TU Solution

QN. 3) In a single through of two dice, find the probability that sum of two faces is 7 or 11.

Solution: Let the event of getting sum 7 and sum 11 be denoted by A and B respectively.

when two dice were through together number of outcomes n(S) = 36

Then, A = {(3,4), (4, 3), (5, 2), (2, 5), (1, 6), (6, 1)}

and B = {(5, 6), (6, 5)}

n(A) = 6, n(B) = 2

Now, P(A or B) = P(A) + P(B)

P(A ∪ B) = n(A) + n(B) = 6 + 2 = 8 = 2
n(S) n(S) 36 36 36 9

QN. 4) Differentiate between probability and non-probability sampling. Solution: The method of
selecting samples from populations is of fundamental importance in the theory of sampling and
depends very much on the objectives and available resources. Sampling methods can be classified
into following two parts:

a) Probability Sampling (Random Sampling)
Probability sampling is that method of drawing a portion (part) of population so that all
possible samples of fixed size have the same probability of being selected. In the
probability sampling each sample unit has equal chance of being selected.

b) Non-probability Sampling (Non-random Sampling)
If the selection of individual from the population to form a sample is not depending on the
chance but depends on the judgment or convenience of the investigator, then the sampling
is said to be non-probability sampling. The non-probability sampling is a procedure of
selecting a sample without the use of probability or randomization.

QN. 5) The year of the origin of the straight-line y = 35 + 2x is trend equation of profits in lakhs
of rupees in 2008.

Estimate profit for the year 2015.

Solution: Now, x = 2015 – 2008 = 7
Then, the estimated profit of the year 2015 is y = 35 + 2×7 = 35 + 14 = Rs.49 lakhs.

QN. 6) Prepare the regret table from the given conditional profit table.

Strategies Decision Alternative 18
150 120 17 90
160 90 100
15 150 160 130 140
16 150 160 170 180
17 150 160 170
18 150
50

Solution: Regret Table Decision Alternative 18
16 17 60
Strategies 30 60 60
15 0 30 30
10 0 0
15 0 20 10
16 10
17 20
18 30

QN. 7) The following calculation were based on the life of refrigerators of two companies:

Company A Company B

Average life 8 years 6 years

Standard deviation 12 years 8 years

Which company's refrigerator show greater constancy in terms of life?

Solution: Let the life of company A and company B be respectively denoted by X and Y.

Here, X̅ = 8, σX = 12, ̅Y = 6, σy = 8

Now, CV (A) = σX × 100% = 12 × 100% = 150%
̅X 8

CV(B) = σY × 100% = 8 × 100% = 133.33%
Y̅ 6

Since, CV(A) > CV(B). Therefore, company B has greater consistency.

QN.8) On the basis of the given information find the regression coefficient Y on X.
ΣXY = 750, ΣX2 = 2085, ΣY2 = 285, ΣX = 135, ΣY = 45, N = 9

a. Solution: Regression coefficient Y on X NΣXY − ΣXΣY
NΣX2 − (ΣX)2
bYX =

= 9 × 750 − 135 × 45
9 × 2085 − (135)2

= 6750 − 6075
18765 − 18225

51

= 675
540

= 1.25

QN.9) The coefficient of correlation between 10 pairs of values of demand and supply was

found to be 0.8. Test the significance of the result.

Solution:

Here, r = 0.8 and N = 10

Now,

P. E. = 0.6745 1 − r2
√N

= 0.6745 1 − 0.82
√10

= 0.6745 1 − 0.64
3.16

= 0.6745 × 0.1139 = 0.077

Again,
0.6.× P. E. = 0.6 × 0.077 = 0.0462
Since, r>0.6×P.E., so r is significant.

QN.10) The first four moments about mean of the distribution are:
μ1 = 0, μ2 = 16, μ3 = −30, μ4 = 40
Test the normality of the distribution.

Solution: Now,

β2 = μ4 = 40 = 40 = 0.15625
(μ2)2 162 256

Since, β2 = 0.15625 < 3, Platykurtic.

Group B

Descriptive question Answer (Attempt any five question)

11. a) Two machines M1 and M2 had the following units of three commodities and the prices of
three commodities in the three different cities of the country.

Supply matrix commodities Price matrix commodities

X YZ X YZ
C1 18 20 16
A = M1 (5400 60 9700) P = C2 (16 22 14)
M2 50 C3 20 24 16

52

To which city each merchant should supply the commodities in order to get the maximum receipt?

Solution: AP = (5400 60 7900) 18 20 16
50 (16 22 14)
24 16
20

= (792000 + 960 + 1800 1000 + 1320 + 2160 800 + 840 + 11142400)
+ 800 + 1400 800 + 1100 + 1680 640 + 700 +

C1 C2 C3

= M1 (32936200 4480 23406800)
M2 3580

Therefore, both machines M1 and M2 gives maximum receipts in city 2.

b) An Aeroplan has 40 seats for passenger. Passenger travelling in economics class can take 20kgs
of baggage each and business can take 60Kgs baggage each. The air craft can carry only 1200
Kgs. Find the number of passengers of each kind by using Cramer's rule.

Solution: Let the number of passengers of economic class be x and business class be y. Then

x + y = 40 …………… (i)

20x + 60y = 1200 (ii)

Now, D = |210 610| = 60 − 20 = 40

Dx = |124000 610| = 2400 − 1200 = 1200

Dy = |210 124000| = 1200 − 800 = 400

x = Dx = 1200 = 30
D 40

y = Dy = 400 = 10
D 40

Thus, economics class passengers are 30 and business class are 10.

QN.12) The following distribution represent yearly income of 2500 employees of an industrial
concern in thousand rupees. Employees earning 2 lakhs or more have to pay 15% tax to the
government. Find the average income of the employees and amount of tax to be paid to the
government.

Income 70 -100 100 - 130 130 - 160 160 - 190 190 - 220 220 - 250 250 - 280
(000'Rs)

No. of 50 150 200 400 900 500 300

Employees

53

Solution:

Calculation of average income

Income No. of Employees Mid-value (x) fx
(000'Rs) (f)

70 – 100 50 85 4250

100 – 130 150 115 17250

130 – 160 200 145 29000

160 – 190 400 175 70000

190 – 220 900 205 184500

220 - 250 500 235 117500

250 - 280 300 265 79500

N = 2500 Σfx=5022000

Now, Average income

X̅ = Σfx = 5022000 = 2008.8
N 2500

Therefore, average income of the employees is Rs. 2008800

Again, Number of employees earning 2 lakhs and more are 600 + 500 + 300 = 1400
Total income of employees who earn more than 2 lakhs = 600 × 205 + 500 × 235 + 300×265

= Rs 320,000

Income tax = Rs320000 × 15 = Rs48000
100

Therefore, amount of tax paid to the government is Rs.48000000.

QN. 13) The following is the net profit of two companies in millions of rupees, which company
shows greater consistency in net profit. Justify.

Year 2006 2007 2008 2009 2010 2011 2012

Company A 15 16 20 16 22 20 24

Company B 14 15 19 20 21 22 22

Solution: Calculation of coefficient of variance

54

Year Profit of Profit of X2 Y2

2006 Company A (X) Company B (Y)
2007
2008 15 14 225 196
2009
2010 16 15 256 225
2011
2012 20 19 400 361

16 20 256 400

22 21 484 441

20 22 400 484

24 22 576 484
ΣX =133 ΣY= 133 ΣX2=2597 ΣY2= 2591

Now, ̅X = ΣX = 133 = 19
N 7

σX = √ΣNX2 − (ΣNX)2 = √25797 − (1733)2 = √371 − 361 = √10 = 3.16

and, ̅Y = ΣY = 133 = 19
N 7

σY = √ΣNY2 − (ΣNY)2 = √25791 − (1733)2 = √370.14 − 361 = √9.14 = 3.02

Again, CV of Company A

σX × 100% = 3.16 × 100% = 16.63%
X̅ 19

Again, CV of Company B

σY × 100% = 3.02 × 100% = 15.89%
Y̅ 19

Since, CV of company B 15.89% < CV of company A = 16.63.

Therefore, company B has greater consistency in net profit.

