Business Statistics Solved Paper FBISE 2018 2nd Annual ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2018 2nd Annual 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 2nd Annual 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, Business Statistics 2018 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 2nd Annual 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)The data which have not undergone through any statistical process are called:
 A. Primary DataB. Discrete DataC. Secondary DataD. Qualitative Data 
      
(ii)A measure computed from sample data is called:
 A. ParameterB. StatisticC. StatisticsD. Data 
      
(iii)Cumulative frequency distribution is graphically presented by:
 A. OgiveB. HistogramC. Bar ChartD. Pie Chart 
      
(iv)The number of observations falling in a particular class is known as:
 A. 𝐶𝑙𝑎𝑠𝑠 𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦B. Class MarkC. Class LimitD. Mid Point 
      
(v)If ∑(X-11)=6, ∑(X-30)=19 & ∑(X-17)=0 then mean of X is:
 A. 11B. 30C. 17D. 19 
      
(vi)Sum of deviations taken from mean is:
 A. PositiveB. ZeroC. NegativeD. Minimum 
      
(vii)If Laspayr’s index = 116 & Paasche’s index = 110 then Fisher’s index is=?
 A. 110.96B. 116.00C. 113.69D. 112.96 
      
(viii)In chain base method the base period is:
 A. FixedB. ConstantC. Not FixedD. Zero 
      
(ix)n(n-1)(n-2)……3.2.1 is equal to:
 A. ∑nB. ∑(n)(n-1)C. 𝑛!D. n(n-1) 
      
(x)Probability of a sure event is equal to:
 A.0B. 1C. -1D. 0.5 
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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) Define sample and population.

Answer:

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) Define 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.

(iii) Define histogram and historigram.

Answer

Histogram

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.

Historigram

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.

(iv) Describe any three qualities of a good statistical table.

Answer

A good statistical table must contain at least the following components:

i Table Number  ii Title of the Table  iii Caption  iv Stub  v Body  vi Head Note  vii Footnote

(v) Find arithmetic mean given that x=10 +5U; ∑fu = -40; ∑f=125

Solution

Here ∑fU = -40, ∑f = 125, A = 10 & h = 5

    \[ \overline X\;=A+\frac{\sum fu}{\sum f}\times h \]

    \[ \overline X\;=10+\frac{-40}{125}\times5=8.4 \]

(vi) For a moderately skewed distribution mode = 60 and median = 30. Find mean.

Solution

Mode = 3 Median – 2 Mean

60 = 3(30) – 2(Mean)

60 = 90 – 2(Mean)

2(Mean) = 90 – 60

2(Mean)  = 30

Mean = 30/2

Mean = 15

(vii) The logarithm of five values of x are: 1.8062, 1.2304, 1.6532, 1.5798, 1.4314. Find X̄.

Solution

Log of XAntilog Value of X or X
1.806264
1.230417
1.653245
1.579838
1.431427
 ∑X =191

    \[ A.M\ \bar{X} = \frac{\sum X}{n} = \frac{191}{5} = 38.2\ \]

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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:

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

Solution

YearsL.R

    \[ \mathbf{Chain\ Index =}\frac{\mathbf{Previous\ Chain\ x\ Current\ L.R}}{\mathbf{100}}\mathbf{\ } \ \]


1100100 
2120

    \[ \frac{\mathbf{100(120)}}{\mathbf{100}}\mathbf{= 120}\ \]

3102

    \[ \frac{\mathbf{120(102)}}{\mathbf{100}}\mathbf{= 122.4}\ \]

4118

    \[ \frac{\mathbf{122.4(118)}}{\mathbf{100}}\mathbf{= 144.43}\ \]

5112

    \[ \frac{\mathbf{144.43(112)}}{\mathbf{100}}\mathbf{= 161.7616}\ \]

(x) Define simple event and composite event.

Answer

Simple Event

An event of sample space contains only one outcome is called simple event. For example getting 6 or 5 or 3 in a throwing of single die is a simple event.

Compound Event

An event of sample space contains at least two outcomes is called compound event these may be denoted by any letter A to Z except S. For example If event consists of the sum of two dice is ‘’5’’ than it consists of four outcomes i.e., (1, 4), (2, 3), (3, 2), (4, 1) and this is considered to be a compound event

(xi) A pair of dice is rolled find the probability that both faces are same:

Solution

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

Event:

Same Faces η(A) = 6

All Possible Outcomes

1, 12, 13, 14, 15, 16, 1
1, 22, 23, 24, 25, 26, 2
1, 32, 33, 34, 35, 36, 3
1, 42, 43, 44, 45, 46, 4
1, 52, 53, 54, 55, 56, 5
1, 62, 63, 64, 65, 66, 6

Probabilities

\left( \mathbf{i} \right)\mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{6}}{\mathbf{36}}\mathbf{= 0.167}\

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 for the following data:

Weight35–3940–4445–4950–5455–5960–64
Frequency310211514

Solution

MarksfClass BoundariesXfXC.F
35–39334.5—39.5371113
40–441039.5—44.54242013
45–492144.5—49.54798734
50–541549.5—54.55278049
55–59154.5—59.5575750
60–64459.5—64.56224854
 ∑f=54  ∑fx=2603 

    \[ \left( \mathbf{i} \right)\mathbf{\ A.M}\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fX}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{2603}}{\mathbf{54}}\mathbf{= 48.2037}\ \]

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

    \[ \widetilde{\mathbf{X}}\mathbf{= 44.5 +}\frac{\mathbf{5}}{\mathbf{21}}\left( \frac{\mathbf{54}}{\mathbf{2}}\mathbf{- \ 13} \right)\mathbf{= 47.83}\ \]

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

    \[ \mathbf{Mode\ }\widehat{\mathbf{X}}\mathbf{= 44.5 + \ }\frac{\mathbf{21 - 10}}{\left( \mathbf{21 - 10} \right)\mathbf{+ (21 - 15)}}\mathbf{x}\mathbf{5}\ \]

    \[ \mathbf{Mode\ }\widehat{\mathbf{X}}\mathbf{= 44.5 +}\frac{\mathbf{55}}{\mathbf{17}}\mathbf{= 47.73}\ \]

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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 =

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

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

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

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

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

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

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

Q.5 Three coins are tossed. Find the following probabilities:

(a) At most one head appear

(b) Same faces appear

(c) Head on the first coin

Solution:

    \[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{= \ }\mathbf{2}^{\mathbf{3}}\mathbf{= 8}\ \]

All possible outcomes

HHHHHTHTHTHH
TTTTTHTHTHTT

Events:

    \[ \left( \mathbf{i} \right)\mathbf{At\ most\ one\ head = \ \eta}\left( \mathbf{A} \right)\mathbf{= \ 4}\ \]

    \[ \left( \mathbf{ii} \right)\mathbf{Same\ faces\ appear = \ \eta}\left( \mathbf{B} \right)\mathbf{= \ 2}\ \]

    \[ \left( \mathbf{iii} \right)\mathbf{Head\ on\ the\ first\ coin = \ \eta}\left( \mathbf{C} \right)\mathbf{= \ 4}\ \]

Probability:

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

    \[ \left( \mathbf{ii} \right)\mathbf{P}\left( \mathbf{B} \right)\mathbf{= \ }\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{2}}{\mathbf{8}}\mathbf{= 0.25}\ \]

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

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