Business Statistics Solved Paper FBISE 2015 ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2015 ICOM II, MCQS, Short Questions, Extensive Questions
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Business Statistics Solved Paper FBISE 2015 ICOM II, MCQS, Short Questions, Extensive Questions

In this blog post, I am going to discuss the paper of Business Statistics Solved Paper FBISE 2015 ICOM II, MCQS, Short Questions, Extensive Questions . Topics included are Introduction to Statistics, Averages, Index Numbers, Probability.

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

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MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.

(i)A measure computed from sample data is:
 A. ParameterB. StatisticC. StatisticsD. Data 
      
(ii)Number of Farz in five prayers is an example of:
 A. Discrete VariableB. ConstantC. AttributeD. None of these 
      
(iii)A sector diagram is also called:
 A. Bar DiagramB. HistogramC. Pie DiagramD. Historigram 
      
(iv)In a relative frequency distribution, the total of relative frequencies is equal to:
 A. 100B. 1C. ∑fD. 0 
      
(v)For a certain distribution, if ∑(x – 20) = 25, : ∑(x – 25) = 15 and ∑(x – 35) = 0, then X̅  is equal to:
 A. 20B. 25C. -35D. 35 
      
(vi)The sum of deviation is zero, when deviations are taken from:
 A. MedianB. MeanC. ModeD. Geometric Mean 
      
(vii)If Laspayer’s Index = 110, Paasche’s Index = 108, then Fisher ideal Index is:
 A. 110B. 109C. 100D. None of these 
      
(viii)What is called a number that measures a relative change in a single variable with respect to a base?
 A. Good Index NumberB. Quantity Index NumberC. Simple Index NumberD. Composite Index Number 
      
(ix)Two coins are tossed. Probability of getting head on first coin is:
 A. 2/4B. 1C. 0D. 4 
      
(x)Two events A and B are said to be mutually exclusive if:
 A.

    \[A \cup B=\varnothing\]

B.

    \[ A \cap B = S\ \]

C.

    \[A \cap B = \varnothing\]

D.

    \[ A \cap B = 1\ \]

 

Short Questions

SECTION-B (Marks 24)

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

(i) Define Descriptive & Inferential Statistics?

Answer: Descriptive statistics deals with collection and presentation of data in various forms, such as tables, graphs and diagrams and findings averages and other measures of data.

Inferential statistics deals with the testing of hypothesis and inference about population parameter is called Inferential Statistics.

(ii) Define Histogram, Historigram and Tabulation.

Answer

Histogram is a graph of frequency distribution in which class boundaries with suitable width is taken on X-axis whereas respective frequencies are taken on Y-axis. In Histogram relative frequencies are shown in the shape of adjacent rectangular bars.

Whereas graph of Time Series or historical series is called historigram.

 Tabulation is the arrangement and presentation of the data in the form of columns and rows.

(iii) Represent the following data by Pie-Charts?

DistrictsLahoreMultanRawalpindiGujrat
Area50115135165
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Solution

DistrictsLahoreMultanRawalpindiGujratSum
Area50115135165465
Sector

    \[ \frac{\mathbf{50}}{\mathbf{465}}\mathbf{\ x\ 360 = 39}\ \]

    \[ \frac{\mathbf{115}}{\mathbf{465}}\mathbf{\ x\ 360 = 89}\ \]

    \[ \frac{\mathbf{135}}{\mathbf{465}}\mathbf{\ x\ 360 = 104}\ \]

    \[ \frac{\mathbf{165}}{\mathbf{465}}\mathbf{\ x\ 360 = 128}\ \]

  360
Business Statistics Solved Paper FBISE 2015 ICOM II, MCQS, Short Questions, Extensive Questions
Pie Chart

(iv) In a moderately skewed distribution, the value of median is 42.8 and the value of mode is 40, find mean.

Solution

Mode = 3 Median – 2 Mean

40 = 3(42.8) – 2(Mean)

40 = 128.4 – 2(Mean)

2(Mean) = 128.4 – 40

2(Mean)  = 88.4

Mean = 88.4/2

Mean = 44.2

Answer  Price Relative is used in fixed base method of Index Number in which each current price is divided by base period price which remains fixed. The formula for price relative is given below:

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

  where Pn is current period price & Po is base period price.

Link Relative is used in chain index in which each period price is divided by previous time period price. The formula for Link relative is given below:

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

    \[ where\ Pn\ is\ current\ period\ price\ \&\ Pn - 1\ is\ previous\ time\ period\ price.\ \]

(vi) Given W= 20, 25, 30, 40 and I = 100, 105, 110 & 120. Find weighted average of relative Index Number

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Answer

WIWI
201002000
251052625
301103300
401204800
115 12725
∑W= ∑WI=

    \[ \mathbf{Weighted\ Index = \ }\frac{\mathbf{\sum WI}}{\mathbf{\sum W}}\mathbf{= \ }\frac{\mathbf{12725}}{\mathbf{115}}\mathbf{= \ 110.65}\ \]

(vii) The deviation about X = 180 are 4, 11, -8, -12, 7, 9, 16, 9, 13, 15. Calculate Arithmetic Mean.

