Solved Paper Statistics I FBISE 2017

Solved Paper Statistics I FBISE 2017

Solved Paper Statistics I FBISE 2017, Dive into a comprehensive solution guide to the FBISE Statistics I 2017!. This blog post provides detailed explanations and step-by-step solutions for key topics like measures of central tendencydispersion, 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 FBISEBISERWPBISELHRMUDUPU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Table of Contents

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Solved Paper Statistics I FBISE 2017

MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)Census collects the:
 A. Primary dataB. Secondary dataC. Fictitious dataD. Official data 
      
(ii)The word “Statistics” has been derived from the German Word:
 A. StatistaB. StatusC. StatistikD. Statistique 
      
(iii)A variable which can take any possible value in an interval is called:
 A. Discrete VariableB. Continuous VariableC. Qualitative VariableD. Finite Variable 
      
(iv)The graph of Time Series data is called:
 A. HistorigramB. Pie-ChartC. HistogramD. Ogive 
      
(v)Total of relative frequency is called:
 A. HalfB. OneC. 100D. Quarter 
      
(vi)The sum of deviation from mean is always:
 A.LeastB. MaximumC. OneD. Zero 
      
(vii)Which of the following averages is not affected by extreme values:
 A.Arithmetic MeanB. MedianC. ModeD. G.M 
      
(viii)If mean of 5 values is 10, then the sum of the values will be:
 A.2B. 15C. 25D. 50 
      
(ix)The variance of the values 7,7,7,7,7,7 is:
 A.42B. 7C. ZeroD.

    \[ \sqrt{7}\ \]

 
      
(x)If X and Y are two independent random variables, Var(x)=4 and Var(y)=9, then Var(2x + y) is:
 A.13B. 17C. 25D. 26 
      
(xi)In a symmetrical distribution, if Q1 = 6 and Q3 = 18 then median is:
 A. 12B. 15C. 24D. Zero 
      
(xii)The empirical relationship between mean, median and mode is: Mode =:
 A. 3Mean – 2MedianB. 2Mean – 3 MedianC. 3Median – 2 MeanD. 2Median – 3Mean 
      
(xiii)The link relatives are the percentage ratios of current year price and:
 A. Previous year quantityB. base year quantityC. Next year priceD. Previous Year Price 
      
(xiv)Which index number helps the Government to formulate economic policies and determine the wages of employees:
 A. Whole sale Price IndexB. Consumer Price IndexC. Quantity IndexD. Simple Index 
      
(xv)The dependent variable is also called:
 A. RegressorB. Explanatory variableC. PredictorD. Response Variable 
      
(xvi)The value of correlation coefficient  lies between:
 A. 0 and 1B. -1 and 0C. -1 and 1D. -2 and 2 
      
(xvii)Increased demand of soft drink in summer and woolen clothes in winter season is:
 A. Seasonal VariationB. Secular VariationC. Cyclical VariationD. Random Variation 

Short Questions

(i) Differentiate between primary and secondary data.

Answer:

Primary Data

First hand, newly collected, ungrouped data is called primary data or data which is not collected by someone previously is called primary data. For example, fresh data obtaining during research survey is an example of primary data.

Secondary Data

Second hand, previously collected, grouped data is called secondary data or data which is collected by someone previously is called secondary data. For example, data obtained from college record is an example of secondary data.

(ii) Differentiate between discrete and continuous variable.

Answer:

Discrete Variable

A variable in which data has some specific values within a given range is called discrete variable. In discrete variable, data is countable. For example, Number of students in a class, Number of houses in a street, number of children in a family etc.

Continuous Variable

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Solved Paper Statistics I FBISE 2017

A variable in which data has any values within a given range is called continuous variable. In continuous variable, data is measurable. For example, age of persons, speed of car, temperature, height, etc.

