Statistics I HSSC I FBISE Solved Paper 2023, MCQS, Short Questions, Extensive Questions

Statistics I HSSC I FBISE Solved Paper 2023, MCQS, Short Questions, Extensive Questions

In this post, we are going to discuss, Statistics I HSSC I FBISE Solved Paper 2023, MCQS, Short Questions, Extensive Questions. Dive into the world of statistics with our comprehensive solutions to the Statistics I paper, covering key topics such as Introduction to Statistics, Measures of Central Tendency and Dispersion, Index Numbers, Correlation & Regression, and Time Series Analysis. Perfect for students and professionals alike, our detailed explanations will help you master these fundamental concepts. Explore more resources on statistical, economical, accounting, and finance topics at bcfeducation.com.

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Table of Contents

Statistics I HSSC I FBISE Solved Paper 2023, MCQS, Short Questions, Extensive Questions

MCQS

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Fill the relevant bubble against each question according to curriculum:

1The branch of statistics that is concerned with procedures for obtaining valid conditions is called:
A)Descriptive StatisticsB)Inferential Statistics
C)Theoretical StatisticsD)Applied Statistics
2Issuing a National identity card is an example of:
A)CensusB)Registration
C)SamplingD)Investigation through enumerators
3The process of systematic arrangement of data into rows and columns is called:
A)ClassificationB)Tabulation
C)Frequency DistributionD)Array
4An Ogive is also called:
A)Frequency PolygonB)Frequency Curve
C)HistogramD)Cumulative Frequency Polygon
5The modal letter(s) of the word :STATISTICS” is/are:
A)SB)T
C)S,TD)I,T
6If X̅ =10 and Y = 2X + 5, then Y̅ = __________:
A)10B)15
C)20D)25
7The sum of squared deviations from mean is always:
A)NegativeB)Maximum
C)MinimumD)Zero
8Geometric mean of 2, 4, 6, 8, 64 is:
A)7B)7.55
C)16.8D)8.5
9For normal distribution, approximately 68% of the values are included by an interval:
A)X̅±SB)X̅±2S
C)X̅±3SD)X̅±4S
10If Var(X) = 2, then Var(3X +4) = _______:
A)10B)15
C)18D)20
11The reversal test is satisfied by:
A)Laspeyre’s IndexB)Paasche’s Index
C)Fisher’s IndexD)Un-weighted Index Number
12The index number is given by (∑pnqn/∑poqn)x100 is called:
A)The Laspeyre’s indexB)The Paasche’s index
C)The value indexD)Wholesale price index
13The price relative is the percentage ratio of current year price and:
A)Current year quantityB)Base year quantity
C)Current year priceD)Base year price
14In method of least square, the sum of errors will be:
A)Less than zeroB)Greater than zero
C)ZeroD)Not equal to zero
15The regression line always passing through:
A)(a, Y̅)B)(b, Y̅)
C)(a,b)D)(X̅,Y̅)
16When two variables move in same direction, the correlation will be:
A)PositiveB)Negative
C)ZeroD)Neutral
17A decline in death rate due to advancement of Science is an example of:
A)Seasonal VariationB)Secular Variation
C)Cyclical VariationD)Random Variation

Total Marks Sections B and C: 68

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Note: Answer any fourteen parts from section “B” and any two questions from Section “C”. Write your answers neatly and legibly. Statistical table will be provided on demand.

Section B- (Marks 42)

Short Questions

(i) Distinguish between primary and secondary data.

Answer

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

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

(ii) Differentiate between discrete variable and continuous variable.

Answer:

Discrete Variable

Discrete variable is a variable in which a data has some specific value within a given range or we can say that discrete variable has variable that is countable. For example, number of persons, number of cars, number of students etc.

Continuous Variable

Continuous variable is a variable in which a data has any value within a given range or we can say that continuous variable has variable that is measurable. For example, height of students, speed of car, length of wood etc.

(iii) Make class boundaries and find missing frequencies of the following frequency distributions.

