Course Contents:

Measure of Location: Mean, Median, Mode, Partition values: Quartiles, Deciles, Percentiles, Selection of measure of location
Measure of Dispersion: Range, Inter quartile Range, Quartile Deviation, Standard deviation, Variance, Coefficient of variation,
Shape of the Distribution: Shape of the distribution by using Mean and Median, Five number summary, Box and whisker plot.
Project Work: Descriptive statistics and numerical measures of tourism and hospitality data by using computer software

Table Of Contents

Practical Problems

Practical Problems

The following are the prices of shares of a company from Sunday to Friday.

Day	S	M	T	W	Th	F
Price (Rs)	200	210	208	160	220	250

Compute the average price by (i) direct method (ii) short-cut method

Solution:

By Using Direct Method (Direct Method)

Here,

Σx = 200 + 210 + 208 + 160 + 220 + 250
∴ Σx = 1248

n = 6
Arithmetic Mean (x̄) = ?

Now,

Arithmetic Mean (x̄) = Σx / n
or, x̄ = 1248 / 6
∴ x̄ = 208

Hence, the arithmetic mean is Rs 208.

By Using Short-cut Method

Here,
Let us assume 208 as assume mean (a=208)

Calculation of A.M

Day	Price (Rs) (X)	d = x – 208 (a)
S M T W Th F	200 210 208 160 220 250	-8 2 0 -48 12 42
n = 6	Σx = 1248	Σd = 0

Here,

Assumed mean (a)= 208
Sum of deviations (Σd) = 0
Σx = 1248
n = 6
Arithmetic Mean (x̄) = ?

Now,

x̄ = a + (Σd /n)
or, x̄ = 208 + (0/6)
or, x̄ = 208 + 0
∴ x̄ = 208

Hence, average price is Rs 208.

a) Find the arithmetic mean from the following frequency table.

Height (in cms)	40	45	50	52	60	70
No. of plants	7	8	9	6	4	2

By Using Direct Method

x	f	fx
40 45 50 52 60 70	7 8 9 6 4 2	280 360 450 312 240 140
	N = 36	Σfx = 1782

Now,

Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 1,782 / 36
∴ x̄ = 49.5

By Using Short-cut Method

Height (in cms)	No of Plants (f)	d = x – 50	fd
40 45 50 52 60 70	7 8 9 6 4 2	-10 -5 0 2 10 20	-70 -40 0 12 40 40
	N = 36		Σfd = -18

Arithmetic Mean (x̄) = a + (Σfd / N)
or, x̄ =50 + (-18 / 36)
∴ x̄ = 49.5

b) Calculate the arithmetic mean from the following data (i) direct method (ii) shortcut method.

Marks	5	10	15	20	25
No. of Students	4	6	8	5	2

Solution

By using Direct Method

x	f	fx
5 10 15 20 25	4 6 8 5 2	20 60 120 100 50
	N = 25	Σfx = 350

Now,
Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 350 / 25
∴ x̄ = 14

By Using Short-cut Method

Marks	No of students (f)	d = x – 15	fd
5 10 15 20 25	4 6 8 5 2	-10 -5 0 5 10	-40 -30 0 25 20
	N = 25		Σfd = -25

Arithmetic Mean (x̄) = a + (Σfd / N)
or, x̄ = 15 + (-25 / 25)
∴ x̄ = 14

c) From the following data, find the missing frequency when arithmetic mean is 15.8.

Central value	10	12	14	16	18	20
Frequency	2	6	?	21	8	6

Solution

Let the missing frequency be f₁.

Computation of Missing Frequency

X	f	fx
10 12 14 16 18 20	2 6 f₁ 21 8 6	20 72 14f₁ 336 144 120
	N = 43 + f₁	Σfx = 692 + 14f₁

Here,
N = 43 + f₁
Σfx = 692 14f₁

x̄ = 15.8 ()

Now,

Arithmetic Mean (x̄) = Σfx / N
or, 15.8 = [(692 + 14f₁ )/ (43 + f₁)]
or, 679.4 + 15.8f₁ = 692 + 14f₁
or, 15.8f₁ – 14f₁ = 692 – 679.4
or, 1.8 f₁ = 12.6
or, f₁ = 12.6 / 1.8
∴ f₁ = 7

a) The number of students in different examination and the percentage results of the two universities A and B, is give below, Find out which is a better university.