(iii) In a music competition, students are asked to rate the music on five points scale A, B, C, D, E where A represents the maximum enjoyment and E represents the minimum enjoyment. The ratings are:

A, D, A, D, E, B, C, D, A, B, B, C, E, A, C, E, C, A, B, E, D, E, B, A, B, E, E, C, B, A

Construct the frequency distribution for the above ratings

Answer:

Ratings (X)Frequency (f)
A7
B7
C5
D4
E7
 ∑f = 30

(iv) Write down the properties of a good average.

Solution:

(i) Deviation of X values from mean is always equal to zero.

    \[ \mathbf{\sum(X -}\overline{\mathbf{X}}\mathbf{)\ = 0\ or\ \sum f(X -}\overline{\mathbf{X}}\mathbf{)\ = 0}\ \]

(ii) Sum of squared deviation from mean is always less than the sum of squared deviation from arbitrary origin.

    \[ \mathbf{\sum(X -}\overline{\mathbf{X}}\mathbf{)\ ²\ < \ \sum(X - A)\ ²\ or\ \sum f(X -}\overline{\mathbf{X}}\mathbf{)\ ²\ < \ \sum f(X - A)\ ²}\ \]

(iii) Combine mean of multiple distributions can be calculated through following formula:

    \[ \mathbf{\ }\overline{\mathbf{X}}\mathbf{c}\mathbf{=}\frac{\mathbf{n1}\overline{\mathbf{X}}\mathbf{1}\mathbf{+}\mathbf{n2}\overline{\mathbf{X}}\mathbf{2}\mathbf{+}\mathbf{n3}\overline{\mathbf{X}}\mathbf{3}\mathbf{+}\mathbf{\ldots\ldots nk}\overline{\mathbf{X}}\mathbf{k}}{\mathbf{n1}\mathbf{+}\mathbf{n2}\mathbf{+}\mathbf{n3}\mathbf{+}\mathbf{\ldots\ldots nk}}\ \]

(v) The average marks obtained by three sections of first year class are given below: Find the combine mean of the class.

SectionsNumber of studentsMeans
A4568
B4258
C3852

Solution:

    \[ \overline{\mathbf{X}}\mathbf{c = \ }\frac{\mathbf{n}\mathbf{1\ }\overline{\mathbf{X}}\mathbf{1 + n}\mathbf{2}\overline{\mathbf{X}}\mathbf{2 + n}\mathbf{3}\overline{\mathbf{X}}\mathbf{3}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 + n}\mathbf{3}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{c = \ }\frac{\left( \mathbf{45} \right)\left( \mathbf{68} \right)\mathbf{+}\left( \mathbf{42} \right)\left( \mathbf{58} \right)\mathbf{+}\left( \mathbf{38} \right)\left( \mathbf{52} \right)}{\mathbf{45 + 42 + 38}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{c =}\frac{\mathbf{3060 + 2436 + 1976}}{\mathbf{125}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{c =}\frac{\mathbf{7472}}{\mathbf{125}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{c = 59.776}\ \]

(vi) For a frequency distribution of X: D=X-40, ∑fD=150, ∑f=50. Calculate arithmetic mean.

Answer:

Data

A = 40, ∑f = 50, ∑fD = 150

    \[ \overline{\mathbf{X}}\mathbf{= A +}\frac{\mathbf{\sum fD}}{\mathbf{\sum f}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{= 40 +}\frac{\mathbf{150}}{\mathbf{50}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{= 43}\ \]

(vii) A variable Y is determined from a variable X by an equation, Y = 10 – 4X and X = -3, -2, -1, 0, 1, 2, 3, 4, 5. Find Ȳ and show that Ȳ = 10 – 4X̄

Solution

XY = 10 – 4XY
-310 – 4(-3) = 2222
-210 – 4(-2) = 1818
-110 – 4(-1) = 1414
010 – 4(0) = 1010
110 – 4(1) = 66
210 – 4(2) = 22
310 – 4(3) = -2-2
410 – 4(4) = -6-6
510 – 4(5) = -10-10
∑X = 9 ∑Y = 54

    \[ \overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{9}}{\mathbf{9}}\mathbf{= 1}\ \]

    \[ \overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{54}}{\mathbf{9}}\mathbf{= 6}\ \]

    \[ \overline{\mathbf{Y}}\mathbf{\  = \ 10\ -\ 4}\overline{\mathbf{X}}\ \]

    \[ \mathbf{6\  = \ 10\ -\ 4(1)}\ \]

    \[ \mathbf{6\  = 6}\ \]

(viii) Define Mean deviation and variance.