QN.14) (a) A company has two plants to manufacture Scooters. Plant I manufacture 80% of
Scooters and plant II manufacture 20% of Scooters. At plant I, 85 out of 100 Scooters rated as
standard quality. At plat II, only 65 out of 100 Scooters are rated as standard quality. What is the
probability that Scooter come from plant II, if it is known that the Scooter is standard quality?

55

Solution: Let the event of manufacturing from plat I be denoted by A and that of plat II be B;
and standard quality of Scooter be denoted by D.

Then, P(A) = 0.8, P(B) = 0.2, P(D/A) = 0.85 and P(D/B) = 0.65

Now, P(B/D) = P(B)P(D/B)
P(A)P(D/A) + P(B)P(D/B)

= 0.8 × 0.2 × 0.65 0.65 = 0.13 = 0.13 = 0.1663
0.85 + 0.2 × 0.68 + 0.13 0.81

Therefore, the probability that Scooter come from plant II is 16.65%.

(b) In certain distribution, the first four moments about an arbitrary point were 1, 3, 7 and, 21. Test
the skewness of the distribution.

Solution: Now we have μ1′ = 1, μ′2 = 2, μ′3 = 7 and μ′4 = 21

The central moments can be calculated as follows:

μ1 = 0

μ1 = μ′2 − (μ1′ )2 = 3 − 1 = 2

μ3′ = μ′3 − 3μ1′ μ′2 + 2(μ13) = 7 − 3 × 1 × 3 + 2 × 1 = 0

μ4 = μ4′ − 4μ1′ μ3′ + 6μ′2(μ1′ )2 − 3(μ1′ )4 = 21 − 4 × 1 × 7 + 6 × 3 × 1 − 3 × 1 = 8

For skewness, β1 = μ23 = 0 = 0
(μ2)3 23

Since, β1 = 0, the distribution is symmetrical.

QN. 15) (a) Why it is necessary to analyze time series data? Discuss various components of time
series.

(b) Calculate the moving average for the following data, assuming the length of business cycle as
3 years.

Year 2004 2005 2006 2007 2008 2009 2010 2011 2012

Sales in '000' 35 50 60 65 70 80 90 92 95

Solution: Calculation of three yearly moving average:

Year Sales in '000' 3-yearly moving 3-yearly moving
total total

2004 35 48.33

2005 50 145

56

2006 60 175 58.33
2007 65 195 65
2008 70 215
2009 80 240 71.67
2010 90 262 80
2011 92 277
2012 95 87.33
92.33

QN. 16) (a) What do you mean by understand by classification of data? What are its objectives?
Classify the given data Sturge's rule.

Solution: The classification of data is the process of arranging data into homogeneous (similar)
groups according to their common characteristics. Raw data cannot be easily understood, and it is
not fit for further analysis and interpretation. Arrangement of data helps users in comparison and
analysis. For example, the population of a town can be grouped according to sex, age, marital
status, etc.

The primary objectives of data classification are:
i) To consolidate the volume of data in such a way that similarities and differences can be

quickly understood. Figures can consequently be ordered in sections with common traits.
ii) To aid comparison.
iii) To point out the important characteristics of the data at a flash.
iv) To give importance to the prominent data collected while separating the optional elements.
v) To allow a statistical method of the materials gathered.

Sturge's has given a formula which can be used to get a rough idea about the number of classes to
be formed. The formula is k = 1 + 3.3222logN

where k is the number of classes and N is the number of observations.

Here number of observations N = 50

Number of class intervals k = 1 + 3.3222log50 = 1 + 3.3222 × 1.7 = 6.65≈ 7

Length of class interval = L − S = 203 − 30 = 273 = 24.72 ≈ 25
k 7 7

Classification data using Sturge's rule

Class Interval Tally Marks Frequency

30 – 55 ||| 3

55 – 80 |||| 4

57

80 – 105 |||| 6
105 – 130 |||| |||| 9
130 – 155 |||| |||| |||| 12
155 – 180 |||| |||| ||| 11
180 - 205 |||| | 5
Total 50

b) What do you understand by sampling distribution? Difference between point and interval
estimate.

Solution: The probability distribution of the values of a statistic, can be defined as a sampling
distribution of a statistics. If select a number of independent random samples from any population
and calculate some statistic from each sample, we shall get a series of values of statistics. All these
values of statistic associated with its relatively frequency would describe the sampling distribution
of statistics.

If from the sample values, a single value is calculated as an estimate of unknown population
parameter, the procedure is termed as point estimation. A single value of the sample statistics that
is used to estimate a population parameter called a point estimate. For example, if we wish to
estimate the population mean μ, the average percentage mark of all the students of 12 class
students of 10 + 2 college of Sindhuli. We take a random sample of n students and calculate the
sample mean x̅ ; then x̅ is a point estimator.

A range of values used to estimate a population parameter is called an interval estimate. There are
two ways for indicating error by the extent of its range and by the probability of the true
population parameter lying within that range. In this case, the decision maker has a better idea of
the reliability of his estimate. For example, if the average salary per month of all professors of TU
lies between Rs.9000 to Rs.15000, then TU management can estimate the exactly salary that fall
within this interval, such estimate is an interval estimate.

Group C (2 × 15 = 30)
Analytical Answer Question

QN. 17) From the following bi-variate table find out whether there exists any association between
security prices and dividends and test the significance of the result. Also estimate the amount
dividend when price of security is Rs.150.

Security Price (in Rs) Annual dividend in (Rs)

6 - 8 8 - 10 10 - 12 12 - 14 14 - 16 16 - 18

70 – 80 21 - - - -
80 – 90 - -
31 - 1

58

90 – 100 22 1 1 - -

100 – 110 -2 2 3 - -

110 – 120 -1 3 2 2 -

120 - 130 -1 3 3 3 2

130 - 140 -- 3 1 4 1

Solution: Let X and Y denotes the mid-point of security price and annual dividend respectively.
X − 105 Y − 11
Then, dx = 10 and dy = 2

Now, correlation coefficient NΣfdxdy − (Σfdx)(Σfdy)
√NΣfdx2 − (Σfdx)2 √NΣfdy2 − (Σfdy)2
r =

= 50 × 101 − 34 × 16
√50 × 190 − (34)2 √50 × 110 − (16)2

= 5050 − 544
√9500 − 1156√5500 − 256

= 4506
√8344√5244

= 4506
91.35 × 72.42

= 4506
6615.567

59

= 0.68

Again,

P. E. = 0.6745 1 − r2
√N

= 0.6745 1 − (0.68)2
√100

= 0.6745 1 − 0.4624
10

= 0.6745 0.5376
10
= 0.036

0.6× P.E. = 0.6× 0.036 = 0.021
Since, r > 0.6 × r, the value of r is significance.

Regression coefficient y on x

b = NΣfdxdy − ΣfdxΣfdy × k
NΣfdx2 − (Σfdx)2 h

= 50 × 101 − 34 × 16 × 2
50 × 190 − (34)2 10

= 5050 − 544 × 1
9500 − 1156 5

= 4506 × 1
8344 5

= 0.54 × 0.2

= 0.11

̅X = A + Σfdx × h = 105 + 34 × 10 = 105 + 6.8 = 111.8
N 50

Y̅ = B + Σfdy × k = 11 + 16 × 2 = 11 + 0.64 = 11.64
N 50

The regression equation Y on X is

Y − Y̅ = byx(X − ̅X)

⇒ Y − 11.64 = 0.11(X − 111.8)

60

⇒ Y − 11.64 = 0.11X − 12.3
⇒ Y = 0.11X − 12.3 + 11.64
⇒ Y = 0.11X − 0.66

Finally, when x = 150
Y = 0.11 × 150 − 0.66 = 15.84

Therefore, the amount of dividend when price of security Rs.150 is Rs.15.84.

