Solved Paper Statistics I FBISE 2011

Solved Paper Statistics I FBISE 2011, Dive into a comprehensive solution guide to the FBISE Statistics I 2011 paper! This blog post provides detailed explanations and step-by-step solutions for key topics like measures of central tendency, dispersion, data presentation, index numbers, correlation, regression, and time series. Whether you’re preparing for exams or reinforcing your understanding, this post is tailored to simplify concepts and help you excel in statistics. This topic is equally important for the students of statistics across all the major Boards and Universities such as FBISE, BISERWP, BISELHR, MU, DU, PU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Solved Paper Statistics I FBISE 2011

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

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Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)	The data which have already been collected by someone are called_______ data.
	A	Secondary	B	Primary
	C	Array	D	None of these

(ii)	The number 143.9500 rounded off to the nearest tenth (one decimal place) is:
	A	143.9	B	144.0
	C	143.0	D	144

(iii)	Weight of the earth is:
	A	Discrete variable	B	Qualitative variable
	C	Continuous Variable	D	Difficult to tell

(iv)	The budgets of two families can be compared by:
	A	Sub divided rectangles	B	Pie-diagram
	C	Both A & B	D	Histogram

(v)	A relative frequency distribution presents frequencies in terms of:
	A	Fractions	B	Whole Numbers
	C	Percentage	D	None of these

(vi)	If X̅ = 25, which of the following will be minimum?
	A	∑(X – 27)²	B	∑(X – 25)²
	C	∑(X – 22)²	D	∑(X + 25)²

(vii)	Harmonic mean for any two positive numbers a and b is:
	A	$\frac{2}{a + b}\$	B	$\frac{\mathbf{2}\mathbf{ab}}{\mathbf{a}\mathbf{+}\mathbf{b}}\$
	C	$\frac{a + b}{2}\$	D	$\frac{a + b}{2ab}\$

(viii)	To compare the variation of two or more than two series, we use:
	A	Mean absolute variation	B	Variance
	C	Coefficient of variation	D	Corrected standard deviation

(ix)	In a symmetrical distribution if Q1 = 4, Q3 = 12 then median is:
	A	4	B	6
	C	8	D	Zero

(x)	The first three moments of a distribution about the mean X̅ are 1, 4 and 0. The distribution is:
	A	Symmetrical	B	Skewed to left
	C	Skewed to the right	D	Normal

(xi)	The distribution is positively skewed if:
	A	Arithmetic Mean < Mode	B	Arithmetic Mean > Mode
	C	Arithmetic Mean > Median	D	Both B and C

(xii)	_________method is uses quantities consumed in the base period when computing a weighted index:
	A	Laspeyre’s	B	Paasche’s
	C	Fisher’s	D	None of these

(xiii)	Prices used in the construction of consumer price index number are______prices.
	A	Retail	B	Wholesale
	C	The fixed	D	None of these

(xiv)	The mean of 10 observations is 10. All observations are increased by 10%. The mean of the increased observations shall be:
	A	20	B	11
	C	10	D	100

(xv)	If all the actual and estimated values of y are the same on the regression line, the sum of squares of errors will be:
	A	0	B	11
	C	10	D	100

(xvi)	The value of the coefficient of correlation lies between:
	A	0 and 1	B	0 and -1
	C	-1 and +1	D	-2 and +2

(xvii)	In time series a business cycle has:
	A	One phase	B	Two phases
	C	Four Phases	D	Three Phases

Short Question

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SECTION-B

Q.2 Attempt any fourteen parts.

(i) Differentiate between descriptive & Inferential Statistics.

Answer:

Descriptive Statistics

Descriptive statistics is a branch of statistics in which data is presented in the form of tables, graphs and charts. Measures of central tendency and dispersion is also a part of descriptive statistics. The aim of descriptive statistics is to present the data in informative way.

Example: Teacher gathering data of student’s marks, calculates averages, variation and presenting in graph.

Inferential Statistics

Inferential statistics is also a branch of statistics in which sample data is used to predict about population. Inferential statistics is used to predict the future trend and fluctuations. Key functions of inferential statistics is to make hypothesis, regression analysis etc.