Solution

Dispersion is a measure of scatteredness or dispersement of the data. Broadly there are two types of dispersion one is called absolute dispersion which measures the dispersion in absolute terms and another is relative dispersion which measures the dispersion in relative terms.

Mean deviation and standard deviation both are absolute measures of dispersion.

Formulas for Mean Deviation from mean and variance are given below:

    \[ \mathbf{M.D\ from\ Mean =}\frac{\mathbf{\sum}\left| \mathbf{X -}\overline{\mathbf{X}} \right|}{\mathbf{n}}\ \]

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

(ix)For a frequency distribution of X, it is given that Mean = 50, Mode = 45 and variance = 64. Find coefficient of variation and coefficient of skewness

Solution:

Data

Mean = 50, Mode = 45, Variance = 64, Standard Deviation = 8

Coefficient of Variation

    \[ \mathbf{C.V = \ }\frac{\mathbf{S}}{\overline{\mathbf{X}}}\mathbf{\times 100}\ \]

    \[ \mathbf{C.V =}\frac{\mathbf{8}}{\mathbf{50}}\mathbf{\times 100 = 16}\ \]

Karl Pearson Coefficient of Skewness

    \[ \mathbf{C.\ of\ Skewness = \ }\frac{\mathbf{Mean - Mode\ }}{\mathbf{S.D}}\ \]

    \[ \mathbf{C.\ of\ Skewness = \ }\frac{\mathbf{50 - 45\ }}{\mathbf{16}}\mathbf{= 0.625}\ \]

(x) If Mean = 75, Mode= 70, using empirical relation, find the value of Median.

Answer:

Data

Mean = 75, Mode = 70

Mode = 3Median – 2Mean

70 = 3Median – 2(75)

70 = 3Median – 150

3Median = 70 + 150

3Median = 120

Median = 120/3

Median = 40

(xi) Differentiate between fixed and chain base method.

Answer:

In fixed base method, base period remains fixed whereas in chain base method base period does not remain fixed. On the other hand, in fixed base method, we calculate price relative whereas in chain base method, we calculate link relative and then we calculate chain index. Both methods can be used in simple and composite index. Formulas for Price and Link relatives are given below:

    \[ \mathbf{Price\ Relative = \ }\frac{\mathbf{Pn}}{\mathbf{Po}}\mathbf{\  \times \ 100}\ \]

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

(xii) Compute the chain indices from the following:

Year199019911992199319941995
Price101215202530

Answer:

YearPriceLink RelativeChain Index
199010100100
199112(12/10) x 100 = 120(100 x 120)/100 = 120
199215(15/12) x 100 = 125(120 x 125)/100 = 150
199320(20/15) x 100 = 133.34(150 x 133.34)/100 = 200
199425(25/20) x 100 = 125(200 x 125)/100 = 250
199530(30/25) x 100 = 120(250 x 120)/100 = 300

(xiii) Define Consumer Price Index and write down the major groups included in CPI.

Solution

Consumer Price Index

Consumer Price Index numbers are used 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.

Major Groups Included in CPI

Food & Beverages

Housing

Transportation

Healthcare

Education

Energy

Clothing etc.

(xiv) If Laspeyre’s index is 120 and Paasche’s index is 130, then find Fisher’s index number.