QN. 18) A firm manufactures two types of electronic items E1 and E2. The profit contribution per
unit of E1 and E2 are Rs. 1600 and Rs. 2400 respectively. Both E1 and E2 make use of two
essential components, a motor and transformer. Each unit of E1 requires 3 units of motors and 2
transformers and each unit of E2 requires 2 motors and 4 transformers. The total supply of
components per month is restricted to 210 motors and 300 transformers. E2 is an export model
requiring a voltage stabilizer, which has supply restriction to 65 units. Formulate the above
problem in a mathematical form describing the objective and limitation of the problem. Solve the
formulated problem by graphical method with an objective of maximization of profit.

Solution: Here we have given

Items Motors Transformers Restriction Profit (Rs.)

E1 3 2 - 1600

E2 2 4 65 2400

Restriction 210 300

Let x units of E1 and y units of E2 of should be produce to maximize the profit.
Mathematical model of the given problem
Objective function
Maximize: Z = 1600x + 2400x
subject to
3x + 2y ≤ 210 …………. (i)
2x + 4y ≤ 300 ………… (ii)
x ≤ 65 …………….. (iii)
x ≥ 0, y ≥ 0 ………… (iv)
Now, corresponding equation of (i) is 3x + 2y = 210
when x = 0, y = 105

61

when y = 0, x = 70
Corresponding equation of (ii) is 2x + 4y = 300
when x = 0, y = 75
when y = 0, x = 150
Corresponding equation of x = 65
As, x ≥ 0 and y ≥ 0 shows that feasible reason lies on the first quadrant.

Calculation of maximum value

Vertices Z = 1600x + 2400y Result
O(0, 0)
A(65, 0) Z = 1600× 0 + 2400 × 0 = 0
B(65, 7.5)
C(30, 60) Z = 1600× 65 + 2400 × 0 = 104000
D(0, 75)
Z = 1600× 65 + 2400 × 7.5 = 122000

Z = 1600× 30 + 2400 × 60 = 192000 Maximum

Z = 1600× 0 + 2400 × 75 = 180000

The maximum value of is Rs. 192000 at (30, 60).

62

QN. 19) What are index number? Why are they called economics barometer? You are required to
prove from the following data that Fisher's index number is an ideal number.

Commodities 2010 Price/ unit Expenditure 2011 Price/ unit Expenditure
A 50 600 60 840
B 40 840 80 2400
C 30 900 30 1500
D 20 600 20 1000
E 60 360 70 2100
F 80 640 80 2400

Solution: Calculation of Fisher's Index number

Com 2010 2011
modit Expn.
Price Expn. Quant. Price (p1q1) Quant p1q0 p0q1
ies (p0) (Poqo) (q0) (p1) 840 . (q1) 720 3000
50 12 60 2400 14
A 600 80 1500
30 1000
B 40 840 21 20 2100 30 1680 1200
70 2400
C 30 900 30 80 10240 50 900 1500

D 20 600 30 50 600 1000

E 60 360 6 30 4200 1800

F 80 640 8 30 6400 2400

Total 3940 14500 10900

Here, ∑p0q0 = 3940, ∑p1q0 = 14500, ∑p0q1 = 10900, ∑p1q1 = 10240

Time reversal test

P01(F) × P10(F) = √ΣΣpp10qq00 × Σp1q1 × Σp0q1 × Σp0q0
Σp0q1 Σp1q1 Σp1q0

= √134954000 × 10240 × 10900 × 3940 = 1
10900 10240 14500

Hence, Fisher's index satisfies the time reversal test.

Factor reversal test

63

P01(F) × Qo1(F) = √ΣΣpp10qq00 × Σp1q1 × Σq1p0 × Σq1p1
Σp0q1 Σq0p0 Σq0p1

= √134954000 × 10240 × 10900 × 10240
10900 3940 14500

= 10240
3940

Qo1(FV)01==V0ΣΣ1pp10qq10 = 10240
3940
∴ P01(F) ×

Hence, Fisher's index satisfies the factor reversal test.

Since Fisher's index number satisfies both time reversal test and factor reversal test, thus it is
an ideal index number.

64

Unit wise Solution

Chapter - 1

Introduction to Statistics and Data Collection

QN. 1) (2077, New) What are the various methods of collecting primary data? Explain with
suitable example.
Solution: The are required for a statistical inquiry are sometimes collected by the investigator
himself and his agents. These data collected for the first time are original in character and they are
called primary data. There are various methods for collecting primary data. The following are
commonly used for the collection of primary data;

1. Direct personal interviews.
In this method the investigators himself goes to the spot and makes direct personal contact
with the people from whom the data are to be collected. These people are called
informants. He asks them various questions and collects first hand data from them. For
examples, if an investigator wants to make an enquiry on the wages of the workers of a
factory, he will visit the factory, meet the workers and get the information from them.

2. Indirect oral interviews.
In this method, the investigator does not make any contact with the persons who are
already concerned with the problem under study. The required data are obtained by
interviewing third persons who are in touch with the field of inquire and can give the
desire information. Usually some enumerators are appointed to have the required
information, put these questions to them and collected the data needed. For example, when
some information about the living conditions of the workers of a factory is wanted, no
contact is made with workers. The information may be obtained by interrogating some
officials or the owner of the factory.

3. Informational from local correspondents.
In this method data are not collected by the investigators or enumerators. Some agents
known as correspondents are employed in various places. The collected information in
their own ways and send them to the central office where the data are processed and
analyzed. This method is generally used by newspapers to get information about important
events in politics or sports or about riots, accidents, strikes, etc.

4. Mailed questionnaire method.
In this method a questionnaire relating to the enquiry is prepared. It consists a list of
questions and some space for the answer to be filled in the respondents. Copies of the
questionnaire are sent by post to various it by post within a specified time. This method is
suitable when the area of inquire is vast spread far and wide, the information are literate
and they know the importance of such enquiries.

5. Schedule to be filled in by enumerators.
Just like a questionnaire a schedule also contains a set of questions. While the
questionnaire is sent by post and filled in by the respondents, the schedule is sent through
the enumerators who fills it up. He explains the purpose of his visit and ask them questions
contained in the schedule and records or write the answers of the questions. Population
census of Nepal by this method.

QN. 2) (2077, Old) Define primary data and describe the methods collecting primary data.
Answer: Primary data are those that are collected for the first time. Primary data is the data that is
collected for the first time through personal experiences or evidence, particularly for research. It is
also described as raw data or first-hand information. The mode of assembling the information is
costly, as the analysis is done by an agency or an external organization, and needs human

65

resources and investment. The investigator supervises and controls the data collection process
directly. The data is mostly collected through observations, physical testing, mailed
questionnaires, surveys, personal interviews, telephonic interviews, case studies, and focus groups,
etc.

The following are commonly used for the collection of primary data;
1. Direct personal interviews.
In this method the investigators himself goes to the spot and makes direct personal contact
with the people from whom the data are to be collected. These people are called
informants. He asks them various questions and collects first hand data from them. For
examples, if an investigator wants to make an enquiry on the wages of the workers of a
factory, he will visit the factory, meet the workers and get the information from them.
2. Indirect oral interviews.
In this method, the investigator does not make any contact with the persons who are
already concerned with the problem under study. The required data are obtained by
interviewing third persons who are in touch with the field of inquire and can give the
desire information. Usually some enumerators are appointed to have the required
information, put these questions to them and collected the data needed. For example, when
some information about the living conditions of the workers of a factory is wanted, no
contact is made with workers. The information may be obtained by interrogating some
officials or the owner of the factory.
3. Informational from local correspondents.
In this method data are not collected by the investigators or enumerators. Some agents
known as correspondents are employed in various places. The collected information in
their own ways and send them to the central office where the data are processed and
analyzed. This method is generally used by newspapers to get information about important
events in politics or sports or about riots, accidents, strikes, etc.
4. Mailed questionnaire method.
In this method a questionnaire relating to the enquiry is prepared. It consists a list of
questions and some space for the answer to be filled in the respondents. Copies of the
questionnaire are sent by post to various it by post within a specified time. This method is
suitable when the area of inquire is vast spread far and wide, the information are literate
and they know the importance of such enquiries.
5. Schedule to be filled in by enumerators.
Just like a questionnaire a schedule also contains a set of questions. While the
questionnaire is sent by post and filled in by the respondents, the schedule is sent through
the enumerators who fills it up. He explains the purpose of his visit and ask them questions
contained in the schedule and records or write the answers of the questions. Population
census of Nepal by this method.

