Business Statistics Solved Paper FBISE 2023 Annual ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2023 Annual ICOM II, MCQS, Short Questions, Extensive Questions

In this blog post, I am going to discuss the paper of Business Statistics Solved Paper FBISE 2023 Annual ICOM IIMCQS, Short Questions, Extensive Questions, topics included are Introduction to StatisticsAverages, Index Numbers, Probability. Solved paper of Business Statistics Paper 2012 & solved paper of Business Statistics 2013Business Statistics 2015Business Statistics 2016Business Statistics 2016 SupplementaryBusiness Statistics 2017Business Statistics 2017 2nd AnnualBusiness Statistics 2018Business Statistics 2018 2nd Annual are already published on the website. Stay Connected for other boards solutions such as BISELHRBISERWP etc.

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Business Statistics Solved Paper FBISE 2023 Annual ICOM II, MCQS, Short Questions, Extensive Questions

MCQS

Q 1: Fill the relevant bubble against each question. 

1A constant can assume:
A)More than one valueB)Only one value
C)Different ValueD)No value at all
2Issuing a new N.I.C is an example of:
A)RegistrationB)Sampling
C)CensusD)Fictitious data
3Data obtained by internet source are:
A)Raw dataB)Private data
C)Secondary dataD)Primary data
4A pie diagram is represented by a:
A)RectangleB)Square
C)TriangleD)Circle
5In symmetrical distribution mean, median and mode are always:
A)EqualB)Different
C)NegativeD)Zero
6If mean of 10 observations is 20, then their sum will be equal to:
A)20B)200
C)2D)0.5
7If Y = -75 – 25X and X̅=3, then Y̅=?
A)150B)25
C)-150D)0
8Base year weighted index number is also known as:
A)Fisher’s IndexB)Paasche’s Index
C)Marshall’s IndexD)Laspeyre’s Index
9Index for base period is:
A)OneB)Fix
C)100D)More than 100
10When three dice are rolled then total number of possible outcome will be:
A)6B)12
C)18D)216

Short Questions

SECTION-B (Marks 24)

Q.2: Answer any eight parts from Section ‘B’ and any two questions from Section ‘C’ on the separately provided answer book. Write your answers neatly and legibly. (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) Name any three sources of primary data.

Answer:

  • Through Direct Questionnaire
  • Through Email
  • Through Telephone
  • Through Public Social Media Poll

(iii) Find the class boundaries, midpoint, and width for each class.

  • 293 – 353
  • 11.8 – 14.7
  • 3.13 – 3.93 

Answer:

ClassesClass BoundariesMid-PointWidth
293 – 353292.5 – 353.532360
11.8 – 14.711.75 – 14.7513.252.9
3.13 – 3.93 3.125 – 3.9353.530.8

(iv) What is a pie-chart?

Answer

A pie chart also called a circular chart is a statistical graphic chart that is divided into slices to illustrate numerical proportion. Total area of pie chart is about to 360 degree. Certain variable has multiple proportion can be shown graphically through pie chart.

(v) Arithmetic mean of 20 values is 25, by adding 4 more values the mean becomes 30. Find the four values if the ratio between these values is 1:2:3:4.

Answer

    \[ \bar{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\ \]

    \[ \mathbf{25 =}\frac{\mathbf{\sum X}}{\mathbf{20}}\ \]

    \[ \mathbf{\sum X = 25\ }\mathbf{\times}\mathbf{\ 20 = 500}\ \]

    \[ \bar{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\ \]

    \[ \mathbf{30 =}\frac{\mathbf{\sum X}}{\mathbf{24}}\ \]

    \[ \mathbf{\sum X = 30\ }\mathbf{\times}\mathbf{\ 24 = 720}\ \]

Sum of four added values = 720 – 500 = 220

Ratio of four values = 1:2:3:4

Sum of ratios = 1+2+3+4=10

    \[ 1st\ value = \frac{220}{10}\ (1) = 22\ \]

    \[ 2nd\ value = \frac{220}{10}\ (2) = 44\ \]

    \[ 3rd\ value = \frac{220}{10}\ (3) = 66\ \]

    \[ 4th\ value = \frac{220}{10}\ (4) = 88\ \]

(vi) The logarithms of 3 values of x are 1.7076, 1.6812, and 1.6532. Find the mean of x values.

