., \(\frac{600}{4}\) = 150

Draw AB || x-axis, meeting the curve at B.
Draw BC ⊥ x-axis. then the abscissa of C is 72
∴ Upper quartile = 72 marks

(iii) From the graph, we have
No. of students scoring above 65 marks are (200 – 126) = 74

(iv) Minimum marks required to qualify are 92
⇒ The students from 191 to 200 (not the 190th student) qualify for merit scholarship. From point F on the y-axis indicating 190, draw a horizontal line meeting the Ogive at G. From G, draw a vertical line meeting x-axis at FI which indicates 91 marks.
Since the 190th student does not qualify for merit scholarship, so all students next to him who score more than 91 mark qualify for the scholarship.
Hence, the minimum marks required to qulify for merit scholarship = 92.

Question 14.
In a school the weekly pocket money of 50 students is as follows.

Weekly pocket money in Rs.	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	2	8	12	14	8	6

Draw an histogram and a frequency polygon on the same graph. Find the mode from this graph.
Solution:
The frequency distribution table is as under:

Weekly pocket money (in Rs.)	Class mark	No. of students
40-50 50-60 60-70 70-80 80-90 90-100	45 55 65 75 85 95	2 8 12 14 8 6

Draw the histogram by taking ‘weekly pocket money along .Y-axis and no. of students along r-axis.

The frequency polygon is obtained by joining the consecutive mid-points of the upper sides (tops) of the adjacent rectangle of the histogram by means of line segments.

Now, join the two vertices A and B of the highest rectangle, diagonally to the upper corners C and D of the adjacent rectangles on either side of the highest rectangles, by line segments AC and BD. Let AC and BD intersect at L.

From L, draw the perpendicular LM on the horizontal line-the x-axis. Then, the abscissa of the point M, i.e.; OM = 72.5 determines the mode.
Hence, the required mode is Rs. 72.50

Question 15.
The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.

Marks obtained	5	6	7	8	9	10
No. of students	3	9	6	4	2	1

Solution:

Marks obtained (xi)	No. of students (fi)	Cumulative frequency (c,f)	fi xi
5	3	3	15
6	9	12	54
7	6	18	42
8	4	22	32
9	2	24	18
10	1	25	10
Total	Σfi = 25		Σfi xi =171

Mean
\(\bar{x}\) = \(\frac{\Sigma f_i}{x_i} \Sigma f_i\) = \(\frac { 171 }{ 25 }\)
\(\bar{x}\) = 6.84
Number of terms = 25 (odd)
Mean = \(\left(\frac{25+1}{2}\right)\)th term = \(\left(\frac{26}{2}\right)\)th term = 13th term
∴ Median = 7
Mode = Marks with maximum frequency is 6
∴ Mode = 6

Question 16.
The mean of the following distribution is 52 and the frequency of class interval 30-40 is ‘f’. Find ‘f’.

Class Interval	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	5	3	f	7	2	6	13

Solution:

Class interval	Frequency (fi)	Class mark (xi)	fixi
10-20	5	15	75
20-30	3	25	75
30-40	f	35	35 f
40-50	7	45	315
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
	Σfi = 36 + f		Σfixi = 1940 + 35f

Now, \(\bar{x}\) = \(\frac{\Sigma f_i x_i}{\Sigma f_i}\)
⇒ 52 = \(\frac{1940+35 f}{36+f}\) ⇒ 52(36 + f) = 1940 + 35f
⇒ 1872 + 52f = 1940 + 35f ⇒ 52f – 35f = 1940 – 1872
⇒ 17f = 68
⇒ f = \(\frac{68}{17}\)
⇒ f = 4

Question 17.
The monthly income of a group of 320 employees in a company is given below :

Monthly Income in Rs.	No. of Employees
6000-7000	20
7000-8000	45
8000-9000	65
9000-10000	95
10000-11000	60
11000-12000	30
12000-13000	5

Draw an ogive of the given distribution on a graph sheet taking 2 cm = Rs. 1000 on one axis and 2 cm = 50 employees on the other axis.
From the graph to determine.
(i) The median wage
(ii) the number of employees whose income is below Rs. 8500.
(iii) If the salary of a senior employee is above Rs. 11,500, find the number of senior employees in the company.
(iv) the upper quartile.
Solution:

Monthly Income in Rs. (c)	No. of Employees (f)	c.f.
6000-7000 7000-8000 8000-9000 9000-10000 10000-11000 11000-12000 12000-13000	20 45 65 95 60 30 5	20 65 130 225 285 315 320
Total	320

