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

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

In this post, I am going to discuss Business Statistics Solved Paper FBISE 2012 ICOM II, MCQS, Short Questions, Extensive Questions of the topics of Introduction to Statistics, Averages, Index Numbers and Probability is discussed and solved. Business Statistics Solved Paper 2013 is also posted.

Table of Contents

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

MCQS

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Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)The Word Statistics may have been derived from the German Word:
 A. StatusB. StatistikC. StatistiqueD. Statista 
      
(ii)The data, which have not undergone any statistical treatment are:
 A. Primary DataB. Secondary DataC. Discrete DataD. Continuous Data 
      
(iii)If the data are classified according to their time of occurrence then this procedure is called:
 A. Multi-way ClassificationB. Chronological ClassificationC. Geographical ClassificationD. Spatial Classification 
      
(iv)The total of relative frequencies is always equal to:
 A. – 1B. 0.5C. 1D. 100 
      
(v)In a symmetrical distribution mean, median and mode are always:
 A. NegativeB. ZeroC. DifferentD. Equal 
      
(vi)Median is a/an:
 A. Calculating averageB. Positional AverageC. Exact AverageD. None of these 
      
(vii)The most suitable average for index number is:
 A. Arithmetic MeanB. Geometric MeanC. MedianD. Mode 
      
(viii)The index number is called?
 A. Economic BarometerB. Statistical BarometerC. Mathematical BarometerD. Physical Barometer 
      
(ix)The probability of an event always lies between:
 A. -1 and 0B. -1 and 1C. 0 and +1D. -0.5 and 0.5 
      
(x)The probability of drawing red card out of 52 cards is:
 A.1/2B. 1/4C. 1/13D. None of these 

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) Write down the applications of Statistics in Business.

Answer:

Statistics is applied in several areas of business given below:

  • Decision Analysis
  • Descriptive Analysis
  • Predictive Analysis
  • Management Analysis
  • Production Analysis
  • Risk Analysis
  • Sales Analysis
  • Quality Assurance Analysis
  • Financial Analysis
  • & many others.

(ii) Differentiate between Primary & secondary data.

Answer:

First hand, newly collected, ungrouped data is called primary data or data which is not collected by someone previously is called primary data.

Second hand, previously collected, grouped data is called secondary data or data which is collected by someone previously is called secondary data.

(iii) The following data show the number of absent students during the month of November 2011 from I COM class:

3, 4, 5, 6, 7, 1, 0, 2, 3, 4, 5, 7, 8, 7, 2, 1, 5, 6, 7, 8, 9, 10, 6, 7, 3 Make a frequency distribution taking class intervals as one.

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Solution

X012345678910
f12232335211

(iv) What is meant by graph?

Solution

A diagram which shows the relationship of normally two variables is called graph. For example Bar graph, Line graph, Pie Chart, Histogram are common examples of graphs.

(v) The following data indicate the size of shoes sold at a store during the given week. Find model size of shoes:

Size of Shoes4.5  55.5  66.5  77.5  8
No. of Pairs Sold257202517154

Solution

XffX
4.529
5525
5.5738.5
620120
6.525162.5
717119
7.515112.5
8432
 ∑f = 95∑fX = 618.5

    \[ \overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{= \ }\frac{\mathbf{618.5}}{\mathbf{95}}\mathbf{= 6.5}\ \]

(vi) Write down the desirable 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.

(vii) In a certain distribution mean is 45 and median is 50 then find mode. Also give the assumption you take regarding the shape of distribution.

Solution

Mode = 3 Median – 2 Mean

Mode = 3(50) – 2(45)

Mode = 150 – 90

Mode = 60

So Mean = 45, Median = 50 and Mode = 60

Mean < Median < Mode

It is negatively Skewed distribution

(viii)Differentiate between Fixed base & 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.

