Business Statistics Solved Paper FBISE 2022 Annual ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2022 Annual ICOM II, MCQS, Short Questions, Extensive Questions
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In this post, I am going to discuss the paper of Business Statistics Solved Paper FBISE 2022 Annual ICOM II, MCQS, Short Questions, Extensive Questions. Topics included are Introduction to Statistics, Averages, Index Numbers, Probability. Topics discovered in this solved paper are also beneficial for the students of BCOM, ADC, BBA and other disciplines in which business statistics is included in the syllabus. Stay connected for more papers. Links of solved papers of Business Statistics and Principles of Accounting are given below.

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For other Solved Papers of Business Statistics and Principles of Accounting, please follow links below:

Solved Paper Business Statistics

Solved Papers Principles of Accounting II

Solved by Iftikhar Ali, M.Sc Economics, MCOM Finance Lecturer Statistics, Finance and Accounting

Table of Contents

Business Statistics Solved Paper FBISE 2022 Annual ICOM II, MCQS, Short Questions, Extensive Questions

MCQS

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Q 1: Fill the relevant bubble against each question. 

1The word “statistics” may have been derived from a Latin word:
A)StatistikB)Status
C)StatistaD)Statistique
2A data undergone through any statistical treatment at-least once is called:
A)Primary DataB)Secondary Data
C)Ungrouped DataD)Discrete Data
3The graph of frequency distributions is called:
A)HistogramB)Ogive
C)HistorigramD)Pie-Chart
4The process of systematic arrangement of data into rows and columns is called:
A)ClassificationB)Tabulation
C)GroupingD)Manipulation
5Sum of deviations from mean is always:
A)ZeroB)Positive
C)NegativeD)One
6Which of the following average is the most suitable for computing chain indices?
A)MeanB)Median
C)ModeD)Geometric Mean
7Price relative is the percentage ratio of current year price and:
A)Previous year priceB)Base year price
C)Preceding year priceD)Next year price
8Laspeyre’s index number is also called:
A)Base year weighted indexB)Current year weighted index
C)Ideal Index numbersD)Good index number
9A card is drawn from pack of 52 playing cards. What is probability of king of hearts?
A)1/52B)4/52
C)13/52D)2/52
10If 4 coins are tossed, the possible outcomes of sample space are:
A)4B)8
C)16D)32
Business  Statistics Solved Paper FBISE 2022 Annual ICOM II, MCQS, Short Questions, Extensive Questions
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Short Questions

SECTION-B (Marks 24)

Q.2: Attempt any eight parts. The answer to each part should not exceed 3 to 4 line. (8 x 3=24)

(i) Differentiate between population and sample.

Answer

Population Vs Sample

Whole group under discussion is called population for example the whole strength of college student or whole population of a certain district etc.

Whereas part of population is called sample for example 30 students out of 100 students or 300 persons out of district population etc.

(ii) The following are the number of children per family in a town obtained from sample survey.

5,8,2,1,5,2,4,6,3,3,4,7,5,6,4,2,1,3,5,4,6,4,5,7,3,5,4,4,5,3,2

Make a discrete frequency distribution fro number of children taking one as class interval.

Solution:

Xf
12
24
35
47
57
63
72
81
 ∑f = 31

(iii) Calculate median and mode from the following discrete frequency distribution.

X123456
f369762

Solution:

XfC.F
133
269
3918
4725
5631
6233
 ∑f =n= 33 

Median

16.5 falls in C.F of 18 so X = 3 is median

Mode

Maximum frequency is 9 so X = 3 is Mode

(iv) A student obtained the following marks in the following subjects:

SubjectsUrduEnglishMathematicsAccountingCommerce
Marks (X)7065809072
Weights (W)21453

Find the weighed arithmetic mean.

Solution:

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SubjectsMarks (X)Weights (W)WX
Urdu702140
English65165
Mathematics804320
Accounting905450
Commerce723216
  ∑W  =15∑WX  =1191

    \[ \mathbf{Weighted\ Mean =}\frac{\mathbf{\sum WX}}{\mathbf{\sum W}}\mathbf{=}\frac{\mathbf{1191}}{\mathbf{15}}\mathbf{= 79.4}\ \]

(v) For a frequency distribution, it is given that D = X – 500, ∑fD =80, ∑f = 10. Find Arithmetic Mean.

Solution:

Data

A= 500, ∑f = 10, ∑fD =80

    \[ A.M\ \overline{X} = A + \ \frac{\sum fD}{\sum f}\ \]

    \[ A.M\ \overline{X} = 500 + \ \frac{80}{10} = 508\ \]

(vi) If arithmetic mean is 40 and median is 36 of a frequency distribution. Using empirical relationship find the value of mode.

Solution:

Mode = 3 Median – 2 Mean

Mode = 3(36) – 2(40)

Mode = 108 – 80

Mode = 28

(vii) Differentiate between fixed base and chain base method.

Answer

In fixed base method, base period remains fixed whereas in chain base method base period does not remain fixed. On the other hand in fixed base method, we calculate price relative whereas in chain base method, we calculate link relative and then we calculate chain index.