66

Chapter – 2
Classification and Presentation of Data

QN.1) (2077, New) Draw the pie diagram from the following information.

Items Expenditure %

Family A Family B

Food 60 55

Clothing 15 17

Education 10 8

Electricity 5 6

Rent 5 6

Miscellaneous 2 3

Solution: Calculation of central angles for pie diagram

Items Family A Family B

Expenditure Central angle Expenditure Central angle

(%) (Degree) (%) (Degree)

Food 60 216 55 208.4

Clothing 15 54 17 64.4

Education 10 36 8 30.3

Electricity 8 28.8 6 22.7

Rent 5 18 6 22.7

Misc. 2 7.2 3 11.4

Total 100 360 95 359.9

Working Note:

Central Angle of Food = (Expenditure %) × 360o = 60 × 360o = 216o
100 100

Central Angle of Clothing = (Expenditure %) × 360o = 15 × 360o = 54o
100 100

Central Angle of Education = (Expenditure %) × 360o = 10 × 360o = 36o
100 100

67

and so on.

68

Chapter - 3
Measure of Central Tendency
Brief Question Answer
QN. 1) (2078) Find the combined mean from the following data:

Group A B C

Numbers 200 250 300

Mean 25 10 15

Let the number of group A and group B, and group C be X1, X2 and X3 respectively.

Here, n1 = 200, n2 = 250, n3 = 300, X̅ 1 = 25, X̅ 2 = 10, X̅ 3 = 15

̅X123 = X̅1n1 + ̅X2n2 + ̅ 3 3 = 200 × 25 + 250 × 10 + 300 × 15 = 12000 = 16
n1 + n2 + 3 200 + 250 + 300 750

Therefore, combined mean of is 16

QN. 2) (2077) What is the average weekly wages for the workers of two companies combined
together?

Companies A Companies B

Numbers of workers 250 210

Average weekly 1000 1250
wages (Rs.)

Solution: Let the income of company A and company B be X1 and X2 respectively.

Here, n1 = 250, n2 = 210, ̅X1 = 1000, ̅X2 = 210

̅X12 = ̅X1n1 + ̅X2n2 = 1000 × 250 + 1250 × 210 = 512500 = 1114.13
n1 + n2 250 + 210 460

Therefore, combined weekly wages of two companies is Rs. 1114.13.

QN.3) (2076) If the arithmetic mean of the following data is 33, find the value of x: x + 10, 30, x +
20, 4x + 5, 40.

Solution: Here number of terms (N) = 5

Mean(̅X) = ΣX
N

⇒ 33 = x + 10 + 30 + x + 20 + 4x + 5 + 40
5

69

⇒ 165 = 6x + 105

⇒ 6x = 165 − 105

⇒ x = 60 = 10
6

QN.4) (2075) The mean monthly salary of 80 male and 20 female employees of a company are Rs.
1560 and Rs. 1260 respectively. What is the mean salary of all employees?

Solution: Here give
Mean salary of male (̅X1) = Rs. 1560, female (̅X2) = Rs. 1260,
Number of male (n1) = 80, number of female (n2) = 20

Now, Combined mean

̅X12 = ̅X1n1 + ̅X2n2 = 1560 × 80 + 1260 × 20 = 150000 = 1500
n1 + n2 80 + 20 100

Therefore, mean salary of all employees is Rs. 1500.

QN. 5) (2074) A person drives a car between two place X and Y. On his outward journey he
consumes 8km per liter of petrol, on his return journey he covers 12km per liter. Find the average
of his mileage assuming the distance between the two places to be 100km.

Solution: Since the distance travelled is constant, i. e. distance between two places is 100 km.
Thus, the appropriate average is Harmonic Mean.

H. M. = n = 2 = 2 × 24 = 48 = 9.6
+ 3+2 5
1 1 1 + 1
X1 X2 8 12

Therefore, average mileage is 9.6km.

QN. 6) (2073) Find the most repeated value by empirical relationship if X̅= 42.2 and median (Md)
= 41.9.

Solution: Now, we have
Mode = 3 median -2 mean = 3×41.9 – 2×42.2 = 41.3.

QN. 7) (2073) A man travelled by a car for 3 days. He covered 480 km each day. On first day he
drove for 10 hours at 48 km/hr. On the second day, drove for 12 hours at 40 km/hr., and on the last
day he drove for 15 hours at 32 km/hr. What is his average speed?

Solution: Since distance travelled is constant, i. e. 480km each day, the appropriate average is
Harmonic Mean.

H. M. = 1 + n + 1 = 1 + 3 + 1 = 2 × 480 = 48 = 38.92
X1 1 X3 48 1 32 10 + 12 + 15 37
X2 40

Therefore, average speed is 38.92 km/hr.

70

QN.8) (2072) The average monthly salary of 10 male staffs and 5 female staffs of a manufacturing
company are Rs. 20,000 and Rs.18,000 respectively. Find the average monthly salary of all staffs
taken together.

Solution: Here give

Mean salary of male (̅X1) = Rs. 20,000, female (X̅2) = Rs. 18,000,
Number of male (n1) = 10, number of female (n2) = 5

Now, Combined

̅X12 = ̅X1n1 + ̅X2n2 = 20000 × 10 + 18000 × 5 = 290000 = 19,333.333
n1 + n2 10 + 5 15

Therefore, mean salary of all staffs is Rs. 19333.333.

QN.9) (2072) From the batch of 13 students who had appear in the examination, 4 students were
failed. The marks of passed students were 43, 57, 45, 61, 75, 64, 53, 50, and 40. Calculate the
median marks of all students.

Solution: Let the marks of failed students in ascending order be x1, x2, x3, x4.

Now, arranging the marks of all students in ascending order, we have x1, x2, x3, x4, 40, 43, 45, 50,
53, 57, 61, 64, 75.

Number of observations (n) = 13

Median (Md) = (N + 1)th item = (132+ 1)th item = 7th item = 45
2

Therefore, median mark of all students is 45.

QN. 10) (2072) If mean = 40, and mode = 30, find median.

Solution: we know that

Mode = 3 median -2 mean
or 30 = 3 median – 2×40

or 30 + 80 = 3 median

∴ Median = 110 = 36.67
3

QN. 11) What will be the value median of a moderately asymmetrical distribution? If mean and
mode are 30 and 24 respectively?

Solution: Here we have given Mean = 30, mode = 24

we know that

Mode = 3 median -2 mean
or 24 = 3 median – 2×30
or 24 + 60 = 3 median

71

∴ Median = 84 = 28
3

Descriptive Question Answer

QN. 12) (2078) Before and after the implementation of the economic program to uplift the
economic condition of a community following information were found.

Monthly income Prior to the plan After the plan
(in Rs’00) No. of families No. of families

4–6 10 8

6–8 70 65

8 – 10 35 37

10 – 12 20 15

12 – 14 10 5

14 – 16 3 3

16 - 18 2

a) Find the highest income of the poorest 40% of the population before and after the plan.
b) Find the lowest income of richest 40% of the population before and after plan.
c) Obtain the limits of income of middle 50% of families before and after the plan.

Solution:

Monthly income Prior to the plan After the plan
(in Rs’00)
No. of families(f) cf No. of families(f) cf

4–6 10 10 8 8

6–8 70 80 65 73

8 – 10 35 115 37 110

10 – 12 20 135 15 125

12 – 14 10 145 15 140

14 – 16 3 148 5 145

16 - 18 2 150 5 150

150 150

Prior to the plan

72

For P40, 40N = 40 × 150 = 60
100 100

cf just greater or equal to 60 is 80, so P40 class is 6 - 8.

L= 6, f = 70, cf = 10, h = 2

P40 = L + 40N − cf × h = 6 + 60 − 10 × 2 = 6 + 50 × 2 = 6 + 1.43 = 7.43
100 70 70
f

Therefore, highest income of the poorest 40% of the population before plan is Rs.743.

For P60, 60N = 60 × 50 = 30
100 100

cf just greater or equal to 20 is 80, so P60 class is 6 - 8.

