TU Solution

Table of Contains 1
15
Importance formulae 32
49
Model question -2076 (Set A) 65
67
Model question -2076 (Set B) 69
90
Model Question -2070 127
Chapter – 1: Introduction to Statistics and Data Collection 154
Chapter – 2: Classification and Presentation of Data 184
Chapter – 3: Measure of Central Tendency 202
Chapter – 4: Measure of Dispersion 222
Chapter – 5: Skewness, Kurtosis and Moments 228
Chapter – 6: Simple Correlation and Regression Analysis 231
Chapter– 7: Time Series Analysis 246
Chapter – 8: Index Number 263
Chapter – 9: Probability 278
Chapter – 10: Sampling and Estimation
Chapter – 11: Quantitative Analysis
Chapter – 12: Linear Programming Problem
Chapter – 13: Determinant
Chapter – 14: Matrix

1

Importance Formulae
Measures of Central Tendency

1. Arithmetic Mean Individual Series Discrete Series Continuous Series
a) Simple Arithmetic Mean Σx Σfx X̅ = ΣNfm,
̅X = N ̅X = N
Direct Method where m = mid-value

Short-cut or ̅X = A + Σx ̅X = A + Σfd X̅ = A + Σfd
Assumed Mean N N N
Method
Step Deviation X̅ = A + Σd′ × h ̅X = A + Σfd′ × h ̅X = A + Σfd′ × h
Method N N N

b) Weighted Arithmetic Mean ̅Xw = ΣwX
Σw
c) Combined Arithmetic Mean
X̅12 = n1X̅1 + n2X̅2
2) Geometric Mean n1 + n2
a) Individual Series

GM = Antilog [ΣlongX]
b) Discrete Series

GM = Antilog [ΣflnogX]

c) Continuous Series
GM = Antilog [Σflongm] , where m = mid − value
) Harmonic Mean

a) IHnMdi=vidΣun 1a l Series
b) HDMisc=retΣefnS 1 eries
c) HCMont=inΣufno 1u s, wSehreirees m = mid − value

2

4) Median and Partition Values

Types of Individual Series Discrete Series Continuous Series

Partition Md = (N + 1)th item Md = (N + 1)th item Md = L + N − cf × h
2 2 2 f
Values
Median ( )

Quartiles ( ), Qi = i (N + 1)th item Qi = i (N + 1)th item Qi = L + i (N4 ) − cf × h
i = 1, 2, 3 4 4 f

Deciles( ) Di = i (N1+0 1)th item Di = i (N1+0 1)th item Di = L + i (1N0) − cf ×h
i = 1, 2, …, 9 Pi = i (N1+001)th item Pi = i (N1+001)th item i (1N00f) − cf ×h
Percentile ( )
= 1, 2, … , 99 Pi = L + f

5) Mode

a. Individual Series: Mode is the value which repeats maximum times.
b. Discrete Series: Mode is the variate value which has maximum frequency.
c. Continuous Series:

Mode(Mo) = L + f1 − fo f2 × h
2f1 − fo −

Or

Mode (Mo) = 3Median – 2Mean

Measure of Dispersion

1. Range (R) = L – S, where L = largest value, S = smallest value
L − S
Coefficient of Range = +

2. Inter-quartile Range (I.Q.R.) = Q3 – Q1
3 − 1
Quartile Deviation (Q. D. ) = 2

Coefficient of Q. D. = Q3 − Q1
Q3 + Q1

3. Mean deviation/Average Deviation

Individual Series X̅| Discrete Series ̅X| Continuous Series ̅X|

From Mean M. D. = Σ|X − M. D. = Σf|X − M. D. = Σf|m −
n n n
Where m = mid-value

3

From Median M. D. = Σ|X − Md| M. D. = Σf|X − Md| M. D. = Σf|m − Md|
From Mode n n n
Where m = mid-value

M. D. = Σ|X − Mo| M. D. = Σf|X − Mo| M. D. = Σf|m − Mo|
n n n
Where m = mid-value

Coefficient of Mean Deviation (CMD)

i)From Mean = MD From Mean
Mean

ii) From Median = MD From Median
Median

ii) From Mode = MD From Mode
Mode

4. Standard deviation (σ)

Individual Series Discrete Series Continuous Series

Actual Mean √Σ(X − ̅X)2 or√ΣNX2 − (ΣNX)2 √Σf(X − ̅X)2 √Σf(X − X̅)2
Method or N N N
Direct Method
√ΣNd2 − (ΣNd)2 , where d = X − A √Σfd2 − (ΣNfd)2 √Σfd2 − (ΣNfd)2
Short-cut or N N
Assumed
Mean Method

Step Deviation √ΣfNd′2 − (ΣNfd′)2 × h, where d′ = X − A
Method h

Coefficient of Standard Deviation = σ
̅X

Coefficient of Variation σ × 100%
̅X

5. Combined Standard Deviation

σ12 = √n1s12 + n1d12 + n2s22 + n2d22 , where d1 = X̅1 − X̅12, d2 = X2 − X̅12
n1 + n2

Skewness, Kurtosis and Moments

1. Absolute measurement of Skewness
a) Karl’s Pearson’s Method
i) Spk = Mean − Mode
ii) Skp = Mean − Median
b) Bowley’s Method = 3 + 1 − 2

2. Relative Measurement of Skewness
a) Karl Pearson’s Coefficients of Skewness

4

ii)i)SSpkkp==M3e(Manea−σnM−σoMdeedian) (If mode is not defined)
b) Bowley’s Method
3 + 1 − 2
= Q3 + 1

c) Measure of Skewness based on Moments
μ23
i) β1 = μ32μ3
ii) 1 = √μ3

3. Moment

a) Central Moments

i) Individual Series
Σ(X X̅)r
μr = − , r = 0, 1, 2, 3, … …
n
ii) Discrete and Continuous Series
Σf(X − ̅X)r
μr = n , r = 0, 1, 2, 3, … …

b) Raw moments

i) Individual Series
Σ(X − A)r Σdr
μ′r = n = n , r = 0, 1, 2, 3, … … , d = X − A

ii) Discrete and Continuous Series
Σf(X − A)r Σfdr
μ′r = n = n , r = 0, 1, 2, 3, … … , d = X − A

If there is any common factor h in the variable we use the
Σf(X − A)r Σfd′r X − A
μ′r = n = n × h, r = 0, 1, 2, 3, … … , d′ = h

4. Relation between Central Moments and Raw Moments
μμμμ1′′4′3′2 μ43(μ1′μμ1′′3′2=)μμ211′′0++
μ1 = − μ6′2μ(2′ μ(1′μ)1′ 2)3− 3(μ1′ )
μ2 = −
μ3 = −
μ4 = −

Correlation & Regression

1. Karl Pearson’s Correlation Coefficient σxy
Cov(X, Y) σxσy
r = √Var(x) Var(Y) =

a) Direct Method

r = Σ(X − ̅X)(Y − Y̅)
NσXσY

5

r = √NΣX2 NΣXY − ΣX ΣY − (ΣY)2
− (ΣX)2 √NΣY2

b) Actual mean method

r = Σ(X − ̅X)(Y − Y̅) = Σxy
√Σ(X − X̅)2 Σ(X − Y̅)2 √Σx2 Σy2

c) Shortcut or Actual Mean Method

r = √NΣu2 NΣuv − Σu Σv − (Σv)2 , where u = X − ̅X, v = Y = Y̅
− (Σu)2 √NΣv2

d) Step-deviation Method

r = NΣu′v′ − Σu′ Σv′ , where u′ = X − ̅X , v = Y = Y̅
− (Σu′)2 √NΣv′2 h k
√NΣu′2 − (Σv′)2

Probable Error of Correlation Coefficient

Probable error of correlation coefficient

Let r be the coefficient of correlation between N pairs of observations. Then standard

error of r is given by 1 − r2
√N
S. E. =

The probable error (P.E.) is measure of ascertaining the reliability of coefficient of

correlation. It value is given by (1 − 2)
√
. . ( ) = 0.6745

The probable errors are used for interpreting the coefficient of r as:

i) If r < P.E. of r it is insignificant

ii) If r > 0.6 x P. E. of r, it is significant
iii) ± P.E. of r gives the limit of the coefficient of correlation population.
2. Spearman’s Rank Correlation Coefficient

a) When observations are not repeated
6Σd2
R = 1 − N(N2 − 1) , where d = R1 − R2

b) When observations not repeated Σ(m3 −
12
R = 1 − 6 [Σd2 + m)]

N(N2 − 1)

Where m = number of repeation on the observation.

