Business Statistics Solved Paper FBISE 2024 Annual ICOM II, MCQS, Short Questions, Extensive Questions

Business Statistics Solved Paper FBISE 2024 Annual ICOM II, MCQS, Short Questions, Extensive Questions

In this blog post, we delve into the intricacies of the Business Statistics Solved Paper FBISE 2024 Annual ICOM II, MCQS, Short Questions, Extensive Questions, providing detailed solutions and explanations. Whether you’re a student looking to ace your exams or simply keen on enhancing your statistical knowledge, this guide will serve as a valuable resource. We’ll cover key topics including: Introduction to Statistics, Graphical Presentation of Data, Averages, Index Numbers & Probability. By the end of this post, you will have a thorough understanding of these essential statistical concepts and be well-prepared to tackle similar problems in your studies or professional work. Join us as we unravel the complexities of business statistics and turn them into manageable, solvable components.

Business Statistics Solved Paper FBISE 2024 Annual ICOM II, MCQS, Short Questions, Extensive Questions

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Solved Papers Business Statistics

Business Statistics Solved Paper FBISE 2024 Annual ICOM II, MCQS, Short Questions, Extensive Questions

MCQS

MCQS will be available soon when available

Short Questions

SECTION — B (Marks 24)

Q. 2 Write brief answers of any EIGHT parts. (8 x 3 = 24)

(i) Distinguish between variable and constant.

Answer A variable is a data, numbers, figures which does not remain fix for example, income, consumption, saving etc.; whereas constant is a number or figure which remain fix for example, Avogadro number, gravitational force, mass of earth, value of Pi etc.

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(ii) Give the importance of Statistics in Education.

Answer: Statistics can help education in number of ways:

  1. Assessment and Evaluation: With the help of statistics, student’s performance can be assessed and evaluated.
  2. Educational Research: With the help of statistics, educational research about trends can be done.
  3. Policy making & Decisions: With the help of statistics, decisions and policies about education can be made on time.
  4. Improving Teaching & Learning: With the help of statistical data analysis, teaching methods and learning process can be developed.
  5. Quality Control: Quality control of education can be maintained with the help of statistical data analysis and evaluation.
  6. Prediction: Statistics can predict about future forecast of education and timely decision can be made.

(iii) Differentiate between grouped data and ungrouped data.

Answer

Ungrouped Data

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

Grouped Data

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

(iv) Find [∑f(X – 4)]² given X = 3, 4, 5, 8 and f = 2, 1, 2, 2

Solution:

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Xf(X – 4)f(X – 4)
32-1-2
4100
5212
8248
   ∑f(X – 4)=8

[∑f(X – 4)]² = 8²=64

(v) If sum of 15 values is 300 and by addition of two more values, it becomes 360. Find the new values if the ratio between them is 1:4.

Solution:

∑X1=300, n1 =15, ∑X2=360, n2 = 17

Difference = 360 – 300 = 60

Sum of the ratio = 1+4=5

    \[ \mathbf{1}\mathbf{st\ Value = 60 \times}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{= 12}\  \]

    \[ \mathbf{2}\mathbf{nd\ Value = 60 \times}\frac{\mathbf{4}}{\mathbf{5}}\mathbf{= 48}\  \]

(vi) Deviations from 10.5 of ten items are: -1.3, 2.0, 2.9, 7.5, -4.6, -3.4, 8.2, 9.3, -7.4, 5.6. Calculate the arithmetic mean.

Solution:

D-1.32.02.97.5-4.6-3.48.29.3-7.45.6∑D=18.8

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

    \[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{= 10.5 +}\frac{\mathbf{18.8}}{\mathbf{10}}\mathbf{= 12.38}\  \]

(vii) Give fm = 33, f1=26, f2 =23, h = 10, l =40.5. Find mode.

Solution:

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

    \[ \mathbf{Mode = 40.5 +}\frac{\mathbf{33 - 26}}{\left( \mathbf{33 - 26} \right)\mathbf{+}\left( \mathbf{33 - 23} \right)}\mathbf{\times 10}\ \]

    \[ \mathbf{Mode = 40.5 +}\frac{\mathbf{70}}{\mathbf{17}}\ \]

    \[ \mathbf{Mode = 44.61}\ \]

(viii) If the mode and mean of a moderately asymmetrical series are 16 and 20.2 respectively, compute the value of median.

