Chapter 24 - Measures of Central Tendency (Mean, Median, Quartiles and Mode) Exercise Ex. 24(E)

Question 1

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

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:

1. The median height.
2. The interquartile range.
3. The number of students whose height is above 172 cm.

Solution 1

 

 

Taking Height of student along x-axis and cumulative frequency along y-axis we will draw an ogive.

(i)

 

Through mark for 80, draw a parallel line to x-axis which meets the curve; then from the curve draw a vertical line which meets the x-axis at the mark of 157.5.

 (ii)Since, number of terms = 160

(iii)Through mark for 172 on x-axis, draw a vertical line which meets the curve; then from the curve draw a horizontal line which meets the y-axis at the mark of 145.

The number of students whose height is above 172 cm

= 160 - 144 = 16

Question 2

Draw ogive for the data given below and from the graph determine: (i) the median marks.

(ii) the number of students who obtained more than 75% marks.

Marks	10 - 19	20 -29	30 - 39	40 - 49	50 - 59	60 - 69	70 - 79	80 - 89	90 - 99
No. of students	14	16	22	26	18	11	6	4	3

 

Solution 2

 

Marks	No. of students	Cumulative frequency
9.5 - 19.5	14	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

Scale:

1cm = 10 marks on X axis

1cm = 20 students on Y axis

 

 

(i) 

Through mark 60, draw a parallel line to x-axis which meets the curve at A. From A, draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

 

(ii) Total marks = 100

75% of total marks =  marks

The number of students getting more than 75% marks = 120 - 111 = 9 students.

 

Question 3

The mean of 1, 7, 5, 3, 4 and 4 is m. The numbers 3, 2, 4, 2, 3, 3 and p have mean m-1 and median q. Find p and q.

Solution 3

Mean of 1, 7, 5, 3, 4 and 4 = 

m=4

Now, mean of 3, 2, 4, 2, 3, 3 and p = m-1 = 4-1 = 3

Therefore, 17+p = 3 x n …. Where n = 7

17+p = 21

p = 4

Arranging in ascending order:

2, 2, 3, 3, 3, 3, 4, 4

Mean = 4^th term = 3

Therefore, q = 3

Question 4

In a malaria epidemic, the number of cases diagnosed were as follows:

Date July	1	2	3	4	5	6	7	8	9	10	11	12
Num	5	12	20	27	46	30	31	18	11	5	0	1

 

On what days do the mode and upper and lower quartiles occur?

Solution 4

 

Date	Number	C.f.
1	5	5
2	12	17
3	20	37
4	27	64
5	46	110
6	30	140
7	31	171
8	18	189
9	11	200
10	5	205
11	0	205
12	1	206

 

(i) Mode = 5^th July as it has maximum frequencies.

(ii) Total number of terms = 206

Upper quartile = 

Lower quartile = 

Question 5

The income of the parents of 100 students in a class in a certain university are tabulated below.

Income (in thousand Rs)	0-8	8-16	16-24	24-32	32-40
No. of students	8	35	35	14	8

 

(i) Draw a cumulative frequency curve to estimate the median income.

(ii) If 15% of the students are given freeships on the basis of the basis of the income of their parents, find the annual income of parents, below which the freeships will be awarded.

(iii) Calculate the Arithmetic mean.

Solution 5

 

We plot the points (8, 8), (16, 43), (24, 78), (32, 92) and (40, 100) to get the curve as follows:

 

 

 

 

 

 

 

At y = 50, affix A.

 

Through A, draw a horizontal line meeting the curve at B.

 

Through B, a vertical line is drawn which meets OX at M.

 

OM = 17.6 units

 

Hence, median income = 17.6 thousands

 

 

Question 6

The marks of 20 students in a test were as follows:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19 and 20.

Calculate:

(i) the mean (ii) the median (iii) the mode

Solution 6

Arranging the terms in ascending order:

2, 6, 8, 9, 10, 11, 11, 12, 13, 13, 14, 14, 15, 15, 15, 16, 16, 18, 19, 20

Number of terms = 20

(i) 

(ii) 

(iii) Mode = 15 as it has maximum frequencies i.e. 3

Question 7

The marks obtained by 120 students in a mathematics test is given below:

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

 

Draw an ogive for the given distribution on a graph sheet. Use a suitable scale for your ogive. Use your ogive to estimate:

(i)the median

(ii)the number of students who obtained more than 75% in test.