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

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

(ix) Find Price Relatives for the data given below using average of last three years as base:

Year199019911992199319941995199619971998
Price151620222423251520
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Solution

YearsPrices

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

199015

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{15}}{\mathbf{20}}\mathbf{\ x\ 100}\  \]

=75
199116

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{16}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=80
199220

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{20}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=100
199322

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{22}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=110
199424

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{24}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=120
199523

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{23}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=115
199625

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{25}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=125
199715

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{15}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=75
199820

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{20}}{\mathbf{20}}\mathbf{\ x\ 100}\ \]

=100

    \[ \overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{25 + 15 + 20}}{\mathbf{3}}\mathbf{= \ }\frac{\mathbf{60}}{\mathbf{3}}\mathbf{= 20}\ \]

(x) Two unbiased coins are tossed once. Find the following.

(a) Make a sample space (b) Probability that two head appears

(c) Probability that exact one head appeared

Solution

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

All possible outcomes

HHHTTHTT

    \[ \left( \mathbf{b} \right)\mathbf{Two\ head\ appeares = \ \eta}\left( \mathbf{A} \right)\mathbf{= \ 1}\  \]

    \[  \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}}\mathbf{= 0.25}\ \]

    \[ \left( \mathbf{c} \right)\mathbf{Exactly\ one\ head\ appeared = \ \eta}\left( \mathbf{B} \right)\mathbf{= \ 2}\   \]

    \[  \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{4}}\mathbf{= 0.5}\    \]

(xi) A fair cubical die is rolled once. What is the probability of obtaining?

(a) Six (b) An even number (c) The number greater than 4

Solution

    \[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{= \ }\mathbf{6}^{\mathbf{1}}\mathbf{= 6}\  \]

All possible outcomes = 1, 2, 3, 4, 5, 6

Events

    \[ \left( \mathbf{i} \right)\mathbf{Six = \ \eta}\left( \mathbf{A} \right)\mathbf{= \ 1}\  \]

    \[ \left( \mathbf{ii} \right)\mathbf{An\ even\ number = \ \eta}\left( \mathbf{B} \right)\mathbf{= \ 3}\  \]

    \[ \left( \mathbf{iii} \right)\mathbf{The\ number\ greater\ than\ 4 = \ \eta}\left( \mathbf{C} \right)\mathbf{= \ 2}\  \]

    \[  \mathbf{P}\left( \mathbf{A} \right)\mathbf{= \ }\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{6}}\mathbf{= 0.167}\ \]

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    \[  \mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{3}}{\mathbf{6}}\mathbf{= 0.5}\ \]

    \[ \mathbf{P}\left( \mathbf{C} \right)\mathbf{= \ }\frac{\mathbf{\eta(C)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{6}}\mathbf{= 0.34}\  \]

Business Statistics

Section C (Marks 16)

Extensive Questions

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

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

GroupsFrequency
0–420
5–923
10–1430
15–1921
20–2417
25–2919

Solution

GroupsFrequencyClass BoundariesXfXC.f
0–420-0.5 – 4.5 24020
5–923 4.5 – 9.5716143
10–1430 9.5 – 14.51236073
15–1921 14.5 – 19.51735794
20–2417 19.5 – 24.522374111
25–2919 24.5 – 29.527513130
 130  1805 
 ∑f=  ∑fX= 

    \[ \left( \mathbf{i} \right)\mathbf{\ A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{1805}}{\mathbf{130}}\mathbf{= 13.88}\  \]

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

Model Class = n/2 = 130/2 = 65 falls in C.f of 73 so data is:

L = 9.5, h = 5, f = 30, n/2 =130/2 = 65 & C = 43

    \[   \left( \mathbf{ii} \right)\mathbf{\ Median = 9.5 +}\frac{\mathbf{5}}{\mathbf{30}}\mathbf{(}\frac{\mathbf{130}}{\mathbf{2}}\mathbf{- \ 43)}\  \]

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    \[ \mathbf{Median = 13.17}\  \]

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

Model Class: Maximum frequency is 30 so data for model class is:

L = 9.5, fm = 30, f1 = 23, f2 = 21 & h = 5

    \[ \mathbf{Mode = 9.5 +}\frac{\mathbf{30 - 23}}{\left( \mathbf{30 - 23} \right)\mathbf{+ \ (30 - 21)}}\mathbf{x\ 5 = \ 9.5 +}\frac{\mathbf{7}}{\mathbf{7 + \ 9}}\mathbf{x\ 5}\  \]

    \[ \mathbf{Mode = 9.5 +}\frac{\mathbf{7}}{\mathbf{16}}\mathbf{x\ 5 = \ 9.5 +}\frac{\mathbf{35}}{\mathbf{16}}\mathbf{= 9.5 + 2.1875 = 11.6875}\  \]

Q.4 Construct Index number for 1995 from the following data taking 1990 as base.

(i) Laspeyre’s Method

(ii) Paasche’s method

(iii) Fisher Ideal Method

  Items19901995
PriceQuantityPriceQuantity
A1012012100
B815010130
C12801370
D15602050

Solution

Article19901995 
Price (Po)Quantity (qo)Price   (P1)Quantity (q1)poqop1qop1q1poq1
A10120121001200144012001000
B8150101301200150013001040
C128013709601040910840
D1560205090012001000750
Sum4260518044103630
 ∑poqo =∑p1qo =∑p1q1 =∑poq1 =

    \[ \left( \mathbf{i} \right)\mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ 1995}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{0}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{0}}}\mathbf{x\ 100}\  \]

    \[ \mathbf{Laspeyr}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index}\mathbf{\ 1995}\mathbf{=}\frac{\mathbf{5180}}{\mathbf{4260}}\mathbf{\ x\ 100 = \ 121.59}\  \]

    \[  \left( \mathbf{ii} \right)\mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 1995}\mathbf{=}\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}\mathbf{q}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{0}}\mathbf{q}_{\mathbf{1}}}\mathbf{x\ 100}\   \]

    \[  \mathbf{Paasch}\mathbf{e}^{\mathbf{'}}\mathbf{s\ Index\ 1995}\mathbf{=}\frac{\mathbf{4410}}{\mathbf{3630}}\mathbf{x\ 100 = 121.48}\   \]

    \[ \left( \mathbf{iii} \right)\mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 1995 =}\sqrt{\mathbf{LxP}}\   \]

    \[  \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 1995 =}\sqrt{\mathbf{121.59}\mathbf{x}\mathbf{121.48}}\  \]

    \[  \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index\ 1995 = 121.53}\  \]

Q.5 A marble is drawn at random from a box containing 10 red and 30 white, 20 blue and 15 orange marbles. What is the probability that it is: (i) White (ii) Orange or Red (iii) Not red or blue (iv) Red, White or blue

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Solution

Red marbles = 10

White marbles = 30

Blue marbles = 20

Orange marbles = 15

Total marbels = 10 + 30 + 20 + 15 = 75

Events:

    \[ \left( \mathbf{i} \right)\mathbf{\ White\ marble\ \ \eta}\left( \mathbf{A} \right)\mathbf{= \ 30}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ Orange\ or\ Red\ \ \eta}\left( \mathbf{B} \right)\mathbf{= \ 10 + 15 = 25}\  \]

    \[  \left( \mathbf{iii} \right)\mathbf{\ }\mathbf{Red,\ White\ or\ blue}\mathbf{\ \ \eta}\left( \mathbf{C} \right)\mathbf{= \ 10 + 30 + 20 = 60}\ \]

Probabilities

    \[ \left( \mathbf{i} \right)\mathbf{\ P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{30}}{\mathbf{75}}\mathbf{= \ 0.4}\  \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{\eta(B)}}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{25}}{\mathbf{75}}\mathbf{= \ 0.34}\  \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ P}\left( \mathbf{C} \right)\mathbf{=}\frac{\mathbf{\eta(C)}}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{60}}{\mathbf{75}}\mathbf{= \ 0.8}\  \]

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