(ii) What are the functions of statistics?

Answer:

Statistics play important role across different fields all around the world. Its key functions are given below:

Data Collection: Gathering data through surveys, Questionnaires, Researches.

Data Organization: Organizing raw data into organized form for further analysis.

Data Summarization: Calculate Central Tendencies and dispersion for meaningful insight.

Data Interpretation: Making pattern and trends from the data analysis and draw conclusions.

Data Comparison: Make data comparison through cross tabulation of ANOVA and hypothesis.

Prediction and Forecasting: To predict the future prediction and trends through inferential statistics.

Decision Making: Making timely decision with the help of data.

Quality Control: Ensuring quality control through using data techniques.

Customer Relationship: Statistics is a key tool to know the customer taste, demand and trends to make better relationship with customer.

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(iii) What are the common types of frequency distribution?

Answer:

Grouped frequency distribution
Ungrouped frequency distribution
Cumulative frequency distribution
Relative frequency distribution
Cumulative frequency distribution

(iv)Differentiate between Historigram and Histogram.

Answer:

Histogram is a graph of frequency distribution in which class boundaries with suitable width is taken on X-axis where as 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.

Solution:

Data

Shares	Per unit price	Total price
20	8.20 Per share	164
100	10.90 Per share	1090
50	9.40 Per share	470
200	7.80 per share	1560
Total Shares = 370 = n		∑x = 3284

$Mean\ Cost\ per\ Share = \ \frac{\sum x}{n} = \ \frac{3284}{370} = 8.87\ \$

(vi) The mean and geometric mean of three numbers are 7 and 4, respectively. Find all the three numbers if mean of two numbers is 10.

Solution:

Let the three numbers be a, b, c, and if mean of two numbers is 10, then:

$\frac{a + b}{2} = 10$

$a + b = 20(i)$

$b = 20 - a$

$\frac{a + b + c}{3} = 7$

$a + b + c = 21(ii)$

$c = 21 - a - b$

$\sqrt[3]{abc} = 4$

$abc = 4^{3} = 64$

Take equation i & ii & get c

a + b = 20 (i)

a + b + c = 21 (ii)

C = 1

If c = 1 then ab = 64

ab = 64

⟹a (20−a) =64

⟹20a – a² – 64= 0

⟹a2−20a+64=0

⟹ (a−16) (a−4) =0

⟹ {a,b,c}={16,4,1}

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(vii) What are the 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.

(viii) The first four moments about X = 17.5 of a distribution are 0.3, 74, 45 and 12125. Find whether the distribution is leptokurtic or Platykurtic.

Solution:

$\mathbf{b}\mathbf{2 = \ }\frac{\mathbf{m}\mathbf{4}}{\mathbf{m}\mathbf{2²}}\mathbf{=}\frac{\mathbf{12125}}{\mathbf{74²}}\mathbf{=}\frac{\mathbf{12125}}{\mathbf{5476}}\mathbf{= 2.21\ }\$

$\mathbf{Since\ \beta}\mathbf{2} < 3\ so\ the\ distribution\ is\ platykurtic\$

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(ix) Given: Calculate Median, Mode, Q3 and D7.

Given:

No. of Children	0	1	2	3	4	5
No. of families	4	7	14	22	12	6

Calculate Median, Mode, Q3 and D7.

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Solution:

No. of Children (X)	No. of Families (f)	C.f
0	4	4
1	7	11
2	14	25
3	22	47
4	12	59
5	6	65

$Median = \frac{n + 1}{2} = \frac{65 + 1}{2} = 33\ falls\ in\ c.f\ of\ 47\ so\ median\ is\ 3.\$

$Mode = maximum frequency is 22 so mode is 3.$

$Q3 = \frac{3n + 1}{4} = \frac{3(65) + 1}{4} = 49\ falls\ in\ c.f\ of\ 59\ so\ Q3\ is\ 4.\$

$D7 = \frac{7n + 1}{10} = \frac{7(65) + 1}{10} = 45.6\ falls\ in\ c.f\ of\ 59\ so\ D7\ is\ 4.\$

(x) Given n1=15, n2 = 25, n3 = 40, X̅1=15.8, X̅2=22.5, X̅3=30.2, S1=2.45, S2 = 4.65, S3 = 7.28. Compute combined coefficient of variation.