Solution

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index =}\sqrt{\mathbf{Las.\ Index \times Paas.\ Index\ }}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index =}\sqrt{\mathbf{120 \times 130\ }}\ \]

    \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{'}}\mathbf{s\ Ideal\ Index = 124.89}\ \]

(xv) Differentiate between regression and correlation

Answer:

Correlation

Correlation or Correlation coefficient is a statistical measure that calculates the quantitative relationship between two variables. It is denoted by (r). It is always ranges between -1 to +1 both inclusive. It has five different types namely (i) Positive (ii) Perfectly positive (iii) Negative (iv) Perfectly negative and (v) Zero.

Regression

Regression or regression coefficient is also a statistical measure which calculates the strength of relationship between two or more than two variables in which one is dependent and all other variables are considered as independent variables. Two variable model is called simple regression and model which has more than two variables is called multiple regression. Regression is very important tool used in forecasting or trend analysis.

(xvi) It is given that: Syx=32, Sx = 2.4, Sy = 25, X̅=155, Y̅=7, n = 10. Calculate Regression Coefficients byx and bxy.

Solution:

Data

Here

Covariance of X and Y = Syx=32

Standard Deviation of X variable = Sx = 2.4

Variance of X Variable S²x = 5.76

Standard Deviation of Y variable = Sx = 25

Variance of Y Variable S²y = 625

Mean of X Variable = X̅=155

Mean of Y Variable = Y̅=7 and n = 10

Regression Coefficient of Y on X

    \[ \mathbf{b}{\mathbf{yx}}\mathbf{=}\frac{\mathbf{S}{\mathbf{yx}}}{\mathbf{S}_{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{32}}{\mathbf{5.76}}\mathbf{= 5.56}\ \]

Regression Coefficient of X on Y

    \[ \mathbf{b}{\mathbf{yx}}\mathbf{=}\frac{\mathbf{S}{\mathbf{yx}}}{\mathbf{S}_{\mathbf{y}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{32}}{\mathbf{625}}\mathbf{= 0.0512}\ \]

(xvii) Find the correlation coefficients (r) from the regression coefficients. (i) 0.85 and 0.6 (ii) -0.96 and -0.55

Solution:

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

    \[ \mathbf{r =}\sqrt{\mathbf{0.85 \times 0.6\ }}\ \]

    \[ \mathbf{r =}\sqrt{\mathbf{051\ }}\ \]

    \[ \mathbf{r = 0.714}\ \]

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

    \[ \mathbf{r = -}\sqrt{\mathbf{( - 0.96)( - 0.55)\ }}\ \]

    \[ \mathbf{r = -}\sqrt{\mathbf{0.528\ }}\ \]

    \[ \mathbf{r = - 0.7266}\ \]

(xviii) What are the different components of a time series?

Answer

The factors that are responsible to bring about changes in a time series, also called the components of time series, are as follows:

  1. Secular Trend (T)
  2. Seasonal Movements (S)
  3. Cyclical Movements (C)
  4. Irregular Fluctuations or movements (I)

(xix) Calculate three years moving average for the following time series:

Year19801981198219831984198519861987
Sale100140168120200210170220

Answer:

YearSale3 Year M.T3 Year M.A
1980100
1981140408136
1982168428142.67
1983120488162.67
1984200530176.67
1985210580193.34
1986170600200
1987220
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Extensive Questions

Q.3 (a) Find Arithmetic Mean, Median, Mode Q1 and Q3 of the following frequency distribution:

Max: Load (Short tons)No of cables
118—1262
127—1355
136—14412
145—15317
154—16214
163—1716
172—1803

Solution:

Max: Load (Short tons)No of cables (f)C.BXfxC.F
118—1262117.5—126.51222442
127—1355126.5—135.51316557
136—14412135.5—144.5140168019
145—15317144.5—153.5149253336
154—16214153.5—162.5158221250
163—1716162.5—171.5167100256
172—1803171.5—180.517652859
 ∑f=n=59  ∑fx =8854 