University A

University B

Examination	% result	No. of students	% result	No. of students
BBS MBS BA MA	50 60 50 70	100 50 200 150	65 50 60 50	100 60 150 90

Solution:

For University A

Examination	% result (x)	No. of students (f)	fx
BBS MBS BA MA	50 60 50 70	100 50 200 150	5,000 3,000 10,000 10,500
		N = 500	Σfx = 28,500

Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 28,500 / 500
∴ x̄ = 57

For University B

Examination	% result (x)	No. of students (f)	fx
BBS MBS BA MA	65 50 60 50	100 60 150 90	6,500 3,000 9,000 4,500
		N = 400	Σfx = 23,000

Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 23,000 / 400
∴ x̄ = 57.5

Since, the arithmetic mean of university B is greater than university A by 0.5, University B is a better university.

b) A travelling salesman made five trips in two months. The records of sales is given below:

Trip	No. of days	Sales per day (Rs)
1 2 3 4 5	5 4 3 7 6	800 500 400 600 500
	25	2800

The sales manager calculated the simple arithmetic mean and the salesman calculate the weighted arithmetic mean of the daily sales. One criticizes the result of the other.

i. Find the average daily sales obtained by the sales Manager.
ii. Find the average daily sales obtained by the salesman.
iii. In your opinion, who has used the appropriate average and why?

Solution:

Calculating arithmetic mean as per sales manager (Simple Arithmetic Mean)

Arithmetic Mean (x̄) = Σx / N
or, x̄ = 2800 / 5
∴ x̄ = Rs 560 per day

ii.

Calculating arithmetic mean as per salesman (Weighted Arithmetic Mean)

Trip	No. of days (w)	Sales per day (Rs) (x)	wx
1 2 3 4 5	5 4 3 7 6	800 500 400 600 500	4,000 2,000 1,200 4,200 3,000
	Σw = 25	Σx = 2800	Σwx = 14,400

Weighted Arithmetic Mean x̄ _w (Salesman) = Σwx / Σw
or, x̄ _w = 14,400 / 25
∴ x̄ _w = Rs 576 per day

iii. The sales manager used the simple arithmetic mean, while the salesman used the weighted arithmetic mean. The weighted arithmetic mean gives more weight to the sales figures on days when more sales were made.

Hence, the salesman has used the appropriate method.

a) The quantities of water used by 40 families in a certain locality is as follows:

Quantity of water (in “000” litres)	No. of families
5 – 10 10 – 15 15 – 20 20 – 25 25 – 30	4 12 16 6 2

Compute the arithmetic mean from the above distribution.

Solution

Quantity of water (in “000” litres)	Mid Value (x)	No. of families (f)	fx
5 – 10 10 – 15 15 – 20 20 – 25 25 – 30	7,500 12,500 17,500 22,500 27,500	4 12 16 6 2	30,000 150,000 280,000 135,000 55,000
		N = 40	Σfx = 650,000

Now,
Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 650,000 / 40
∴ x̄ = 16,250

b) Compute the arithmetic mean of the following distribution by (i) direct (ii)Short- cut (iii) Step deviation Method

Production	0-10	10-20	20-30	30-40	40-50
No. of factories	5	8	25	16	6

Solution

i. By using direct Method

Production	Mid value (x)	No. of factories (f)	fx
0-10 10-20 20-30 30-40 40-50	5 15 25 35 45	5 8 25 16 6	25 120 625 560 270
		N = 60	Σfx = 1600

Now,
Arithmetic Mean (x̄) = Σfx / N
or, x̄ =1,600 / 60
∴ x̄ = 26.66

ii. By using short-cut method

Production	Mid Value (x)	No. of factories (f)	d = x – 25	fd
0-10 10-20 20-30 30-40 40-50	5 15 25 35 45	5 8 25 16 6	-20 -10 0 10 20	-100 -80 0 160 120
		N = 60		Σfx = 100