Plot the point (7000, 20), (8000, 65), (9000, 130), (10,000, 225) (11000, 285) (12,000, 315), (13,000, 1320) on the graph and join them with free hand to get an ogive as shown below
Here n = 320

(i) For median wage, Take OP = \(\frac{320}{2}\) = 160
on y-axis
Draw a line PQ || x-axis and from Q,
draw QM ⊥ x-axis, abcissa of M point is 9400
⇒ Median = Rs. 9400

(ii) Take OM’ = 8500 on x-axis. Draw Q’M’ || to y-axis and P’Q’ || x-axis
Where ordinate of P’ is 90
There are approximately 90 employees whose monthly wage is below Rs. 8500

(iii) There are approximately 20 employees whose salary is above Rs. 11500.

(iv) Upper quartile

Mark a point A on y-axis on \(\frac{3N}{4}\) = \(\frac{3 × 320}{4}\) = 240 and draw a line AB || x-axis, then draw BB’ ⊥ x-axis abscissa of B’ is upper quartile i.e., Rs. 10200.

Question 18.
A Mathematics aptitude test of 50 students was recorded as follows :

Marks	50-60	60-70	70-80	80-90	90-100
No. of Students	4	8	14	19	5

Draw a histogram for the above data using a graph paper and locate the mode.
Solution:

Represent marks along x-axis and students on y-axis and draw histogram as shown in the graph. In the highest rectangle. join A to C and B to D which intersect each other at P. From P, draw PQ perpendicular on x-axis meeting it at Q
Q is the mode which is 82.5
Hence, the required mode is 82.5

Question 19.
(a) (i) Using step-deviation method, calculate the mean marks of the following distribution.
(ii) State the modal class.

Class Interval	50-55	55-60	60-65	65-70	70-75	75-80	80-85	85-90
Frequency	5	20	10	10	9	6	12	8

Solution:

C.I.	Xi	Fi	Assumed mean A=67.5 Ui = \(\frac{x_t-\mathrm{A}}{h}\), h = 5	fiui
50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90	52.5 57.5 62.5 67.5 = A 72.5 77.5 82.5 87.5	5 20 10 10 9 6 12 8	-3 -2 -1 0 1 2 3 4	-15 -40 -10 0 9 12 36 32
Total		80		24

(i) Mean = A + \(\frac{\Sigma f_i u_i}{\Sigma f_i}\) × h
= 67.5 + \(\frac{24}{80}\) × 5
= 67.5 + 1.5 = 69
(ii) Modal class = 55 – 60 (∵ frequency of this class is maximum i.e. 20)

Question 20.
Marks obtained by 200 students in an examination are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
Frequency	5	11	10	20	28	37	40	29	14	6

Draw an ogive for the given distribution taking 2 cm = 10 marks on one-axis and 2 cm = 20 students on the other axis. Using the graph, determine
(i) the median marks,
(ii) the number of students who failed if minimum marks required to pass is 40.
(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade one in the examination.
Solution:

Marks	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 11 10 20 28 37 40 29 14 6	5 16 26 46 74 111 151 180 194 200
	N = 200

Plot the point (10, 5) (20, 16) (30, 26) (40, 46) (50, 74) (60, 111) (70, 151) (80, 180) (90, 196) (100, 200) on the graph and join them in free hand to get an ogive as shown.
From the graph it is seen that

(i) Median = \(\left(\frac{n}{2} \text { th }\right)\) term = \(\left(\frac{200}{2} \text { th }\right)\) 100 observation = 57 from graph.
(ii) No. of students who failed = 46
(iii) No. of students who secured grade one = 200 – 188 = 12

Question 21.
Marks obtained by 40 students in a short assessment is given below, where a and b are two missing data.

Marks	5	6	7	8	9
Number of students	6	a	16	13	b

If the mean of the distribution is 7.2, find a and b.
Solution:
Mean of distribution = 7.2

Marks	Number of students	fx
5	6	30
6	a	6 a
7	16	112
8	13	104
9	b	9 b
Total	36 + a + b = 40	246 + 6a + 9b

35 + a + b = 40
a + b = 40 – 35 = 5 …(i)
Mean = 7.2
∴ Total = 7.2 x 40 = 288
246 + 6a + 9b = 288
6a + 9b = 288 – 246 = 42
2a + 3b = 14 …(ii)
From (i) a + b = 5 ⇒ a = 5 – b
∴ 2(5 – b) + 3b = 14 ⇒ 10 – 2b + 3b = 14
⇒ b = 14 – 10 = 4
∴ a = 5 – 4 = 1
a = 1, b = 4

Question 22.
The following distribution represents the height of 160 students of a school.