Mean

    \[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum X}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{8854}}{\mathbf{59}}\mathbf{= 150.06}\ \]

Median

∑f or n = 59, n/2 = 59/2 = 29.5 falls in C.F of 36 so l = 144.5, f = 17, h = 9 & C = 19

    \[ \widetilde{\mathbf{X}}\mathbf{= l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{n}}{\mathbf{2}}\mathbf{- C} \right)\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{9}}{\mathbf{17}}\left( \mathbf{29.5 - 19} \right)\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{9}}{\mathbf{17}}\left( \mathbf{10.5} \right)\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{94.5}}{\mathbf{17}}\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 150.05}\ \]

Mode

Maximum frequency is 17 so fm = 17, f1 = 12, f2 = 14, h = 9 and l = 144.5

    \[ \widehat{\mathbf{X}}\mathbf{= l +}\frac{\mathbf{f}_{\mathbf{m}}\mathbf{-}\mathbf{f}_{\mathbf{1}}}{\left( \mathbf{f}_{\mathbf{m}}\mathbf{-}\mathbf{f}_{\mathbf{1}} \right)\mathbf{+ (}\mathbf{f}_{\mathbf{m}}\mathbf{-}\mathbf{f}_{\mathbf{2}}\mathbf{)}}\mathbf{\times h}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{17 - 12}}{\left( \mathbf{17 - 12} \right)\mathbf{+ (17 - 14)}}\mathbf{\times 9}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{5}}{\mathbf{5 + 3}}\mathbf{\times 9}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 144.5 +}\frac{\mathbf{45}}{\mathbf{8}}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 5.625}\ \]

Quartile 1 Q1

∑f or n = 59, 1n/4 = 59/4 = 14.75 falls in C.F of 19 so l = 135.5, f = 12, h = 9 & C = 7

    \[ \mathbf{Q}_{\mathbf{1}}\mathbf{= l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{1}\mathbf{n}}{\mathbf{4}}\mathbf{- C} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{1}}\mathbf{= 135.5 +}\frac{\mathbf{9}}{\mathbf{12}}\left( \mathbf{14.75 - 7} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{1}}\mathbf{= 135.5 +}\frac{\mathbf{9}}{\mathbf{12}}\left( \mathbf{7.75} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{1}}\mathbf{= 135.5 +}\frac{\mathbf{69.75}}{\mathbf{12}}\ \]

    \[ \mathbf{Q}_{\mathbf{1}}\mathbf{= 141.3125}\ \]

Quartile 3 Q3

∑f or n = 59, 3n/4 = [(3)(59)]/4 = 44.25 falls in C.F of 50 so l = 153.5, f = 14, h = 9 & C = 36

    \[ \mathbf{Q}_{\mathbf{3}}\mathbf{= l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{1}\mathbf{n}}{\mathbf{4}}\mathbf{- C} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{3}}\mathbf{= 153.5 +}\frac{\mathbf{9}}{\mathbf{14}}\left( \mathbf{44.25 - 36} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{3}}\mathbf{= 153.5 +}\frac{\mathbf{9}}{\mathbf{14}}\left( \mathbf{8.25} \right)\ \]

    \[ \mathbf{Q}_{\mathbf{3}}\mathbf{= 153.5 +}\frac{\mathbf{74.25}}{\mathbf{14}}\ \]

    \[ \mathbf{Q}_{\mathbf{3}}\mathbf{= 158.80}\ \]

(b) Calculate variance, standard deviation and coefficient of variation:

Daily WagesFrequency
1—32
3—54
5—710
7—93
9—111

Solution

Daily WagesFrequency (f)Xfxfx²
1—32248
3—5441664
5—710660360
7—93824192
9—1111010100
 ∑f = 20 ∑fx =114∑fx² =724

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

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{=}\frac{\mathbf{724}}{\mathbf{20}}\mathbf{-}\left( \frac{\mathbf{114}}{\mathbf{20}} \right)^{\mathbf{2}}\ \]

