Business Statistics Solved Paper FBISE 2018 ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2018 ICOM II, MCQS, Short Questions, Extensive Questions
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In this blog post, I am going to discuss the paper of Business Statistics Solved Paper FBISE 2018 ICOM II, MCQS, Short Questions, Extensive Questions topics included are Introduction to Statistics, Averages, Index Numbers, Probability. Solved paper of Business Statistics Paper 2012 & solved paper of Business Statistics 2013, Business Statistics 2015, Business Statistics 2016, Business Statistics 2016 Supplementary, Business Statistics 2017, Business Statistics 2017 2nd Annual are already published on the website. Stay Connected for other boards solutions such as BISELHR, BISERWP etc.

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Business Statistics Solved Paper FBISE 2018 ICOM II, MCQS, Short Questions, Extensive Questions

Solved by Iftikhar Ali, M.Sc Economics, M.Com Finance Lecture Statistics, Finance & Accounting

MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)Statistics is always:
 A. FixedB. Aggregate of Facts and FiguresC. Always ContinuousD. Always True 
      
(ii)The data which have not undergone and statistical treatments are:
 A. Discrete dataB. Secondary DataC. Primary DataD. Qualitative Data 
      
(iii)The number of observations falling in a particular class is known as:
 A. MidpointB. Class markC. Class IntervalD. Class Frequency 
      
(iv)In a pie chart, angle of a sector is:
 A. (Component Part/Total) x 100B. (Component Part/Total) x 360⁰C. (Total/Component Part) x 100D. (Total/Component Part) x 360⁰ 
      
(v)The sum of square deviations from mean is:
 A. ZeroB. MinimumC. NegativeD. Maximum 
      
(vi)The empirical relationship among mean, median and mode is:
 A. Mode=3Mean – 2 MedianB. Mode = 3Median – 2 MeanC. Mode = 3Median – 3MeanD. Mode = 2Mean – 3Median 
      
(vii)If all the values are not of equal importance, the index number is called:
 A. SimpleB. CompositeC. Un-WeightedD. Weighted 
      
(viii)Imports and exports of Pakistan is an example of:
 A. Composite Index NumberB. Wholesale price Index NumberC. Volume Index NumberD. Simple Index Number 
      
(ix)The probability of an event always lies between:
 A. 0 and 1B. -1 and 1C. -1 and 0D. 1 and 2 
      
(x)The probability of sure event is:
 A.

    \[ A\ U\ B = \ \varnothing\ \]

B.

    \[ A\ U\ B = \ A\ \]

C.

    \[ A\ \bigcap\ B = \ \varnothing\ \]

D.

    \[ A\ \bigcap\ B = \ S\ \]

 
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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 lines. (8 x 3 = 24)

(i) Differentiate between descriptive and 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 discrete variable by giving examples.

Answer

In discrete variable, data has some specific value within a given range for example Number of persons, Number of Cars, Number of Students in a class etc.

(iii) Define Histogram.

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.

(iv) Define classification and tabulation.

Answer

The process of classifying data into groups is called classification whereas to present data in the form of table or to present into rows and columns is called tabulation.

(v) A student obtained 40, 50, 60, 80 and 45 marks in the subjects of English, Urdu, Maths, Stats and Pakistan Studies, respectively, assuming weights 5, 2, 4, 3 and 1 respectively for above mentioned subjects. Find weighted A-M for subjects.

Answer

SubjectsMarks (X)Weights (W)WX
English405200
Urdu502100
Maths604240
Statistics803240
Pak. Studies45145
  ∑W=15∑WX=825

    \[ \mathbf{Weighted\ Mean =}\frac{\mathbf{\sum WX}}{\mathbf{\sum W}}\mathbf{=}\frac{\mathbf{825}}{\mathbf{15}}\mathbf{= 55}\ \]

(vi) A distribution consists of four components with frequencies 40, 50, 65 and 45 having their means 20, 38, 45 and 52. Find mean of combined distribution.