Height (in cm)	No. of Students
140-145	12
145-150	20
150-155	30
155-160	38
160-165	24
165-170	16
170-175	12
175-180	8

Draw an ogive for the given distribution taking 2 cm = 5 cm of height on one axis and 2 cm = 20 students on the other axis. Using the graph, determine :
(i) The median height.
(ii) The inter quartile range.
(iii) The number of students whose height is above 172 cm.
Solution:
The cumulative frequency table may be prepared as follows :

Height (in cm)	No. of Students (f)	Cumulative frequency (c.f)
140-145	12	12
145-150	20	32
150-155	30	62
155-160	38	100
160-165	24	124
165-170	16	140
170-175	12	152
175-180	8	160

Now, we take height along x-axis and number of students along the y-axis. Now, plot the point (140, 0), (145, 12), (150, 32), (155, 62), (160, 100), (165, 124), (170, 140), (175, 152) and (180, 160). Join these points by a free hand curve to get the ogive.

(i) Here N = 160 ⇒ \(\frac { N }{ 2 }\) = 80
On the graph paper take a point A on the y-axis representing 80. Through A draw horizontal line meeting the ogive at B. From B, draw BC ⊥ x-axis, meeting the x-axis at C.
The abscissa of C is 157.5
So, median = 157.5 cm

(ii) Proceeding in the same way as we have done in above, we have, Q₁ = 152 and Q₃ = 164 – 152 = 12cm
So, inter quartile range = Q₃ – Q₁ = 164 – 152 = 12 cm

(iii) From the ogive, we see that the number of students whose height is less than 172 is 145.
No. of students whose height is above 172 cm = 160- 145 = 15

Question 23.
Find the mode and median of the following frequency distribution :

X	10	11	12	13	14	15
f	1	4	7	5	9	3

Solution:

x	f	Cumulative frequency (c.f)
10	1	1
11	4	5
12	7	12
13	5	17
14	9	26
15	3	29

Here n- 29 which is odd
∴ Median = \(\left(\frac{n+1}{2}\right)\)th term = \(\left(\frac{29+1}{2}\right)\)th term = 15th term = 13
Since, the frequency corresponding to 14 is maximum, so mode = 14.

Question 24.
The median of the following observations 11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47 arranged in ascending order is 24. Find the value of x and hence find the mean.
Solution:
Observation are:
11, 12, 14, (x – 2), (x + 4), (x + 9), 32, 38, 47
n = 9
∴ Median = \(\left(\frac{9+1}{2}\right)^{\text {th }}\) term
i.e. 5^th term = x + 4
∴ Median = x + 4
24 = x + 4
x = 24 – 4
x = 20
Now observation are :
11, 12, 14, (20 – 2), (20 + 4), (20 + 9), 32, 38, 47 i.e. 11, 12, 14, 18, 24,29, 32, 38, 47
∴ Mean = \(\frac{11+12+14+18+24+29+32+38+47}{9}\) = \(\frac{225}{9}\) = 25

Question 25.
Draw a histogram from the following frequency distribution and find the mode from the graph :

Class	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	2	5	18	14	8	5

Solution:

Class	0-5	5-10	10-15	15-20	20-25	25-30
Frequency	2	5	18	14	8	5

∴ Mode =13.6

Steps for calculation of mode
(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.
(ii) Mark the inner corner of adjacent rectangles as C and D.
(iii) Join AC and BD do intersect at K. From K, draw KL perpendicular to x-axis.
(iv) The value of L on x-axis represents the mode.