6

Regression Equation of Y on X
Computational Procedure of Regression Equation of Y on X

The linear regression equation Y on X is defined by Y = a + bX

where −
2 − ( )2
=

and = Σ − Σ


The coefficient b is called regression coefficient and is denoted by . Its values can be

expressed as Cov(X, Y)
Var(X)
bYX =

The linear regression equation X on Y is defined by = 1 + 1

where NΣXY − ΣXΣY
NΣY2 − (ΣY)2
b1 =

and 1 = Σ − 1Σ

The coefficient 1 is called regression coefficient and is denoted by . Its values can

be expressed as ( , )
( )
=

Different Forms of Regression Equations

1. The regression coefficient of Y on X is given by

bYX = NΣXY − ΣXΣY = Cov(X, Y ) = σXY
NΣX2 − (ΣX)2 Var(X) σ2X

or bYX = σXY = σY × σXY = σXY × σY = r. σY
Hence, σ2X σσYY σX2 σXσY σX σX
σX
bYX = r

2. The regression coefficient of X on Y is given by
NΣXY − ΣXΣY Cov(X, Y ) σXY
bXY = NΣY2 − (ΣY)2 = Var(Y) = σ2Y

or bXY = σXY = σX × σXY = σXY × σX = r. σX
Hence, σY2 σσXX σ2Y σXσY σY σY
σY
bXY = r

3. The regression equation Y on X is given by Y = a + bX ……….. (i)
Σ = Σ + Σ

7

Σ = Σ= ++ Σ ̅ .….….….….. (ii)
̅
∴

Subtracting equation (ii) from (i) we get
− ̅ = ( − ̅ )
∴ − ̅ = ( − ̅ )

4. The regression equation X on Y is given by

Σ = 1 + 1 … … … . . ( )
Σ == ̅ ΣΣ= 11 ++1 + 11 ΣΣ +
1 ̅ .….….….….. (ii)

∴

Subtracting equation (ii) from (i) we get
̅ = 1 ( −( ̅ −)
− − = ̅ )
∴ ̅

5. The regression equation Y on X is given by − ̅ = ( − ̅ )


6. The regression equation X on Y is given by − ̅ = ( − ̅ )


Relationship Between Correlation coefficient and Regression Coefficient

bbXYYX×=bYrσXσXY=arnσσdXY bXY = rσX
σY
× rσX
σY = r2

∴ r = √bXY × bYX

Index Number

1. Calculation of Unweighted Index Numbers
i. Simple(unweighted) aggregative method

P01 = ΣP1 × 100 … … … … … … (i)
number ΣP0 '1' with respect
P01 =
where price index for current year to base year '0'

ΣP1 = total price of current year
ΣP0 = total price of base year

Unweighted quantity index numbers

Q01 = Σq1 × 100 … … … … … … (ii)
Σq0
Q01 =
where price index number for current year '1' with respect to base year '0'

Σq1 = total price of current year

8

Σq0 = total price of base year

ii. Simple average of price relative method: p1
Price relative (P) p0
= × 100, where

p1 = price of each commidity for current year
p0 = price of each commidity for base year

Simple average of price relative index
ΣP
P01(AM) = n … … … … … … … … (iii) p1
P01(GM) = p0
where p = × 100
× 100
Antilog [ΣlongP] … … … … … … . . (iv)
where
p = p1
p0
2. Weighted index number

i. Weighted aggregated method

a. Laspeyre's method or base year method

P01(L) = Σp1q0 × 100 … … … … … … . . (v)
ΣP0q0
where
p1 = price of each commidity for current year
p0 = price of each commidity for current year
q0 = quantity of base year

b. Paasche's method or given year method

P01(P) = Σp1q1 × 100 … … … … … … . . ( )
ΣP0q1
where
p1 = price of each commidity for current year
p0 = price of each commidity for current year
q1 = quantity of current year

c. Fisher's Ideal Index Number

Fisher′s index number p01(F) = √pp10qq00 × p1q1 × 100 … … … … … … . . (vii)
p0q1

ii. Weighted average of price relative
=) =pp10ΣΣ×WW1P0…0,…W…=…
P01(A. M. … … … … (viii) to the price relative p
where p weight attached

9

P01(G. M. )==pp10An×ti1lo0g0[ΣWΣlWogP] … … … … … … . . (ix)
where p

Quantity Index Number

i. The quantity index by simple average aggregative method is given by
Σ 1
01 = Σ 0 × 100 … … … … … … … ( )

ii. The quantity index by simple average of quantity relative is given by
wh e r01e(Q .= .q)u=anΣt i t y…re…la…tiv…es…=…qq. 1.0(× 1)00

iii. The Lasperyer's quantity index by weighted aggregative method is given by

Q01(L) = Σq1p0 × 100 … … … … … … . . ( )
Σq0p0

The Paasche's quantity index is given by

Q01(P) = Σq1p1 × 100 … … … … … … . . ( )
Σq0p1

Fisher's ideal quantity index is given by

Q01(F) = √qq01pp00 × q1p1 × 100 … … … … … … . . ( )
q0p1

iv) The quantity index by weighted average of quantity relatives is given by
w h 0e1re=QΣΣ= qq 10…×…1…00…,
… . . ( )
W = q0p0

Value Index

V01 = Σp1q1 × 100 … … … … … … … … (xvi)
Σp0q0

Test of Consistency of Index Number Formulae

a. Time reversal Test

The formula for the index number should be such that the product of two index numbers (omitting

the factor 100) based on the same data with time interchanged should be unity
i.e. 01 × 1 = 1
For Laspeyer,s formula

10

P01(L) = Σp1q0 and P10(L) = Σp0q1
ΣP0q0 ΣP1q1
Σp1q0 Σp1q0
P01(L) × P10(L) = ΣP0q0 × ΣP0q0 ≠ 1 … … … … … … . (xvii)

Thus, the Laspeyer's formula does not satisfies time reversal test.

For Paasche's formula

P01(P) = Σp1q1 and P10(P) ΣΣ=pP10ΣΣqqpP0010qq≠00
P01(P) × ΣP0q1
= Σp1q1 × 1 … … … … … … … … … . (xviii)
P10(P) ΣP0q1

Thus, the Paasch's formula does not satisfies time reversal test.

For Fisher's ideal formula

p01(F) = √pp10qq00 × p1q1 and p10(F) = √pp01qq11 × p0q0
p0q1 p1q0

∴ p01(F) = √pp10qq00 × p1q1 × p0q1 × p0q0 = 1 … … … … … … … … … (xix)
p0q1 p1q1 p1q0

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

b. Factor Reversal Test

The formula for the index number should be such that the product of the two index numbers

(omitting the factor 100) based on the same data with price and quantity interchanged should

be equal to the value indices

. . 01 × 01 = 01

For Laspeyer,s formula

P01(L) = Σp1q0 and Q01(L) = Σq1p0
ΣP0q0 Σq0p0
Σp1q0 Σq1p0 Σp1q1
P01(L) × Q01(L) = ΣP0q0 × Σq0p0 ≠ Σp0q0 = v01 … … … … … … . (xx)

Thus, the Laspeyer's formula does not satisfies factor reversal test.