Solution:

    \[ \overline X1=20,\;\overline X2=38,\;\overline X3=45,\;\overline X4=52,\;n1=40,\;n2=50,\;n3=65,\;n4=45 \]

    \[ \overline Xc=\frac{\overline X1n1+\overline X2n2+\overline X3n3+\overline X4n4}{n1+n2+n3+n4} \]

    \[ \overline Xc=\frac{(20)(40)+(38)(50)+(45)(65)+(52)(45)}{40+50+65+45} \]

    \[ \overline Xc=\frac{800+1900+2925+2340}{200} \]

    \[ \overline Xc=\frac{7965}{200}=39.825 \]

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(vii) Define median and also write it’s any two advantages.

Solution

Median is a second type of average. Its formula for ungrouped data is:

    \[ \mathbf{The\ value\ of\ }\frac{\mathbf{n + 1}}{\mathbf{2}}\mathbf{nd\ item}\ \]

In grouped data, the formula for median is:

    \[ \widetilde{\mathbf{X}}\mathbf{= L +}\frac{\mathbf{h}}{\mathbf{f}}\mathbf{(}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{- \ C)}\ \]

    \[ \mathbf{Advantages\ of\ median:}\ \]

    \[ \left( \mathbf{i} \right)\mathbf{It\ is\ less\ affected\ by\ extreme\ values}\ \]

    \[ \textbf{(ii) It is preferred when data is not symmetrical} \]

    \[ \textbf{(iii) It is easy to calculate & understand} \]

    \[ \textbf{(iv) It has a mathematical formula} \]

(viii) Laspeyre’s Index No= 115, Fisher’s Index No=112.98 Paasch’s Index No=?

Answer

    \[ \mathbf{112.98 = \ }\sqrt{\mathbf{Paasch's}}\mathbf{\ x}\mathbf{\ }\sqrt{\mathbf{115}}\mathbf{\ }\ \]

    \[ \mathbf{112.98 =}\sqrt{\mathbf{Paasch's}}\mathbf{\ x\ 10.7238\ \ }\ \]

    \[ \mathbf{112.98 =}\sqrt{\mathbf{Paasch's}}\mathbf{\ x\ 10.7238\ \ }\ \]

    \[ \sqrt{\mathbf{Paasch's}}\mathbf{=}\frac{\mathbf{112.98}}{\mathbf{10.7238}}\mathbf{= 10.5354}\ \]

    \[ \textbf{Taking square root on both sides} \]

    \[ {\mathbf{(}\sqrt{\mathbf{Paasch's}}\mathbf{)}}^{\mathbf{2}}\mathbf{=}\mathbf{(10.5354)}^{\mathbf{2}}\ \]

    \[ \mathbf{Paasc}\mathbf{h}^{\mathbf{'}}\mathbf{s\ Index = \ 110.994}\ \]

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(ix) Find Index No. from the following data taking average of all five years as base:

Year19701971197219731974
Price9691110

Solution

    \[ \bar{X} = \frac{9 + 6 + 9 + 11 + 10}{5} = \frac{45}{5} = 9\ \]

YearsPrices

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


197014

    \[ \frac{\mathbf{14}}{\mathbf{9}}\mathbf{\ x}\mathbf{100 = 155.55}\ \]

197115

    \[ \frac{\mathbf{15}}{\mathbf{9}}\mathbf{\ x}\mathbf{100 = 166.67}\ \]

197216

    \[ \frac{\mathbf{16}}{\mathbf{9}}\mathbf{\ x}\mathbf{100 = 177.77}\ \]

  
197317

    \[ \frac{\mathbf{17}}{\mathbf{9}}\mathbf{\ x}\mathbf{100 = 188.88}\ \]

197418

    \[ \frac{\mathbf{18}}{\mathbf{9}}\mathbf{\ x}\mathbf{100 = 200}\ \]

 

(x) If one card is drawn from 52 playing cards. Find the probability that it is a king of heart.

