Business Statistics Solved Paper 2024 Lahore Board (BISE Lahore) – Complete Solution

Business Statistics Solved Paper 2024 Lahore Board (BISE Lahore) – Complete Solution
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Business Statistics Solved Paper 2024 Lahore Board (BISE Lahore) – Complete Solution. Get the complete solution of the Business Statistics Lahore Board 2024 paper. Step-by-step solved questions, easy explanations, and exam preparation guide for students.

This post provides a complete and detailed solution to the Business Statistics 2024 paper of the Lahore Board (BISE Lahore). All questions are solved step by step with clear explanations to help students understand key concepts such as measures of central tendency, dispersion, probability, and data analysis. This solved paper is highly useful for I.Com and business students preparing for exams, helping them practice past paper questions and improve their problem-solving skills. For more solutions, stay in touch with bcfeducation.com

Table of Contents

Business Statistics Solved Paper 2024 Lahore Board (BISE Lahore) – Complete Solution

MCQS

Q#QuestionABCD
1Battery life time is variable:QualitativeDiscreteComparableContinuous
2Graph of class boundaries and frequency is:HistogramOgiveHistogriamBar chart
3One dimensional diagram is:Rectangular diagramSquare diagramSimple bar chartPie diagram
4Single value which represent a set of data:SymmetricCentral tendencySkew-symmetricQuartile
5If Σx = 150, X̄ = 10 then n =1050515
6Mode of 2, 7, 10, 15 is:ZeroNo mode215
7The year for which index number is 100 known as:Current yearPrevious yearChain yearBase year
8Consumer price index number is also called:ValueVolumeCost of living indexWholesale price
9Probability of getting red card when a card is drawn from 52:1/261/5426/521/2
10If P(A∩B) = P(A) then A and B are:IndependentDependentMutually exclusiveExhaustive

Q.2 Write short answers to any SIX (6) questions. Marks (6 x 2=12)

i. Define parameter by giving an example.

Answer: A measure computed from a population data is called parameter. For example, a population mean is a parameter. Mean of population µ is a parameter.

ii. Explain the concept of primary data.

Answer: Most original first hand collected data is called primary data or the data without any statistical analysis is called primary data.

iii. Differentiate ungrouped and grouped data.

Answer:

Primary/Ungrouped Data

Most original first hand collected data is called primary/ungrouped data or the data without any statistical analysis is called primary data.

Secondary/Grouped Data

The data which has already collected and statistically analyzed by someone is called secondary/grouped data. The data given in your book and all grouped data is called secondary data.

iv. Define Statistics in Plural Sense.

Answer: In plural sense “Statistics” refers to data collected in systematic manner for definite purpose in any field of study i.e. industrial production, population growth of country etc.

v. Distinguish between histogram and historigram.

Answer:

Histogram

A Histogram is a graphical representation of data using bars to show the frequency distribution of continuous data. Unlike a bar graph, the bars in a histogram touch each other, representing intervals (bins) of data values. It is used to analyze the shape, spread, and central tendency of data.

Historigram

Historigram is a type of chart which is used to present the data of time series in which different variations of time series can be shown.It is also called graph of time series or historical series. In historigram, time is taken on x-axis and variable is taken on y-axis.

vi. What is an array?

Answer: An array is a simple way of arranging data in a systematic order or a set of data arranged in ascending or descending order of magnitude.

vii. Explain the term “Equally Likely Events”.

Answer: If two events A and B have equal chance to happen or occur then the events are called equally likely events. For example, in case of tossing a coin, head has equal chance to appear as tail similarly in case of throwing a die, even number has equal chance as the odd number.

viii. Write down sample space when three coins are tossed.

Solution:

Sample Space η(S)=Nn=23=8\text{Sample Space } \eta(S) = N^n = 2^3 = 8

All Possible Outcomes

H,H,HH,H,TH,T,HT,H,H
T,T,TT,T,HT,H,TH,T,T

ix. A die is rolled. What is the probability that it shows odd numbers?

Solution:

Sample Space=η(S)=Nn\text{Sample Space} = \eta(S) = N^n

Sample Space=η(S)=61=6\text{Sample Space} = \eta(S) = 6^1 = 6

All Possible Outcomes or Sample Points

{1, 2, 3, 4, 5, 6}

Events

Odd Number η(A)=3\text{Odd Number } \eta(A) = 3

Probability

P(A)=η(A)η(S)=36=12P(A) = \frac{\eta(A)}{\eta(S)} = \frac{3}{6} = \frac{1}{2}

Q.3 Write short answers to any Six (6) questions: Marks (6 x 2=12)

i. Given D = X – 2075, ΣfD = 10730, Σf = 500, find arithmetic mean.

Solution:

X‾=A+(∑fD∑f)\bar{X} = A + \left( \frac{\sum fD}{\sum f} \right)

X‾=2075+10730500=2096.46\bar{X} = 2075 + \frac{10730}{500} = 2096.46

ii. Describe four 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.

iii. Write down any two properties of arithmetic mean.