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{= 36.2 - 32.49}\ \]

    \[ \mathbf{Variance\ }\mathbf{S}^{\mathbf{2}}\mathbf{= 3.71}\ \]

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

    \[ \mathbf{S.D\ S =}\sqrt{\mathbf{3.71}}\ \]

    \[ \mathbf{S.D\ S = 1.92}\ \]

    \[ \mathbf{Mean\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{114}}{\mathbf{20}}\mathbf{= 5.7\ }\ \]

Coefficient of Variation

    \[ \mathbf{C.V = \ }\frac{\mathbf{S}}{\overline{\mathbf{X}}}\mathbf{\times 100}\ \]

    \[ \mathbf{C.V =}\frac{\mathbf{1.92}}{\mathbf{5.7}}\mathbf{\times 100 = 33.68}\ \]

Q.4 (a) Compute Chain index from the following price relatives using (i) Mean (ii) geometric mean as an average:

YearCommodities
ABCD
2000817711955
2001625412852
200210487111100
2003937515496
2004604316588

Solution

YearLink Relatives
ABCD
2000817711955
2001(62/81)×100=76.54(54/77)×100=70.12(384/330)×100=116.36(52/55)×100= 94.54
2002(104/62)×100=167.74(87/54)×100=161.12(333/384) × 100=86.71(100/52)×100= 192.30
2003(93/104) × 100=89.42(75/87) × 100=86.20(462/333)×100=138.73(96/100)×100= 96
2004(60/93) × 100=64.51(43/75) × 100=57.33(495/462)×100=107.14(88/96)×100= 91.67
  • Chain Index Using Mean as an average
Year  X̅ = X/n 
2000332/4 =8383
2001357.56/4 = 89.39

    \[ \frac{\mathbf{83}\mathbf{\ \times \ 8}\mathbf{9.39}}{\mathbf{100}}\mathbf{=}\mathbf{74.1937}\ \]

2002607.87/4 = 151.96 

    \[ \frac{\mathbf{74.1937}\mathbf{\ \times \ }\mathbf{151.96}}{\mathbf{100}}\mathbf{=}\mathbf{112.74}\ \]

2003410.35/4 = 102.58 

    \[ \frac{\mathbf{112.74}\mathbf{\ \times \ 102.}\mathbf{58}}{\mathbf{100}}\mathbf{=}\mathbf{115.64}\ \]

2004320.65/4= 80.1625 

    \[ \frac{\mathbf{115.64}\mathbf{\ \times \ }\mathbf{80.1625}}{\mathbf{100}}\mathbf{=}\mathbf{92.699}\ \]

  • Chain Index Using Geometric Mean as an average
Year 

    \[ {\mathbf{G.M(X}\mathbf{1.X}\mathbf{2.X}\mathbf{3)}}^{\mathbf{1/3}}\ \]

 

    \[ \mathbf{C.I = \ }\frac{\mathbf{Chai}\mathbf{n}{\mathbf{n - 1}}\mathbf{\ \times \ Averag}\mathbf{e}{\mathbf{n}}\mathbf{\ L.R}}{\mathbf{100}}\ \]

2000 

    \[ \mathbf{(81 \times 77 \times 119 \times 55)}^{\mathbf{1/4}}\mathbf{= 79.93}\ \]

79.93
2001 

    \[ \mathbf{(76.54 \times 70.12 \times 116.36 \times 94.54)}^{\mathbf{1/4}}\mathbf{= 87.65}\ \]

    \[ \frac{\mathbf{79.93\ \times \ 87.65}}{\mathbf{100}}\mathbf{= 70.05}\ \]

2002 

    \[ \mathbf{(167.74 \times 161.12 \times 86.71 \times 192.30)}^{\mathbf{1/4}}\mathbf{= 145.69}\ \]