Solution:

Mode = 3Median – 2Mean

16 = 3Median – 2(20.2)

3Median = 16 + 40.4

3Median = 56.4

Median = 56.4/3

Median = 18.8

(ix) Compute Fisher’s price index number for the given data: ∑poqo = 35310, ∑p1qo = 41140, ∑p1q1= 46707 and ∑poq1 = 39644.

Solution:

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Index =}\sqrt{\left( \frac{\sum p1qo}{\sum poqo}\mathbf{\times}\frac{\sum p1q1}{\sum poq1} \right)}\mathbf{100}\  \]

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    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Index =}\sqrt{\left( \frac{41140}{35310}\mathbf{\times}\frac{46707}{39644} \right)}\mathbf{100}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Index = 137.26}\  \]

(x) If W = 20, 25, 30, 40, & I = 100, 105, 110, 120. Find weighted average of relative’s Index No.

Solution:

WIWI
201002000
251052625
301103300
401204800
W=115 WI=12725

    \[ \mathbf{Weighted\ Average = \ }\frac{\mathbf{\sum WI}}{\mathbf{\sum W}}\mathbf{=}\frac{\mathbf{12725}}{\mathbf{115}}\mathbf{= 110.65}\  \]

(xi) Solve: (a) 14P11 (b) 20C5

    \[ \mathbf{a.\ }\begin{pmatrix}\mathbf{14} \\\mathbf{P} \\\mathbf{11} \\\end{pmatrix}\mathbf{,\ b.}\begin{pmatrix}\mathbf{20} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{\ }\  \]

Solution:

    \[ \left( \mathbf{a} \right)\mathbf{Permutation =}\frac{\mathbf{n!}}{\mathbf{(n - r)!}}\  \]

    \[ \begin{pmatrix}\mathbf{14} \\\mathbf{P} \\\mathbf{11} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{14!}}{\left( \mathbf{14 - 11} \right)\mathbf{!}}\mathbf{=}\frac{\mathbf{14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{\mathbf{3 \times 2 \times 1}}\  \]

    \[ \begin{pmatrix}\mathbf{14} \\\mathbf{P} \\\mathbf{11} \\\end{pmatrix}\mathbf{= \ }\mathbf{14}\mathbf{,}\mathbf{529}\mathbf{,}\mathbf{715}\mathbf{,}\mathbf{200}\ \]

    \[ \left( \mathbf{b} \right)\mathbf{Combination =}\frac{\mathbf{n!}}{\mathbf{r!(n - r)!}}\  \]

    \[ \begin{pmatrix}\mathbf{20} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{= \ }\frac{\mathbf{n!}}{\mathbf{r!(n - r)!}}\mathbf{=}\frac{\mathbf{20!}}{\mathbf{5!}\left( \mathbf{20 - 5} \right)\mathbf{!}}\  \]

    \[ \frac{\mathbf{20!}}{\mathbf{5!}\left( \mathbf{20 - 5} \right)\mathbf{!}}\mathbf{=}\frac{\mathbf{20!}}{\mathbf{5! \times 15!}}\ \]

    \[ \begin{pmatrix}\mathbf{20} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{20 \times 19 \times 18 \times 17 \times 16 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{\mathbf{5 \times 4 \times 3 \times 2 \times 1 \times 15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}\ \]

    \[ \begin{pmatrix}\mathbf{20} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{20 \times 19 \times 18 \times 17 \times 16}}{\mathbf{5 \times 4 \times 3 \times 2 \times 1}}\ \]

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    \[ \begin{pmatrix}\mathbf{20} \\\mathbf{C} \\\mathbf{5} \\\end{pmatrix}\mathbf{=}\frac{\mathbf{1860480}}{\mathbf{120}}\mathbf{= 15504}\ \]

Section-C (Marks 16)

Extensive Questions

Note: Attempt any two questions. All questions carry equal marks.