Here,
N= 60
a = 25 (supposed)
Σfx = 100
Here,
a = 25 (supposed)
Σfx = 100

Arithmetic Mean = a + (Σfd / N )
or, x̄ = 25 + (100 / 60)
∴ x̄ = 26.66

iii. By using step-deviation method

Production	Mid Value (x)	No. of factories (f)	d = x – 25	d’ = d / 10	fd’
0-10 10-20 20-30 30-40 40-50	5 15 25 35 45	5 8 25 16 6	-20 -10 0 10 20	-2 -1 0 1 2	-10 -8 0 16 12
		N = 60			Σfd’ = 10

Here,
a = 25
Σfd’ = 10
N = 60
h= 10
x̄ = ?

Now,
Arithmetic Mean (x̄) = a + (Σfd’ / N) * h
or, x̄ = 25 + (10 / 60) * 10
∴ x̄ = 26.66

c) Calculate the arithmetic mean from the following data:

Age (in years)	18-21	22-25	26-35	36 -45	46-55
No. of employee	8	32	54	36	20

Solution

By using Direct Method

Age (in years)	Mid value (x)	No. of factories (f)	fx
18-21 22-25 26-35 36-45 46-55	19.5 23.5 30.5 40.5 50.5	8 32 54 36 20	156 752 1,647 1,458 1,010
		N = 150	Σfx = 5,023

Now,
Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 5,023 / 150
∴ x̄ = 33.48

d) The following table presents the weekly wages of the workers in a firm. Calculate the average weekly wage per worker:

Wages (Rs)	100-120	120-140	140-160	160-180	180-200	200-220
Total hours	180	140	156	153	195	143
Average no. of hours worked per worker	15	10	12	9	15	13

Solution

Wages	Mid Value (x)	Workers
100-120 120-140 140-160 160-180 180-200 200-220	110 130 150 170 190 210

e) From the following data of income distribution, calculate the arithmetic mean. It is given that

i. the total income of the person in the highest group is Rs 435
ii. none is earning less than Rs 20

Income (Rs)							80 and
Below –	30	40	50	60	70	80	above
No of persons	16	36	61	76	87	95	5

Solution:

Sequencing:

Income	No of persons
20-30 30-40 40-50 50-60 60-70 70-80 80 and above	16 10 25 15 11 8 5

Calculating Arithmetic Mean

Income	Mid Point	No of persons (f)	fx
20-30 30-40 40-50 50-60 60-70 70-80 80 and above	25 35 45 55 65 75 –	16 10 25 15 11 8 5	400 350 1125 825 715 600 435
		Σf = 100	Σfx = 4,450

Now,
Arithmetic Mean (x̄) = Σfx / N
or, x̄ = 4,450 / 100
∴ x̄ = 44.5

The following are the monthly salaries in rupees of 20 employees of a firm.

130	62	145	118	125	76	151	142	110	98
65	116	100	`103	71	85	80	122	132	95

The firm gives bonus of Rs 10, 15, 20, 25 and 30 for individuals in the respective salary groups exceeding Rs 60 but not exceeding Rs 80, exceeding Rs 80 but not exceeding Rs 100 and so on upto exceeding Rs 140 but not exceeding Rs 160. Find the average bonus paid per employee.

Solution:

Salary Class	Bonus (x)	Tally Bar	f	fx
61-80 81-100 101-120 121-140 141-160	10 15 20 25 30	llll lll lll lll ll	4

a) From the information given below, find
i. Which factory pays larger amount as daily wage?
ii. What is the average daily wage for the workers of the two factory.