Answer:

(1)The sum of the deviations, of all the values of x, from their arithmetic mean, is zero i.e.

∑(X−X‾)=0\sum (X – \bar{X}) = 0

∑f(X−X‾)=0\sum f(X – \bar{X}) = 0

(2) Sum of squares of deviations from arithmetic mean is least i.e. 

∑(X−X‾)2<∑(X−A)2\sum (X – \bar{X})^2 < \sum (X – A)^2

∑f(X−X‾)2<∑f(X−A)2\sum f(X – \bar{X})^2 < \sum f(X – A)^2

iv. If mode is 15 and median = 12. Find mean.

Solution:

Mode = 3 Median – 2 Mean

15= 3(12) – 2(Mean)

15 = 36 – 2(Mean)

2(Mean) = 36-15

2(Mean)  = 21

Mean = 21/2

Mean = 10.5

v. Find the median of: 0, -1, -4, 3, 5, 10, -3, -7, 10, 3.

Solution:

Array the Data

S/No12345678910
Data-7-4-3-103351010
Median=value of (n+12)nd item\text{Median} = \text{value of } \left( \frac{n+1}{2} \right)^{\text{nd}} \text{ item}

Median=value of (10+12)nd item\text{Median} = \text{value of } \left( \frac{10+1}{2} \right)^{\text{nd}} \text{ item}

Median=value of 5.5th item\text{Median} = \text{value of } 5.5^{\text{th}} \text{ item}

Median=5th value+0.5(6th−5th value)\text{Median} = 5^{\text{th}} \text{ value} + 0.5(6^{\text{th}} – 5^{\text{th}} \text{ value})

Median=0+0.5(3−0)\text{Median} = 0 + 0.5(3 – 0)

Median=1.5\text{Median} = 1.5

vi. Describe four advantages of mode.

Answer:

  • It is easy to understand and simple to calculate.
  • It is not affected by extreme large or small values.
  • It can be located only by inspection in ungrouped data and discrete frequency distribution.
  • It can be useful for qualitative data.
  • It can be computed in open-end frequency table.
  • It can be located graphically.

vii. Contrast between simple and composite index numbers.

Answer:

BasisSimple Index NumberComposite Index Number
DefinitionMeasures change in a single itemMeasures change in a group of items
Data UsedOnly one commodityMultiple commodities combined
PurposeShows price/quantity change of one goodShows overall change in a group (e.g., cost of living)
AccuracyLess representativeMore representative of real conditions
ComplexityEasy to calculateMore complex (may use weights)
ExamplesPrice of wheat over timeConsumer Price Index (CPI)

viii. Given Σp1qo = 7052, Σpoqo = 6095, Σpoq1 = 6980, Σp1q1 = 8061. Find Fisher Ideal Index.

Solution:

Fisher’s Index=(∑p1q0∑p0q0)×(∑p1q1∑p0q1)×100\text{Fisher’s Index} = \sqrt{ \left( \frac{\sum p_1 q_0}{\sum p_0 q_0} \right) \times \left( \frac{\sum p_1 q_1}{\sum p_0 q_1} \right) } \times 100

Fisher’s Index=70526095×80616980×100\text{Fisher’s Index} = \sqrt{ \frac{7052}{6095} \times \frac{8061}{6980} } \times 100

Fisher’s Index=56,846,17242,543,100×100\text{Fisher’s Index} = \sqrt{ \frac{56,\!846,\!172}{42,\!543,\!100} } \times 100
Fisher’s Index=133.62\text{Fisher’s Index} = 133.62

ix. Define consumer price index number.

Answer:

“Consumer Price Index numbers are intended to measure the changes in the prices paid by the consumer for purchasing a specified “basket” of goods and services during the current year as compared to the base year”.

Methods to Calculate

Aggregative Expenditure Method

P0n=∑pnq0∑p0q0P_{0n} = \frac{\sum p_n q_0}{\sum p_0 q_0}

Household Budget method Or Family Budget Method

P0n=∑(pnp0)p0q0∑p0q0×100=∑WI∑WP_{0n} = \frac{\sum \left( \frac{p_n}{p_0} \right) p_0 q_0}{\sum p_0 q_0} \times 100 = \frac{\sum W I}{\sum W}

Where I=pnp0×100andW=p0q0\text{Where } I = \frac{p_n}{p_0} \times 100 \quad \text{and} \quad W = p_0 q_0

SECTION – II (Marks: 16) (2×8=16)

4. (a) Following data represents the reported sales for 26 companies in the shoe industry:
32, 36, 54, 38, 17, 41, 22, 33, 22, 31, 21, 18, 46, 36, 11, 31, 29, 12, 23, 51, 12, 13, 37, 33, 27, 26

Construct a frequency distribution using classes with a width of 10 i.e. 10-20, 20-30 etc.