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{Pn}}{\mathbf{Po}}\mathbf{\times}\mathbf{100}\ \]

    \[ \mathbf{Link\ Relative = \ }\frac{\mathbf{Pn}}{\mathbf{Pn - 1}}\mathbf{\ }\mathbf{\times}\mathbf{\ 100}\ \]

(viii) Calculate chain indices from the following prices:

Year199019911992199319941995
Price404548505570

Solution:

YearsPriceLink RelativeChain Index
1990404040
199145 

    \[ \frac{45}{40} \times 100 = 112.5\ \]

 

    \[ \frac{40 \times 112.5}{100} = 45\ \]

199248 

    \[ \frac{48}{45} \times 100 = 106.67\ \]

 

    \[ \frac{45 \times 106.67}{100} = 48\ \]

199350 

    \[ \frac{50}{48} \times 100 = 104.16\ \]

 		 

    \[ \frac{48 \times 104.16}{100} = 50\ \]

 
199455 

    \[ \frac{55}{50} \times 100 = 110\ \]

 		 

    \[ \frac{50 \times 110}{100} = 55\ \]

 
199570 

    \[ \frac{70}{55} \times 100 = 127.27\ \]

 

    \[ \frac{55 \times 127.27}{100} = 70\ \]

 

(ix) If Laspeyre’s index = 120, Fisher’s index = 118, then find Paasche’s index number.

Solution:

    \[ Fisher^{'}s\ Index = \ \sqrt{L \times P}\ \]

    \[ 118 = \ \sqrt{120}\ \times \ \sqrt{P}\ \]

    \[ 118 = \ 10.95\ \times \ \sqrt{P}\ \]

    \[ \sqrt{P} = \frac{118}{10.95}\ \]

    \[ \sqrt{P} = 10.77\ \]

Take Square root on both sides:

    \[ P = {(10.77)}^{2}\ \]

    \[ Paasche^{'}s\ Index = 116\ \]

(x) Describe (a) Random Experiment (b) Sample Space

Answer:

(a) Random Experiment

Random experiment is an experiment that is performed in probability distribution in order to know the probability of a particular event.

(b) Sample Space

All possible outcomes of a random experiment are called sample space. Sample space is denoted by Upper Case Letter S.

(xi) An unbiased dice is rolled once. What is the probability of (i) Even Number (ii) Number greater than 2?

Solution

    \[ \mathbf{S}ample\ Space\ = \ \eta(s) = 6^{1} = 6\ \]

All possible outcomes

1, 2, 3, 4, 5, 6

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Events

    \[ (i)\ Even\ Number\ \eta(A) = 3\ \]

    \[ (ii)\ Greater\ than\ 2\ \eta(B) = 4\ \]

Probabilities

    \[ (i)P(A) = \frac{\eta(A)}{\eta(S)} = \frac{3}{6} = 0.5\ \]

    \[ (ii)P(B) = \frac{\eta(B)}{\eta(S)} = \frac{4}{6} = 0.67\ \]

Extensive Questions

Section C (Marks 16)

Note: Attempt any two questions. All questions carry equal marks. (2×8=16)

Q.3: Find Mean, Median and Mode from the following distribution:

Class5—910—1415—1920—2425—2930—3435—39
Frequency23710521

Solution

Class LimitsFrequency (f)Class BoundariesXfXC.f
5—924.5—9.57142
10—1439.5—14.512365
15—19714.5—19.51711912
20—241019.5—24.52222022
25—29524.5—29.52713527
30—34229.5—34.5326429
35—39134.5—39.5373730
Sum30  625 
 ∑f or n=  ∑fX= 

    \[ \left( \mathbf{i} \right)\mathbf{\ A.M}\overline{\mathbf{X\ }}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{625}}{\mathbf{30}}\mathbf{= 20.83}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ Median = L +}\frac{\mathbf{h}}{\mathbf{f}}\mathbf{(}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C)}\ \]

Model Class = n/2 = 30/2 = 15 falls in C.f of 22 so data is:

L = 19.5, h = 5, f = 10, n/2 =30/2 = 15 & C = 12

    \[ \mathbf{Median = L +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C} \right)\ \]

    \[ \mathbf{Median = 19.5 +}\frac{\mathbf{5}}{\mathbf{10}}\left( \mathbf{15 - \ 12} \right)\ \]

    \[ \mathbf{Median = 19.5 +}\frac{\mathbf{15}}{\mathbf{10}}\ \]

    \[ \mathbf{Median = 21}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{\ Mode = L +}\frac{\mathbf{fm - f}\mathbf{1}}{\left( \mathbf{fm - f}\mathbf{1} \right)\mathbf{+ \ (fm - f}\mathbf{2)}}\mathbf{\times \ h}\ \]

Model Class: Maximum frequency is 10 so data for model class is:

L = 19.5, fm = 10, f1 = 7, f2 = 5 & h = 5

    \[ \mathbf{Mode = 19.5 +}\frac{\mathbf{10 - 7}}{\left( \mathbf{10 - 7} \right)\mathbf{+ \ }\left( \mathbf{10 - 5} \right)}\mathbf{\times \ 5}\ \]

    \[ \mathbf{Mode = \ 19.5 +}\frac{\mathbf{15}}{\mathbf{8}}\ \]

    \[ \mathbf{Mode = 21.375}\ \]

Q.4: Calculate index numbers from the following data for 2005 taking 2000 as base year by: (i) Laspeyre’s Method (ii) Paasche’s Method (iii) Fisher’s Method

Commodities20002005
PriceQuantityPriceQuantity
A520830
B210515
C425630
D8301240
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Solution:

  Items20002005 
Price (Po)Quantity (qo)Price   (P1)Quantity (q1)poqop1qop1q1poq1
A520830100160240150
B21051520507530
C425630100150180120
D8301240240360480320
Sum460720975620
 ∑poqo =∑p1qo =∑p1q1 =∑poq1 =

    \[ (i)Laspeyre^{'}s\ Index\ 2005 = \frac{\sum p_{1}q_{0}}{\sum p_{0}q_{0}} \times \ 100\ \]

    \[ Laspeyre^{'}s\ Index\ 2005 = \frac{720}{460}\ \times \ 100 = \ 156.52\ \]

    \[ (ii)Paasche^{'}s\ Index\ 2005 = \frac{\sum p_{1}q_{1}}{\sum p_{0}q_{1}} \times \ 100\ \]

    \[ Paasche^{'}s\ Index\ 2005 = \frac{975}{620} \times \ 100 = 157.25\ \]

    \[ (iii)Fisher^{'}s\ Ideal\ Index\ 2005 = \sqrt{L \times P}\ \]

    \[ Fisher^{'}s\ Ideal\ Index\ 2005 = \sqrt{156.52 \times 157.25}\ \]

    \[ Fisher^{'}s\ Ideal\ Index\ 2005 = 156.88\ \]

Q.5: Three balls are drawn at random from a bag containing 6 red and 4 black balls. Find the probability that:

(i) All balls are red

(ii) All balls are of same colour

(iii) Two red and one black ball

(iv) No red ball

Solution

Red Balls = 6, Black Balls = 4, N = 6+4 = 10, r = 3

    \[ \mathbf{Sample\ Space\ = \ }\mathbf{\eta}\left( \mathbf{S} \right)\mathbf{= \ }\begin{pmatrix}\mathbf{N} \\\mathbf{C} \\\mathbf{r} \\\end{pmatrix}\mathbf{=}\begin{pmatrix}\mathbf{10} \\\mathbf{C} \\\mathbf{3} \\\end{pmatrix}\mathbf{= 120}\ \]

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Events

    \[ (i)All\ balls\ are\ Red\ \mathbf{\eta}\left( \mathbf{A} \right) = \begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{3} \\\end{pmatrix}\begin{pmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{0} \\\end{pmatrix}\mathbf{= 20 \times 1 = 20}\ \]

    \[ (ii\ a)All\ balls\ are\ Red\ \mathbf{\eta}\left( \mathbf{B} \right) = \begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{3} \\\end{pmatrix}\begin{pmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{0} \\\end{pmatrix}\mathbf{= 20 \times 1 = 20}\ \]

    \[ (ii\ b)All\ balls\ are\ Black\ \mathbf{\eta}\left( \mathbf{C} \right) = \begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{0} \\\end{pmatrix}\begin{pmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{3} \\\end{pmatrix}\mathbf{= 1 \times 4 = 4}\ \]

    \[ (iii)Two\ Red\ One\ Black\ \mathbf{\eta}\left( \mathbf{D} \right) = \begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{2} \\\end{pmatrix}\begin{pmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{1} \\\end{pmatrix}\mathbf{= 15 \times 4 = 60}\ \]

    \[ (iv)No\ Red\ \mathbf{\eta}\left( \mathbf{E} \right) = \begin{pmatrix}\mathbf{6} \\\mathbf{C} \\\mathbf{0} \\\end{pmatrix}\begin{pmatrix}\mathbf{4} \\\mathbf{C} \\\mathbf{3} \\\end{pmatrix}\mathbf{= 1 \times 4 = 4}\ \]

Probabilities

    \[ \left( \mathbf{i} \right)\mathbf{\ P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{20}}{\mathbf{120}}\mathbf{= 0.167}\ \]

    \[ \left( \mathbf{ii\ a} \right)\mathbf{\ P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{B} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{20}}{\mathbf{120}}\mathbf{= 0.167}\ \]

    \[ \left( \mathbf{ii\ b} \right)\mathbf{\ P}\left( \mathbf{C} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{C} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{4}}{\mathbf{120}}\mathbf{= 0.033}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{\ P}\left( \mathbf{D} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{D} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{60}}{\mathbf{120}}\mathbf{= 0.5}\ \]

    \[ \left( \mathbf{iv} \right)\mathbf{\ P}\left( \mathbf{E} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{E} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{4}}{\mathbf{120}}\mathbf{= 0.033}\ \]

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