Question 26.
Find the mean of the following distribution by step deviation method :

Class interval	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	10	6	8	12	5	9

Solution:

C.I.	Frequency (f)	Class mark (x)	d = \(\frac{x-\mathrm{A}}{i}\) [A = 45]	f × d
20-30	10	25	-2	-20
30-40	6	35	-1	-6
40-50	8	45	0	0
50-60	12	55	1	12
60-70	5	65	2	10
70-80	9	75	3	27
	50			23

Mean = A + \(\frac{\Sigma f d}{n}\) × i = 45 + \(\frac { 23 }{ 50 }\) × 10
= 45 + 4.6 = 49.6

Question 27.
The marks obtained by 120 students in a test are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	5	9	16	22	26	18	11	6	4	3

Use suitable scale for ogive to estimate the following :
(i) The median.
(ii) The number of students who obtained more than 75% marks in the test.
(ii) The number of students who did not pass the test if minimum marks required to pass is 40.
Solution:

Marks	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	5 9 16 22 26 18 11 6 4 3	5 14 30 52 78 96 107 113 117 120
	120

(i) Through marks 60, draw a line segment parallel to x-axis which meets the curve at A. From A, draw a line perpendicular to x-axis meeting at B.
∴ Median = 43

(ii) Through marks 75, draw a line segment parallel to y-axis which meets the curve at D. From D, draw a line perpendicular to y-axis which meets y-axis at 110.
Number of students getting more than 75% = 120 – 110 = 10 students.

(iii) Through marks 40, draw a line segment parallel toy-axis which meets the curve at C. From C, draw a line perpendicular to y-axis which meets y-axis at 52.
∴ Number of students who did not pass = 52.

Question 28.
The numbers 6, 8,10,12, 13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x.
Solution:
Arranging the numbers in ascending order
6, 8, 10, 12, 13, x
Mean of the given numbers = \(\frac{6+8+10+12+13+x}{6}\) = \(\frac{49+x}{6}\)
Number of terms (n) = 6 which is even
∴ Median = \(\frac{\left(\frac{n}{2}\right) \text { th term }+\left(\frac{n}{2}+1\right) \text { th term }}{2}\) = \(\frac{3^{r d}+4^{t h}}{2}\) = \(\) = \(\frac{10+12}{2}\) = \(\frac{22}{2}\) = 11
According to statement,
It is given that the mean of 6, 8, 10, 12, 13, x is equal to the median of 6, 8, 10, 12, 13, x
∴ \(\frac{49+x}{6}\) = 11 ⇒ 49 + x = 66 ⇒ x = 66 – 49
x = 17

Question 29.
Calculate the mean of the distribution given below using the short cut method.

Marks	11-20	21-30	31-40	41-50	51-60	61-70	71-80
No. of students	2	6	10	12	9	7	4

Solution:

Marks	Frequency (f)	Mid Value (x)	di = xi – A A = 45.5	Ui = \(\frac{x_i-\mathrm{A}}{10}\)	F × ui
11-20	2	15.5	-30	-3	-6
21-30	6	25.5	-20	-2	-12
31-40	10	35.5	-10	-1	-10
41-50	12	45.5	0	0	0
51-60	9	55.5	10	1	9
61-70	7	65.5	20	2	14
71-80	4	75.5	30	3	12
	Σ f=50				Σfui = 7

A = 45.5
Mean = A + \(\frac{\sum f u_i}{\sum f}\) × 10 = 45.5 + \(\frac{7}{50}\) × 10 = 45.5 + 1.4 = 46.9

Question 30.
(Use a graph paper for this question.) The daily pocket expenses of 200 students in a school are given below:

Marks	Number of students (frequency)
0-5	10
5-10	14
10-15	28
15-20	42
20-25	50
25-30	30
30-35	14
35-40	12

Draw a histogram representing the above distribution and estimate the mode from the graph.
Solution:

Estimated mode = 21.5

Steps for calculation of mode
(i) Mark the end points of the upper corner of rectangle with maximum frequency as A and B.
(ii) Mark the inner corner of adjacent rectangles as C and D.
(iii) Join AC and BD do intersect at K. From K, draw KL perpendicular to x-axis.
(iv) The value of L on x-axis represents the mode.

Question 31.
The marks obtained by 100 students in a Mathematics test are given below :

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of students	3	7	12	17	23	14	9	6	5	4

Draw an ogive for the given distribution on a graph sheet.
Use a scale of 2 cm = 10 units on both axis).
Use the ogive to estimate the:
(i) median.
(ii) lower quartile.
(iii) number of students who obtained more than 85% marks in the test.
(iv) number of students who did not pass in the test if the pass percentage was 35.
Solution:

Marks	Number of students	Cumulative frequency (c.f.)
0-10	3	3
10-20	7	10
20-30	12	22
30-40	17	39
40-50	23	62
50-60	14	76
60-70	9	85
70-80	6	91
80-90	5	96
90-100	4	100

On the graph paper, we plot the following points A (10, 3), B (20, 10), C (30, 22), D, (40, 39), E (50, 62), F (60, 70), G (70, 85), H (80, 91), 1 (90, 96) and V (100, 100). Join all these points by a free hand drawing. The required ogive is shown on the graph paper given below :
Here, number of students (n) = 100, which is even.