Solution

X̅1 = 15.8, n1 = 15, S1=2.45, S1² = 6

X̅2 = 22.5, n2 = 25, S2=4.65, S2² = 21.6225

X̅3 = 30.2, n3 = 40, S3=7.28, S3² = 52.9984

$\overline{X}c = \ \frac{n1\ \overline{X}1 + n2\overline{X}2 + n3\overline{X}3}{n1 + n2 + n3}\$

$\overline{X}c = \ \frac{(15)(15.8) + (25)(22.5) + (40)(30.2)}{15 + 25 + 40}\$

$\overline{X}c = \frac{237 + 562.5 + 1208}{80} = \frac{2007.5}{80}\$

$\overline{X}c = 25.09\ \$

Combined Variance

$S²c = \ \frac{n1\lbrack S^{2}1 + \left( \overline{X}1 - \overline{X}c \right)^{2} + \ n2\lbrack S^{2}2 + \left( \overline{X}2 - \overline{X}c \right)^{2} + n3\lbrack S^{2}3 + \left( \overline{X}3 - \overline{X}c \right)^{2}}{n1 + n2 + n3}\$

$S²c = \ \frac{15\lbrack 6 + (15.8 - 25.09)^{2} + \ 25\lbrack 21.6225 + (22.5 - 25.09)^{2} + 40\lbrack 52.9984 + (30.2 - 25.09)^{2}}{15 + 25 + 40}\$

$S^{2}c = \ \frac{1384.5615 + \ 708.265 + 3164.42}{80}\$

$S²c = \ \frac{5257.2465}{80} = 65.71\$

$S²c = \ \sqrt{65.71} = 8.10\$

$Combined\ C.V = \ \frac{Combined\ Standard\ Deviation}{Combined\ A.M}\$

$Combined\ C.V = \frac{8.10}{25.09}\$

$Combined\ C.V = 0.322\$

(xi) What will be the standard deviation and the variance in each of the following case? (i) 2X (ii) X + 2 (iii) 2X + 4 if Var(X) = 25

Solution:

(i) 2X

$2Var(X)\$

$2(25)\ = \ 50\$

$S.D = \sqrt{50} = 7.07\$

(ii) X + 2

$var(X)\ + \ 2\$

$(25)\ + \ 2\ = \ 27\$

$S.D = \sqrt{27} = 5.19\$

(iii) 2X + 4

$2\ var(X)\ + \ 4\$

$2(25)\ + \ 4\ = \ 54\$

$S.D = \ \sqrt{54} = 7.34\$

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(xii) Compute chain index taking 1920 as base for the following data:

Year	1920	1921	1922	1923	1924	1925	1926	1927	1928	1929	1930
Price Relative	100	116	120	120	137	136	149	156	137	162	149

Solution:

\[ \frac{\mathbf{pn}}{\mathbf{pn - 1}}\mathbf{\times 100}\ \]

\[ \mathbf{ChainIndex}\mathbf{\ = \ }\frac{\mathbf{PreviousChain}\mathbf{\times}\mathbf{CurrentL}\mathbf{.}\mathbf{R}}{\mathbf{100}}\ \]

(xiii) What are the important uses of index number?

Answer:

Index numbers are used in the fields of commerce, meteorology, labour, industrial, etc.
The index numbers measure fluctuations during intervals of time, group differences of geographical position of degree etc.
They are used to compare the total variations in the prices of different commodities in which the unit of measurements differs with time and price etc.
They measure the purchasing power of money.
They are helpful in forecasting the future economic trends.
They are used in studying difference between the comparable categories of animals, persons or items.
Index numbers of industrial production are used to measure the changes in the level of industrial production in the country.
Index numbers of import prices and export prices are used to measure the changes in the trade of a country.
The index numbers are used to measure seasonal variations and cyclical variations in a time series.