	Factory A	Factory B
No. of wage earners:	250	200
Average daily wage:	Rs 20	Rs 25

Solution:

Total Wages = Nx̄
For factory a =

b) The mean weight of 150 students in a certain class is 60 kg. The mean weight of boys in the class is 70 kg and that of girls is 55 kg. Find the number of boys and girls in the class.

c) The pass result of 50 students who took up a class test is given below:

Marks	40	50	60	70	80	90
No. of students	8	10	9	6	4	3

If the average mark of all the 50 students was 51.6, find out the average mark of the students who failed.

d) The mean mark of 100 students found to be 40. Later on it was discovered that a score 53 was misread as 83. Find the correct mean corresponding to the correct score. What will be the mean mark when the wrong score is omitted?

a) Find the median and the two quartiles from the following data:
i. 40, 50, 37, 60, 12, 96, 25
ii. 7, 18, 55,33 ,67,41, 28, 73
iii. Find the 6th decile and 35th percentile of 96, 98, 75, 80, 102, 100, 94, 78

b) In a batch of 13 students, 3 students failed in a test. The marks of 3 failed students and the one student with highest mark are missing. If the marks of remaining 9 students are 83, 46, 71, 64, 90, 58, 43, 62, 88, what is the median of the marks of all 13 students.

c) The following marks have been obtained in two papers of statistics in examination by 10 students. In which paper is the general level of knowledge of the students highest? Give reason.

A:	36	56	41	46	54	59	55	51	52	44
B:	58	54	21	51	59	46	65	31	68	41

a) Find the median and the two quartiles from the following data:

Expenditure (in Rs)	1500	3000	4500	6000	7500	9000	10500
No. of families	3	7	10	5	3	5	2

b) Calculate the median from the following frequency table of marks at a test in statistics.

Marks	5	10	15	20	25	30	35	40	45	50
No. of students	6	13	15	8	25	32	18	12	15	5

Also find the two quartiles, 4th decile and 80th percentile.

a) Find the median and the two quartiles from the following frequency distribution:

Marks	20-30	30-40	40-50	50-60	60-70	70-80
No. of students	3	5	6	8	4	4

b) Calculate the appropriate measure of central tendency from the following distribution and give reason for the choice of the measure

Wages:	Below 20	20-30	30-50	50-70	70-80	80 and above
No. of workers:	5	3	4	8	4	6

Also find the 4th decile and 60th percentile.

c) Calculate the median and the two quartiles from the following frequency distribution

Value:	0-5	5-10	10-20	20-30	30-50
Frequency	5	7	15	20	13

d) Amend the following table and locate the median from the amended data:

Size	Frequency	Size	Frequency
10-16 16-17.5 17.5-20 20-30	10 15 17 25	30-35 35-40 40 and onwards	28 30 40

e) From the following frequency table, calculate the median, two quartiles, 6th decile and 70th percentile.

Marks (less than)	10	20	30	40	50	60	70	80
No. of students:	5	13	20	32	60	80	90	100

f) Calculate the appropriate measure of central tendency from the following distribution and support for your choice of the measure

Monthly income (Rs)	No. of families
Below 100 100-199 200-299 300-399 400-499 500 and above	5 20 40 15 12 8

Also find two quartiles.

a) The monthly expenditure in rupees for a group of families is as follows:

Expenditure (in Rs)	100-200	200-300	300-350	350-400	400-500
No. of families	8	–	20	12	5

Median of the expenditure is known to be Rs. 317.5. Determine the number of families having expenditure between Rs. 200 – Rs 300

b) Income distribution of 100 workers is given below:

Income per day (in Rs)	0-200	200-400	400-600	600-800	800-1000
No of workers:	15	–	30	20	–

MODE

i. When 11 students were asked for their favorite number, the responses were: 7, 10, 72, 7 1600, 4, 1 , 7, 2, 1, 7.
What measure of central tendency is the most appropriate to describe their answers?
Solution:

x	f
1 2 4 7 10 72 1600	2 1 1 4 1 1 1

Maximum frequency is 4 times, whose corresponding variate’s value is 7.
∴ Mode (M_o) = 7

ii. Compute mode from the following:

x	50	100	150	200	250	300	350	400
f	5	14	40	91	150	87	60	38

Solution:

x	f
50 100 150 200 250 300 350 400	5 14 40 91 150 87 60 38

Maximum frequency is 150 times, whose corresponding variate’s value is 250.
∴ Mode (M_o) = 250

Calculate the modal size in the following distribution:

Size (inches)	Below 10	10 – 12	12 – 14	14 – 16	16 – 18	18 – 20
Demand	3	15	27	20	3	2

Will the median fall under the same size?