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

Class Limits3.0—3.94.0—4.95.0—5.96.0—6.97.0—7.98.0—8.9
Frequency13274030164

Solution:

Class LimitsFrequencyC.BXfxC.F
3.0—3.9132.95—3.953.4544.8513
4.0—4.9273.95—4.954.45120.1540
5.0—5.9404.95—5.955.4521880
6.0—6.9305.95—6.956.45193.5110
7.0—7.9166.95—7.957.45119.2126
8.0—8.947.95—8.958.4533.8130
 ∑f=n=130  ∑fx=729.5 

    \[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\  \]

    \[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{729.5}}{\mathbf{130}}\mathbf{= 5.611}\  \]

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

n/2=130/2=65 falls in c.f of 80 so L = 4.95, h = 1, f = 40 & C = 40

    \[ \mathbf{Median = 4.95 +}\frac{\mathbf{1}}{\mathbf{40}}\left( \mathbf{65 - 40} \right)\  \]

    \[ \mathbf{Median = 5.575}\  \]

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

Maximum frequency fm is 40 so L = 4.95, h = 1, f1 = 27 & f2 = 30

    \[ \mathbf{Mode = 4.95 +}\frac{\mathbf{40 - 27}}{\left( \mathbf{40 - 27} \right)\mathbf{+}\left( \mathbf{40 - 30} \right)}\mathbf{\times 1}\  \]

    \[ \mathbf{Mode = 4.95 +}\frac{\mathbf{13}}{\mathbf{23}}\  \]

    \[ \mathbf{Mode = 5.515}\  \]

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Q.4 Compute index numbers of prices from the following data taking 1981 as base using median as an average: (08)

YearPrice
ABC
1981188552
1982227660
1983288066
1984319580

Solution:

YearPrice RelativesMedian
ABC 
1981100100100100
1982(22/18)x100=122.22(76/85)x100=89.411(60/52)x100=115.38115.38
1983(28/18)x100=155.55(80/85)x100=94.11(66/52)x100=126.92126.92
1984(31/18)x100=172.22(95/85)x100=111.76(80/52)x100=153.84153.84

Q.5 Find the probability for each of the following:

(i) An even number appears in a single toss of fair die.

(ii) The sum of 5 appears in a single toss of a pair of fair dice.

(iii) A queen, ace, jack of spades or king of diamonds appears in drawing a single card from a well-shuffled ordinary deck of 52 cards.

(iv) Of getting a number less than 7, if a die is rolled one time.

Solution (i) and (iv):

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

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

All Possible outcomes: {1, 2, 3, 4, 5, 6}

Events:

    \[ \left( \mathbf{i} \right)\mathbf{Even\ Number\ \eta}\left( \mathbf{A} \right)\mathbf{= 3}\  \]

    \[ \left( \mathbf{iv} \right)\mathbf{Number\ less\ than\ 7\ \eta}\left( \mathbf{B} \right)\mathbf{= 6}\  \]

Probability:

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    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{3}}{\mathbf{6}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{= 0.5}\  \]

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

Solution (ii):

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

    \[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{=}\mathbf{6}^{\mathbf{2}}\mathbf{= 36}\  \]

All Possible Outcomes

11 21 31 41 51 61
12 22 32 42 52 62
13 23 33 43 53 63
14 24 34 44 54 64
15 25 35 45 55 65
16 26 36 46 56 66

Sum Table

234567
345678
456789
5678910
67891011
789101112

Events:

    \[ \left( \mathbf{i} \right)\mathbf{Sum\ of\ 5\ \eta}\left( \mathbf{A} \right)\mathbf{= 4}\  \]

Probability

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{4}}{\mathbf{36}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{9}}\ \]

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Solution (iii):

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

    \[ \mathbf{Sample\ Space\ \eta}\left( \mathbf{s} \right)\mathbf{=}\mathbf{52}^{\mathbf{1}}\mathbf{= 52}\  \]

Events:

    \[ \left( \mathbf{i} \right)\mathbf{\ A\ queen,\ ace,\ jack\ of\ spades\ or\ king\ of\ diamonds}\mathbf{\ \eta}\left( \mathbf{A} \right)\mathbf{= 10}\ \]

Probability:

    \[ \mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta}\left( \mathbf{A} \right)}{\mathbf{\eta}\left( \mathbf{S} \right)}\mathbf{=}\frac{\mathbf{10}}{\mathbf{52}}\mathbf{= 0.1923}\  \]

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