ClassesFrequencyCumulative Frequency
0.7312—0.731355
0.7314—0.73157?
0.7316—0.7317?22
0.7318—0.73198?
0.7320—0.7321?35

Solution:

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ClassesFrequencyCumulative Frequency
0.7312—0.731355
0.7314—0.731575+7=12
0.7316—0.731722-12=1022
0.7318—0.7319822+8=30
0.7320—0.732135-30=535

(iv) Calculate Arithmetic mean from the following deviations.

D=X-10-5-303681013

Solution:

D=X-10-5-303681013∑D=32

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

    \[ \mathbf{A.M\ \ }\overline{\mathbf{X}}\mathbf{= 10 +}\frac{\mathbf{32}}{\mathbf{8}}\  \]

    \[ \mathbf{A.M = 14}\  \]

(v) Calculate Geometric Mean and Harmonic Mean for five values of X for the following reciprocal values:

1/X0.20.10.050.040.025

Solution:

1/XX=1/1/x
0.25
0.110
0.0520
0.0425
0.02540
∑(1/X)=0.415

    \[ \mathbf{Harmonic\ Mean =}\frac{\mathbf{n}}{\mathbf{\sum}\left( \frac{\mathbf{1}}{\mathbf{x}} \right)}\  \]

    \[ \mathbf{Harmonic\ Mean =}\frac{\mathbf{5}}{\mathbf{0.415}}\  \]

    \[ \mathbf{Harmonic\ Mean = 12}\  \]

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    \[ \mathbf{G.M =}\mathbf{(5 \times 10 \times 20 \times 25 \times 40)}^{\frac{\mathbf{1}}{\mathbf{5}}}\  \]

    \[ \mathbf{G.M = 15.84}\  \]

(vi) Calculate combined Arithmetic mean for the following two groups.

 BoysGirls
Number of Persons2030
Mean Weight (kg)62.553.4

Solution:

    \[  \overline{\mathbf{X}}\mathbf{c =}\frac{\overline{\mathbf{X}}\mathbf{1}\mathbf{n}\mathbf{1 +}\overline{\mathbf{X}}\mathbf{2}\mathbf{n}\mathbf{2}}{\mathbf{n}\mathbf{1 + n}\mathbf{2}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{c =}\frac{\left( \mathbf{62.5 \times 20} \right)\mathbf{+ (53.4 \times 30)}}{\mathbf{20 + 30}}\  \]

    \[ \overline{\mathbf{X}}\mathbf{c =}\frac{\mathbf{1250 + 1602}}{\mathbf{50}}\  \]

    \[ \overline{\mathbf{X}}\mathbf{c = 57.04}\ \]

(vii) What is meant by range and semi-interquartile range?

Answer: Range and semi-interquartile range both are absolute measures of dispersion. Range calculates the dispersion of class marks and semi interquartile deviation or quartile deviation calculates the dispersion of quartiles. Formulas are given below:

    \[  Range\ = \ Xm\ -\ Xo\ \]

    \[ Quartile\ Deviation\ or\ S.I.Q.R = \ \frac{Q3 - Q1}{2}\  \]

(viii) If lower quartile (Q1) = 40, upper quartile (Q3) = 90 and Median = 60, Compute Quartile Coefficient of Skewness.

Solution:

    \[ \mathbf{Quartile\ Coefficient\ of\ Skewness =}\frac{\mathbf{Q3 + Q1 - 2Median}}{\mathbf{Q3 - Q1}}\  \]

    \[ \mathbf{Quartile\ Coefficient\ of\ Skewness =}\frac{\mathbf{90}\mathbf{+}\mathbf{40}\mathbf{- 2}\mathbf{(60)}}{\mathbf{90 - 40}}\  \]

    \[ \mathbf{Quartile\ Coefficient\ of\ Skewness =}\frac{\mathbf{10}}{\mathbf{50}}\mathbf{= 0.2}\  \]

(ix) Given n =5, ∑x=180, ∑x²=6660. Compute variance, Standard deviation and coefficient of variation.