 

    \[ \frac{\mathbf{70.05\ \times \ 145.69}}{\mathbf{100}}\mathbf{= 102.05}\ \]

2003 

    \[ {\mathbf{(89.42 \times 86.20 \times 138.73}\mathbf{\times 96}\mathbf{)}}^{\mathbf{1/}\mathbf{4}}\mathbf{= 10}\mathbf{0.65}\ \]

 

    \[ \frac{\mathbf{102.05}\mathbf{\ \times \ 1}\mathbf{00.65}}{\mathbf{100}}\mathbf{=}\mathbf{102.71}\ \]

2004 

    \[ {\mathbf{(64.51 \times 57.33 \times 107.14}\mathbf{\times 91.67}\mathbf{)}}^{\mathbf{1/}\mathbf{4}}\mathbf{= 7}\mathbf{7.63}\ \]

 

    \[ \frac{\mathbf{102.71}\mathbf{\ \times \ 7}\mathbf{7.63}}{\mathbf{100}}\mathbf{=}\mathbf{79.73}\ \]

(b) An inquiry into budgets of the middle class families in a city for year 1989 – 1990 was conducted. Construct Consumer Price Index.

The following price relatives are given:

ExpensesFoodRentClothingFuelMisc.
Weights (W)35%15%20%10%20%
Price Relative (I)116120125125150

Solution:

ExpensesWeights (W)Price Relative (I)WI
Food351164060
Rent151201800
Clothing201252500
Fuel101251250
Misc.201503000
 ∑W = 100 ∑WI = 12610

Method to Calculate

Household Budget method or Family Budget Method

    \[ \mathbf{P}_{\mathbf{on}}\mathbf{=}\frac{\mathbf{\sum}\left( \frac{\mathbf{Pn}}{\mathbf{Po}} \right)\mathbf{Poqo}}{\mathbf{\sum Poqo}}\mathbf{\times 100 =}\frac{\mathbf{\sum IW}}{\mathbf{\sum W}}\ \]

Where Price Relative I equals to:

    \[ \frac{\mathbf{Pn}}{\mathbf{Po}}\mathbf{\times 100,\ W = Poqo}\ \]

    \[ \mathbf{P}_{\mathbf{on}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{WI}}{\mathbf{\sum W}}\mathbf{=}\frac{\mathbf{12610}}{\mathbf{100}}\mathbf{= 126.1}\ \]

Q.5 (a) The following data is obtained in a study on the number of absentees (X) and the final Marks (Y) of seven students from a class. (i) Compute Correlation coefficient (r) (ii) Obtain regression line Yon X and estimate final marks when there are 20 absentees.

X621591258
Y85864374589078

Solution:

XYXY
685510367225
28617247396
15436452251849
974666815476
12586961443364
590450258100
878624646084
∑X = 57∑Y = 514∑XY = 3763∑X² = 579∑Y² = 39494

Solution (i)

Correlation Coefficient (r)

Regression Coefficient X on Y

    \[ \mathbf{d}\mathbf{= \ }\frac{\mathbf{\sum}\mathbf{xy - n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{Y}^{\mathbf{2}}\mathbf{- n}\mathbf{(}\overline{\mathbf{Y}}\mathbf{)}\mathbf{²}}\ \]

    \[ \mathbf{d}\mathbf{= \ }\frac{\mathbf{3763 -}\left( \mathbf{7} \right)\left( \mathbf{8.14} \right)\left( \mathbf{73.42} \right)}{\mathbf{39494}\mathbf{- 7}\left( \mathbf{73.42} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{d}\mathbf{= \ }\frac{\mathbf{- 420.4716}}{\mathbf{1}\mathbf{760.5252}}\ \]

    \[ \mathbf{d}\mathbf{=}\mathbf{-}\mathbf{0.2388}\ \]