For Paasche's formula

P01(P) = Σp1q1 and Q10(P) = Σq1p1
ΣP0q1 Σq0p1
Σp1q1 Σq1p1 Σp1q1
P01(P) × Q10(P) = ΣP0q1 × Σq0p1 ≠ Σp0q0 = v01 … … … … … … . (xxi)

Thus, the Paasch's formula does not satisfies factor reversal test.

11

For Fisher's ideal formula

p01(F) = √pp10qq00 × p1 q1 and Q10(F) = √pp10qq11 × p0q0
p0 q1 p1q0

∴ p01(F) × Q10(F) = √pp10 q0 × p1q1 × q1p0 × q1p1
q0 p0q1 q0p0 q0p1

= Σp1q1 = v01 … … … … … … . . (xxii)
Σp0q0

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

Cost of Living index number (or Consumer Price Index Number)

i. Aggregative expenditure method or weighted aggregative method

Cost of living index number = Σp1q1 × 100 … … … … … . (xxiii)
Σp0q0

ii. Family budget method or method of weighted average of price relatives

Cost of living index = ΣpΣWW=Ppp…10 … …… … … … … (xxiv)
where × 100

If w is not given, then we take w=p0q0

When the geometric mean is given, then cost of living index is given by
Cost of living index = Antilog (ΣΣWWP) … … … … … … … … (xxv)

When index number is given, the cost of living index is given by

Cost of living index = ΣIW … … … … … … … … (xxvi)
ΣW

Base shifting

Recast index number of any year = Index numbr of old base × 100 … … … … … . (xxvii)
Index number of new base

Real wages (or income) index numbers = Real wage of current year × 100 … … … … … . (xxviii)
Real wage of base year
1
Purchasing power of money = Price Index … … … … (xxix)

Real wages = Money or Nominal wages × 100 … … … … (xxx)
Price Index

Probability

12

Probability of an Event A is total no. favourable cases getting an event A
total no. of all possible outcomes
P(A) =

Empirical probability

Empirical probability, also known as experimental probability, refers to a probability that is based
on historical data. In other words, empirical probability illustrates the likelihood of an event
occurring based on historical data.

Axioms of probability
Axiom 1) (Positivizes):- p(A)≥0 for any subset of S. That is, the probability assigned to each
events is positive or zero.

Axiom 2) (Certainty):- P(S) = 1 i. e. the probability of sample space is always 1.

Axiom 3(Addition rule):- If 1, 2, 3, … … are finite or infinite sample space of mutually
exclusive events of S, then

P(A1 ∪ A2 ∪ A3 … . . ) = P(A1) + P(A2) + P(A3) … . . = ∑n∞=i ( ) ∑ =1 ( )

Permutation and Combination in Probability

Permutation:

The arrangement in the specific order of n different objects taken r at a time is called the

permutation. It is denoted by P(n, r) or or , .
The number of permutation of n distinct objects taken r at a time is given by
!
= ( − )!

Combination:

The selection of rr)doifrfe r e n t o b j e c, t s f r o(m )th. e n different objects is called the combination. It is
denoted by C(n,

The number of combination of n things taken r at a time is given by
!
= ( − )! !

Some theorems on Probability
Theorem 5.1: Let be an empty event then P( )= 0 for any sample space S.

Theorem 5.2: If A and B are events in a sample space S and
⊆ , h ( ) ≤ ( ).

Theorem 5.3: Let A be any event in the sample space S then,
0 ≤ P( ) ≤ 1.

Theorem 5.4: If A and B be any two events in a sample space S then,
P(A - B) =P(A) -P(A∩B)

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Theorem 5.5: Let A and B be the two events in the sample S then,
( ∪ ) = ( ) + ( ) − ( ∩ )

Theorem 5.7: If A and A' are two complementary events in then,

sample space S, then P(A) + P(A') = 1

Conditional Probability
If A and B be two evens in the sample space S and P(A)> 0, then the conditional probability of

the event B given the event A, denoted by P(B/A), is defined as:
( ∩ )
( / ) = ( )

Similarly, the coitional probability of the event A given the event B is given by:

( / ) = ( ∩ )
( )

Independent and dependent events

Independent Events

Two events A and B are said to be independent events if the probability of occurrence and non-

occurrence of A does not effect the probability of the occurrence of B. If A and B are two

independents of sample space S then,
( ∩ ) = ( ). ( )

Dependents events
Two events A and B are said to be dependent events if the probability of occurrence and non-
occurrence of A effect the probability of the occurrence of B. If A and B are two dependents of
sample space S then,
P(A ∩ B) = P(A). P(A)P(B/A)

Quantitative Analysis

Maximax = Maximum of the maximum pay off value
Maximin = Maximum of the minimum pay off value
Minimax Regret = Minimum of the maximum regret value
Expected Monetary Value (EMV) = Σ(Pay off value × probability)
Expected Opportunity Loss (Regret value × Probability)
Expected Profit with Perfect Information (EPPI) = Σ(Diagonal pay-off value × probability)
Expected value of Perfect Information (EVPI) = EPPI – Maximum EMV
Realism = α × max profit + (1 – α) × min payoff
where, α = coefficient of optimistic, 1- α = coefficient of pessimistic

14

Model Question -2076 (Set A) [ × = ]
Group – A: Brief Question Answer

QN. 1) The mean of 200 items was 50. Later it was found that two items were wrongly
taken as 92 and 8 instead of 192 and 88. Find the correct mean.

Solution: Here given

Number of items (n) = 200
mean (̅X) = 50

Incorrect values = 92 and 8

Correct values = 192 and 8

Now, we have

̅X = ∑X
n

50 = ∑X
200

∴ ΣX = 10000

Now, Correct ΣX = 10000 − 92 − 8 + 192 + 88 = 10180

Again, Correct mean = Correct ∑X = 10180 = 50.9
n 200

QN.2) In the batch of 15 students, 3 students failed in an examination. The marks of
passed 12 students were; 9, 6, 7,8, 4, 5, 8, 10, 9, 7, 5, 7. What was the median mark of the
15 students?

Solution: Here given

Number of students = 15

Passed students = 12

Failed students = 3

Now, arranging the marks of passed students, we have

4, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9, 10

Now, the median of all 15 students is;

(N + 1)th item = 15 + 1 item = 8th item
2 2

∴ Median mark of all 15 students = 7.

QN.3) In the moderately skewed frequency distribution, the mean is 10 and its median is 9,
if the coefficient of variation is 20%. Find he Pearson's coefficient of skewness of the
distribution.

Solution: Now, we have

15

Coefficient of variation = σ × 100%
̅X
σ
⇒ 20% = 10 × 100%

⇒ 200 = σ
100

∴σ=2

Again, Pearson's coefficient of skewness is

X̅ − Md = 10 − 9 = 1 = 1.5
σ 2 2

QN. 4) List out the various methods of collecting primary and secondary data.

Solution: The following are the commonly used for the collection of primary data:

1) Direct personal interviews.
2) Indirect oral interviews.
3) Information from local correspondents.
4) Mailed questionnaire method.
5) Schedules to be filled in by enumerators.

The methods of collecting secondary data are as follows:
1) Data available on the internet.
2) Government and nongovernment agencies.
3) Public libraries.
4) Educational Institutions.
5) Commercial information sources.

QN. 5) Calculate the lower and upper quartiles from the following marks distributions:
Marks Below 25 25 - 29 30 - 34 35 - 39 40 - 44 44 above

Students 5 12 22 25 17 9

Solution: Calculation of Lower and Upper quartiles

Marks Students (f) cf

Below 24.5 55

24.5 – 29.5 12 17

29.5 – 34.5 22 39

34.5 – 39.5 25 54

39.5 – 44.5 17 71

Above 44.5 9 80

N = 80

For, Lower quartile = n th item = 80 th item = 20 th item
4 4

16

cf just greater and equal to 20 is 39, the lower quartile class is 29.5 – 34.5

Thus, L = 29.5, cf = 17, f = 22, h = 5

Now, the lower quartile is

Q1 = L + N − cf × h
2 f

= 29.5 + 20 − 17 × 5
22

= 29.5 + 3 × 5
22

= 29.5 + 0.68

= 30.18

QN. 6) If regression coefficient of y on x (byx) = −0.61 and regression coefficient y on x
(bxy) = −053, calculate the coefficient of correlation and interpret the result

Solution: Now, correlation coefficient

r = √byx × bxy = √−0.61 × −0.53 = −0.5686

QN. 7) The standard deviation of symmetric distribution is 5. what must be the value of
fourth moment about mean in order that the distribution be mesokurtic?