(iii)the number of students who did not pass in the test if the pass percentage was 40.

(iv)the lower quartile

Solution 7

Marks	No. of students	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

 

(i) 

Through mark 60.5, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 43

(ii) Number of students who obtained up to 75% marks in the test = 110

Number of students who obtained more than 75% marks in the test = 120 - 110 = 10

(iii) Number of students who obtained less than 40% marks in the test = 52 (from the graph; x=40, y=52)

(iv) Lower quartile = Q₁ = 

Question 8

Using a graph paper, draw an ogive for the following distribution which shows a record of the width in kilograms of 200 students.

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

Use your 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 student fall

(iii) The number of students who are

(a) underweight

(b) overweight,

if 55.70 kg is considered as standard weight.

Solution 8

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
75-80	9	200

(i) Number of students weighing more than 55 kg = 200-44 = 156

Therefore, percentage of students weighing 55 kg or more

(ii) 30% of students = 

Heaviest 60students in weight = 9 + 21 + 30 = 60

weight = 65 kg ( from table)

(iii) (a) underweight students when 55.70 kg is standard = 46 (approx) from graph

(b) overweight students when 55.70 kg is standard = 200- 55.70 = 154 (approx) from graph

Question 9

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 9

Marks obtained(x)	No. of students (f)	c.f.	fx
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	25		171

Number of terms = 25

(i) Mean = 

(ii) 

(iii) Mode = 6 as it has maximum frequencies i.e. 6

Question 10

The mean of the following distribution is 52 and the frequency of class interval 30-40 is 'f'. Find f.

C.I	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Freq	5	3	f	7	2	6	13

Solution 10

C.I.	Frequency(f)	Mid value (x)	fx
10-20	5	15	75
20-30	3	25	75
30-40	f	35	35f
40-50	7	45	315
50-60	2	55	110
60-70	6	65	390
70-80	13	75	975
Total	36+f		1940+35f

 

Question 11

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

Monthly Income (thousands)	No. of employees
6-7	20
7-8	45
8-9	65
9-10	95
10-11	60
11-12	30
12-13	5

Draw an ogive of the given distribution on a graph paper taking 2 cm = Rs 1000 on one axis and 2 cm = 50 employees on the other axis. From the graph determine :

(i) the median wage.

(ii) number of employees whose income is below Rs 8500.

(iii) if salary of a senior employee is above Rs 11,500, find the number of senior employees in the company.

(iv) the upper quartile.

Solution 11

Monthly Income (thousands)	No. of employees (f)	Cumulative frequency
6-7	20	20
7-8	45	65
8-9	65	130
9-10	95	225
10-11	60	285
11-12	30	315
12-13	5	320
Total	320

 

Number of employees = 320

(i) 

Through mark 160, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = Rs 9.3 thousands

(ii) The number of employees with income below Rs 8500 = 95 (approx from the graph)

(iii) Number of employees with income below Rs 11500 = 305 (approx from the graph)

Therefore number of employees (senior employees) = 320-305 =15

(iv) Upper quartile = 

Question 12

A mathematics aptitude test of 50 students was recorded as follows:

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

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

Solution 12

 

(i)Draw the histogram

(ii) In the highest rectangle which represents modal class draw two lines AC and BD intersecting at P.

(iii) From P, draw a perpendicular to x-axis meeting at Q.

(iv) Value of Q is the mode = 82 (approx)

Question 13

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

Marks	No. of students
0-10	5
10-20	11
20-30	10
30-40	20
40-50	28
50-60	37
60-70	40
70-80	29
80-90	14
90-100	6

Draw an ogive of the given distribution on a graph paper taking 2 cm = 10 marks on one axis and 2 cm = 20 students on the other axis. Using the graph:

(i) the median marks.