(i) Let P be the point on Y-axis representing frequency = \(\frac { n }{ 2 }\) = \(\frac { 100 }{ 2 }\) = 50.
Through P, draw a horizontal line to meet the ogive at point Q. Through Q, draw a vertical line to meet the X-axis at T. The abscissa of the point T represents 45 marks.
Hence, median marks is 45.

(ii) Let R be the point on Y-axis representing frequency = \(\frac { n }{ 4 }\) = \(\frac { 100 }{ 4 }\) = 25.
Through R, draw a horizontal line to meet the ogive at point S. Through S, draw a vertical line to meet the X-axis at N. The abscissa of the point N represents 31 marks.
Hence, the lower quartile = 31 marks.

(iii) 85% marks = 85% of 100 = 85 marks
Let the point M on X-axis represents 85 marks.
Through M, draw a vertical line to meet the ogive at the point O. Through O draw a horizontal line to meet the Y-axis at point J. The ordinate of point J represents 95 students on Y-axis.
∴ Number of students who obtained more than 85% in the test = 100 – 95 = 5.

(iv) 35% marks = 35% of 100 = 35
Let the point K on X-axis represents 35 marks.
Through K, draw a vertical line to meet the ogive at the point L. Through L, draw a horizontal line to meet the Y-axis at point U. The ordinate of point U represents 30 students on Y-axis.
Hence, the number of students, who did not pass in the test is 30.

Question 32.
Marks obtained by 30 students in a class assessment of 5 marks is given below (next page):

Marks	0	1	2	3	4	5
No. of students	1	3	6	10	5	5

Calculate the mean, median and mode of the above distribution.
Solution:

Marks	0	1	2	3	4	5
No. of students	1	3	6	10	5	5
Cumulative Frequency	1	4	10	20	25	30

Mean = \(\frac{\sum f x}{\sum f}\)
= \(\frac{0 \times 1+1 \times 3+2 \times 6+3 \times 10+4 \times 5+5 \times 5}{1+3+6+10+5+5}\)
= \(\frac { 90 }{ 30 }\) = 3
∴ 3 is the mean
There are a total of 30 observations in the data
The median is the arithmetic mean of \(\left(\frac{n}{2}\right)^{\text {th }}\) and \(\left(\frac{n}{2}+1\right)^{\text {th }}\) observation in case of even number of observations = Arithmetic mean of \(\left(\frac{30}{2}\right)^{\mathrm{th}}\) and \(\left(\frac{30}{2}+1\right)^{\text {th }}\)
= Arithmetic mean of 15^th and 16^th observation will be the median
∴ Median = \(\frac{3+3}{2}\) = 3
Frequency is highest for the observation x_i = 3
Mode = 3

Question 33.
Calculate the mean of the following distribution:

Class Interval	0-10	10-20	20-30	30-40	40-50	50-60
Frequency	8	5	12	35	24	16

Solution:
Consider the following distribution:

Class interval	Frequency	Class mark  xi	fi xi
0-10	Fi	5	40
10-20	8	15	75
20-30	5	25	300
30-40	12	35	1225
40-50	35	45	1080
50-60	24	55	880
Total	fi = 100		fi xi =3600

Mean = \(\frac{\sum f_i x_i}{\sum f_i}\) = \(\frac { 3600 }{ 100 }\) = 36

Question 34.
The weight of 50 workers is given below:

Weight in Kg	50-60	60-70	70-80	80-90	90-100	100-110	110-120
No. of Workers	4	7	11	14	6	5	3

Draw an ogive of the given distribution using a graph sheet. Take 2 cm = 10 kg on one axis and 2 cm = 5 workers along the other axis. Use a graph to estimate the following:
(i) the upper and lower quartiles.
(ii) if weighing 95 kg and above is considered overweight find the number of workers who are overweight.
Solution:
The cumulative frequency table of the given distribution table is as follows:

Weith in kg	Number of workers	Cumulative frequency
50-60 60-70 70-80 80-90 90-100 100-110 110-120	4 7 11 14 6 5 3	4 11 22 36 42 47 50