Solution:

Here, σ2 = 3, ∴ μ2 = σ2 = 32 = 9

For the distribution to be mesokurtic, β2 = 3
μ4
⇒ μ22 = 3

⇒ μ4 = 3
92

∴ μ4 = 3 × 81 = 243

QN.8) A bag contains 20 balls numbers from 1 to 20. One ball is drawn at random. Find
the probability that the number of the drawn ball be multiple of (i) 3 or 7 (ii) 3 or 5.

Solution:

Let the event of getting multiple of 3, 5, and 7 be denoted by A, B, and C respectively.

Now, we have

A = {3, 6, 9, 12, 15, 18}, B = {5, 10, 15, 20} and C = {7, 14}
A ∪ C = {3, 6, 7, 9, 12, 14, 15, 18} and n(A ∪ B) = 8

A ∪ B = {3, 5, 6, 9, 10, 12, 15, 18, 20} and n(A ∪ B) = 9

then, n(S) = 20, n(A) = 6, n(B) = 4, and n(C) = 2

17

i) P(A or C) = n(A ∪ C) = 8 = 2
n(S) 20 5

ii) P(A or B) = n(A ∪ B) = 9
n(S) 20

QN.9) Calculate the price index number from the following data by simple aggregative
method.

Commodities A B C D E

Price in 2075 125 105 260 150 250

Price in 2076 125 155 250 160 300

Solution: Calculation of Price Index Number

Commodities Price in 2075 (p0) Price in 2076 (p1)

A 125 125

B 105 155

C 260 250

D 150 160

E 250 300

Σpo = 890 Σp1 = 990

Now, the price index number (P01) = Σp1 × 100 = 990 × 100 = 111.24
Σp0 890

QN. 10) The following table shows the pay-off matrix related to the demand and strategy
of a business person. What should be the decision if he/she use i) maxi-max criterion ii)
maxi-min criterion?

Strategies D1 Demand D3
300 D2 80
S1 200 60
S2 80 100 20
S3 120
60

Solution: Calculation of maximum and minimum pay-off

18

Demand Pay-off

Strategies D1 D2 D3 Maximum Minimum
300 100 80 300 80
S1 200 120 60 200 60
S2

S3 80 60 20 80 20

Therefore, according to both maxi-max and maxi-min criterion he/she must choose
strategy S1.

Group – B: Descriptive Question Answer [ × = ]

QN. 11) List the five-number summary and prepare a box-and -whisker plot from the
following information. also comment the nature of the frequency distribution.

Class size 20-30 30-40 40 - 50 50 - 60 60 - 70 70 – 80
25 35 40 50
Frequency 10 12

Solution: Calculation of Q1, Md, Q3

Class size Mid-value (X) Frequency (f) Cumulative frequency (c. f.)
25 10 10
20 – 30 35 12 22
30 – 40 45 25 47
40 – 50 55 35 82
50 – 60 65 40 122
60 – 70 75 50 172
70 - 80
N = 172
N 172
For Q1, 4 = 4 = 43.

The cf just greater or equal to 43 is 47, so the corresponding quartile class is 40 – 50.

Here, L = 40, c. f. = 22, f = 25, h = 10

Now, Q1 = L + N − cf × h
4 f

= 40 + 43 − 22 × 10
25

= 40 + 21 × 10
25

= 40 + 8.4

= 48.4

19

For Md, N = 172 = 86
2 2

c. f. just greater or equal to 86 is 122. So, the median class is 60 – 70.

Here, L = 60, c. f. = 82, f = 40, h = 10

Now, Md = L + N − cf × h
2 f

= 60 + 86 − 82 × 10
40

= 60 + 4 × 10
40

= 60 + 1

= 61

For Q3, 3N = 3 × 172 = 129.
4 4

c. f. just greater or equal to the 129 is 172. So, the corresponding quartile class is 70 – 80.

Here, L = 70, c. f. = 122, f = 50, h = 10

Now, Q3 = L + 3N − cf × h
4 f

= 70 + 129 − 122 × 10
50

= 70 + 7 × 10
50

= 70 +1.4

= 71.4

Therefore, five-number summary is

smallest number =25, first quartile = 48.4, median = 61, third quartile = 71.4 and greatest
number = 75.

20

Box-and-whisker plot

QN. 12) Solve the following system of equations by using determinant method.

2x + 5y – z = -3, 4x + 3y +2z =1, & x + 2y +3z = -5

Solution:
2 5 −1

Now, D = |4 3 2 |
12 3

= 2(9 − 4) − 5(12 − 2) − 1(8 − 3)
= 10 − 50 − 5
−45

−3 5 −1
Dx = | 1 3 2 |

−5 2 3
= −3(9 − 4) − 5(3 + 10) − 1(2 + 15)
= −15 − 65 − 17
= −97

2 −3 −1
Dy = |4 1 2 |

1 −5 3
= 2(3 + 10) + 3(12 − 2) − 1(−20 − 1)
= 26 + 30 + 21
= 77

2 5 −3
Dz = |4 3 1 |

1 2 −5
= 2(−15 − 2) − 5(−20 − 1) − 3(8 − 3)

21

= −34 + 105 − 15

= 56

∴ x = Dx = − 97 = 97
D −45 45

y = Dy = 77 = −77
D −45 45

z = Dz = 56 = −56
D −45 45

QN. 13) Calculate Fisher index number for the following data and show that it satisfies

i) Time reversal test and ii) Factor reversal test.
Solution:

Calculation of Fisher's Index Number

Commodities 2018 Quantity 2019 Quantity p0q0 p1q0 p0q1 p1q1
A Price (q0) Price (q1) 720 464 696
(p0) 60 (p1) 58 480
8 12

B 4 110 4 125 440 440 500 500

C 6 70 8 70 420 560 420 560

D 12 40 14 38 480 560 456 532

Total 30 280 38 291 1820 2280 1840 2288

Here, ∑p0q0 = 1820, ∑p1q0 = 2280, ∑p0q1 = 1840, ∑p1q1 = 2288

Now Fisher′s index number p01(F) = √pp10qq00 × p1q1 × 100
p0q1

= √12828200 × 2288 × 100
1840

= √1.2527 × 1.2435 × 100
= 1.2481 × 100
= 124.88

Time reversal test

22

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

= √12282800 × 2288 × 1840 × 1820 = 1
1840 2288 2280

Hence, Fisher ideal index satisfies the time reversal test.

Factor reversal test

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

= √21828200 × 2288 × 1840 × 2288
1840 1820 2280

= 2288
1820

V01 = Σp1q1 = 2288
Σp0q0 1820

∴ P01(F) × Qo1(F) = V01
Hence, Fisher's ideal index satisfies the factor reversal test.