(ii) 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 13

Marks	No. of students	Cumulative frequency
0-10	5	5
10-20	11	16
20-30	10	26
30-40	20	46
40-50	28	74
50-60	37	111
60-70	40	151
70-80	29	180
80-90	14	194
90-100	6	200

Number of students = 200

(i) 

Through mark 100, draw a parallel line to x-axis which meets the curve at A, From A draw a perpendicular to x-axis meeting it at B.

The value of point B is the median = 57 marks (approx)

(ii) The number of students who failed (if minimum marks required to pass is 40)= 46 (approx from the graph)

(iii) The number of students who secured grade one in the examination = 200 - 188 = 12 (approx from the graph)

Question 14

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

If mean of the distribution is 7.2, find a and b.

Solution 14

 

Question 15

Find the mode and the median of the following frequency distribution.

 

Solution 15

Since the frequency for x = 14 is maximum.

So Mode = 14.

 

According to the table it can be observed that the value of x from the 13^th term to the 17^th term is 13.

So the median = 13.

Question 16

The median of the 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 16

 

Question 17

The number 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 17

 

Question 18

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

Pocket expenses (in Rs)	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40
No. of students (frequency)	10	14	28	42	50	30	14	12

 

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

Solution 18

Histogram is as follows:

 

 

 

In the highest rectangle which represents modal class draw two lines AC and BD intersecting at E. 

From E, draw a perpendicular to x-axis meeting at L.

Value of L is the mode. Hence, mode = 21.5

Question 19

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 the axes.

Use the ogive to estimate :

(i) Median

(ii) Lower quartile

(iii) Number of students who obtained more than 85% marks in the test.

(iv) Number of students failed, if the pass percentage was 35.

Solution 19

 

Marks	Number of students (Frequency)	Cumulative Frequency
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

 

The ogive is as follows:

 

 

 

 

 

 

 

Question 20

The mean of following numbers is 68. Find the value of 'x'.

45, 52, 60, x, 69, 70, 26, 81 and 94.

Hence, estimate the median.

Solution 20

 

 

Question 21

The marks of 10 students of a class in an examination arranged in ascending order is as follows:

13, 35, 43, 46, x, x + 4, 55, 61, 71, 80

If the median marks is 48, find the value of x. Hence, find the mode of the given data.

Solution 21

Here the number of observations i. e is 10, which is even.'

 

 

So, the given data is 13, 35, 43, 46, 46, 50, 55, 61, 71, 80.

In the given data, 46 occurs most frequently.

∴ Mode = 46 

Question 22

The daily wages of 80 workers in a project are given below.

Wages	400- 450	450- 500	500- 550	550-600	600-650	650-700	700- 750
No.of workers	2	6	12	18	24	13	5

 

Use a graph paper to draw an ogive for the above distribution. (Use a scale of 2 cm = Rs. 50 on x - axis and 2 cm = 10 workers on y - axis). Use your ogive to estimate.

i. the median wages of the workers.

ii. the lower quartile wage of workers.

iii. the number of workers who earn more than Rs. 625 daily.

Solution 22

The cumulative frequency table of the given distribution is as follows:

Wages (Rs.)	Upper limit	No. of workers	C.f.
400-450	450	2	2
450-500	500	6	8
500-550	550	12	20
550-600	600	18	38
600-650	650	24	62
650-700	700	13	75
700-750	750	5	80

The ogive is as follows:

 

 

Number of workers = n = 80

1) Median = term = 40^th term, draw a horizontal line which meets the curve at point A.

Draw vertical line parallel to y axis from A to meet x axis at B.

The value of point B is 605.

2) Lower quartile (Q1)= term=20^th term = 550

3) Through mark of point 625 on x axis draw a vertical line which meets the graph at point C Then through point C, draw a horizontal line which meets the y axis at the mark of 50.

Thus, the number of workers that earn more than Rs. 625 daily = 80 - 50 = 30

Question 23

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 23

i. The frequency distribution table is as follows:

 

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)	fx
0-10	2	5	10
10- 20	5	15	75
20-30	8	25	200
30-40	4	35	140
40-50	6	45	270
	Sf = 25		Sf = 695

 

 

iii. Here the maximum frequency is 8 which is corresponding to class 20 - 30.

 Hence, the modal class is 20 - 30.