Solution:

Size (inches) (x)	Demand(f)
0 – 10 10 – 12 12 – 14 14 – 16 16 – 18 18 – 20	3 15 27 20 3 2

Here, maximum frequency is 27 so mode lies between 12 – 14.

Where,

f_m= 27
f₁= 15
f₂= 20
L = 12
h = 2

We have,

M_o = L + [( f_m – f₁) / (2 f_m – f₁ – f₂)] * h
or, M_o = 12 + [( 27 – 15) / ( 2 * 27 – 15 – 20)] * 2
∴ M_o = 13.26 inches

Computation of Median

Size (x)	Demand (f)	C.f
Below 50 10 – 12 12 – 14 14 – 16 16 – 18 18 – 20	3 15 27 20 3 2	3 18 45 65 68 70

We have,
Median (Md) = N / 2
= 70 / 2
= 35

The cumulative frequency just greater than 35 is 45 and its corresponding value is 12 – 14 which is the required median.

Yes, the median falls under the same size.

Calculate the modal value from the following data:

Income (Rs) (less than)	100	200	300	400	500	600
No. of Persons	8	22	35	60	67	70

Solution:

Income (x)	No. of persons (f)
Less than 100 100 – 200 200 – 300 300 – 400 400 – 500 500 – 600	8 14 13 25 7 3

Here, maximum frequency is 25.

so, we have,
f_m= 25
f₁= 13
f₂= 7
L = 300
h = 100

Now,

M_o = L + [( f_m – f₁) / (2 f_m – f₁ – f₂)] * h
or, M_o = 300+ [( 25 – 13) / ( 2 * 25 – 13 – 7)] * 100
∴ M_o = Rs 340

Determine the mode using empirical relation between mean, median and mode.

x	20	21	22	23	24	25	26	27	28	29
f	6	9	4	2	10	8	7	5	1	3

Solution:

x	f	fx	cf
20 21 22 23 24 25 26 27 28 29	6 9 4 2 10 8 7 5 1 3	120 189 88 46 240 200 182 135 28 87	6 15 19 21 31 39 46 51 52 55
	N = 55	Σfx = 1315

Mean = Σfx / N
= 1315 / 55
= 23.909

Median = N / 2
= 55 / 2
= 27.5

The cf greater than 27.5 is 31 and it’s corresponding value 24 is the required median.

Since the maximum frequency is repeated once only.
Mode ( M_o) = 3 median – 2 mean
or, M_o= 3 * 27.5 – 2 *23.909
∴ Mode ( M_o) = 24.18

You are given the following incomplete frequency distribution. It is known that the total frequency is 1000 and that the median is 413.11. Estimate by calculation, the missing frequencies and find the value of the mode.

Value (x)	300 – 325	325 – 350	350 – 375	375 – 400	400 – 425	425 – 450	450 – 475	475 – 500
Frequency (f)	5	17	80	?	326	?	88	9

Solution:

Value (x)	f
300 – 325 325 – 350 350 – 375 375 – 400 400 – 425 425 – 450 450 – 475 475 – 500	5 17 80 f₁ 326 f₂ 88 9

Let f₁ and f₂be the frequencies corresponding to the classes 375 – 400 and 425 – 450 respectively.

Computation of missing frequencies

Value (x)	f	cf
300 – 325 325 – 350 350 – 375 375 – 400 400 – 425 425 – 450 450 – 475 475 – 500	5 17 80 f₁ 326 f₂ 88 9	5 22 102 102 + f₁ 428 + f₁ 428 + f₁ + f₂ 516 + f₁ + f₂ 525 + f₁ + f₂
	N = 525 + f₁ + f₂

From the table,
Total frequency = 525 + f₁ + f₂
or, 1000 = 525 + f₁ + f₂
∴ f₁ + f₂ = 475 ——– Suppose, equation (i)

By the given value of median= 413.11, which lies in the class 400 – 425.