Solution:

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{6660}}{\mathbf{5}}\mathbf{-}\left( \frac{\mathbf{180}}{\mathbf{5}} \right)^{\mathbf{2}}\  \]

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{= 1332 - 1296}\  \]

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{= 36}\ \]

    \[ \mathbf{Standard\ Deviation\ S =}\sqrt{\frac{\mathbf{\sum}\mathbf{x}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)^{\mathbf{2}}\mathbf{\ }}\  \]

    \[ \mathbf{Standard\ Deviation\ S =}\sqrt{\mathbf{36}\mathbf{\ }}\  \]

    \[ \mathbf{Standard\ Deviation\ S = 6}\  \]

    \[ \mathbf{Mean =}\frac{\mathbf{\sum x}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{180}}{\mathbf{5}}\mathbf{= 36}\  \]

    \[ \mathbf{C.V =}\frac{\mathbf{S.D}}{\mathbf{Mean}}\mathbf{\times}\mathbf{100}\  \]

    \[ \mathbf{C.V =}\frac{\mathbf{6}}{\mathbf{36}}\mathbf{\times}\mathbf{100}\ \]

    \[ \mathbf{C.V = 16.67}\ \]

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(x) Differentiate between un-weighted and weighted index number.

Answer:

Un-weighted Index Number

Un-weighted index number is a measure of composite index number. In un-weighted index number, weights or quantities are not given. Unweighted index numbers are calculated through following methods:

  • Aggregative Index
  • Average of Relative Method

Weighted Index

Weighted index is also a measure of composite index numbers in which weights or quantities are given. Weighted index can be calculated through number of methods given below:

  • Laspeyre’s Method
  • Paasche’s Method
  • Fisher’s Ideal Index Method
  • Marshal Edgesworth Method
  • Walsh Method

(xi) Compute index number taking 2010 as base:

Year201020112012201320142015
Price101415202533

Solution:

YearPriceIndex Number taking 2010 as base
201010 

    \[  \left( \frac{\mathbf{10}}{\mathbf{10}} \right)\mathbf{100\ = \ 100}\ \]

201114 

    \[  \left( \frac{\mathbf{14}}{\mathbf{10}} \right)\mathbf{100\ = \ 140}\ \]

201215 

    \[ \left( \frac{\mathbf{15}}{\mathbf{10}} \right)\mathbf{100\ = \ 150}\  \]

201320 

    \[ \left( \frac{\mathbf{20}}{\mathbf{10}} \right)\mathbf{100\ = \ 200}\  \]

201425 

    \[ \left( \frac{\mathbf{25}}{\mathbf{10}} \right)\mathbf{100\ = \ 250}\  \]

201533 

    \[ \left( \frac{\mathbf{33}}{\mathbf{10}} \right)\mathbf{100\ = \ 330}\  \]

 

(xii) If Laspeyre’s index number is 150 and Fisher’s index is 147, calculate the Paasche’s index number.

Solution:

    \[  \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Index =}\sqrt{\mathbf{L \times P}}\ \]

    \[ \mathbf{147 =}\sqrt{\mathbf{150 \times P}}\  \]

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Taking square root on both sides:

    \[ \mathbf{(147)}^{\mathbf{2}}\mathbf{=}\left( \sqrt{\mathbf{150}} \right)^{\mathbf{2}}\mathbf{\times}\left( \sqrt{\mathbf{P}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{21609 = 150 \times P}\  \]

    \[ \mathbf{P =}\frac{\mathbf{21609}}{\mathbf{150}}\mathbf{= 144.06}\  \]

    \[ \mathbf{Paasche}\mathbf{s}^{\mathbf{'}}\mathbf{s\ Index = 144.06}\  \]

(xiii) Compute two regression coefficients byx and bxy of following data: n=10, ∑Dx=12, ∑Dy=-5, ∑DxDy=390,∑Dx²=2830, ∑Dy²=91

Solution:

Regression Coefficient y on x:

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{\sum DxDy -}\frac{\mathbf{(\sum Dx)(\sum Dy)}}{\mathbf{n}}}{\mathbf{\sum}\mathbf{Dx}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum Dx} \right)^{\mathbf{2}}}{\mathbf{n}}}\  \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{390}\mathbf{-}\frac{\mathbf{(}\mathbf{12}\mathbf{)(}\mathbf{- 5}\mathbf{)}}{\mathbf{10}}}{\mathbf{2830}\mathbf{-}\frac{\left( \mathbf{12} \right)^{\mathbf{2}}}{\mathbf{10}}}\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{39}\mathbf{6}}{\mathbf{28}\mathbf{15.6}}\  \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\mathbf{0.140}\  \]

Regression Coefficient x on y:

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{\sum DxDy -}\frac{\mathbf{(\sum Dx)(\sum Dy)}}{\mathbf{n}}}{\mathbf{\sum}{\mathbf{D}\mathbf{y}}^{\mathbf{2}}\mathbf{-}\frac{\left( \mathbf{\sum D}\mathbf{y} \right)^{\mathbf{2}}}{\mathbf{n}}}\  \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{390 -}\frac{\mathbf{(}\mathbf{12}\mathbf{)(}\mathbf{- 5}\mathbf{)}}{\mathbf{10}}}{\mathbf{91}\mathbf{-}\frac{\left( \mathbf{- 5} \right)^{\mathbf{2}}}{\mathbf{10}}}\  \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\frac{\mathbf{396}}{\mathbf{88.5}}\  \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{=}\mathbf{4.47}\ \]

(iv) The two regression lines are Ŷ=25+0.83x and X̂=40+0.97y are given, identify the two regression coefficients and compute the correlation coefficient (r).

Solution:

Regression Coefficient Y on X

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= 0.83}\ \]

Regression Coefficient X on Y

    \[ \mathbf{b}_{\mathbf{xy}}\mathbf{= 0.97}\ \]

Correlation Coefficient

    \[ \mathbf{r}{\mathbf{xy}}\mathbf{=}\sqrt{\left( \mathbf{b}{\mathbf{yx}} \right)\mathbf{(}\mathbf{b}_{\mathbf{xy}}\mathbf{)}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\left( \mathbf{0.83} \right)\mathbf{(0.97)}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\mathbf{0.8051}}\ \]

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{= 0.8972}\ \]

(xv) Write down the properties of correlation coefficient (r).

Answer:

1. Symmetric Property

Correlation between X variable and Y variable and Y Variable and X variable have same meaning or:

    \[ \mathbf{r}{\mathbf{xy}}\mathbf{=}\mathbf{r}{\mathbf{yx}}\ \]

2. Free from unit of measurement

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Correlation coefficient is free from unit of measurement means if data is given in kilograms or length, the answer will not be in kilograms or length.

3. Independent to Origin and Scale

Correlation Coefficient is independent to origin and scale means:

    \[ r_{xy} = r_{uv}\ where\ U = \ \frac{X \pm A}{h}\ and\ V = \frac{Y \pm B}{h}\ \]

4. Range of Correlation Coefficient

Correlation Coefficient always lies between -1 to +1 both inclusive.

5. Geometric Mean of Regression Coefficients

Correlation Coefficient is a geometric mean of two regression coefficients.

    \[ \mathbf{r}_{\mathbf{xy}}\mathbf{=}\sqrt{\mathbf{byx\ \times bxy}}\  \]

6. Relation with Covariance

If X and Y are two independent variables then Cov(X, Y) = 0 but it does not mean that if Cov (X, Y) = 0 then X and Y are definitely independent.

(xvi) If rxy =0.60, U=(X – 50)/10, V = (Y – 60)/5, then what is the value of rxy and ruv?

Answer:  It is the property of correlation coefficient that it is independent to origin and scale so rxy is equals to ruv and its value is 0.60.

(xvii) Describe seasonal variation with example in time series.

Answer: Seasonal variation is a variation of time series that occurs repeatedly when season changes. For example, in summer, the demand for cold drinks increases and when winter comes it reduces and this happens again and again.

(xviii) If the least square line fitted to the data for the years 1960-65 (both inclusive) with the origin at the middle of 1962 and 1963 is Ŷ = 75+0.85x, the unit of X is being half year, then find the trend values for 1960 to 1965.