QN. 14) Fit straight line trend by the method of least square to the data given below. Also, find the
trend values and predict the sales for the year 2022.
Year 2013 2014 2015 2016 2017 2018 2019
Sales (000 units) 15 16 17 16 19 23 25

Solution: Let the trend line be y = a + b x, where x = X - Middle-year
Calculation of trend line

Year(X) Sales(y) x = X-2016 x2 xy Trend values
Yc = 18.714 + 1.643x

2013 15 -3 9 -45 yc = 18.714 + 1.643 × −3 = 13.785

2014 16 -2 4 -32 yc = 18.714 + 1.643 × −2 = 15.428

2015 17 -1 1 -17 yc = 18.714 + 1.643 × −1 = 17.071

2016 16 0 0 0 yc = 18.714 + 1.643 × −0 = 18.714

23

2017 19 1 1 19 yc = 18.714 + 1.643 × 1 = 20.384
2018 23 2 4 46 yc = 18.714 + 1.643 × 2 = 22
2019 25 3 9 75 yc = 18.714 + 1.643 × 3 = 23.643
131 0
28 46

Σy = 131, ΣX = 0, Σx2 = 28, Σxy = 46
Σy 131 Σxy 92
Now, a = n = 7 = 18.714 and b = Σx2 = 28 = 1.643

Thus, the trend line is y = a + bx

y= 18.714 + 1.643 x

Again, trend values

when x = -3, y = 18.714 + 1.643 *-3= 13.785

when x = -2, y = 18.714+ 1.643*-2 = 15.428
Estimated sales for 2022, yc = 18.714+1.643*6=28.572

Therefore, required estimated sales for 2022 is Rs.28572

QN. 15) A factory manufactures three types of white sheet of papers I, II, and III and distributes
them into two markets A and B. The sales of papers during one year are given below.

Markets I Papers III
12,000 II 23,000
A 14,000 13,000
B 5,000
22,000

a) If unit sales prices of papers I, II, and, III are Rs.5, Rs.4, Rs.3 respectively, find the total

revenue in each market with the help of matrix algebra.

b) If the unit costs of above papers are Rs.3, Rs.2, and Rs.1 respectively, find the gross profit.

Solution: Let the matrix X, Y and Z respectively denotes the matrices of distribution of papers in

three markets, Sales prices of three types of papers, and cost of per unit of papers.

Then, 5 3
[4] [2]
X = [1124000000 5000 1233000000] , Y = 3 and Z = 1
22000

a) Total revenue in each market 5
[4]
XY = [1124000000 5000 1233000000] 3
= 22000
[114997000000]
Therefore, total revenue in market A is Rs.149000 and market B is Rs.197000.

b) Total cost of papers in each market 3
[2]
XZ = [1124000000 5000 1233000000] 1
22000

= [6999000000]

24

Again, gross profit = total revenue – total cost
=[114997000000] − [9699000000] = [8908000000]
Therefore, gross profit = Rs.80000 + Rs.98000 = Rs. 178,000

QN. 16) The following is the profit of two companies in millions of rupees, find which company
shows the greater consistency in the net profit. Justify your answer with statistical evidence.

Company A 18 19 23 19 25 23 27
Company B 17 18 22 23 24 25 25

Solution: Calculation of Variance

Company A X2 Company B
X 324 Y Y2
18 361
19 529 17 289
23 361
19 625 18 324
25 529
23 729 22 484
27 ∑ X2 = 3458
∑ X=154 23 529

24 567

25 625

25 625
∑ Y=154 ∑ Y2 =3443

Now, ∑X 154
n 7
Mean of company A (̅X) = = = 22

Standard deviation of A (σ) = √∑nX2 − (∑nX)2

= √34758 − (1574)2

= √494 − 484
= √10

= 3.16

25

C. V. of company A = σ × 100% = 3.16 × 100% = 14.37%
X̅ 22
Again,
∑Y 154
Mean of company B = n = 7 = 22

Standard deviation of B (σ) = √∑nY2 − (∑nY)2

= √3443 − (1574)2
7

= √491.85 − 484
= √7.857

= 2.8

C. V. of company A = σ × 100% = 2.8 × 100% = 12.74%
̅X 22

Since the C.V. of company B is less than company B. So, Company B has consistency in net
profit.

Group – B: Analytical Question Answer [ × = ]

QN.17) A family income and its percentage expenditure on food for 100 families gave the

following bivariate frequency distribution. Find out if there exists any relationship between family

income and expenditure on food and interpret the result. Also test the significance of the result.

Estimate the percentage expenditure on food when family income = Rs.90,000.

Solution: Let X denote the mid-point of family income and Y denotes mid-point of food

expenditure then − 65
X 5 Y – 25
dx = and dy = 5

26

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

= √100 100 × −152 − 0 × 0 − 0
× 1390 − 0 √100 × 100

= −15200 × 100
√139000 √100

= −15200
372.83 × 100

= −15200
37283

= −0.41

Now, P. E. = 0.6745 1 − r2
√N

= 0.6745 1 − 0.412
√100

= 0.6745 1 − 0.1681
10

= 0.6745 0.8319
10

= 0.056

Again,

0.6× P.E. = 6× 0.056 = 0.336
Since, |r| > 6 × P. E. , the value of r is significance.

Regression coefficient y on x

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

= 100 × −152 − 0 × k
100 × 1390 − 0 h

27

= −15200
139000

= −15200
139000

= − 0.11

̅X = A + Σfdx × h = 65 + 0 = 65
N
Σfdy
Y̅ = B + N × k = 25 + 0 = 25

The regression equation Y on X is

Y − Y̅ = byx(X − ̅X)

⇒ Y − 25 = −0.11(X − 65)

⇒ Y =-0.11X+7.11+25

⇒ Y = 90 − 0.11X

Finally, when x = 90
Y = 32.11 − 0.11 × Rs. 90 = 22.21

Percentage expenditure on food when family income is Rs.90,00 is 22.21%.

18. The following distribution shows the frequency distribution of weakly expenditure of foods of

students in certain locality of Kathmandu Metropolitan City. Describe the various characteristic
feature of the frequency distribution. Also, comment the nature of the distribution.

Expenditure 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 100 - 110
(Rs.00)

Number of 10 15 28 33 40 35 25 14
students

Solution: Calculation of various Sums

Expenditure No. of Students (f) Mid-value d′ = X − 75 fd' fd'2
(x) 10 - 40
160
30 – 40 10 35 -4 135
112
40 – 50 15 45 -3 - 45 33
0
50 – 60 28 55 -2 - 56 35
100
60 – 70 33 65 -1 - 33 126

70 – 80 40 75 00

80 – 90 35 85 1 35

90 - 100 25 95 2 50

100 - 110 14 105 3 42

28

Total 200 560 -4 -47 701

Here, N = 200, ΣX = 560, ∑d′ = −4, Σfd′ = −47, Σfd′2 = 701

Now, ∑ fd′
N
X̅ = A + × h

= 75 + −47 × 10
200

= 75 − 2.35

= 72.65

Therefore, average expenditure is Rs.7265

σ = √∑ fd′2 − (∑Nfd′)2 × h
N

= √720001 − (−20407)2 × 10

= √4.9 − 0.055 × 10

= √3.45 × 10
= 1.857 × 10
= 18.57

QN. 19) What do you mean by linear programming problem? A person requires minimum 10, 12,

and 12 units of Chemicals A, B, and C respectively. A liquid product contains 5, 2, and 1 unit of
A, B, and C per jar. A dry product contains 1, 2, and 4 units of A, B, and C per carton. If the liquid

product cost Rs.30 per jar and dry product Rs.20 per carton. Formulate this problem in linear

programming model. How many of each product should be purchased in order to minimum the to

cost the requirements. Also, find the minimum cost.

Solution:
Linear Programming Problem: Linear programming is defined as that branch of mathematics

which deals with the optimization (maximization or minimization) of a linear function of a

number of various subjects to a number of conditions on the variables, in the form of linear

inequalities in the variables of the optimization function.

Here given

29

Let x units of dry product, and y units of liquid product should purchase to minimize the cost.

Then the mathematical model of given problem is

Minimize: C = 30x + 20y

subject to constraints:
5x + y ≥ 10
2x + 2y ≥ 12
x + 4y ≥ 12
and x ≥ 0, y ≥ 0

Let associated equations of the given constraints be

5x + y = 10 ………….. (i)

2x + 2y = 12 ……….. (ii)

x + 4y = 12 …………. (iii)

From (i), we get

if x = 0, y = 10

if y = 0, x = 2

Thus, the line (i) passes through the points (0, 10) and (2, 0).

From (ii)

if x = 0, y = 6

if y = 0, x = 6

Thus, line (ii) is passing through the points (0, 6) and (6, 0).