Regression Coefficient Y on X (b) is calculated in (ii)

    \[ \mathbf{r = \pm}\sqrt{\mathbf{b \times d}}\ \]

    \[ \mathbf{r = -}\sqrt{\mathbf{( - 3.65)( - 0.2388)}}\ \]

    \[ \mathbf{r = -}\sqrt{\mathbf{0.87162}}\ \]

    \[ \mathbf{r = -}\mathbf{0.9336\ Negative\ Relation}\ \]

Solution (ii)

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

    \[ \mathbf{b}\mathbf{= \ }\frac{\mathbf{\sum}\mathbf{xy - n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}\mathbf{- n}\mathbf{(}\overline{\mathbf{X}}\mathbf{)}\mathbf{²}}\ \]

    \[ \mathbf{Mean}\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{X}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{57}}{\mathbf{7}}\mathbf{= 8.14}\ \]

    \[ \mathbf{Mean}\overline{\mathbf{Y}}\mathbf{=}\frac{\mathbf{\sum}\mathbf{Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{514}}{\mathbf{7}}\mathbf{=}\mathbf{73.42}\ \]

    \[ \mathbf{b}\mathbf{= \ }\frac{\mathbf{\sum}\mathbf{xy - n}\overline{\mathbf{X}}\overline{\mathbf{Y}}}{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}\mathbf{- n}\left( \overline{\mathbf{X}} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{b}\mathbf{= \ }\frac{\mathbf{3763}\mathbf{-}\left( \mathbf{7} \right)\left( \mathbf{8.14} \right)\left( \mathbf{73.42} \right)}{\mathbf{579}\mathbf{-}\mathbf{7}\left( \mathbf{8.14} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{b}\mathbf{= \ }\frac{\mathbf{- 420.4716}}{\mathbf{115.1828}}\ \]

    \[ \mathbf{b}\mathbf{=}\mathbf{-}\mathbf{3.65}\ \]

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

    \[ \mathbf{a}\mathbf{=}\mathbf{73.42}\mathbf{-}\left( \mathbf{- 3.65} \right)\left( \mathbf{8.14} \right)\ \]

    \[ \mathbf{a}\mathbf{=}\mathbf{103.131}\ \]

    \[ \widehat{\mathbf{y}}\mathbf{\  = \ a\  + \ bx\ }\ \]

    \[ \widehat{\mathbf{y}}\mathbf{= \ }\mathbf{103.131 -}\mathbf{\ }\mathbf{3.65}\mathbf{x}\ \]

Final marks when there are 20 absentees

    \[ \widehat{\mathbf{y}}\mathbf{= \ 103.131 - \ 3.65x}\ \]

    \[ \widehat{\mathbf{y}}\mathbf{= \ 103.131 - \ 3.65}\mathbf{(20)}\ \]

    \[ \widehat{\mathbf{y}}\mathbf{=}\mathbf{1}\mathbf{03.131 -}\mathbf{73}\ \]

    \[ \widehat{\mathbf{y}}\mathbf{=}\mathbf{30.131}\ \]

Approximately 30 marks if absentees are 20

b. Find 4-qurarters centered moving averages from the following time series data:

YearsQuarters
IIIIIIIV
2005160165163161
2006170167172171
2007172169167170
2008175177172170

Solution:

Year & QuarterValues4 Quarter Moving Total (C.1)2 Values Moving Total (C.2)4 Quarter centered Moving Average (C.2/8)
2005   (I)160   
  
(ii)165
  649
(iii)163 1308163.5
  659  
(iv)161 1320165
  661  
2006 (i)170 1331166.375
  670  
(ii)167 1350168.75
  680  
(iii)172 1362170.25
  682  
(iv)171 1366170.75
  684  
2007 (i)172 1363170.375
  679  
(ii)169 1357169.625
  678  
(iii)167 1359169.875
  681  
(iv)170 1370171.25
  689  
2008 (i)175 1383172.875
  694  
(ii)177 1388173.5
  694  
(iii)172 
  
(iv)170
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