Plot the points (60, 4), (70, 11), (80, 22), (90, 36), (100, 42), (110, 47) and (120, 50). Then draw a free-hand curve (ogive) as shown below:

∴ N = 50

(i) Upper quartile = \(\left(\frac{3 \times N}{4}\right)^{\text {th }}\) term
= \(\left(\frac{3 \times 50}{4}\right)^{\text {th }}\) term = 37.5^th term = 92.5 kg
Lower quartile = \(\left(\frac{\mathrm{N}}{4}\right)^{\text {th }}\) term
= \(\left(\frac{\mathrm{50}}{4}\right)^{\text {th }}\) term = 12.5^th term = 71.5 kg

(ii) Through 95 kg mark, draw a vertical line that meets graph at point P. Through point P, draw a horizontal line which meets axis for c.f. at point A and A = 39.
⇒ Weights of 39 workers are 95 kg or below it.
∴ Number of workers who are overweight = 50 – 39 = 11

Question 35.
The mean of following number is 68. Find the value of ‘x’. 45, 52, 60, x, 69, 70, 26, 81 and 94. Hence estimate the median.
Solution:
Mean = \(\frac{\text { Sum of all observations }}{\text { Total number of observation }}\)
∴ 68 = \(\frac{45+52+60+x+69+70+26+81+94}{9}\)
∴ 68 = \(\frac{497+x}{9}\)
∴ 612 = 497 + x
∴ x = 612 – 497
∴ x = 115
Arrange the numbers in ascending order 26, 45, 52, 60, 69, 70, 81, 94, 115
Since the number of observations’ is odd, the median is \(\left(\frac{n+1}{2}\right) \text { th }\) observation
⇒ The mrdian is the \(\left(\frac{9+1}{2}\right)\) = 5th observation
Hence, the median is 69.

Question 36.
The table shows the distribution of the scores obtained by 160 shooters in a shooting competition. Use a graph sheet and draw an ogive for the distribution. (Take 2 cm = 10 scores on the X-axis and 2 cm = 20 shooters on the Y-axis).

Scores	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70-80	80-90	90-100
No. of shooters	9	13	20	26	30	22	15	10	8	7

Use your graph to estimate the following :
(i) The median.
(ii) The interquartile range.
(iii) The number of shooters who obtained a score of more than 85%.
Solution:

Scores	f	c.f.
0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100	9 13 20 26 30 22 15 10 8 7	9 22 42 68 98 120 135 145 153 160
	N = 160	c.f.

(i) Median = \(\left(\frac{n}{2}\right)^{\mathrm{th}}\) term = \(\left(\frac{160}{2}\right)^{\text {th }}\) term = 80^th term
Through mark 80 on y-axis, draw a horizontal line which meets the ogive drawn at point Q. Through Q, draw a vertical line which meets the x-axis at the mark of 43.
∴ Median = 43

(ii) Since the number of terms = 160
Lower quartile (Q₁) = \(\left(\frac{160}{4}\right)^{\text {th }}\) term = 40^th term = 28
Upper quartile (Q₃) = \(\left(\frac{3 \times 160}{4}\right)^{\text {th }}\) term = 120^th term = 60
∴ Inter-Quartile range = Q₃ – Q₁
= 60 – 28 = 32
∴ Inter-Quartile range is 32

(iii) Since 85% scores = 85% of 100 = 85
Through mark for 85 on x-axis, draw a vertical line which meets the ogive drawn at point B. Through the point B, draw a horizontal line which meets the y-axis at the mark of 150.
∴ The number of shooters who obtained more than 85% score
= 160 – 150 = 10
So, the number of shooters who obtained more than 85% score is 10.

Question 37.
The histogram below represents the scores obtained by 25 students in a Mathematics mental test. Use the data to :
(i) Frame a frequency distribution table.
(ii) To calculate mean.
(iii) To determine the Modal class.

Solution:
(i)

Class interval	Frequency
0-10	2
10-20	5
20-30	8
30-40	4
40-50	6

(ii)

Class interval	Frequency (f)	Mean value (x)	f x
0-10 10-20 20-30 30-40 40-50	2 5 8 4 6	5 15 25 35 45	10 75 200 140 270
	Σf = 25		Σfx = 695

∴ Mean = \(\frac{\sum f x}{\sum f}\) = \(\frac{695}{25}\) = 27.8

(iii) Modal class = class with maximum frequency = 20-30