L= 6, f = 70, cf = 10, h = 2

P60 = L + 60N − cf × h = 6 + 30 − 10 × 2 = 6 + 20 × 2 = 6 + 0.57 = 6.57
100 70 70
f

Therefore, lowest income of the richest 40% of the population before plan is Rs.657.

For P25, 25N = 25 × 50 = 12.5
100 100

cf just greater or equal to 12.5 is 80, so P25 class is 6 - 8.

L= 6, f = 70, cf = 10, h = 2

P40 = L + 40N − cf × h = 6 + 12.5 − 10 × 2 = 6 + 2.5 × 2 = 6 + 0.07 = 6.07
100 70 70
f

For P75, 75N = 75 × 50 = 37.5
100 100

cf just greater or equal to 37.5 is 80, so P75 class is 6 - 8.

L= 6, f = 70, cf = 10, h = 2

P75 = L + 75N − cf × h = 6 + 37.5 − 10 × 2 = 6 + 27.5 × 2 = 6 + 0.77 = 6.77
100 70 70
f

Therefore, minimum and maximum income of middle 50% of families before the plan is
Rs.Rs.607 and Rs.677 respectively.

After the plan

For P40, 40N = 40 × 50 = 20
100 100

cf just greater or equal to 20 is 73, so P40 class is 6 - 8.

73

L= 6, f = 65, cf = 8, h = 2

P40 = L + 40N − cf × h = 6 + 20 − 8 × 2 = 6 + 12 × 2 = 6 + 0.37 = 6.37
100 65 65
f

Therefore, highest income of the poorest 40% of the population before plan is Rs.637.

For P60, 60N = 60 × 50 = 30
100 100

cf just greater or equal to 20 is 73, so P60 class is 6 - 8.

L= 6, f = 65, cf = 8, h = 2

P60 = L + 60N − cf × h = 6 + 30 − 8 × 2 = 6 + 22 × 2 = 6 + 0.68 = 6.68
100 65 65
f

Therefore, lowest income of the richest 40% of the population before plan is Rs.668.

For P25, 25N = 25 × 50 = 12.5
100 100

cf just greater or equal to 12.5 is 80, so P25 class is 6 - 8.

L= 6, f = 65, cf = 8, h = 2

P25 = L + 25N − cf × h = 6 + 12.5 − 8 × 2 = 6 + 4.5 × 2 = 6 + 0.14 = 6.14
100 65 65
f

For P75, 75N = 75 × 50 = 37.5
100 100

cf just greater or equal to 37.5 is 80, so P75 class is 6 - 8.

L= 6, f = 70, cf = 8, h = 2

P75 = L + 75N − cf × h = 6 + 37.5 − 8 × 2 = 6 + 29.5 × 2 = 6 + 0.91 = 6.91
100 65 65
f

Therefore, minimum and maximum income of middle 50% of families before the plan is
Rs.Rs.614 and Rs.691 respectively.

QN.13) (2077)

Find the missing frequencies of the following frequencies table when mean value is Rs. 45 and
total number of workers is 30.

Wages (Rs. '00) 20 - 30 30 - 40 40 - 50 50 – 60 60 - 70

Workers 3 - 10 7 -

Solution: Let the frequencies be a and b. Then

74

Wages workers(f) mid-value(X) fX

20 – 30 3 25 75

30 – 40 a 35 35a

40 – 50 10 45 450

50 - 60 7 55 385

60 – 70 b 65 65b

20 + a + b 910+35a + 65b

Here ΣfX = 910+ 35a + 65b, N = 20 + a + b

Now, mean(̅X) = ΣfX
N

or 45 = 910 + 35a + 65b
30

or 1350 = 910 + 35a + 65b

or 1350 − 910 = 35a + 65b

or 440 = 35a + 65b … … … … … (i)

and N = 20 + a + b

or 30 = 20 + a + b

or 10 = a + b

or a = 10 – b

putting the value of a in equation (i), we obtain

440 = 35(10 – b) + 65b

or 440 = 350 – 35b + 65b

or 30b = 90

or b = 3

and a = 10 – 3 = 7

Therefore, missing frequencies are 3 and 7.

QN. 14) (2075)

Classify the following score in class interval 0 – 9, 10 – 19, 20 – 29, …… etc. Calculate the model
value of the construct frequency table; 24, 12, 37, 49, 30, 6, 28, 10, 13, 18, 12, 19, 2, 33, 4, 25, 42,

22, 16, 29, 2, 21, 23, 31, 21, 34, 20, 27, 23, 35.

Solution: Frequency Distribution Table

75

Class Interval Tally Marks Frequency
0-9 |||| 4
10 - 19 |||| || 7
20 - 29 |||| |||| | 11
30 - 39 |||| 5
40 - 49 ||| 3
Total 30

Calculation of Model value

Class Interval Frequency (f)

-0.5 – 9.5 4

9.5 – 19.5 7

19.5 – 29.5 11

29.5 – 39.5 5

39.5 – 49.5 3

Total 30

Here, maximum frequency is 11, thus the model class is 19.5 – 29.5.

L = 19.5, f1 = 11, f0 = 7, and f2 = 5

Now, Mode = L + f1 − f0 f2 × 10
2f1 − f0 −

= 19.5 + 2 × 11 − 7 − 5 × 10
11 − 7

= 19.5 + 4 × 10
10

= 19.5 + 4

= 23.5

Therefore, model value is 23.5.

QN.15) Define arithmetic mean and its properties. Following are the income distribution in
thousands of 50 families.

76

63, 52, 65, 78, 79, 52, 81, 41, 77, 70, 96, 104,
94, 76, 78, 79, 40, 76, 69, 67, 50, 59, 115, 49,
56, 82, 74, 65, 50, 110, 61, 79, 69, 79, 81, 72,
73, 103, 66, 64, 60, 72, 86, 84, 53, 51, 80, 68,
42, 90.

Finance minister want to know the limit of the income of middle 60% observed families.

Solution: Calculation of income of middle 60% families

Income Frequency (f) Cumulative frequency (c.f.)

40 – 50 4 4

50 – 60 8 12

60 – 70 11 23

70 – 80 14 37

80 – 90 6 43

90 – 100 3 46

100 - 110 2 48

110 - 120 2 50

N = 50

For first P20,

20N = 20 × 50 = 10
100 100

cf just greater or equal to 10 is 12, so P20 class is 50 - 60.

L= 50, f = 8, cf = 4, h = 10

P20 = L + 20N − cf × h
100
f

= 50 + 10 − 4 × 10
8

= 50 + 6 × 10
8

= 50 + 7.5

= 57.5

Again,
For first P80,

77

80N = 80 × 50 = 40
100 100

cf just greater or equal to 40 is 43, so P20 class is 80 - 90.

L= 80, f = 6, cf = 37, h = 10

P80 = L + 80N − cf × h
100
f

= 80 + 40 − 37 × 10
6

= 80 + 3 × 10
6

= 80 + 5

= 85

Thus, minimum, and maximum income of middle 60% family is Rs. 57,500 and Rs. 85,000
respectively.

QN. 16) (2074) The daily expenditure of 100 students in mobile phone is given below:

Expenditure (in Rs.) 0 – 20 20 – 40 40 – 60 60 – 80 80 - 100

Number of students 15 - 30 20 -

Find the missing frequencies given that model expenditure is Rs.50

Solution: Let the missing frequencies be a and b. Then

Expenditure (in Rs.) Number of students

0 – 20 15

20 – 40 a

40 – 60 30

60 – 80 20

80 - 100 b

N = 65 + a + b

Here, model expenditure is Rs. 50. The model class is 40 – 50,

L = 40, f1 = 30, f0 = a, f2 = 20, h = 20

Now, Mode = L + f1 − f0 f2 × 20
2f1 − f0 −

78

⇒ 50 = 40 + 2 × 30 − a 20 × 20
30 − a −

⇒ 50 − 40 = 30 − a × 20
40 − a

⇒ 10 = 30 − a × 20
40 − a

⇒ 40 − a = 60 − 2a

⇒ a = 20

Again, we have N = 100
⇒ 65 + a + b = 100
⇒ 65 + 20 + b = 100
⇒ 85 + b = 100
⇒ b = 15

Therefore, a = 20 and b = 15.