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(xiv) Given the following information: Compute Base year weighted & current year weighted price index numbers.

$\sum p1qo = 41140,\ \sum poqo = 35310,\ \sum poq1 = 40048,\ \sum_{}^{}{p1q1 = 46707.}\$

Solution

$Base\ year\ weighted\ Index = \ \frac{\sum p1qo}{\sum poqo}\ \times \ 100\$

$Base\ year\ weighted\ Index = \ \frac{41140}{35310}\ \times 100 = 116.51\$

$Current\ year\ weighted\ Index = \ \frac{\sum p1q1}{\sum poq1}\ \times 100\$

$Current\ year\ weighted\ Index = \ \frac{46707}{40048} \times \ 100 = 116.62\$

(xv) What is meant by regression?

Answer:

Regression calculates the relationship between one dependent variables and one or more independent variables. Regression shows the effect of one unit change in an independent variable on the dependent variable.

(xvi) Given n = 10, ∑Dx = – 8, ∑Dy = 0, ∑D²x = 66, ∑D²y = 99, ∑DxDy = 72. Find r, bxy, byx.

Solution:

$rxy = \ \frac{\sum DxDy - \ \frac{(\sum Dx)(\sum Dy)}{n}}{\sqrt{\left\lbrack \left{ \sum Dx^{2} - \frac{(\sum Dx)^{2}}{n} \right}\left{ \sum Dy^{2} - \frac{(\sum Dy)^{2}}{n} \right} \right\rbrack}}\$

$rxy = \ \frac{72 - \ \frac{( - 8)(0)}{10}}{\sqrt{\left\lbrack \left{ 66 - \frac{( - 8)^{2}}{10} \right}\left{ 99 - \frac{(0)^{2}}{10} \right} \right\rbrack}}\$

$rxy = \ \frac{72}{\sqrt{59.6\ x\ 99}} = \ \frac{72}{76.81} = 0.93\$

$byx = \ \frac{\sum DxDy - \ \frac{(\sum Dx)(\sum Dy)}{n}}{\left( \sum Dx^{2} - \frac{(\sum Dx)^{2}}{n} \right)}\$

$byx = \ \frac{72 - \ \frac{( - 8)(0)}{10}}{66 - \frac{( - 8)^{2}}{10}}\$

$byx = \ \frac{72}{59.6}\$

$byx = \ \frac{72}{76.81}\$

$byx = 1.20\$

$bxy = \ \frac{\sum DxDy - \ \frac{(\sum Dx)(\sum Dy)}{n}}{\left( \sum Dy^{2} - \frac{(\sum Dy)^{2}}{n} \right)}\$

$bxy = \ \frac{72 - \ \frac{( - 8)(0)}{10}}{99 - \frac{(0)^{2}}{10}}\$

$bxy = \ \frac{72}{99}\$

$bxy = \ \frac{72}{76.81}\$

$bxy = 0.7272\$

(xvii) Differentiate between Signal and Noise

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Answer:

A variation in any time series caused by sequence of systematic components is called “Signal” whereas A variation in any time series caused by sequence of unsystematic components is called “Noise”

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(xviii) Fit a straight line to the following results for the years 1951-58 both inclusive ∑x = 0, ∑y = 2861.1, ∑x² = 168, ∑xy = 2215.5. Also compute trend values.

Solution:

Data:

∑x = 0, ∑y = 2861.1, ∑x² = 168, ∑xy = 2215.5, n = 8

Ŷ = a + bx

$b = \ \frac{\sum XY}{\sum X^{2}}\&\ a = \ \frac{\sum Y}{n}\$

$a = \ \frac{\sum Y}{n} = \frac{2861.1}{8} = 357.6375\$

$b = \ \frac{\sum XY}{\sum X^{2}} = \frac{2215.5}{168} = 13.1875\$

Ŷ = a + bx

Ŷ = 357.6375 + 13.1875x

Calculation of Trend Values

Year	X	Ŷ = 357.6375 + 13.1875x
1951	-7	Ŷ = 357.6375 + 13.1875(-7) = 265.325
1952	-5	Ŷ = 357.6375 + 13.1875(-5) = 291.7
1953	-3	Ŷ = 357.6375 + 13.1875(-3) = 318.075
1954	-1	Ŷ = 357.6375 + 13.1875(-1) = 344.45
1955	+1	Ŷ = 357.6375 + 13.1875(1) = 370.825
1956	+3	Ŷ = 357.6375 + 13.1875(3) = 397.2
1957	+5	Ŷ = 357.6375 + 13.1875(5) = 423.575
1958	+7	Ŷ = 357.6375 + 13.1875(7) = 449.95