Here,

L = 400
f = 326
h = 25
cf = 102 + f₁
Md = 413.11
N/ 2 = 1000 / 2 = 500

Now,

Md = L + [{(N / 2) – cf} / f ]* h
or, 413.11 = 400 +[{ 500 – 102 + f₁}/ 326 ]* 25
or, 13.11 = [398 + f₁ / 326 ]* 25
∴ f₁ = 227

From the equation (i), We have,

f₁ + f₂ = 475
or, 227 + f₂ = 425
∴ f₂ = 248

Now, calculating the value of mode:
Since the highest frequency is 326 which corresponding class is 400 – 425.

Where,

f_m= 326
f₁= 227
f₂= 248
L = 400
h = 25

We have,

M_o = L + [( f_m – f₁) / (2 f_m – f₁ – f₂)] * h
or, M_o = 400 + [( 326 – 227) / ( 2 * 326 – 227- 248)] * 25
∴ M_o = 413.98

Calculate mean, median and mode from the following data of the heights in inches of a group of students.

61, 62, 63, 61, 61, 64, 64, 60, 65, 63, 64, 65, 66, 64

Now suppose that a group of students whose heights are 60, 66, 59, 68, 67 and 70 inches is added to the original group. Find the mean, median and mode of the combined group.

Solution:

x	f	cf	fx
59 60 61 62 63 64 65 66 67 68 70	1 2 3 1 2 4 2 2 1 1 1	1 3 6 7 9 13 15 17 18 19 20	59 120 183 62 126 256 130 132 67 68 70
	N = 20		Σfx = 1273

Mean = Σfx / N
= 1273 / 20
= 63.65

Median (Md) = N / 2
= 20 / 2
= 10 th

the c.f greater than 10 is 13 and it’s corresponding value 64 is the required median.

Mode (M_o)= 3 Median – 2 mean
or, M_o= 3 * 64 – 2 * 63.65
∴ M_o = 64.7

A locality with 16 schools has the following distribution of average number of teachers in different income groups.

Income (Central Value)	600	800	1000	1200	1400
No. of schools	2	3	5	4	2
Average no. of teachers	20	26	21	21	9

Find the mean, median and modal income of all the teachers.

Solution:

A cement company sells his production in different cities through the appointed dealers. The sales of his production in the last year is given in the following tabular form:

Sales (in ’00’ bags)	Number of dealers
0 – 500 500 -1000 1000 – 1500 1500 – 2000 2000 – 2500 2500 and more	40 48 60 52 35 22

The annual general meeting of the company decided to give award of Rs 5,000 to each dealer whose sakes are more than the most usual sales. Calculate the total amount of money that would be given to the dealers.

Solution:

The following table represents the marks of 100 students.

Marks	0 – 20	20 – 40	40 – 60	60 – 80	80 – 100
No. of students	14	–	27	–	15

If the mode value is 48 find the missing frequencies and the mean marks of all 100 students.

Solution:

Marks	Mid value	No. of students (f)
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100	10 30 50 70 90	14 a 27 b 15

In the given data, the mode value is 48 which lies in the class 40-60. So,

We have,

f_m= 27
f₁= a
f₂= b
L = 40
h = 20

We have,

M_o = L + [( f_m – f₁) / (2 f_m – f₁ – f₂)] * h
or, 48 = 40+ [( 27 – a) / ( 2 * 27 – a – b)] * 20
or, 8 = (27 – a) / (54-a-b) * 20
or, 432 – 8a -8b = 540 – 20a
or, 20a – 8a – 8b = 540 – 432
∴ 12a – 8b = 108 ——— Suppose, equation (i)

Again, We have total 100 students

14+ a + 27 + b + 15 = 100
or a + b = 44 —– Suppose, equation (ii)

Solving equation i and ii

a + b = 44
∴a = 44 – b

12 a – 8b = 108
or, 12 * (44 – b) – 8b = 108
or, 528 – 12b – 8b = 108
or, 528 – 108 = 20b
∴ b = 21
Again,
a = 44 – b
or, a = 44 – 21
∴ a = 23

Calculating mean of all 100 students

Marks	Mid value	No. of students (f)	fx
0 – 20 20 – 40 40 – 60 60 – 80 80 – 100	10 30 50 70 90	14 23 27 21 15	140 690 1350 1470 1350
		N = 100	Σfx = 5000

Mean = Σfx / N
= 5000 / 100
= 50

(i) For moderately asymmetrical distribution, the arithmetic mean = 28 and the media = 25. Find the mode of this distribution.