QN. 17) (2073) The median expenditure of 1000 families were found to be Rs. 650 and the
distribution was as below:

Expenditure (in Rs.) 400 – 500 500 – 600 600 – 700 700 – 800 800 - 900
Number of families 50 - 450 - 100

Calculate the missing frequencies and find the mean of the distribution.

Solution: Let the missing frequencies be a and b. Then

Expenditure (in Rs.) Number of students(f) Cumulative frequency (c.f.)

400 – 500 50 50

500 – 600 a 50 + a

600 – 700 450 500 + a

700 – 800 b 500 + a + b

800 - 900 100 600 + a + b

Here median expenditure is Rs. 650. The median class is 600 – 700,

L = 600, f = 450, c. f. = 50 + a, h = 100

Median(Md) = L + N − cf × h
2 f

⇒ 650 = 600 + 500 − (50 + a) × 100
450

79

⇒ 650 − 600 = 500 − 50 − a × 100
450

⇒ 50 = 450 − a × 100
450

⇒ 1 = 450 − a
2 450

⇒ 450 = 900 − 2a

⇒ 2a = 450

⇒ a = 225

Again, we have N = 1000
⇒ a + b + 600 = 1000
⇒ 225 + b + 600 = 1000
⇒ b + 600 = 775
⇒ b = 175

Therefore, missing frequencies are 225 and 175.

Again, calculation of mean

Expenditure (in Number of Mid- fX
Rs.) students(f) value(X)
22500
400 – 500 50 450 123750
500 – 600 225 550 292500
600 – 700 450 650 131250
700 – 800 175 750 85000
800- 900 100 850 SfX = 655000
N = 1000

Mean(X̅) = ΣfX = 655000 = 655
N 1000

Therefore, mean expenditure is Rs. 655

QN.18) (2072) A frequency distribution of 100 students is given below. Frequencies
corresponding to two groups are missing from the table. The median is known to be 49.5
marks.

Marks 0 - 19 20 - 39 40 - 59 60 - 79 80 - 99

80

No of students 14 - 26 - 16
a) Find the missing frequencies.

b) Calculate the limit of marks obtained by middle 60% students?

Solution: Let the missing frequencies be a and b. Then

Marks No of students (f) Cumulative frequencies (c. f)

- 0.5 – 19.5 14 14

19.5 – 39.5 a 14+a

39.5 – 59.5 26 40 + a

59.5 – 79.5 b 40 + a + b

59.5 – 99.5 16 56 + a + b

Here median mark is Rs. 49.5. The median class is 39.5 – 59.5,

L = 39.5, f = 26, c. f. = 14 + a, h = 20

Median(Md) = L + N − cf × h
2 f

⇒ 49.5 = 39.5 + 50 − (14 + a) × 20
26

⇒ 49.5 − 39.5 = 50 − 14 − a × 20
26

⇒ 10 = 36 − a × 20
26

⇒ 1 = 36 − a
2 26

⇒ 26 = 72 − 2a

⇒ 2a = 72 − 26

⇒ a = 23

Again, we have N = 100
⇒ a + b + 56 = 100
⇒ 23 + b + 56 = 100
⇒ b + 79 = 100
⇒ b = 21

81

Therefore, missing frequencies are 23 and 21.

b) Calculation of the limit of marks obtained by middle 60% students

Marks No of students (f) Cumulative frequencies (c. f)

- 0.5 – 19.5 14 14
19.5 – 39.5 23 37
39.5 – 59.5 26 63
59.5 – 79.5 21 86
79.5 – 99.5 16 100

N = 100

For first P20,

20N = 20 × 100 = 20
100 100

cf just greater or equal to 20 is 35, so P20 class is 19.5 – 39.5.

L= 19.5, f = 23, cf = 14, h = 20

P20 = L + 20N − cf × h
100
f

= 19.5 + 20 − 14 × 20
23

= 19.5 + 6 × 20
23

= 19.5 + 5.21

= 24.72

Again,

For first P80,

80N = 80 × 100 = 80
100 100

cf just greater or equal to 80 is 86, so P20 class is 59.5 – 79.5.

L= 59.5, f = 21, cf = 63, h = 20

82

P80 = L + 80N − cf × h
100
f

= 59.5 + 80 − 63 × 20
21

= 59.5 + 17 × 20
21

= 59.5 + 16.19

= 75.69

Thus, minimum, and maximum marks of middle 60% family is 24.72 and 75.69
respectively.

QN.19) (2072)

The following table shows the distribution of 100 families according to their expenditure
per week.

Expenditure 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50
27 - 15
No of families 14 -

If median = 25, find mean and mode.

Solution: Let the missing frequencies be a and b. Then

Expenditure No of families (f) Cumulative frequencies (c. f.)

0 - 10 14 14

10 - 20 a 14 + a

20 - 30 27 41 + a

30 - 40 b 41 + a + b

40 - 50 15 56 + a + b

Here median mark is 25. The median class is 20 – 30,

L = 20, f = 27, c. f. = 14 + a, h = 10

Median(Md) = L + N − cf × h
2 f

⇒ 25 = 20 + 50 − (14 + a) × 10
27

⇒ 25 − 20 = 50 − 14 − a × 10
27

83

⇒ 5 = 36 − a × 10
27

⇒ 1 = 36 − a
2 27

⇒ 27 = 72 − 2a

⇒ 2a = 72 − 27

⇒ a = 22.5 ≈ 23

Again, we have N = 100
⇒ a + b + 56 = 100
⇒ 23 + b + 56 = 100
⇒ b + 79 = 100
⇒ b = 21

Calculation of mean and mode

Expenditure No of families (f) Mid-value (X) fX
70
0 - 10 14 5 345
650
10 - 20 23 15 735
720
20 - 30 27 25 ΣfX = 2500

30 - 40 21 35

40 - 50 15 45

N = 100

Now, Mean(̅X) = ΣfX = 2500 = 25
N 100

Again, maximum frequency is 27, so model class is 20 – 30.

Here, L = 20, f1 = 27, f0 = 23, f2= 21, h = 10

Now, Mode = L + f1 − f0 f2 × 10
2f1 − f0 −

= 20 + 2 × 27 − 23 21 × 10
27 − 23 −

= 20 + 54 4 44 × 10
−

84

= 20 + 40
10

= 20 + 4

= 24

Therefore, mean = 25 and mode = 24.

QN. 20) (2072 old) Find the geometric and harmonic mean from the following data:

Class interval 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60

Frequencies 30 75 70 60 15

Solution: Calculation of Geometric and Harmonic mean

Class Frequency Mid- logX f logX 1 f 1
X X
interval (f) value (X) 0.0667

10 – 20 30 15 1.1761 35.283 0.04 2.001

20 – 30 75 25 1.3979 104.8425 0.0286 3.000

30 – 40 70 35 1.5440 108.08 0.0222 2.002

40 – 50 60 45 1.6532 99.192 0.0182 1.332

50 - 60 15 55 1.7404 26.106 0.273

250 373.5035 8.608

Here, N = 250, ΣflogX = 256.6775, Σf 1 = 8.608
X

Now, G. M. = Antilog (ΣflNogX)

= Antilog (3732.550035) = Antilog(1.4940)

= 31.19

Again, H. M. = n = 250 = 29.04
8.608
Σf 1
X

Therefore, G. M. = 31.19 and H. M. = 29.04.

QN.21) (2072 old)
The weekly expenditure of 1000 families is given below.

85

Expenditure (in'00') 40 - 59 60 - 79 80 - 99 100 - 119 120 - 139

No of families 50 - 500 - 50

If median of the distribution is Rs. 8,700. Calculate the missing frequencies. Also calculate

the mode of the distribution.