(xix) Find 4-years moving averages cantered from the following data:

Year	1990	1991	1992	1993	1994	1995	1996
Y	45	88	65	59	95	81	788

Solution:

Year	Values	4 Years Moving Total	2 Values Moving Total	4 Years Centered Moving Average
1990	45
1991	88
		257
1992	65		564	70.5
		307
1993	59		607	75.875
		300
1994	95		1323	165.375
		1023
1995	81
1996	788

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Extensive Questions

Q.3 Calculate: (i) Mean Deviation from Median, (ii) Coefficient of Variation. From the following frequency distribution:

Max: Load (Short tons)	No of cables
9.3—9.7	2
9.8—10.2	5
10.3—10.7	12
10.8—11.2	17
11.3—11.7	14
11.8—12.2	6
12.3—12.7	3
12.8—13.2	1

Solution:

Max: Load (Short tons)	No of cables (f)	Class Boundaries	X	fX	fX²	C.f	\|X-median\|	f\|X-median\|
9.3—9.7	2	9.25—9.75	9.5	19	180.5	2	1.57	3.14
9.8—10.2	5	9.75—10.25	10	50	500	7	1.07	5.35
10.3—10.7	12	10.25—10.75	10.5	126	1323	19	0.57	6.84
10.8—11.2	17	10.75—11.25	11	187	2057	36	0.07	1.19
11.3—11.7	14	11.25—11.75	11.5	161	1851.5	50	0.43	6.02
11.8—12.2	6	11.75—12.25	12	72	864	56	0.93	5.58
12.3—12.7	3	12.25—12.75	12.5	37.5	468.75	59	1.43	4.29
12.8—13.2	1	12.75—13.25	13	13	169	60	1.93	1.93

	60			665.5	7413.75			34.34
	∑f=			∑fX=	∑fX²=			∑f\|X-median\| =

$A.M\ \overline{X} = \frac{\sum fx}{\sum f}\ = \frac{665.5}{60}\ = 11.09\$

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

$Median = 10.75 + \frac{0.5}{17}(30 - 19)\$

$Median = 10.75 + \frac{5.5}{17} = 11.07\$

$\frac{n}{2} = \frac{60}{2} = 30\ so\ model\ class\ is\ 10.75 - 11.25\$

$S.D = \ \sqrt{\left\lbrack \frac{\sum fx^{2}}{\sum f} - \ \left( \frac{\sum fx}{\sum f} \right)^{2} \right\rbrack}\$

$S.D = \sqrt{\left\lbrack \frac{7413.75}{60} - \ \left( \frac{665.5}{60} \right)^{2} \right\rbrack}\$

$S.D = \sqrt{123.5625 - 122.9881}\$

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$S.D = 0.757\$

$(i)M.D\ through\ Median = \frac{\sum f|X - median|}{median}\ \$

$M.D\ through\ Median = \frac{34.34}{11.07}\ \$

$M.D\ through\ Median = 3.10\$

$(ii)C.V = \left( \frac{S.D}{mean} \right)100\ \$

$C.V = \frac{0.757}{11.09} \times 100\ = 6.82\$

Q.4 Construct Consumer Price Index of 1980 on the basis of 1978 using: (i) Aggregative Method (ii) Family Budget Method

Articles	Quantity Consumed	Unit of Price	Prices (Rs.)
Articles	Quantity Consumed	Unit of Price	1978	1980
Rice	6 maunds	Per Seer	6	6.50
Wheat	10 maunds	Per maund	35	40
Grain	3 maunds	Per maund	60	90
Pulses	5 maunds	Per maund	120	144
Ghee	5 Seers	Per Seer	8	10
Sugar	1 maund	Per maund	240	300