Solution

D = X -180411-8-12791691315∑D =64

A = 180, n = 10, ∑D = 64

    \[ \overline{\mathbf{X}}\mathbf{= A + \ }\frac{\mathbf{\sum D}}{\mathbf{n}}\mathbf{= 180 + \ }\frac{\mathbf{64}}{\mathbf{10}}\mathbf{= 180 + 6.4 = 186.4}\ \]

(viii) Describe the qualities of a good average?

Answer

(1) It should be easy to calculate and simple to understand.

(2) It should be clearly defined by a mathematical formula.

(3) It should not be affected by extreme values.

(4) It should be based on all the observations.

(5) It should be capable of further mathematical treatment.

(6) It should have sample stability.

(ix) If three coins are tossed what is the probability of getting: (a)  At least two heads (b) Two tails

Solution

    \[ \textbf{Sample Space =}\mathbf{\eta}\left( \mathbf{s} \right)\mathbf{=}\mathbf{2}^{\mathbf{2}}\mathbf{= 4}\ \]

All possible outcomes

HHHHHTHTHTHH
TTTTTHTHTHTT

Events

    \[ \left( \mathbf{i} \right)\mathbf{\ At\ least\ two\ heads\ \eta}\left( \mathbf{A} \right)\mathbf{=}\mathbf{4}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ Two\ Tails\ \eta}\left( \mathbf{B} \right)\mathbf{=}\mathbf{3}\ \]

Probabilities

    \[ \left( \mathbf{i} \right)\mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{8}}\mathbf{= 0.5}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{8}}\mathbf{= 0.375}\ \]

(x) Suppose P(A) = ¼, P(B) = 1/3 and P(AUB)= ½. Determine P = (A \cap B)

Solution

Since A & B are not mutually exclusive events therefore

    \[ \mathbf{P}\left( \mathbf{AUB} \right)\mathbf{= \ P}\left( \mathbf{A} \right)\mathbf{+ \ P}\left( \mathbf{B} \right)\mathbf{- \ P(A \cap B)}\ \]

    \[ \frac{\mathbf{1}}{\mathbf{2}}\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+ \ }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{- \ P(A \cap B)}\ \]

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    \[ \mathbf{P}\left( \mathbf{A \cap B} \right)\mathbf{= \ }\frac{\mathbf{1}}{\mathbf{4}}\mathbf{+ \ }\frac{\mathbf{1}}{\mathbf{3}}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\ =}\frac{\mathbf{1}}{\mathbf{12}}\ \]

(xi) Given X = 10 + 5U, ∑fU = 308, ∑f = 7. Find A.M

Solution

Here ∑fU = 308, ∑f = 7, A = 10 & h = 5

    \[ \overline{\mathbf{X}}\mathbf{= A + \ }\frac{\mathbf{\sum fu}}{\mathbf{\sum f}}\mathbf{x\ h = 10 + \ }\frac{\mathbf{308}}{\mathbf{7}}\mathbf{x\ 5 = 10 + 220 = 230}\ \]

Extensive Questions

Section C (Marks 16)

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

Q.3 Calculate Mean, Median Mode from the following data:

Max. loadNo. of Cables
9.3—9.72
9.8—10.25
10.3—10.712
10.8—11.217
11.3—11.714
11.8—12.26
12.3—12.73
12.8—13.21

Solution

Max. loadNo. of CablesClass BoundariesXfXC.f
9.3—9.729.25–9.759.5192
9.8—10.259.75–10.2510507
10.3—10.71210.25–10.7510.512619
10.8—11.21710.75–11.251118736
11.3—11.71411.25–11.7511.516150
11.8—12.2611.75–12.25127256
12.3—12.7312.25–12.7512.537.559
12.8—13.2112.75–13.25131360
Sum60  665.5 
 ∑f=  ∑fX= 
      

    \[ \left( \mathbf{i} \right)\mathbf{\ A.M}\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{665.5}}{\mathbf{60}}\mathbf{= 11.09}\ \]

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

Model Class = n/2 = 60/2 = 30 falls in C.f of 36 so data is:

L = 10.75, h = 0.5, f = 17, n/2 =60/2 = 30 & C = 19

    \[ Median\;=L+\frac hf\left(\frac n2-C\right) \]

    \[ Median\;=10.75+\frac{0.5}{17}\left(\frac{60}2-19\right) \]

    \[ Median\;=10.75+\frac{5.5}{17} \]

    \[ Median\;=11.07 \]

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

Model Class: Maximum frequency is 17 so data for model class is:

L = 10.75, fm = 17, f1 = 12, f2 = 14 & h = 0.5

    \[ Mode\;=10.75+\frac{(17-12)}{(17-12)+(17-14)}\times0.5 \]

    \[ Mode\;=10.75+\frac5{5+3}\times0.5 \]

    \[ Mode\;=10.75+\frac{2.5}8 \]

    \[ Mode\;=11.06 \]

Q.4 Construct Price Index number for 2000 on the basis of 1990.