Solution:

Arithmetic Mean (AM) = 28
Median (Md) = 25
Mode (M_o) = ?

We have,

Mode (M_o) = 3 Median – 2 mean
= 3 * 25 – 2 * 28
= 19

ii. In a moderately asymmetrical distribution, the values of mode and mean are 32.1 and 35.4 respectively. Find the median value.

Solution:

Mode (M_o) = 32.1
Mean = 35.4

M_o = 3 Median – 2 Mean
or, 32.1 = 3 Md – 2 * 35.4
∴ Md = 34.3

iii. If M be the median and m be the mode of the numbers 10, 70, 20, 40, 70, 90, find the arithmetic mean of M and m be the mode of the numbers 10, 70, 20, 40, 70, 90, find the arithmetic mean of M and m.

Solution:

Arranging the data in ascending order

10, 20, 40, 70, 70, 90

Md lies between 40 and 70

Md = (40 +70) / 2
= 110 / 2
= 55

.Arithmetic mean = (M + m) / 2
= (55 + 70 ) / 2
= 125 / 2
= 62.5

Geometric Mean and Harmonic Mean

(a) Calculate G.M and H.M of the following data.

Variable value (x)

Solution:

log G = 1/n Σlog X
= 1/5 (5+7+9+12+18)
=10^10.2

ii.

X	10	20	30	40	55
f	2	4	8	3	2

Solution:

We have,

G = Anti log of [1/n Σflog X]

Calculation of Geometric Mean

x	f	log X	flogx	1/x	f (1/x)
10 20 30 40 55	2 4 8 3 2	1 1.3010 1.4771 1.6020 1.74036	2 5.204 11.8168 4.806 3.48072	0.10 0.50 0.0333 0.025 0.	0.20 0.20 0.2664 0.075 0.05454
	N=19		Σflogx =27.30752		Σf (1/x) = 0.79594

Geometric Mean = Anti log of [1/19 (27.30752 )]
= 10^1.4372
= 27.3676

Harmonic Mean = N / Σf(1/x)
= 19 / 0.79594
= 23.87

iii.

C.I	5-14	15-24	25-34	35-44	45-54
f	5	7	10	4	2

Solution:

Calculation of Geometric Mean

C.I	f	mid value (x)	log X	flogX
5-14 15-24 25-34 35-44 45-54	5 7 10 4 2	9.5 19.5 29.5 39.5 49.5	0.9777 1.2900 1.4698 1.5965 1.6946	4.8885 9.03 14.698 6.386 3.3892
	N = 28			ΣFlogX = 38.3917

Geometric Mean (GM) = Antilog [ΣFlogX / N]
= Antilog [38.3917 / 28]
= Antilog [1.371132143]
= 23.5034

Calculation of Harmonic Mean

C.I	f	mid value (x)	1/ x	f(1/x)
5-14 15-24 25-34 35-44 45-54	5 7 10 4 2	9.5 19.5 29.5 39.5 49.5	0.1052 0.0512 0.0338 0.0253 0.0202	0.526 0.9984 0.9971 0.9993 0.9999
	N = 28			Σf(1/X) = 4.5207

Harmonic mean (HM) = [ N / Σf(1/X) ]
= 28 / 4.5207
=6.1937

(b) A cyclist pedals from his house to his college at a speed of 10km per hour and back from the college to his house at 15km per hour. Find the average speed.

(c) You make a trip which entrails travelling 900km by train at a speed of 60 km / hour, 3000km by boat at an average of 25km/hour, 400 km by plane at 350km/hour and finally 15km by taxi at 25km/hour, what is your average speed for the entire distance?

(d) If HM, AM, and GM of a set of 5 observations are 10.2, 16 & 14 respectively. Comment upon these values.