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Solution:

Unit of Price	Article	Quantity Consumed (qo)	Prices (Rs.)		P1qo	poqo = W	I=(P1/Po) × 100	WI
Unit of Price	Article	Quantity Consumed (qo)	1978(Po)	1980(P1)	P1qo	poqo = W	I=(P1/Po) × 100	WI
Per Seer(6×40)	Rice	240	6	6.5	1560	1440	108.33333	156000
Per maund	Wheat	10	35	40	400	350	114.28571	40000
Per maund	Grain	3	60	90	270	180	150	27000
Per maund	Pulses	5	120	144	720	600	120	72000
Per Seer	Ghee	5	8	10	50	40	125	5000
Per maund	Sugar	1	240	300	300	240	125	30000

					3300	2850	742.61905	330000
					*∑p1qo=*	*∑poqo=*	*∑I=*	*∑WI=*

$\left( \mathbf{i} \right)Aggregative\ Index\ 1980 = \frac{\sum P1qo}{\sum Poqo}\ \times \ 100\$

$Aggregative\ Index\ 1980 = \ \frac{3300}{2850}\ \times \ 100\$

$Aggregative\ Index\ 1980 = 115.78\ Approx\$

$\left( \mathbf{ii} \right)Family\ budget\ Method\ 1980 = \frac{\sum WI}{\sum W}\ \$

$Family\ budget\ Method\ 1980 = \ \frac{330000}{2850}\ \$

$Family\ budget\ Method\ 1980 = 115.78\ Approx\$

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Q.5 The following table shows the ages (X) and systolic blood pressure (Y) of 12 women..

(X)	(Y)
56	147
42	125
72	160
36	118
63	149
47	128
55	150
49	145
38	115
42	140
68	152
60	155

Compute Regression line of Y on X & X on Y
Compute Coefficient of Correlation between X & Y.
Show that
$r = \ \sqrt{byx\ .bxy}\$

Solution:

(X)	(Y)	X²	Y²	XY
56	147	3136	21609	8232
42	125	1764	15625	5250
72	160	5184	25600	11520
36	118	1296	13924	4248
63	149	3969	22201	9387
47	128	2209	16384	6016
55	150	3025	22500	8250
49	145	2401	21025	7105
38	115	1444	13225	4370
42	140	1764	19600	5880
68	152	4624	23104	10336
60	155	3600	24025	9300

628	1684	34416	238822	89894
∑X =	∑Y =	∑X² =	∑Y² =	∑XY =

$\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{x}}{\mathbf{n}}\mathbf{= \ }\frac{\mathbf{628}}{\mathbf{12}}\mathbf{=}\mathbf{52.34}\$

$\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{y}}{\mathbf{n}}\mathbf{= \ }\frac{\mathbf{1684}}{\mathbf{12}}\mathbf{=}\mathbf{140.34}\$

$\mathbf{Sx}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{x}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\$

$\mathbf{Sx}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{34416}}{\mathbf{12}}\mathbf{-}\mathbf{\ }\left( \mathbf{52.34} \right)^{\mathbf{2}} \right\rbrack}\$

$\mathbf{Sx =}\sqrt{\mathbf{2868 - 2739.47}}\$

$\mathbf{Sx}\mathbf{=}\sqrt{\mathbf{128.53}}\$

$\mathbf{Sx}\mathbf{=}\mathbf{11.33}\$

$\mathbf{Sy}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{y}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{y}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\$

$\mathbf{Sy}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{238822}}{\mathbf{12}}\mathbf{-}\mathbf{\ }\left( \mathbf{140.34} \right)^{\mathbf{2}} \right\rbrack}\$