(i) Base Year Weighted Method

(ii) Current Year Weighted Method

(iii) Fisher Ideal Index

  Items19902000
PriceQuantityPriceQuantity
A21048
B4556
C58610
D320325

Solution

Article19902000 
Price (Po)Quantity (qo)Price   (P1)Quantity (q1)poqop1qop1q1poq1
A2104820403216
B455620253024
C5861040486050
D32032560607575
Sum140173197165
 ∑poqo =∑p1qo =∑p1q1 =∑poq1 =

    \[ \left( \mathbf{i} \right)\mathbf{Base\ Year\ Weighted\ }\mathbf{Index\ 2000}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{0}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{0}}}\mathbf{x\ 100}\ \]

    \[ \mathbf{Base\ Year\ Weighted\ }\mathbf{Index\ 2000}\mathbf{=}\frac{\mathbf{173}}{\mathbf{140}}\mathbf{\ x\ 100 = \ 123.57}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{Current\ Year\ Weighted\ }\mathbf{Index\ 2000}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{1}}}\mathbf{x\ 100}\ \]

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    \[ \mathbf{Current\ Year\ Weighted\ }\mathbf{Index\ 2000}\mathbf{=}\frac{\mathbf{197}}{\mathbf{165}}\mathbf{x\ 100 = 119.39}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2000 =}\sqrt{\mathbf{LxP}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2000 =}\sqrt{\mathbf{123.57}\mathbf{x}\mathbf{119.39}}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 2000 = 121.46}\ \]

Q.5 (a) A class contains 10 men and 20 women of which half of the men and half of the women have brown eyes. Find probability that a person chosen at random is a man or has brown eyes.

Solution

 Brown EyesDo Not Have Brown EyesTotal
Man5510
Women101020
Total151530

    \[ \mathbf{Here\ P}\left( \mathbf{Man} \right)\mathbf{= \ P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{10}}{\mathbf{30}}\mathbf{,\ }\mathbf{P}\left( \mathbf{Brown\ Eyes} \right)\mathbf{= \ P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{15}}{\mathbf{30}}\mathbf{\ }\&\ P\left( \mathbf{A \cap B} \right)\mathbf{=}\frac{\mathbf{10}}{\mathbf{30}}\ \]

    \[ \mathbf{P}\left( \mathbf{Man\ or\ has\ Brown\ Eyes} \right)\mathbf{= P}\left( \mathbf{AUB} \right)\mathbf{= \ P}\left( \mathbf{A} \right)\mathbf{+ \ P}\left( \mathbf{B} \right)\mathbf{- \ P(A \cap B)}\ \]

    \[ \mathbf{P}\left( \mathbf{Man\ or\ has\ Brown\ Eyes} \right)\mathbf{= P}\left( \mathbf{AUB} \right)\mathbf{= \ }\frac{\mathbf{10}}{\mathbf{30}}\mathbf{+ \ }\frac{\mathbf{15}}{\mathbf{30}}\mathbf{- \ }\frac{\mathbf{10}}{\mathbf{30}}\mathbf{=}\frac{\mathbf{15}}{\mathbf{30}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{= 0.5}\ \]

 (b) A dice is rolls. Find the probability of getting complete square.

Solution

    \[ \textbf{Sample Space η(S) = 6² = 36} \]

    \[ \textbf{Event: A Complete Square η(A) = 7} \]

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Sum Table

1, 1 = 22, 1 = 33, 1 = 44, 1 = 55, 1 = 66, 1 = 7
1, 2 = 32, 2 = 43, 2 = 54, 2 = 65, 2 = 76, 2 = 8
1, 3 = 42, 3 = 53, 3 = 64, 3 = 75, 3 = 86, 3 = 9
1, 4 = 52, 4 = 63, 4 = 74, 4 = 85, 4 = 96, 4 = 10
1, 5 = 62, 5 = 73, 5 = 84, 5 = 95, 5 = 106, 5 = 11
1, 6 = 72, 6 = 83, 6 = 94, 6 = 105, 6 = 116, 6 = 12

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta}\mathbf{(}\mathbf{A}\mathbf{)}}{\mathbf{\eta}\mathbf{(}\mathbf{S}\mathbf{)}})\textbf{=}(\mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{7}}{\mathbf{36}}\ \]

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Introduction to Statistics Basic Important Concepts

Measures of Central Tendency, Arithmetic Mean, Median, Mode, Harmonic, Geometric Mean

Business Statistics Solved Paper FBISE 2012 ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2013 ICOM II, MCQS, Short Questions, Extensive Questions

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