Similarly, from (iii)

if x = 0, y = 12

if y = 0, x = 3

Thus, line (iii) is passing through the pints (0, 12), and (3, 0).

Plotting above points in the graph, we have

Calculation of Minimum Value

Vertices C = 30x + 20y Remark

A(12, 0) C = 13 × 12 + 20 × 0 = 156

B(4, 2) C = 13 × 4 + 20 × 2 = 92 Minimum

C(1, 5) C = 13 × 1 + 20 × 5 = 113

D(0, 10) C = 13 × 0 + 20 × 10 = 200

Therefore, minimum value cost is Rs.92 at the point B(4, 2).

30

31

Model Question -2076 (Set B) [ × = ]
Group – A: Brief Question Answer

QN. 1) The mean marks in Statistics of 100 students in a class was 72. The mean marks of 70 boys
was 75. Find out the mean marks of the girls in the class.

Solution: Here given

Number of students (n) = 100

Number of boys (n1) = 70
Mean marks of all students (X12) = 72
Mean marks of boys (X̅1) = 75
Mean marks of girls (X̅2) =?
Now, number of girls (n2) = n − n1 = 100 − 70 = 30

we have,

X̅12 = X̅1n1 + ̅X2n2
n1 + n2

⇒ 72 = 70 × 75 + ̅X2 × 30
70 + 30

⇒ 72 = 70 × 75 + ̅X2 × 30
100

⇒ 7200 = 5250 + ̅X × 30

⇒ 7200 − 5250 = ̅X × 30

⇒ 1950 = ̅X × 30

⇒ ̅X = 1950 = 65
30

Therefore, mean marks of girls is 65.

QN. 2) The difference between the upper quartile and lower quartile of certain frequency

distribution is 4 and their sum is 16. Calculate 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 coefficient of quartile deviation

32

Q3 − Q1 = 4 = 0.25
Q3 + Q1 16

QN. 3) In a single throw of two dice, what is the probability getting same numbers on both dice?

Solution: Number of outcomes when a two dice is thrown n(S) = 36

Events of getting numbers = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}

Therefore, the probability of getting same number is

6 = 1
36 6

QN. 4) The personal director of Nepal Drug Limited recorded the average percentages absentee

rates for each quarter of the four years periods are 55, 67.5, 62.5 and 53, find the seasonal indices.

Solution: Now averages of averages

55 + 67.5 + 62.5 + 53 = 59.5
4

Seasonal index for first quarter = I quaterly avarage = 55 × 100 = 92.44
Avergae of averages 59.5

Seasonal index for second quarter = II quaterly avarage = 67.5 × 100 = 113.45
Avergae of averages 59.5

Seasonal index for third quarter = III quaterly avarage = 62.5 × 100 = 105.04
Avergae of averages 59.5

Seasonal index for first quarter = IV quaterly avarage = 53 × 100 = 89.08
Avergae of averages 59.5

QN. 5) The coefficient of variation a symmetric distribution is 9% and mean of the distribution is
40. Find the value of standard deviation and variance.

Solution:

Now, coffecient of variance = σ × 100%
̅X

9% = σ × 100%
40

360 = σ × 100

σ = 360 = 3.6
100

and variance = σ2 = 3.62 = 12.96

QN.6) What do you mean by five-number summary? What is its application in statistics?

33

Answer: A set consists of five values: the most extreme values in the data set (the maximum and
minimum values), the lower and upper quartiles, and the median is known as the five-number
summary. These values are presented together and ordered from lowest to highest: minimum
value, lower quartile (Q1), median value (Q2), upper quartile (Q3), maximum value. These values
have been selected to give a summary of a data set. A five-number summary is especially useful in
descriptive analyses or during the preliminary investigation of a large data set.
QN.7) Reconstruct the following index number by shifting the base year as 2053.

Year 2049 2050 2051 2052 2053 2054 2055
Index Number 100 115 126 134 147 155 163

Solution: we have Index numbr of old base
Index number of new base
Reconstruct index number by shifting the base year = × 100

Now, 100
147
Index number of 2049 by shifting the base year 2053 = × 100 = 68.03

and so on.

Calculation of index numbers with new base as 2063

Year Old index number New index number(with base 2063)

2049 100 68.03

2050 115 78.23

2051 126 85.71

2052 134 91.16

2053 147 100

2054 152 102.04

2055 163 110.88

QN.8) From the following pay-off table, find the best strategy if i) Maximax criteria is applied ii)
Maximin criteria is applied.
Solution:

N1 N2 Pay-off Table Minimum pay-off
S1 200 50 N3 Maximum pay-off 40
S2 100 60 40 200 30
S3 40 30 30 100 10
10 40

34

i) Strategy S1 ii) Strategy S3

QN. 9) For eight pair of observations on two variables sales (X) and price (Y), the following result

were obtained,
ΣX = 156, ΣY = 132, ∑X2 = 4162, ∑Y2 = 2434, ΣXY = 2884

Find out if there is any relationship between sales and price.

Solution: Now, correlation coefficient between sales and price

r = √NΣX2 NΣXY − ΣXΣY − (ΣY)2
− (ΣX)2 √NΣY2

= √8 × 8 × 2884 − 156 × 132 − (132)2
4162 − (156)2 √8 × 2434

= 2480
√8960 √2048

= 2480
4283.7

= 0.5789
Therefore, positive relation between sales and price.

QN. 10) Find the adjoint matrix of the matrix given below: [ × = ]
[13 −72]

Solution: Now, cofactor of 1 = 7
cofactor of -2 = -3
cofactor of 3 = 2
cofactor of 7 = 1
Adjoint matrix of given matrix is [−73 12]

Group – B: Descriptive Question Answer

QN. 11 The average wages, standard deviation, and number of workers of two factories are
given below:

Average weekly wages Factory A Factory B
Standard Deviation Rs.4600 Rs.4900
Number of workers Rs.50 Rs.40
100 80

Calculate the mean and variance of weekly wages of all workers taken together. Which factory has
greater variability in the distribution of weekly wages? Justify your result with appropriate
statistical tool.
Solution: Let X and Y denotes the weekly wages of factory A and factory B respectively.

35

Then, ̅X = 4600, Y̅ = 4900, σX = 50, σY = 40, n1 = 100, n2 = 80

Now, average weekly wages of all workers

X̅12 = ̅Xn1 + ̅Yn2
n1 + n2

̅X12 = 4600 × 100 + 4900 × 80
100 + 80

̅X12 = 460000 + 392000
180

̅X12 = 852000
180

̅X12 = 4733.33

Again,

d1 = ̅X − ̅X12 = 4600 − 4733.33 = −133.33
and d2 = ̅Y − ̅X12 = 4900 − 4733.33 = 166.67

variance of weekly wages,

σ122 = 100(502 + 133.332) + 80(402 + 166.672)
100 + 80

σ122 = 2027688.89 + 2350311.11
100 + 80

σ122 = 4378000
180

σ122 = 24322.22

Coefficient of Variance of factory A

σX × 100% = 50 × 100% = 1.09%
X̅ 4600

Coefficient of Variance of factory B

σY × 100% = 40 × 100% = 0.82%
̅Y 4900

Since, C.V. of factory A is greater than factory, so factory A has greater variability.

QN. 12) Differentiate between "Census" and "sampling" method of data collection. Why sampling

method suitable to collect data from large population?

Solution: When we have to collect primary data for a statistical investigation, we must first decide

about the method to be used. Broadly speaking, there are two methods of collecting primary data;

a) Census method b) sampling method.

36

a) Census Method: In this method information is collected from each and every unit of the
population associated with the subject matter under inquiry. For example, when we have to
study the smoking habits of 10,000 adults in a city, we take into account each one of them and
obtain the required information. This is the way in which population census of a country in
conducted.
b) Sampling Method: In this method only a part of the population is taken into account and
conclusion is drawn from the information obtained in this way. For example, in order to
study to smoking habit of 10,000 adults, we may select 1000 persons out of them and find
out their habit. On the basis of the information obtained from them we can draw the
conclusion about the habit of the entire population.
Difference between census method and sampling method.