$\mathbf{Sy =}\sqrt{\mathbf{19901.84 - 19695.31}}\$

$\mathbf{Sy}\mathbf{=}\sqrt{\mathbf{206.53}}\$

$\mathbf{Sy}\mathbf{=}\mathbf{14.37}\$

$\mathbf{r}\mathbf{=}\frac{\frac{\mathbf{\sum}\mathbf{xy}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum}\mathbf{x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum}\mathbf{y}}{\mathbf{n}} \right)}{\mathbf{Sx}\mathbf{.}\mathbf{Sy}}\$

$\mathbf{r}\mathbf{=}\frac{\frac{\mathbf{89894}}{\mathbf{12}}\mathbf{-}\left( \frac{\mathbf{628}}{\mathbf{12}} \right)\left( \frac{\mathbf{1684}}{\mathbf{12}} \right)}{\left( \mathbf{11.33} \right)\mathbf{(}\mathbf{14.37}\mathbf{)}}\$

$\mathbf{r}\mathbf{=}\frac{\mathbf{7491.17}\mathbf{-}\mathbf{7345.3956}}{\mathbf{162.8121}}\$

$\mathbf{r}\mathbf{=}\frac{\mathbf{145.7744}}{\mathbf{162.8121}}\$

$\mathbf{r}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89}\$

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Positive Relation

$\mathbf{bxy}\mathbf{=}\mathbf{r}\left( \frac{\mathbf{\sigma x}}{\mathbf{\sigma y}} \right)\$

$\mathbf{bxy}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89}\left( \frac{\mathbf{11.33}}{\mathbf{14.37}} \right)\$

$\mathbf{bxy}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89}\left( \mathbf{0}\mathbf{.}\mathbf{79} \right)\$

$\mathbf{bxy}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{7031}\$

$\mathbf{byx}\mathbf{=}\mathbf{r}\left( \frac{\mathbf{\sigma y}}{\mathbf{\sigma x}} \right)\$

$\mathbf{byx}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89}\left( \frac{\mathbf{14.37}}{\mathbf{11.33}} \right)\$

$\mathbf{byx}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{89}\left( \mathbf{1.268} \right)\$

$\mathbf{byx}\mathbf{=}\mathbf{1.12852}\$

Regression Equation Y on X

$\mathbf{(Y\ - \ }\overline{\mathbf{Y}}\mathbf{) = \ byx\ (X\ - \ }\overline{\mathbf{X}}\mathbf{)}\$

$(Y\ -\ 140.34)\ = \ 1.12852(X\ -\ 52.34)\$

$(Y\ -\ 140.34)\ = \ 1.12852X-\ 59.07\$

$Y\ \ = \ 1.12852X-\ 59.07 + \ 140.34\$

$Y\ \ = \ 1.12852X\ + \ 81.27\$

$\widehat{\mathbf{Y}}\mathbf{\ = \ 81.27\ + \ 1.12852x}\$

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Regression Equation X on Y

$(X\ - \ \overline{X}) = \ bxy\ (Y\ - \ \overline{Y})\$

$(X\ -\ 52.34)\ = \ 0.7031(Y\ -\ 140.34)\$

$(x\ -\ 52.34)\ = \ 0.7031Y-\ 98.67\$

$X\ \ = \ 0.7031Y-\ 98.67 + \ 52.34\$

$X\ \ = \ 0.7031Y\ -46.33\$

$\widehat{\mathbf{X}}\mathbf{\ = \ - 46.33\ + \ \ 0.7031}\mathbf{y}\$

$r = \ \sqrt{byx\ .bxy}\$

$r = \ \sqrt{(1.12852).(0.7031)}\$

$r = \ \sqrt{0.79346}\$

$\mathbf{r = \ 0.89}\$

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Solved Paper Statistics I FBISE 2011

Table of Contents

Solved Paper Statistics I FBISE 2011

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

Short Question

Q.2 Attempt any fourteen parts.

(i) Differentiate between descriptive & Inferential Statistics.

(ii) What are the functions of statistics?

(iii) What are the common types of frequency distribution?

(iv)Differentiate between Historigram and Histogram.

(vi) The mean and geometric mean of three numbers are 7 and 4, respectively. Find all the three numbers if mean of two numbers is 10.

(vii) What are the desirable qualities of a good average?