Solution:

Log Values of XX = Antilog of Log X
1.707651
1.681248
1.653245
 ∑X= 144

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{144}}{\mathbf{3}}\mathbf{= 48}\  \]

(vii) If l=28, fm = 25, f1 = 20, f2 = 18 and h = 7. Compute mode.

Solution:

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

    \[  \mathbf{Mode\ }\widehat{\mathbf{X}}\mathbf{=}\mathbf{28}\mathbf{+ \ }\frac{\mathbf{25 - 20}}{\left( \mathbf{25 - 20} \right)\mathbf{+ (}\mathbf{25 - 18}\mathbf{)}}\mathbf{\times}\mathbf{7}\ \]

    \[  \mathbf{Mode\ }\widehat{\mathbf{X}}\mathbf{=}\mathbf{28}\mathbf{+}\frac{\mathbf{35}}{\mathbf{1}\mathbf{2}}\mathbf{=}\mathbf{30.91}\ \]

(viii) Given ∑(X – 10) = 2.8, n=5, Calculate arithmetic mean.

Solution

    \[  \overline{\mathbf{X}}\mathbf{= A +}\left( \frac{\mathbf{\sum D}}{\mathbf{n}} \right)\mathbf{= 10 +}\frac{\mathbf{2.8}}{\mathbf{5}}\mathbf{= 10.56}\ \]

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(ix) Given that ∑poqo = 1430, ∑p1qo = 1760, ∑p1q1 = 2110, and ∑poq1 = 1715. Find Fisher’s Ideal Price Index.

Solution:

    \[  Fisher^{'}s\ Index = \sqrt{\frac{\sum p1qo}{\sum poqo} \times \frac{\sum p1q1}{\sum poq1}}\ \times 100\ \]

    \[  Fisher^{'}s\ Index = \sqrt{\frac{1760}{1430} \times \frac{2110}{1715}}\ \times 100\ \]

    \[  Fisher^{'}s\ Index = 123.05\ \]

(x) Given po = 5, 4, 3 and qo = 70,75,80. Find ∑W.

Solution:

poqoW=poqo
570350
475300
380240
  ∑W=890

(xi) Solve the following:

    \[  (a)\begin{pmatrix}9 \\P \\4 \\\end{pmatrix}\ (b)\begin{pmatrix}13 \\C \\0 \\\end{pmatrix}\  \]

Solution:

    \[  (a)\begin{pmatrix}9 \\P \\4 \\\end{pmatrix}\ \]

    \[ \frac{n!}{(n - r)!} = \frac{9!}{(9 - 4)!}\ \]

    \[ \frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{5 \times 4 \times 3 \times 2 \times 1} = 3024\ \]

    \[  (b)\begin{pmatrix}13 \\C \\0 \\\end{pmatrix}\ \ \]

    \[ \frac{n!}{(n - r)!r!} = \frac{13!}{(13 - 0)!0!}\ \]

    \[ \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1} = 1\ \]

Extensive Questions

Business Statistics Solved Paper FBISE 2023 Annual ICOM II, MCQS, Short Questions, Extensive Questions

Section C (Marks 16)

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

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

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

Classes0—2020—4040—6060—80
Frequency20304010

Solution

Class LimitsFrequency (f)XfXC.f
0—20201020020
20—40303090050
40—604050200090
60—801070700100
Sum100 3800 
 f or n= ∑fX= 

    \[ \left( \mathbf{i} \right)\mathbf{\ A.M}\overline{\mathbf{X}\mathbf{\ }}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{3800}}{\mathbf{100}}\mathbf{=}\mathbf{38}\  \]