S.N. Basis of Census Sampling
Difference
The extensive enquiry is The limited enquiry is
1. Nature of conducted at each and every conducted as only a few units
enquiry unit of the population. of the population are studied.

2. Economy More Time, Money, and Less Time, Money, and
Labour It requires a large Labour Relatively less money,
amount of money, time, and time, and labour are required.
labour.

3. Suitability It is more suitable if the It is more suitable if the
population is heterogeneous population is homogeneous in
in nature. nature.

4. Reliability and The results are quite reliable The results of the sampling

Accuracy and accurate under the census method are less reliable

method. because a high degree of

accuracy is not achieved.

5. Organisation It is very difficult to organise The sampling method is

and Supervision and supervise the census comparatively easy to

method. organise and supervise.

6. Verification Under this method, the results Under this method, the results

of the investigation cannot be can be tested by taking out

verified. another small sample.

7. Nature of The census method is an old The sampling method is a new
method method of investigation and and practicable method. It is a
not a very scientific method. scientific method.

37

QN. 13) (a) Solve the following linear programming problem using graphical method.
Maximize: Z = 30x + 50y
subject to constraints: x + y ≤ 30, x + 2y ≤ 40, x ≥ 0, y ≥ 0
Solution: Corresponding equations of given inequalities are
x + y = 30 …………….. (i)
x + 2y = 40 ………….. (ii)
Now, from (i), we get
If x = 0, y = 30
If y = 0, x = 30
From (ii), we get
If x = 40, y = 20
If y = 0, x = 40

Vertices Calculation of Maximum Value Remark
O(0, 0) Minimum
A(30, 0) P = 30x + 50y
B(20, 10) P = 30 × 0 + 50 × 0 = 0 Maximum
C(0, 20)
P = 30 × 30 + 50 × 0 = 900
P = 30 × 20 + 50 × 10 = 1100
P = 30 × 0 + 50 × 20 = 1000

Therefore, minimum value cost is Rs.1100 at the point B(20, 10).
b) A manufacturing company has 1,000 employees. 10% of the employees earn less than Rs.500
per day, 200 earn between Rs.500 and Rs.999, 30% earn between Rs.1000 and Rs.1499, 250

38

employees earn between Rs.1500 and rs.1999 and rest earn Rs.2000 and above. Calculate the
suitable average wage. Also, give the reason for your choice of average.
Solution:

Calculation of Median

Earning (Rs.) No. of Employees c. f.

Less than 499.5 100 100

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

ow, Median = N th item = 1000 th item = 500
2 2

c.f. just greater or equal to 500 is 600, so the median class is 1000 – 15000.

then L = 1000, f = 300, c.f. = 300, h = 500

Median(Md) = L + N − cf × h
2 f

= 999.5 + 500 − 300 × 500
300

= 999.5 + 200 × 500
300

= 1000 + 333.333

=13332.83

As, we have given open ended class interval, so the median is suitable average.

QN. 14) Calculate the index number by using suitable formula for 1985 based on 1980
from the following information.

Year Product X Product Y Product Z
Price
Price Quantity Price Quantity 2 Quantity
10 4 5
1980 4 54 3 8 5

1985 10 40 8

Solution: Calculation Fisher's Index Number

2010 2011

39

Commodities p0 q0 P1 q1 p0q0 p0q1 p1q0 p1q1
X 4 54 10 40 216 160 540 400
Y 3 10 8 8 30 24 80 64
Z 2 5 4 5 10 10 20 20

256 194 640 484

Here,
Σp0q0 = 256, Σp0q1 = 204, Σp1q0 = 640, Σp1q1 = 484

P01(F) = √ΣΣpp10qq00 × Σp1q1 × 100
Σp0q1

P01(F) = √624506 × 484 × 10
194

= √2.5 × 2.4948× 100
= 249.74

15. a) Prove the following by using properties of determinants.

111
| a b c | = (a − b)(b − c)(c − a)(a + b + c)
a3 b3 c3

Solution: LHS

111
|a b c|
a3 b3 c3

010 b| [Applying, C2, C3 − C2]
=| a−b b c − b3 C1 → C1 − → C3
a3 − b3 b3 c3 −

= (a − b)(c − b) | 010
1 b 1|
a2 + ab + b2 b3 c2 + cb + b2

= −(a − b)(b − c)(−1)[(c2 + bc + b2) − (a2 + ab + b2)]

= (a − b)(b − c)[c2 + bc + b2 − a2 − ab − b2]

= (a − b)(b − c)[(c − a)(c + a) + b(c − a)]

= (a − b)(b − c)(c − a)(a + b + c)

= LHS

b) Solve the following equations by using matrix method

40

- x + 3y = 5, 2x – 4y = 0
Solution:

The matrix form of the given system is

(−21 −34) (yx) = (05)
which can be written as AX = B … … … … (i)
A = (−21 −34) , X = (yx) and B = (50)

Now,

|A| = |−21 −34| = 4 − 6 = −2
Since, |A| ≠ 0, so the system has unique solution.

Adj(A) = (−−42 −−13)

∴ A−1 = 1 Adj(A) = 1 (−−24 −−13) = 1 (42 31)
|A| −2 2

From equation (i), we have

X = A−1B

⇒ X = 1 (42 31) (50)
2

⇒ X = 1 (2100 + 00)
2 +

⇒ X = 1 (1200)
2
⇒ (yx) = (150)

∴ x = 10, y = 5

QN. 16) From the following data compute Bowley's coefficient of skewness and interpret the
result.

Income (00 Rs.) Below 200 200 - 400 400 - 600 600 - 800 800 - 1000 1000 & above
No. of families 25 40 80 75 20 16
Solution:

Computation of Bowley's Coefficient of Skewness

Income (00 Rs.) No of families (f) c. f.
Below 200 25 25
200 – 400 40 65

41

400 – 600 80 145
600 – 800 75 220
800 – 1000 20 240
1000 & above 16 256
N = 256

For, Q1 = N th item = 256 th item = 64
4 4

c.f. just greater or equal to 64 is 65, so the median class is 200 - 400.

then L = 200, f = 40, c.f. = 25, h = 200

First Quartile(Q1) = L + N − cf × h
4 f

= 200 + 64 − 25 × 200
40

= 200 + 39 × 200
40

= 200 + 195

=395

Median = N th item = 256 th item = 128
2 2

c.f. just greater or equal to 128 is 145, so the median class is 400 - 600.

then L = 400, f = 80, c.f. = 145, h = 200

Median(Md) = L + N − cf × h
2 f

= 400 + 128 − 65 × 200
80

= 400 + 63 × 200
80

= 400 + 157.5

= 557.5

For, Q3 = 3N th item = 3 × 256 th item = 192
4 4

c.f. just greater or equal to 192 is 220, so the median class is 600 - 800.

42

then L = 600, f = 75, c.f. = 145, h = 200

Third Quartile(Q3) = L + 3N − cf × h
4 f

= 600 + 192 − 145 × 200
75

= 600 + 47 × 200
75

= 600 + 125.33

= 725.33

Bowley′s Coefficient of Skewness (SKB) = Q3 + Q1 − 2Md
Q3 − Q1

= 725.33 + 395 − 2 × 557.5
725.33 − 395

= 5.33
330.33

= 0.016

Therefore, the frequency is positively skewed. [ × = ]
Group – C: Analytic Question Answer

QN. 17) The following table presents the annual trends of net profit of two different companies
seeking investment for their development project. As an investment advisor, in which company
would you suggest to invest money. Justify your answer by using necessary statistical tools.