Solution:

YearXTrend Values Ŷ = 75+0.85x
1960-2.5Ŷ = 75+0.85(-2.5)=72.875
1961-1.5Ŷ = 75+0.85(-1.5)=73.725
1962-0.5Ŷ = 75+0.85(-0.5)=74.575
1963+0.5Ŷ = 75+0.85(0.5)=75.425
1964+1.5Ŷ = 75+0.85(1.5)=76.275
1965+2.5Ŷ = 75+0.85(2.5)=77.125

(xix) Estimate the trend values by semi average method for 1970 to 1975.

YearSemi-TotalSemi-Average
1970  
1971720240
1972  
1973  
1974990330
1975  
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Solution:

YearSemi-TotalSemi-AverageTrend Values Y̑
1970  240 – 30 = 210
1971720240240
1972  240 + 30 = 270
1973  270 + 30 = 300
1974990330330
1975  330 + 30 = 360

Increase of trend in 3 years = 330 – 240 = 90

Increase of trend in one year = 90/3 = 30

Section C Marks 26

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

Extensive Questions

Q.3 a. Find Arithmetic Mean, Median and Mode from the following data:        (06)

Q.3 a. Find Arithmetic Mean, Median and Mode from the following data:        (06)

Classes10-1415-1920-2425-2930-3435-39
Frequency35101264

Solution:

ClassesFrequencyClass BoundariesXfxC.F
10—1439.5–14.512363
15—19514.5–19.517858
20—241019.5–24.52222018
25—291224.5–29.52732430
30—34629.5–34.53219236
35—39434.5–39.53714840
∑f =n= 40∑fx = 1005

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

    \[ \mathbf{Mean =}\frac{\mathbf{1005}}{\mathbf{40}}\mathbf{= 25.125}\ \]

    \[ \mathbf{Median = l +}\frac{\mathbf{h}}{\mathbf{f}}\mathbf{(}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{- c)}\ \]

n/2=40/2=20 falls in c.f of 30 so l=24.5, h=5,f=12,n/2=20, c=18

    \[ \mathbf{Median = 24.5 +}\frac{\mathbf{5}}{\mathbf{12}}\mathbf{(20 - 18)}\ \]

    \[ \mathbf{Median = 25.33}\  \]

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

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Maximum frequency is 12 so l = 24.5, fm =12, f1=10, f2=6 and h = 5

    \[ \mathbf{Mode = 24.5 +}\frac{\mathbf{12 - 10}}{\left( \mathbf{12 - 10} \right)\mathbf{+ (12 - 6)}}\mathbf{\times 5}\  \]

    \[  \mathbf{Mode = 24.5 +}\frac{\mathbf{10}}{\mathbf{8}}\mathbf{= 25.75}\ \]

b. Calculate Coefficient of skewness by Karl Pearson’s Methods.  (07)

b. Calculate Coefficient of skewness by Karl Pearson’s Methods.  (07)

Marks20—2425—2930—3435—3940—4445—49
Frequency14811156

Solution:

ClassesFrequencyClass BoundariesXfxfx²
20—24119.5–24.52222484
25—29424.5–29.5271082916
30—34829.5–34.5322568192
35—391134.5–39.53740715059
40—441539.5–44.54263026460
45—49644.5–49.54728213254
∑f =n= 45∑fx = 1705∑fx² = 66365

    \[ \mathbf{Karl\ Person}\mathbf{s}^{\mathbf{'}}\mathbf{s\ C.Skewness =}\frac{\mathbf{Mean - Mode}}{\mathbf{S.D}}\ \]

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

    \[ \mathbf{Mean =}\frac{\mathbf{1705}}{\mathbf{45}}\mathbf{= 37.88}\ \]