Solution: Let the missing frequencies be a and b. Then

Expenditure No of families Cumulative frequencies (c. f.)
(in'00') (f) 50
50
39.5 – 59.5

59.5 – 79.5 a 50 + a
79.5 – 99.5 500 550 + a
99.5 – 119.5 b 550 + a + b

119.5 – 139.5 50 600 + a + b

Here median mark is Rs. 8,700. The median class is 79.5 – 99.5,

L = 79.5, f = 500, c. f. = 50 + a, h = 20

Median(Md) = L + N − cf × h
2 f

⇒ 87 = 79.5 + 500 − (50 + a) × 20
500

⇒ 87 − 79.5 = 500 − 50 − a × 50
500

⇒ 7.5 = 450 − a × 20
500

⇒ 7.5 = 450 − a
25

⇒ 187.5 = 450 − a

⇒ a = 450 − 187.5

⇒ a = 262.5 ≈ 263

Again, we have N = 1000
⇒ a + b + 600 = 100
⇒ 263 + b + 600 = 100
⇒ b + 863 = 1000
⇒ b = 137

86

Missing frequencies are 263 and 137 No of families (f)
50
Again, 263
500
Calculation of mode 137
50
Expenditure (in'00')
39.5 – 59.5
59.5 – 79.5
79.5 – 99.5
99.5 – 119.5
119.5 – 139.5

Again, maximum frequency is 500, so model class is 79.5 – 99.5.

Here, L = 79.5, f1 = 500, f0 = 263, f2= 137, h = 20

Now, Mode = L + f1 − f0 f2 × 20
2f1 − f0 −

= 79.5 + 2 × 500 − 263 137 × 20
500 − 263 −

= 79.5 + 237 400 × 20
1000 −

= 79.5 + 4740
600

= 79.5 + 7.9

= 87.40

Therefore, model expenditure is Rs. 8740.

QN. 22) (2071) A manufacturing company has 1000 employees 10% of the employees earn less
than Rs. 500 per day, 200 earn between Rs. 500 and Rs. 999, 30% earn Rs. 1000 and Rs. 1499;
250 employees earn between Rs. 1500 and Rs. 1999 and earn Rs. 2000 and above. Calculate the
suitable average wage. Give the reason for your choice of average.

Solution: Here total employees (n) = 1000

10% of 1000 = 100, 30% of 1000 = 300

The given distribution has open ended classes, so the median is suitable average.

Calculation of median

Income (in Rs.) No of employees (f) cumulative frequency (c. f.)

less than 499.5 100 100

87

499.5 – 999.5 200 300
999.5 -1499.5 300 600
1499.5 – 1999.5 250 850
above 1999.5 150 1000
N = 1000

For median, N = 1000 = 500
2 2

c.f. just greater or equal to 500 is 600, So median median class is 999.5 – 1499.5.

Here, l = 999.5, f = 300, c.f. = 300, h = 500, Md =?

Median(Md) = L + N − cf × h
2 f

= 999.5 + 500 − 300 × 500
300

= 999.5 + 200 × 500
300

= 999.5 + 333.333

= 1332.83

Therefore, average wage is Rs.1332.83 per day.

QN. 23 (2071 old)

From the following distribution of marks of 250 students of a campus, find the
minimum pass marks if only 20% students have failed, and also find the minimum marks
obtained by the top 25% of the students.

Marks 0 - 20 20 – 40 40 – 50 50 - 60 60 – 80 80 - 100
50 75 45 30 25
No of students 25

Solution:

Calculation of P20 and P25

Marks No of students (f) Cumulative frequency (c. f.)
25 25
0 – 20 50 75
20 – 40 75 150
40 – 50 45 195
50 – 60 30 225
60 – 80 25 250
80 - 100

88

N = 250

For P20, 20N = 20 × 250 = 50
100 100

c.f. just greater or equal to 50 is 70, So P20 lies in class class interval 20 - 40.

Here, l = 20, f = 25, c.f. = 25, h = 10, P20 =?

P20 = L + 20N − cf × h
100
f

= 20 + 50 − 25 × 10
25

= 20 + 25 × 10
25

= 20 + 10

= 30

Therefore, minimum pass mark is 30.

Again,

For P75, 75N = 75 × 250 = 187.5
100 100

c.f. just greater or equal to 187.5 is 195, So P75 lies in class class interval 50 - 60.

Here, l = 50, f = 45, c.f. = 150, h = 10, P75 =?

P75 = L + 75N − cf × h
100
f

= 50 + 187.5 − 150 × 10
45

= 50 + 37.5 × 10
45

= 50 + 8.33

= 58.33

Therefore, minimum pass marks obtained by top 25% students is 58.33.

89

Chapter – 4

Measure of Dispersion

Brief Question Answer

QN. 1) (2078) Coefficient of variation of a distribution is found to be 20% and variance of
distribution is 36. Find the mean value of the distribution.

Solution: Here coefficient of variation (C. V.) = 20%

Variance of distribution (σ2) = 36

Now, C. V. = σ 100%
̅X

⇒ 20% = 6 100%
X̅

⇒ ̅X = 6 100
20

∴ ̅X = 30

QN.2) (2077 old) Calculate the quartile deviation when the values when the values of lower
quartile and upper quartile are 12 and 18 respectively. Also, calculate the coefficient of quartile
deviation.

Solution: Here given

lower quartile (Q1) = 12 and upper quartile (Q1) = 18

Now, Quartile deviatin = Q3 − Q1 = 18 − 12 = 3
2 2

and coeccicient of Q. D. = (Q3 − Q1) = 18 − 12 = 3 = 0.1
Q3 + Q1 18 + 12 30

QN. 3) (2077) Coefficient of variation of a distribution is found to be 20% and variance of
distribution is 49. What is the mean value of the distribution?

Solution: Here coefficient of variation (C. V.) = 20%

Variance of distribution (σ2) = 49

Now, C. V. = σ 100%
̅X

⇒ 20% = 7 100%
X̅

⇒ ̅X = 7 100
20

90

∴ ̅X = 35

QN. 4) (2077) The mean and standard deviation of 100 items are found to be 40 and 10
respectively. If at the time of calculation two items are wrongly taken as 30 and 70 instead of 3
and 27, find the correct mean and correct standard deviation.

Solution: Here number of observations (N) = 100
mean(̅X) = 40, standard deviation (σ) = 10

wrong values = 30 and 70

correct values = 3 and 27

Now, we have

X̅ = ∑X
N

40 = ∑X
100

⇒ ΣX = 400

Now, Correct ΣX = 4000 − 30 − 70 + 3 + 27 = 3930

∴ Correct mean = Correct ∑X = 3930 = 39.3
N 100

Again,

σ = √ΣNX2 − (ΣNX)2

⇒ 10 = √1Σ0X02 − (14000)2

⇒ 100 = ΣX2 − 0.16
100

⇒ 100 + 0.16 = ΣX2
100

⇒ 100.16 = ΣX2
100

⇒ ΣX2 = 10016

Correct ΣX2 = 10016 − 302 − 702 + 32 + 272 = 4954

∴ Correct σ = √ΣNX2 − (ΣNX)2

91

= √4190504 − (3190.03)2

= √49.54 − 0.154
= √49.386
= 7.03

QN.5) (2076) The coefficient of variance of symmetrical distribution is 9% and the mean of the
distribution is 40. Find the value of variance.

Solution: Here given C. V. = 9% and mean (̅X) = 40

variance (σ2) =?

Now, we have

C. V. = σ 100%
̅X

⇒ 9% = σ 100%
40

⇒σ = 40 × 9
100

σ = 3.6

∴ σ2 =12.96

QN. 6) (2076) Find the coefficient of variance from the following distribution.
N = 9, ΣX = 135, ΣX2 = 2085

Solution:

X̅ = ΣX = 135 = 15
N 9

and, variance (σ2) = ΣX2 − (ΣNX)2 = 2085 + (1935)2 = 231.67 − 225 = 6.67
N 9

or σ = 2.58

∴ coefficient of variance (C. V. ) = σ 100% = 2.58 × 100% = 17.22%
X̅ 15

QN. 7) (2075) From the following distribution, determine the quartile deviation.

Hight (inches) 30 35 40 45 50
8
No of students 5 15 20 12

Solution: Calculation of quartile deviation

92

Hight (inches) No of students (f) Cumulative frequency (c.f.)
30 5 5
35 15 20
40 20 40
45 12 52
50 8 60

N = 60

Now, first quartile (Q1) = (N + 1) th item
4

= (60 + 1) th item
4

= (61) th item
4

= 15.25th item

c.f. just greater or equal to 15.25 is 20 and its corresponding value is 35.