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

Model Class = n/2 = 100/2 = 50 falls in C.f of 50 so data is:

L = 20, h = 20, f = 30, n/2 =100/2 = 50 & C = 20

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

    \[  \mathbf{Median =}\mathbf{20}\mathbf{+}\frac{\mathbf{20}}{\mathbf{3}\mathbf{0}}\left( \mathbf{50}\mathbf{- \ }\mathbf{2}\mathbf{0} \right)\ \]

    \[  \mathbf{Median}\mathbf{=}\mathbf{20}\mathbf{+}\frac{\mathbf{600}}{\mathbf{3}\mathbf{0}}\ \]

    \[  \mathbf{Median =}\mathbf{40}\ \]

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

Model Class: Maximum frequency is 40 so data for model class is:

L = 40, fm = 40, f1 = 30, f2 = 10 & h = 20

    \[ \mathbf{Mode =}\mathbf{40}\mathbf{+}\frac{\mathbf{40 -}\mathbf{30}}{\left( \mathbf{40 -}\mathbf{30} \right)\mathbf{+ \ }\left( \mathbf{40 -}\mathbf{1}\mathbf{0} \right)}\mathbf{\times}\mathbf{\ }\mathbf{20}\  \]

    \[ \mathbf{Mode}\mathbf{= \ 4}\mathbf{0}\mathbf{+}\frac{\mathbf{200}}{\mathbf{40}}\  \]

    \[ \mathbf{Mode =}\mathbf{45}\  \]

Q.4: Construct Fisher’s Price Index number for 2002 using 2001 as base…..

Q.4: Construct Fisher’s Price Index number for 2002 using 2001 as base.

  Items20012002
PriceQuantityPriceQuantity
A6427075290
B4012445144
C1813021137
D5818568200

Solution:

  Items20012002 
Price (Po)Quantity (qo)Price   (P1)Quantity (q1)poqop1qop1q1poq1
A642707529017280202502175018560
B40124451444960558064805760
C18130211372340273028772466
D581856820010730125801360011600
Sum35310411404470738386
 ∑poqo =∑p1qo =∑p1q1 =∑poq1 =

    \[  Fisher^{'}s\ Index\ 2002 = \sqrt{\frac{\sum p1qo}{\sum poqo} \times \frac{\sum p1q1}{\sum poq1}}\ \times 100\ \]

    \[  Fisher^{'}s\ Index\ 2002 = \sqrt{\frac{41140}{35310} \times \frac{44707}{38386}}\ \times 100\ \]

    \[ Fisher^{'}s\ Index\ 2002 = 116.48\  \]

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Q.5: If a die is rolled one time, find these probabilities…..

Q.5: If a die is rolled one time, find these probabilities.

(i) Of getting a 4

(ii) Of getting a number less than 7

(iii) Of getting a number greater than 3 or an odd number

(iv) Of getting a number greater than 3 and an odd number

Solution:

    \[  Sample\ Space\ \eta(S) = \ N^{n} = 6^{1} = 6\ \]

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

Events:

    \[ (i) = Getting\ 4\ \eta(A) = 1\  \]

    \[  (ii) = Getting\ a\ number\ less\ than\ 7\ \eta(B) = 6\ \]

    \[ (iii) = Number\ greater\ than\ 3\ or\ an\ odd\ \ \eta(C) = 5\  \]

    \[  (iv) = Number\ greater\ than\ 3\ and\ an\ odd\ \ \eta(D) = 1\ \]

Probabilities:

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

    \[  (ii)P(B) = \frac{\eta(B)}{\eta(S)} = \frac{6}{6} = 1\ \]

    \[  (iii)P(C) = \frac{\eta(C)}{\eta(S)} = \frac{5}{6} = 0.833\ \]

    \[  (iv)P(D) = \frac{\eta(D)}{\eta(S)} = \frac{1}{6} = 0.167\ \]

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