Solution:

    \[ Sample\;Space\;\eta(s)\;=52 \]

Event

    \[ \left( \mathbf{i} \right)\mathbf{\ King\ of\ Heart\ \eta}\left( \mathbf{A} \right)\mathbf{=}\mathbf{1}\ \]

Probability

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{52}}\mathbf{= 0.01923}\ \]

(xi) In how many ways, the word “Committee” can be arranged?

Solution

In word “Committee”, M, T and E appear 2 times each whereas C, O and I appears one time each and total letters are 9, so the permutation applied:

    \[ \frac{\mathbf{9!}}{\mathbf{2!2!2!1!1!1!}}\mathbf{=}\frac{\mathbf{9}\mathbf{x}\mathbf{8}\mathbf{x}\mathbf{7}\mathbf{x}\mathbf{6}\mathbf{x}\mathbf{5}\mathbf{x}\mathbf{4}\mathbf{x}\mathbf{3}\mathbf{x}\mathbf{2}\mathbf{x}\mathbf{1}}{\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}}\mathbf{=}\frac{\mathbf{9}\mathbf{x}\mathbf{8}\mathbf{x}\mathbf{7}\mathbf{x}\mathbf{6}\mathbf{x}\mathbf{5}\mathbf{x}\mathbf{4}\mathbf{x}\mathbf{3}}{\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{2}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}\mathbf{x}\mathbf{1}}\mathbf{=}\frac{\mathbf{181440}}{\mathbf{4}}\mathbf{= 45360}\ \]

Extensive Questions

Section C (Marks 16)

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

Q.3 Calculate arithmetic mean by using:

(a) Direct method

(b) Indirect/Shortcut Method

(b) Step-deviation/Coding method

MarksNo. of Students
30–391
40–493
50–5911
60–6921
70–7943
80–8932
90–999

Solution

MarksfClass BoundariesXfXD=X-64.5fD

    \[ U=\frac{X-64.5}{10} \]

fU
30–39129.5–39.534.534.5-30-30-3-3
40–49339.5–49.544.5133.5-20-60-2-6
50–591149.5–59.554.5599.5-10-110-1-11
60–692159.5–69.564.51354.50000
70–794369.5–79.574.53203.510430143
80–893279.5–89.584.5270420640264
90–99989.5–99.594.5850.530270327
Sum120  8880  1140 114
 ∑f=  ∑fX= ∑fD=  ∑fU=

    \[ \left( \mathbf{i} \right)\mathbf{\ Direct\ Method\ A.M}\ \]

    \[ \mathbf{\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{8880}}{\mathbf{120}}\mathbf{= 74}\ \]

    \[ \left( \mathbf{i}\mathbf{i} \right)\mathbf{Indirect\ or\ Shortcut\ Method}\mathbf{\ }\ \]

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

    \[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{=}\mathbf{64.5}\mathbf{+}\frac{\mathbf{1140}}{\mathbf{120}}\mathbf{= 74}\ \]

    \[ \left( \mathbf{ii}\mathbf{i} \right)\mathbf{Step - deviation}\mathbf{\ or\ coding}\mathbf{\ Method\ }\ \]

    \[ \mathbf{A.M}\mathbf{\ }\overline{\mathbf{X}}\mathbf{= A + \ }\frac{\mathbf{\sum fU}}{\mathbf{\sum f}}\mathbf{\times}\mathbf{h}\ \]

    \[ \mathbf{\ A.M}\overline{\mathbf{X}}\mathbf{= 64.5 + \ }\frac{\mathbf{114}}{\mathbf{120}}\mathbf{\times}\mathbf{10}\ \]

    \[ \mathbf{\ A.M}\overline{\mathbf{X}}\mathbf{= 64.5 + \ }\frac{\mathbf{1140}}{\mathbf{120}}\mathbf{= 64.5 + 9.5 = 74}\ \]

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Q.4 Calculate Chain Index numbers for the following data. Use Median as an average.