(viii) The first four moments about X = 17.5 of a distribution are 0.3, 74, 45 and 12125. Find whether the distribution is leptokurtic or Platykurtic.

(ix) Given: Calculate Median, Mode, Q3 and D7.

(x) Given n1=15, n2 = 25, n3 = 40, X̅1=15.8, X̅2=22.5, X̅3=30.2, S1=2.45, S2 = 4.65, S3 = 7.28. Compute combined coefficient of variation.

(xi) What will be the standard deviation and the variance in each of the following case? (i) 2X (ii) X + 2 (iii) 2X + 4 if Var(X) = 25

(xii) Compute chain index taking 1920 as base for the following data:

(xiii) What are the important uses of index number?

(xiv) Given the following information: Compute Base year weighted & current year weighted price index numbers.

(xv) What is meant by regression?

(xvi) Given n = 10, ∑Dx = – 8, ∑Dy = 0, ∑D²x = 66, ∑D²y = 99, ∑DxDy = 72. Find r, bxy, byx.

(xvii) Differentiate between Signal and Noise

(xviii) Fit a straight line to the following results for the years 1951-58 both inclusive ∑x = 0, ∑y = 2861.1, ∑x² = 168, ∑xy = 2215.5. Also compute trend values.

(xix) Find 4-years moving averages cantered from the following data:

Extensive Questions

Q.3 Calculate: (i) Mean Deviation from Median, (ii) Coefficient of Variation. From the following frequency distribution:

Q.4 Construct Consumer Price Index of 1980 on the basis of 1978 using: (i) Aggregative Method (ii) Family Budget Method

Q.5 The following table shows the ages (X) and systolic blood pressure (Y) of 12 women..

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Table of Contents

Solved Paper Statistics I FBISE 2011

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

Short Question

Q.2 Attempt any fourteen parts.

(i) Differentiate between descriptive & Inferential Statistics.

(ii) What are the functions of statistics?

(iii) What are the common types of frequency distribution?

(iv)Differentiate between Historigram and Histogram.

(v) A given stock was purchased at the following prices at various times, 20 shares at Rs. 8.20 a share, 100 shares at Rs 10.90 a share, 50 shares at Rs. 9.40 a share and 200 shares at Rs. 7.80 a share. Find the mean cost per share.

(vi) The mean and geometric mean of three numbers are 7 and 4, respectively. Find all the three numbers if mean of two numbers is 10.

(vii) What are the desirable qualities of a good average?

(viii) The first four moments about X = 17.5 of a distribution are 0.3, 74, 45 and 12125. Find whether the distribution is leptokurtic or Platykurtic.

(ix) Given: Calculate Median, Mode, Q3 and D7.

(x) Given n1=15, n2 = 25, n3 = 40, X̅1=15.8, X̅2=22.5, X̅3=30.2, S1=2.45, S2 = 4.65, S3 = 7.28. Compute combined coefficient of variation.

(xi) What will be the standard deviation and the variance in each of the following case? (i) 2X (ii) X + 2 (iii) 2X + 4 if Var(X) = 25

(xii) Compute chain index taking 1920 as base for the following data:

(xiii) What are the important uses of index number?

(xiv) Given the following information: Compute Base year weighted & current year weighted price index numbers.

(xv) What is meant by regression?

(xvi) Given n = 10, ∑Dx = – 8, ∑Dy = 0, ∑D²x = 66, ∑D²y = 99, ∑DxDy = 72. Find r, bxy, byx.

(xvii) Differentiate between Signal and Noise

(xviii) Fit a straight line to the following results for the years 1951-58 both inclusive ∑x = 0, ∑y = 2861.1, ∑x² = 168, ∑xy = 2215.5. Also compute trend values.

(xix) Find 4-years moving averages cantered from the following data:

Extensive Questions

Q.3 Calculate: (i) Mean Deviation from Median, (ii) Coefficient of Variation. From the following frequency distribution:

Q.4 Construct Consumer Price Index of 1980 on the basis of 1978 using: (i) Aggregative Method (ii) Family Budget Method

Q.5 The following table shows the ages (X) and systolic blood pressure (Y) of 12 women..

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