Business Statistics Solved Paper FBISE 2017 ICOM II, MCQS, Short Questions, Extensive Questions

In this blog post, I am going to discuss the paper of Business Statistics Solved Paper FBISE 2017 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 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 2017 ICOM II, MCQS, Short Questions, Extensive Questions

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

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MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.

(i)	The data obtained from college record is
	A. Primary	B. Secondary	C. Grouped	D. Qualitative

(ii)	A characteristic that varies from one object to another is called:
	A. Attribute	B. Variable	C. Constant	D. Statistic

(iii)	The size of class is also called:
	A. Class Interval	B. Class Frequency	C. Class Mark	D. Class Boundary

(iv)	A graph of Time Series is called:
	A. Ogive	B. Histogram	C. Historigram	D. Scatter diagram

(v)	Total angle of pie chart is:
	A. 270	B. 300	C. 320	D. 360

(vi)	Step deviation method is used for calculation of:
	A. Median	B. Mode	C. Mean	D. Quartiles

(vii)	If mean of 10 observations is 20, then their sum will be equal to:
	A. 100	B. 200	C. 10	D. 20

(viii)	In chain base method, the base period is:
	A. Fixed	B. 1^st Year	C. Last year	D. Previous

(ix)	If all the values are of equal importance, the index numbers are called?
	A. Weighted	B. Unweighted	C. Simple	D. Value

(x)	When two dice are rolled the total number of possible outcomes will be:
	A. 36	B. 6	C. 16	D. 216

Short Questions

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) What is classification?

Answer:

In statistics, classification is a process of grouping items or data into classes according to their attributes or characteristics.

(iii) The following data shows the number of students absent during the month of September 2015 from 2^nd year Statistics class.

5,3,7,4,6, 1,4,2,3,5,8,4,7, 1,5,2,6,8,7,9,3,10,6,5,0

Make a frequency distribution taking one as class interval.

Solution:

X	f
0	1
1	2
2	2
3	3
4	3
5	4
6	3
7	3
8	2
9	1
10	1
	*∑f = 25*

(iv) What are different methods for collection of Primary data?

Answer:

Direct Questionnaire
Through SMS
Through Telephone
Through Social Media Platforms
Through Poll

(v) What is Histogram?

Answer:

Histogram is a graphical representation of a frequency or probability distribution of the data. In Histogram, we take range of values on x-axis whereas frequency of specific values is taken on y-axis.

(vi) **Given that x = 10 + 5u, ∑fu= 46, ∑f = 125. Find arithmetic mean.**

Solution:

$\overline X=A+\frac{\sum fu}{\sum f}\times h$

$\overline X=10+\frac{46}{125}\times5=11.84$

(vii) Define median.

Answer:

In measures of central tendency, Median is a type of average that calculates central point of the data. It is also called positional average. Specific data takes a part in the calculation of median. Its formulas for ungrouped and grouped data are given below:

The formula for ungrouped data:

$Median\ \widetilde{X} = The\ Value\ of\ \left( \frac{n + 1}{2} \right)nd\ item.\$

The formula for grouped data:

$Median\ \widetilde{X} = L + \frac{h}{f}\left( \frac{n}{2} - C \right)\$

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(viii) Distinguish between simple and composite index numbers.

Answer:

In simple index number price or quantity of a single product is taken whereas in composite index, the price or quantity is taken related to multiple products.

(ix) If Laspeyer’s index is 120 and Paasche’s index is 110, then find Fisher’s index.

Solution:

$\mathbf{Fisher's\ Ideal\ Index\ = \ }\sqrt{\mathbf{Laspeyer's\ Index\ \ \times \ Paasche's\ Index}}\$

$\mathbf{Fisher's\ Ideal\ Index\ = \ }\sqrt{\mathbf{120\ \ \times \ 110}}\mathbf{\ = 114.89}\$

(x) Define random experiment.

Answer:

Random experiment is an experiment that is performed in probability distribution in order to know the probability of a particular event.

(xi) How many different permutations can be formed from the word “PAKPATTAN”

Solution

Total letters (n) = 9

Letter P = 2, Letter A = 3, Letter K = 1, Letter T = 2, Letter N = 1

$\mathbf{P}\mathbf{ermutaton =}\frac{\mathbf{8}\mathbf{!}}{\mathbf{3}\mathbf{!}\mathbf{2!2!1!1!}}\mathbf{= \ }\frac{\mathbf{8}\mathbf{\times}\mathbf{7}\mathbf{\times}\mathbf{6}\mathbf{\times}\mathbf{5}\mathbf{\times}\mathbf{4}\mathbf{\times}\mathbf{3}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{1}}{\mathbf{3}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{1}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{1}\mathbf{\times}\mathbf{2}\mathbf{\times}\mathbf{1}\mathbf{\times}\mathbf{1}\mathbf{\times}\mathbf{1}}\mathbf{=}\frac{\mathbf{40320}}{\mathbf{24}}\mathbf{= 1680}\$

Extensive Questions

Section C (Marks 16)

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

Q.3 Determine median and mode from the following frequency distribution.