YearABCD
1958949882122
1959988397105
1960968795107
1961948297112

Solution

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

\mathbf{Chain\ Index = \ }\frac{\mathbf{Previous\ Chain\ x\ Current\ Average}}{\mathbf{100}}\mathbf{\ }\

YearABCDMedianChain Index
1958949882122(82+94)/2 =8888
1959

    \[ \frac{\mathbf{98}}{\mathbf{94}}\mathbf{x}\mathbf{100}\ \]

  =104.25

    \[ \frac{\mathbf{83}}{\mathbf{98}}\mathbf{x}\mathbf{100}\ \]

  =84.69

    \[ \frac{\mathbf{97}}{\mathbf{82}}\mathbf{x}\mathbf{100}\ \]

  =118.29

    \[ \frac{\mathbf{105}}{\mathbf{122}}\mathbf{x}\mathbf{100}\ \]

  =86.86		(86.86 + 104.25)/2 =95.555(88 x 95.555)/100 = 84
1960

    \[ \frac{\mathbf{96}}{\mathbf{98}}\mathbf{x}\mathbf{100}\ \]

  =97.95

    \[ \frac{\mathbf{87}}{\mathbf{83}}\mathbf{x}\mathbf{100}\ \]

  =104.81

    \[ \frac{\mathbf{95}}{\mathbf{97}}\mathbf{x}\mathbf{100}\ \]

  =97.93

    \[ \frac{\mathbf{107}}{\mathbf{105}}\mathbf{x}\mathbf{100}\ \]

  =101.90		(97.95 + 101.90)/2 =99.925(84 x 99.925)/100 = 83.93
1961

    \[ \frac{\mathbf{94}}{\mathbf{96}}\mathbf{x}\mathbf{100}\ \]

  =97.917

    \[ \frac{\mathbf{82}}{\mathbf{87}}\mathbf{x}\mathbf{100}\ \]

  =94.25

    \[ \frac{\mathbf{97}}{\mathbf{95}}\mathbf{x}\mathbf{100}\ \]

  =102.10

    \[ \frac{\mathbf{112}}{\mathbf{107}}\mathbf{x}\mathbf{100}\ \]

  =104.67		(97.917 + 102.10)/2 =100(83.93 x 100)/100 = 83.93

Q.5 From a pack of playing cards, a card is drawn at random, find the probability that the drawn card is:

(i) Ace card

(ii) A red card

(iii) A king or a queen

(iv) A jack of clubs

Solution

    \[ Sample\;Space\;=\eta(S)=52 \]

Events

    \[ \left( \mathbf{i} \right)\mathbf{\ Ace\ card\ \eta}\left( \mathbf{A} \right)\mathbf{=}\mathbf{4}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{\ A\ red\ card\ \eta}\left( \mathbf{B} \right)\mathbf{=}\mathbf{26}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{\ A\ King\ or\ a\ Queen\ \eta}\left( \mathbf{C} \right)\mathbf{=}\mathbf{8}\ \]

    \[ \left( \mathbf{iv} \right)\mathbf{\ A\ Jack\ of\ Clubs\ \eta}\left( \mathbf{D} \right)\mathbf{=}\mathbf{1}\ \]

Probabilities

    \[ \left( \mathbf{i} \right)\mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{4}}{\mathbf{52}}\mathbf{= 0.0769}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{26}}{\mathbf{52}}\mathbf{= 0.5}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{P}\left( \mathbf{C} \right)\mathbf{=}\frac{\mathbf{\eta(C)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{8}}{\mathbf{52}}\mathbf{= 0.153}\ \]

    \[ \left( \mathbf{iv} \right)\mathbf{P}\left( \mathbf{D} \right)\mathbf{=}\frac{\mathbf{\eta(D)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{52}}\mathbf{= 0.019}\ \]

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