∴ Q1 = 35

Again, third quartile (Q1) = 3(N + 1) th item
4

= 3(60 + 1) th item
4

= 3(61) th item
4

= 45.75th item

c.f. just greater or equal to 45.75 is 52 and its corresponding value is 45.

∴ Q1 = 45

∴ Quartile deviatin = Q3 − Q1 = 45 − 35 = 5
2 2

QN.8) (2074) The mean and standard deviation of a series of 20 items are found to be 15
and 10 respectively. If and additional item of 5 included in the series, find the revised value
of mean and standard deviation.

Solution: Number of items (N) = 20
mean (X̅) = 15 and standard deviation (σ) = 10

Now, we have

93

X̅ = ∑X
N

15 = ∑X
20

⇒ ΣX = 300

Again,

σ = √ΣNX2 − (ΣNX)2

⇒ 10 = √ΣX2 − (1250)2
20

⇒ 100 = ΣX2 − 0.5625
20

⇒ 100 + 0.5625 = ΣX2
20

⇒ 100.5625 = ΣX2
20

⇒ ΣX2 = 2011.25

Again, additional item 5 is included

Number of items (N) = 21

ΣX = 300 + 5 = 305

∴ ̅X = ∑X = 305 = 14.52
N 21

ΣX2 = 2011.25 + 52 = 2036.5625

σ = √ΣNX2 − (ΣNX)2

= √20362.15625 − (142.152)2

= √96.98 − 0.4761
= √96.50
= 9.8
Therefore, revised mean = 14.52 and standard deviation = 9.8

94

QN.9) (2073) From the data relating to the performance of 2 workers, state who is more is
more consistence worker.

Worker A Worker B

Mean time taken to complete job 20 hrs 35 hrs

standard deviation 45

Solution: Let time taken by worker A be X and worker B be Y. Then

X̅ = 20, Y̅ = 35, σX = 4, σY = 5

C. V. (X) = σX 100% = 4 100% = 20%
̅X 20

C. V. (Y) = σY 100% = 5 100% = 14.29%
̅Y 35

As, CV(X) = 20% > CV(Y) = 14.29%

Therefore, worker B is more consistence worker.

QN.10) (2072) The mean and coefficient of variance of a certain data set are 12 and 25%
respectively. Calculate the value of standard deviation and variance of data.

Solution: Here given ̅X = 12 and CV(X) = 25%

Now, we have CV(X) = σ 100%
̅X

25% = σ 100%
12

σ = 25 × 12
100

∴σ=3

and variance σ2 = 32 = 9

QN. 11) (2071) The difference between the upper quartile and lower quartile of a certain
frequency distribution is 4 and their sum is 16. Calculate the quartile deviation and its
coefficient.

Solution: Here given Q3 − Q1 = 4, and Q3 + Q1 = 16

Now, Quartile deviation = Q3 − Q1 = 4 = 2
2 2

and Coeffcient C. V. = (Q3 − Q1) = 4 = 0.25
Q3 + Q1 16

QN. 12) (2071) The average price and standard deviation of the price of Mansuli rice per
kg. for the last seven days in two markets M1 and M1 are recorded below:

Average price (Rs. ) S.D. of price (Rs.)

95

Market M1 90 7

Market M1 94 6

Which market shows greater variation in price?

Solution: Let price in market M1 be X and price in market M2 be Y. Then

X̅ = 90, Y̅ = 94, σX = 7, σY = 6

C. V. (X) = σX 100% = 7 100% = 7.78%
X̅ 90

C. V. (Y) = σY 100% = 6 100% = 6.38%
̅Y 94

As, CV(X) = 7.78% > CV(Y) = 6.38%

Therefore, the market M1 shows greater variation in price.

Descriptive Question Answer

QN.13) (2078)

Following two samples describe the age of students in regular morning MBS program and
evening MBS program of National College. If homogeneity of the class is a positive factor
for learning which of the two program will be easier to teach.

Evening MBS 24 30 28 23 25 22 26 27 28 25
Morning MBS 28 27 34 33 29 27 28 29 33 26
Solution: Let the age of students in evening program and morning program be denoted by
X and Y respectively.

Calculation of Coefficient of Variance

Evening MBS Morning MBS
X X2
24 576 Y Y2
30 900
28 784 28 784
23 529
25 625 27 729
22 484
26 676 34 1156
27 729
28 784 33 1089
25 625
ΣX = 258 ΣX2 = 6712 29 841

27 729

28 784

29 841

33 1089

26 676

ΣY = 294 ΣY2=8718

For Evening program

Mean (X̅) = ΣX = 258 = 25.8
N 10
96

Standard deviation (σX) = √ΣNX2 − (ΣNX)2 = √617012 − (21508)2

= √671.2 − 665.64

= √5.56

= 2.36

C. V. = σX 100% = 2.36 100% = 9.15%
̅X 25.8

For morning MBS

Mean (̅Y) = ΣY = 294 = 29.4
N 10

And, Standard deviation MBS(σY) = √ΣY2 − (ΣNY)2 = √817018 − (21904)2
N

= √871.8 − 864.36

= √7.44

= 2.73

C. V. = σY 100% = 2.73 100% = 9.29%
Y̅ 29.4

Since CV of evening MBS is less than CV of morning MBS. Thus, evening MBS program
is easier to teach than morning program.

QN. 14) (2077, Back) The mean mark of 60 students is found to be 35. However, two
frequencies are missing in the group of marks distribution 20 -30 and 40 – 50. Find the
missing frequencies from the following data. Also, calculate the standard deviation of the

mark distribution.

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 – 50 50 - 60

No of students 4 6- 20 - 10

Solution: Let the missing frequencies be a and b. Then

Marks No of students(f) Cumulative frequencies (c. f.)
0 – 10 4 4
10 – 20 6 10
20 – 30 a
30 – 40 20 10 + a
30 + a

97

40 - 50 b 30 + a + b
40 + a + b
50 - 60 10
fX fx2
N = 40 + a + b 20 100
90 1350
Here median mark is Rs.35. The median class is 30 – 40. 250 6250
700 24500
L = 30, f = 20, c. f. = 10 + a, h = 10, N = 60

Median(Md) = L + N − cf × h
2 f

⇒ 35 = 30 + 30 − (10 + a) × 10
20

⇒ 35 − 30 = 30 − 10 − a × 10
20

⇒ 5 = 20 − a × 10
20

⇒ 5 = 20 − a
2

⇒ 10 = 20 − a

⇒ a = 20 − 10

⇒ a = 10

Again, we have N = 60
⇒ a + b + 40 = 60
⇒ 10 + b + 40 = 60
⇒ b + 50 = 60
⇒ b = 10

Missing frequencies are 10 and 10.

Again, Calculation of standard deviation

Marks No of students(f) Mid-value
0 – 10 4 (X)
5

10 – 20 6 15

20 – 30 10 25
30 – 40 20 35

98

40 - 50 10 45 450 20250
50 - 60 10 55 550 30250
N = 60
ΣfX =2060 ΣfX2 =82700

Now, Standard deviation(σ) = √ΣfNX2 − (ΣNfX)2

= √8267000 − (260060)2

= √1378.33 − 1178.55

= √199.78
= 14.13

QN. 15) (2076)

Two samples (each of size 10) of ages of students of BBS program and MBS program of
Tribhuvan University are described below:

Age of students of BBS (in year) 4 0 8 3 5 2 6 7 8 5
Age of students of BBS (in year) 6 3 9 8 7 9 3 4 7 8
If homogeneity in age of the class in a positive factor in teaching and learning process,
which of the two groups will be easier to teach?

Solution: Calculation of standard deviation

Age of students (in year) X2 Y2

BBS (X) MBS (Y) 16 36
0 9
46 64 81
9 64
03 25 49
4 81
89 36 9
49 16
38 64 49
25 64
57 Σ X2 =292 Σ Y2 =458

29

63

74

87

5 8
Σ X =48 ΣY =64

For BBS Program

99