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

Maximum frequency is 15 so l = 39.5, fm =15, f1=11, f2=6 and h = 5

    \[ \mathbf{Mode = 39.5 +}\frac{\mathbf{15 - 11}}{\left( \mathbf{15 - 11} \right)\mathbf{+ (15 - 6)}}\mathbf{\times 5}\ \]

    \[ \mathbf{Mode = 39.5 +}\frac{\mathbf{20}}{\mathbf{13}}\mathbf{= 41.03}\ \]

    \[ \mathbf{S.D =}\sqrt{\frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{-}\left( \frac{\mathbf{\sum fx}}{\mathbf{\sum f}} \right)^{\mathbf{2}}}\  \]

    \[ \mathbf{S.D =}\sqrt{\frac{\mathbf{66365}}{\mathbf{45}}\mathbf{-}\left( \mathbf{37.88} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{S.D =}\sqrt{\mathbf{39.8756}}\ \]

    \[ \mathbf{S.D = 6.3147}\ \]

    \[ \mathbf{Karl\ Person}\mathbf{s}^{\mathbf{'}}\mathbf{s\ C.Skewness =}\frac{\mathbf{Mean - Mode}}{\mathbf{S.D}}\ \]

    \[ \mathbf{Karl\ Person}\mathbf{s}^{\mathbf{'}}\mathbf{s\ C.Skewness =}\frac{\mathbf{37.88 - 41.03}}{\mathbf{6.3147}}\mathbf{= - 0.4988}\ \]

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Q.4 a. Construct chain indices using Geometric Mean as an average (08)

Q.4 a. Construct chain indices using Geometric Mean as an average (08)

YearPrices
WheatRiceGhee
19901203020
19911323224
19921403830
19931444040
19941504550

Solution:

YearLink RelativesGeometric Mean
WheatRiceGhee 
19901203020 

    \[ \left( \mathbf{120 \times 30 \times 20} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 41.60}\  \]

1991(132/120)100=110(32/30)100=106.66(24/20)100=120 

    \[ \left( \mathbf{110 \times 106.66 \times 120} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 112.07}\ \]

1992(140/132)100=106.06(38/32)100=118.75(30/24)100=125 

    \[ \left( \mathbf{106.06 \times 118.75 \times 125} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 116.33}\ \]

1993(144/140)100=102.85(40/38)100=105.26(40/30)100=133.33 

    \[ \left( \mathbf{102.85 \times 105.26 \times 133.33} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 113.01}\  \]

1994(150/144)100=104.16(45/40)100=112.5(50/40)100=125 

    \[ \left( \mathbf{104.16 \times 112.5 \times 125} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 113.56}\  \]

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YearGeometric MeanChain Index
  
1990 

    \[ \left( \mathbf{120 \times 30 \times 20} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 41.60}\  \]

41.60
1991  

    \[ \left( \mathbf{110 \times 106.66 \times 120} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 112.07}\  \]

 

    \[ \frac{\left( \mathbf{41.60 \times 112.07} \right)}{\mathbf{100}}\mathbf{= 46.62}\  \]

1992  

    \[ \left( \mathbf{106.06 \times 118.75 \times 125} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 116.33}\  \]

 

    \[ \frac{\left( \mathbf{46.62 \times 116.33} \right)}{\mathbf{100}}\mathbf{= 54.23}\ \]

1993  

    \[ \left( \mathbf{102.85 \times 105.26 \times 133.33} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 113.01}\  \]

 

    \[ \frac{\left( \mathbf{54.23 \times 113.01} \right)}{\mathbf{100}}\mathbf{= 61.28}\  \]

1994  

    \[ \left( \mathbf{104.16 \times 112.5 \times 125} \right)^{\frac{\mathbf{1}}{\mathbf{3}}}\mathbf{= 113.56}\  \]

 

    \[ \frac{\left( \mathbf{61.28 \times 113.56} \right)}{\mathbf{100}}\mathbf{= 69.58}\  \]

b. Compute the consumer price index number by aggregative expenditure method      (05)

b. Compute the consumer price index number by aggregative expenditure method      (05)

ItemQuantity Consumed in Base Year (qo)Base year price (po)Current year price (pn)
Rice60150200
Wheat803540
Pulses10120150
Ghee20225250
Sugar15100110

Solution:

Item(qo)(po)(p1)(p1qo)(poqo)
Rice60150200120009000
Wheat80354032002800
Pulses1012015015001200
Ghee2022525050004500
Sugar1510011016501500
    ∑p1qo=23350∑poqo=19000

    \[ \mathbf{Aggregative\ Index =}\frac{\mathbf{\sum p}\mathbf{1}\mathbf{qo}}{\mathbf{\sum poqo}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index =}\frac{\mathbf{23350}}{\mathbf{19000}}\mathbf{\times 100 = 122.89}\ \]

Q.5 a. Calculate regression line Y on X and Correlation Coefficient (r) from the following data: (06)

Q.5 a. Calculate regression line Y on X and Correlation Coefficient (r) from the following data: (06)

X10987643
Y812710896
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Solution:

XYXY
1088010064
91210881144
87566449
7107049100
68483664
49361681
3618936
∑X=47∑Y=60∑XY=416∑X²=355∑Y²=538

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum XY}}{\mathbf{n}}\mathbf{-}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{Sx \times Sy}}\ \]

Where

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{47}}{\mathbf{7}}\mathbf{= 6.714}\ \]

    \[ \overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{60}}{\mathbf{7}}\mathbf{= 8.57}\ \]

    \[ \mathbf{Sx =}\sqrt{\frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum X}}{\mathbf{n}} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{Sx =}\sqrt{\frac{\mathbf{355}}{\mathbf{7}}\mathbf{-}\left( \mathbf{6.714} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{Sx = \ }\sqrt{\mathbf{5.6364}}\  \]

    \[ \mathbf{Sx = 2.374}\ \]

    \[ \mathbf{Sy =}\sqrt{\frac{\mathbf{\sum}\mathbf{Y}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\left( \frac{\mathbf{\sum Y}}{\mathbf{n}} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{Sy =}\sqrt{\frac{\mathbf{538}}{\mathbf{7}}\mathbf{-}\left( \mathbf{8.57} \right)^{\mathbf{2}}}\  \]

    \[ \mathbf{Sy = \ }\sqrt{\mathbf{3.412}}\ \]

    \[ \mathbf{Sy = 1.847}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{\sum XY}}{\mathbf{n}}\mathbf{-}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{Sx \times Sy}}\ \]

    \[ \mathbf{r =}\frac{\frac{\mathbf{416}}{\mathbf{7}}\mathbf{-}\left( \mathbf{6.714} \right)\mathbf{(8.57)}}{\mathbf{2.374 \times 1.847}}\ \]

    \[ \mathbf{r =}\frac{\mathbf{1.88952}}{\mathbf{4.384}}\mathbf{= 0.4310}\ \]

Regression Line Y on X

Ŷ=a+bx

Where

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= r}\left( \frac{\mathbf{sy}}{\mathbf{sx}} \right)\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= 0.4310}\left( \frac{\mathbf{1.847}}{\mathbf{2.374}} \right)\ \]

    \[ \mathbf{b}_{\mathbf{yx}}\mathbf{= 0.3353}\ \]

    \[ \mathbf{a =}\overline{\mathbf{Y}}\mathbf{- b}\overline{\mathbf{X}}\ \]

    \[ \mathbf{a = 8.57 - 0.3353(6.714)}\ \]

    \[ \mathbf{a = 6.32}\ \]

Ŷ=a+bx

Ŷ=6.32 + 0.3353x

b. Smooth the variation of following data with the help of 4-years centered moving average method.    (07)

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b. Smooth the variation of following data with the help of 4-years centered moving average method.    (07)

Year20082009201020112012201320142015
Profit (in Rs.000)100120150160190210350415

Solution

Year & QuarterValues4 Years Moving Total2 Values Moving Total4 Years centered Moving Average
2008100   
     
2009120   
  530  
2010150 1150143.75
  620  
2011160 1330166.25
  710  
2012190 1620202.5
  910  
2013210 2075259.375
  1165  
2014350   
     
2015 415   
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Statistics I Solved Paper FBISE 2008

Statistics I Solved Paper FBISE 2009

Statistics I Solved Paper FBISE 2010

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