Solution: Let the annual profit of company A and B be respectively denoted by Y1 and Y2.
Suppose, X1 = X – 2010

Calculation of various sums

Year X1 X12 Net Profit (Company A) Net profit (Company B)
(X) Y1 Y12 X1Y1 Y2 Y22 X1Y2
2007 -3 9 16 256 - 48 16 256 - 48
2008 -2 4 32 1024 - 64
2009 -1 1 40 1600 - 40 16 256 -32
2010 0 0 24 576 0
2011 1 1 40 1600 40 22 484 - 22
2012 2 4 32 1024 64
2013 3 9 88 7744 264 36 1296 0
Total 0 28 272 13824 216
40 1600 40

44 1936 88

48 2304 144

222 8132 170

43

Here, N = 7, ∑X1 = 0, ∑X12 = 28, ∑Y1 = 272, ∑Y12 = 13824, ∑X1Y1 = 216
∑Y2 = 222, ∑Y22 = 8132, ∑X1Y2 = 170

Now, For Company A

Averageprofit(̅Y1) = ∑Y1 = 272 = 38.86
N 7

Standard deviation (σ1) = √∑NY12 − (∑NY1)2

= √138724 − (2772)2

= √1974.86 − 1503.10

= √471.75

= 21.72

Coefficient of Variance = σ1 × 100% = 21.72 × 100% = 55.89%
̅Y1 38.86

Regression Coefficient = ∑X1Y1 = 216 = 7.71
∑X12 28

Again, For Company B

Averageprofit(̅Y2) = ∑Y2 = 222 = 31.71
N 7

Standard deviation (σ2) = √∑NY22 − (∑NY2)2

= √81732 − (2272)2

= √1161.71 − 1005.52

= √156.19

= 12.5

Coefficient of Variance = σ2 × 100% = 12.5 × 100% = 39.42%
Y̅2 31.71

44

Regression Coefficient = ∑X1Y2 = 170 = 6.07
∑X12 28

Since, both C.V. and regression coefficient of company A is greater than
company B. Therefore, based on the growth rate of net profit company A is better but
based on consistency in profit company B is better.

QN. 18) From the following bi-variate frequency table, find out if there exists any
relationship between advertisement expenditure (in 00 Rs.) and sales revenue (in 000 Rs.)
and test the significance of the result. Also estimate the sales revenue when advertisement
expenditure is Rs.40,000.

Advertisement Sales Revenue (in 000 Rs. )

Expenditure 0 - 50 50 - 100 100 - 150 150 - 200 200 - 250

(in 00 Rs.) 12 6 8 - -
5 1
0 – 40 2 18 4 2 4
40 – 80 2 1
80 – 120 - 8 10 2 3
120 – 160
160 - 200 - 1 10

--1

Solution: Let X denote the mid-point of advertise expenditure and Y denotes mid-point of sales

revenue then =x − 100 y − 125
40 50
dx , and dy =

Calculation of Various Sums

45

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

= 100 × 94 − (−56) × (−32)
√100 × 172 − (−56)2 √100 × 136 − (−32)2

= 7608
√17200 − 3136√13600 − 1024

= 7608
√14064√12576

= 7608
118.59 × 112.14

= 7608
13298.68

= 0.572

Now, P. E. = 0.6745 1 − r2
√N

= 0.6745 1 − (0.572)2
√100

= 0.6745 1 − 0.3272
10

= 0.6745 0.6728
10

= 0.045
Again,

0.6× P.E. = 0.6× 0.045 = 0.027
Since, r > 0.6 × r, the value of r is significance.

Regression coefficient y on x

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

= 100 × 94 − (−32) × (−56) × 50
100 × 172 − (−56)2 40

= 9400 − 1792 × 5
17200 − 3136 4

46

= 7608 × 5
14064 4

= 0.54 × 1.25

= 0.68

X̅ = A + Σfdx × h = 100 + (− 15060) × 40 = 100 − 22.4 = 77.6
N

̅Y = B + Σfdy × h = 125 + (− 13020) × 50 = 125 − 16 = 109
N

The regression equation Y on X is
Y − Y̅ = byx(X − ̅X)
⇒ Y − 109 = 0.68(X − 77.6)
⇒ Y − 109 = 0.68X − 52.77
⇒ Y = 0.68X − 52.77 + 109
⇒ Y = 0.68X + 56.23

Finally, when x = 400
Y = 0.68 × 400 + 56.23 = 328.23

Therefore, sales revenue when expenditure in advertisement Rs.40,00 is Rs.32823.

QN. 19) Under the employment promotion program, it is proposed to allow the sale of newspaper

on the business during peak hours. A newspaper boy has the following probability of selling

magazine.

No of copies sold 10 11 12 13 14

Probability 0.10 0.15 0.20 0.25 0.30

Cost per copy of magazine is Rs.30 and sales price per copy is Rs.50. He cannot return unsold
copies where salvage value is zero.
a) Calculate the expected monitory value (EMV) for each strategy.
b) How many copies should be ordered?
c) Compute expected profit with perfect information (EPPI)
d) Also calculate expected value of perfect information (EVPI).
Solution:

a. Payoff table
Since the company would not sell less than 10 bottles and not more than 14 bottles of keys, it
has six strategies (alternatives) such as S1 = 10, S2 = 11, S3 = 12, S4 = 13, S5 =
14. Similarly there are three possible state of nature i.e. the demand of keys such as
N1 = 10, N2 = 11, N3 = 12, N4 = 13, N5 = 14.
Here, SP= Rs.50, CP= Rs. 30, MP = Rs.20

State of nature (demand of keys)

47

Strategies N1=10 N2=11 N3=12 N4=13 N5=14
S1=10 200 200 200 200 200
S2=11 170 220 220 220 220
S3=12 140 190 240 240 240
S4=13 110 160 210 260 260
S5=14 80 130 180 230 280

b. Expected monetary value (EMV) for each strategy.

Payoff Values (Rs.) Expected payoff (Rs)

Strategies Prob. Decision Alternatives Decision Alternative
(1)
S1=10 S2=11 S3=12 S4=13 S5=14 S1=10 S2=11 S3=12 S4=13 S5=14

(2) (3) (4) (5) (6) (1x2) (1x3) (1x4) (1x5) (1x6)

D1=10 0.10 200 170 140 110 80 20 17 14 11 8

D2=11 0.15 200 220 190 160 130 30 33 28.5 24 19.5

D3=12 0.20 200 220 240 210 180 40 44 48 42 36

D4=13 0.25 200 220 240 260 230 50 55 60 65 57.5

D5=14 0.30 200 220 240 260 280 60 66 72 78 84

Expected Monetary value (EMV)(Rs) 200 215 222.5 220 205

Therefore, the optimal quantity that can maximize the expected profit is 12 units.

d. Expected profit with prefect information;
EPPI = Rs. (0.10 × 200 + 0.15× 220 + 0.20 × 240 + 0.25 ×260 +0.30× 280)

= Rs. (20 + 33 + 48 + 65 + 84) = Rs.250
e. Expected value of the perfect information (EVPI)

= EPPI – EPV
= Rs.250 - Rs.222.50
= Rs.27.50

48

Model Question -2070 [ × = ]
Group – A: Brief Question Answer

QN.1) The mean of 200 items was found to be 80, letter it was found that 61 and 45 were
misread as 16 and 15. Find the correct mean.

Solution: Here given

Number of items (n) = 200
mean (̅X) = 80

Incorrect values = 61 and 45

Correct values = 16 and 15

Now, we have

̅X = ∑X
n

80 = ∑X
200

∴ ΣX = 16000

Now, Correct ΣX = 16000 − 61 − 45 + 16 + 15 = 15925

Again, Correct mean = Correct ∑X = 15925 = 79.63
n 200

QN. 2) The following result were obtained:

Coefficient of variation = 50% Karl Person's coefficient of skewness = 0.5

Standard deviation = 2

Find mean and mode.

Solution: Now, we have

Coefficient of variation = σ × 100%
X̅

⇒ 50% = 2 × 100%
X̅

⇒ ̅X = 2 × 10 = 4
50

Again, Person′s coefficient of skewness = X̅ − M0
σ

⇒ 0.5 = 4 − M0
2

⇒ M0 = 4 − 1 = 3

49