Age Last Birthday	15–19	20–24	25–29	30–34	35–39	40–44
No of Persons	4	20	38	24	10	4

Solution

Marks	f	Class Boundaries	X	C.F
15—19	4	14.5–19.5	17	4
20—24	20	19.5–24.5	22	24
25—29	38	24.5–29.5	27	62
30—34	24	29.5–34.5	32	86
35—39	10	34.5–39.5	37	96
40—44	4	39.5–44.5	42	100
Sum	100
	∑f=n

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

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Model Class = n/2 = 100/2 = 50 falls in C.F of 62 so data is:

L = 24.5, h = 5, f = 38, n/2 =100/2 = 50 & C = 24

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

$\mathbf{Median = 24.5 +}\frac{\mathbf{5}}{\mathbf{38}}\left( \mathbf{50 - \ 24} \right)\mathbf{\ }\mathbf{\ }\$

$\mathbf{Median = 27.92}\$

$\left( \mathbf{ii} \right)\mathbf{\ Mode = L +}\frac{\mathbf{fm - f}\mathbf{1}}{\left( \mathbf{fm - f}\mathbf{1} \right)\mathbf{+ \ (fm - f}\mathbf{2)}}\mathbf{x\ h}\$

Model Class: Maximum frequency is 38 so data for model class is:

L = 24.5, fm = 38, f1 = 20, f2 = 24 & h = 5

$\mathbf{Mode = 24.5 +}\frac{\mathbf{38 - 20}}{\left( \mathbf{38 - 20} \right)\mathbf{+ \ (38 - 24)}}\mathbf{\times \ 5}\$

$\mathbf{Mode = \ 24.5 +}\frac{\mathbf{18}}{\mathbf{32}}\mathbf{\times \ 5\ = 27.31}\$

Q.4 Compute Chain indices from the following data of prices.

Year	Wheat	Sugar	Ghee	Cotton
1965	16	40	200	50
1966	20	50	250	60
1967	18	70	240	80
1968	25	80	300	100

Solution

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

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

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Year	Wheat	Sugar	Ghee	Cotton	Median	Chain Index
1965	16	40	200	50	(40+50)/2=45	45
1966	$\frac{\mathbf{20}}{\mathbf{16}}\mathbf{\times 100}\$ =125	$\frac{\mathbf{50}}{\mathbf{40}}\mathbf{\times 100}\$ =125	$\frac{\mathbf{250}}{\mathbf{200}}\mathbf{\times 100}\$ =125	$\frac{\mathbf{60}}{\mathbf{50}}\mathbf{\times 100}\$ =120	(125+125)/2 =125	(45 x 125)/100 = 56.25
1967	$\frac{\mathbf{18}}{\mathbf{20}}\mathbf{\times 100}\$ = 90	$\frac{\mathbf{70}}{\mathbf{50}}\mathbf{\times 100}\$ = 140	$\frac{\mathbf{240}}{\mathbf{250}}\mathbf{\times 100}\$ = 96	$\frac{\mathbf{80}}{\mathbf{60}}\mathbf{\times 100}\$ = 133	(96+133)/2=114.5	(56.25 x 114.5)/100 =64.40
1968	$\frac{\mathbf{25}}{\mathbf{18}}\mathbf{\times 100}\$ =139	$\frac{\mathbf{80}}{\mathbf{70}}\mathbf{\times 100}\$ =114	$\frac{\mathbf{300}}{\mathbf{240}}\mathbf{\times 100}\$ =125	$\frac{\mathbf{100}}{\mathbf{80}}\mathbf{\times 100}\$ =125	(125+125)/2=125	(64.40 x 125)/100 = 80.5

Q.5 Show that in a single throw with two dice, the chance of throwing more than 7 is equal to that of throwing less than 7 and hence find probability of throwing exactly 7.

Solution

Sample Space

$\mathbf{\eta}\left( \mathbf{s} \right)\mathbf{=}\mathbf{N}^{\mathbf{n}}\mathbf{=}\mathbf{6}^{\mathbf{2}}\mathbf{= 36}\$

All Possible Outcome

1, 1	2, 1	3, 1	4, 1	5, 1	6, 1
1, 2	2, 2	3, 2	4, 2	5, 2	6, 2
1, 3	2, 3	3, 3	4, 3	5, 3	6, 3
1, 4	2, 4	3, 4	4, 4	5, 4	6, 4
1, 5	2, 5	3, 5	4, 5	5, 5	6, 5
1, 6	2, 6	3, 6	4, 6	5, 6	6, 6

Sum Table

2	3	4	5	6	7
3	4	5	6	7	8
4	5	6	7	8	9
5	6	7	8	9	10
6	7	8	9	10	11
7	8	9	10	11	12

Events

$\left( \mathbf{i} \right)\mathbf{\ More\ than\ 7\ \eta}\left( \mathbf{A} \right)\mathbf{=}\mathbf{15}\$

$\left( \mathbf{ii} \right)\mathbf{\ Less\ than\ 7\ \eta}\left( \mathbf{B} \right)\mathbf{=}\mathbf{15}\$

$\left( \mathbf{iii} \right)\mathbf{\ Exactly\ 7\ \eta}\left( \mathbf{C} \right)\mathbf{=}\mathbf{6}\$

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Probabilities

$\left( \mathbf{i} \right)\mathbf{P}\left( \mathbf{A} \right)\mathbf{=}\frac{\mathbf{\eta(A)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{15}}{\mathbf{36}}\mathbf{= 0.417}\$

$\left( \mathbf{ii} \right)\mathbf{P}\left( \mathbf{B} \right)\mathbf{=}\frac{\mathbf{\eta(B)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{15}}{\mathbf{36}}\mathbf{= 0.417}\$

$\left( \mathbf{iii} \right)\mathbf{P}\left( \mathbf{C} \right)\mathbf{=}\frac{\mathbf{\eta(C)}}{\mathbf{\eta(S)}}\mathbf{=}\frac{\mathbf{6}}{\mathbf{36}}\mathbf{= 0.167}\$