Solution:

Class IntervalFrequency
10 – 206
20 – 307
30 – 409
40 – 502
50 – 602
Total∑f = 26

(b) Draw a frequency polygon from the following data:

C.I.5–910–1415–1920–2425–2930–3435–39
f51118221594

Solution:

Frequency Polygon

5. (a) For the following data, find arithmetic mean by coding method:

Marks30–3940–4950–5960–6970–79
f8871908620

Solution:

MarksfXD = X-64.5U = D/10fU
30–39834.5-30-3-24
40–498744.5-20-2-174
50–5919054.5-10-1-190
60–698664.5000
70–792074.510120
∑f =391∑fU =-368
A.M. X‾=A+∑fU∑f×h\text{A.M. } \bar{X} = A + \frac{\sum fU}{\sum f} \times h

A.M. X‾=64.5+−368391×10\text{A.M. } \bar{X} = 64.5 + \frac{-368}{391} \times 10

A.M. X‾=64.5+−3680391\text{A.M. } \bar{X} = 64.5 + \frac{-3680}{391}

A.M. X‾=64.5−9.411\text{A.M. } \bar{X} = 64.5 – 9.411

A.M. X‾=55.089\text{A.M. } \bar{X} = 55.089

(b) If D = X – 112, calculate median of X from the following data:

D-210123
f143045657268

Solution:

DffD
-214-28
13030
0450
16565
272144
368204
∑f = 294∑fD =415
A.M. X‾=A+∑fD∑f\text{A.M. } \bar{X} = A + \frac{\sum fD}{\sum f}

A.M. X‾=112+415294\text{A.M. } \bar{X} = 112 + \frac{415}{294}

A.M. X‾=113.411\text{A.M. } \bar{X} = 113.411

6. (a) Given the prices of four commodities. Construct price index numbers by simple average method taking (i) 2016 as base and (ii) average of all year aggregate as base:

YearCommodity
ABCD
2016817711955
2017625412852
201810487111100
2019937515496

Solution:

(i) 2016 as base

YearCommodity Prices & Price Relatives 
AP.R (A)BP.R (B)CP.R (C)DP.R (D)Median
201681(81/81)x100=10077(77/77)x100=100119(119/119)x100=10055(55/55)x100=100100
201762(62/81)x100=76.5454(54/77)x100=70.12128(128/119)x100=107.5652(52/55)x100=94.54(76.54+94.54)/2 = 85.54
2018104(104/81)x100=128.3987(87/77)x100=112.98111(111/119)x100=93.27100(100/55)x100=181.81(112.98+128.39)/2 = 120.685
201993(93/81)x100=114.8175(75/77)x100=97.40154(154/119)x100=129.4196(96/55)x100=174.54(114.81+129.41)/2 =122.11

(ii) average of all year aggregate as base:

First of all, we have to calculate arithmetic mean for all year prices of commodities:

YearCommodity
ABCD
2016817711955
2017625412852
201810487111100
2019937515496
Sum340293512303
Average A.M(340/4) = 85(293/4) = 73.25(512/4) = 128(303/4) = 75.75

Now, we are going to calculate price relatives and average of price relatives:

YearCommodity Prices & Price Relatives 
AP.R (A)BP.R (B)CP.R (C)DP.R (D)Median
201681(81/85)x100=95.2977(77/73.25)x100=105.11119(119/128)x100=92.9655(55/75.75)x100=72.60(92.96+95.29)/2 = 94.125
201762(62/85)x100=72.9454(54/73.25)x100=73.72128(128/128)x100=10052(52/75.75)x100=68.64(72.94+73.72)/2 = 73.33
2018104(104/85)x100=122.3587(87/73.25)x100=118.77111(111/128)x100=86.71100(100/75.75)x100=132(118.77+122.35)/2 = 120.56
201993(93/85)x100=109.4175(75/73.25)x100=102.38154(154/128)x100=120.3196(96/75.75)x100=126.73(109.41+120.31)/2 =114.86

 (b) A fair die is tossed twice. Find the probabilities that the sum of numbers appearing is: (i) At most 5 (ii) At least 10

Solution:

Sample Space=η(S)=Nn\text{Sample Space} = \eta(S) = N^n

Sample Space=η(S)=62=36\text{Sample Space} = \eta(S) = 6^2 = 36

All Possible outcomes or sample points

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

Sum Table

234567
345678
456789
5678910
67891011
789101112

Events

(i) Sum is at-most 5η(A)=10\text{(i) Sum is at-most 5} \quad \eta(A) = 10

[2,3,3,4,4,4,5,5,5,5]

(ii) Sum is at-least 10η(B)=6\text{(ii) Sum is at-least 10} \quad \eta(B) = 6

[10,10,10,11,11,12]

Probability

P(A)=η(A)η(S)=1036=0.2777P(A) = \frac{\eta(A)}{\eta(S)} = \frac{10}{36} = 0.2777

P(B)=η(B)η(S)=636=0.1666P(B) = \frac{\eta(B)}{\eta(S)} = \frac{6}{36} = 0.1666

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