Solved Paper Statistics I FBISE 2015 and 2015 2nd Annual

Solved Paper Statistics I FBISE 2015 and 2015 2nd Annual

Solved Paper Statistics I FBISE 2015 and 2015 2nd Annual, Dive into a comprehensive solution guide to the FBISE Statistics I 2015 and 2015 2nd Annual 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 FBISEBISERWPBISELHRMUDUPU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

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Table of Contents

Solved Paper Statistics I FBISE 2015 and 2015 2nd Annual

MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)A specific character of sample is called:
 A.VariableB. ConstantC. ParameterD. Statistic 
      
(ii)A variable that assumes only some selected values in a range is called:
 A.Continuous VariableB. Quantitative VariableC. Qualitative VariableD. Discrete data 
      
(iii)Color of flower is an example of:
 A.Quantitative VariableB. Symmetric VariableC. Skewed VariableD. Qualitative Variable 
      
(iv)The grouped data is called:
 A.Primary dataB. Raw dataC. Secondary DataD. Difficult to tell 
      
(v)In classification, the data is arranged according to:
 A.percentagesB. DifferencesC. SimilaritiesD. Ratios 
      
(vi)A sector diagram is also called:
 A.Bar DiagramB. HistogramC. Pie DiagramD. Historigram 
      
(vii)The measure of central tendency listed below is:
 A.The raw scoreB. The MeanC. The RangeD. Standard Deviation 
      
(viii)Mode of the series 0, 0, 0, 2, 2, 3, 3, 8, 10 is:
 A.2B. 0C. 3D. No Mode 
      
(ix)The population mean µ is called:
 A.Discrete variableB. ParameterC. Continuous VariableD. Sampling Unit 
      
(x)In symmetrical distribution Q3 – Q1 = 20, Median = 15, Q3 is equal to:
 A.5B. 25C. 20D. 15 
      
(xi)The Standard deviation of -5, -5, -5, -5 is:
 A. 0B. +5C. -5D. -25 
      
(xii)The measures of dispersion can never be:
 A. NegativeB. ZeroC. PositiveD. Equal to 2 
      
(xiii)When index number is calculated for several variable, it is called:
 A. Composite IndexB. Wholesale price IndexC. Volume IndexD. Simple Index 
      
(xiv)Paasche’s price index number is also called:
 A. Base year weightedB. Current year weightedC. Simple Aggregative indexD. Consumer price Index 
      
(xv)Index number having upward bias is:
 A. Laspeyre’s IndexB. Paasche’s IndexC. Fisher’s IndexD. Marshall Edgeworth Index 
      
(xvi)If regression line of ŷ = 5 then value of regression coefficient of y on x is:
 A. 0.5B. 0C. 1D. 5 
      
(xvii)A business cycle has:
 A. One stageB. Four StagesC. Three stagesD. Two stages 
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Short Questions

(i) Differentiate between Variable & 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 etc.

(ii) Name the sources of primary data.

Answer:

The most common methods or sources of collecting primary data are:

  • Questionnaires
  • Surveys
  • Interviews
  • Polls
  • E-mail & Telephone

(iii) Distinguish between Histogram and Historigram.

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.

(iv)Write down the 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.

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(v) u=(x-170)/5 , ∑fu=100 and ∑f=200. Find arithmetic mean.

Solution:

    \[  \overline{\mathbf{X}}\mathbf{= A +}\frac{\mathbf{\sum fu}}{\mathbf{\sum f}}\mathbf{\ }\mathbf{\times}\mathbf{\ h\ }\ \]

    \[  \overline{\mathbf{X}}\mathbf{= 170 +}\frac{\mathbf{100}}{\mathbf{200}}\mathbf{\ }\mathbf{\times}\mathbf{\ 5\ }\ \]

    \[  \overline{\mathbf{X}}\mathbf{=}\mathbf{170 +}\frac{\mathbf{500}}{\mathbf{200}}\ \]

    \[  \overline{\mathbf{X}}\mathbf{=}\mathbf{170 + 2.5}\ \]

    \[  \overline{\mathbf{X}}\mathbf{=}\mathbf{172.5}\  \]

(vi) Define Geometric mean.

Answer:

It is the type of average which is typically used in investment portfolio. There are two different formulas for ungrouped & grouped data:

For Ungroup Data

    \[ G.M = Antilog\ \left( \frac{\sum\log x}{n} \right)\ \]

For Group Data

    \[ G.M = Antilog\ \left( \frac{\sum f\log x}{\sum f} \right)\ \]

(vii) Find unbiased sample standard deviation of the scores 30, 35, 40..

Solution

X
30900
351225
401600
∑X =105∑X² =3725

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{X}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{3725}}{\mathbf{3}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{105}}{\mathbf{3}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D =}\sqrt{\mathbf{1241.67 - 1225}}\ \]

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\mathbf{16.67}}\ \]

    \[ \mathbf{S.D}\mathbf{=}\mathbf{4.08}\ \]

(viii) ∑x = 180, ∑x² =6660 and n = 5, Find coefficient of variation.

Solution

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

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}{\mathbf{n}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{\sum}\mathbf{X}}{\mathbf{n}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\left\lbrack \frac{\mathbf{6660}}{\mathbf{5}}\mathbf{-}\mathbf{\ }\left( \frac{\mathbf{180}}{\mathbf{5}} \right)^{\mathbf{2}} \right\rbrack}\ \]

    \[ \mathbf{S.D}\) \textbf{=} \(\sqrt{\mathbf{1332 - 1296}}\ \]

    \[ \mathbf{S.D}\mathbf{=}\sqrt{\mathbf{36}}\ \]

    \[ \mathbf{S.D}\mathbf{=}\mathbf{6}\ \]

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

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

    \[ \mathbf{C.V}\mathbf{= \ 16.67}\ \]

Solved Paper Statistics I FBISE 2015 and 2015 2nd Annual
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(ix)The first four moments about the arithmetic mean of a distribution are 0, 4, 6 and 48. Find β2.

Solution

    \[ \mathbf{\beta}\mathbf{2 = \ }\frac{\mathbf{\mu}\mathbf{4}}{\mathbf{\mu}\mathbf{2²}}\mathbf{= \ }\frac{\mathbf{48}}{\mathbf{4²}}\mathbf{=}\frac{\mathbf{48}}{\mathbf{16}}\mathbf{= 3\ \ }\ \]

(x) Define the term skewness.

Answer:

Skewness in statistics represents an imbalance and asymmetry from the mean of a data distribution. In positive skewness mean > median > mode whereas in negative skewness mean < median < mode. In other words  if the frequency curve has a longer tail to right, the distribution is said to be positively skewed.If the frequency curve has a longer tail to left, the distribution is said to negatively skewed.

(xi) ∑poqn = 1000 and ∑pnqn = 1360. Find current year weighted index.

Solution:

    \[ \mathbf{Current\ Year\ Weighted\ Index =}\frac{\mathbf{\sum pnqn}}{\mathbf{\sum poqn}}\mathbf{\ \times \ 100}\ \]

    \[ \mathbf{Current\ Year\ Weighted\ Index}\mathbf{= \ }\frac{\mathbf{1360}}{\mathbf{1000}}\mathbf{\ \times \ 100 = 136\ }\ \]

(xii) What is cost of living index number?

Answer:

Cost of living index is a statistical measure to calculate the living cost with respect to time, place etc. It measures the change of consumer basket goods and services with respect to time and place. It is an important tool to measure the inflation or deflation or purchasing power.

(xiii) X̅=1, Y̅ = 8 and b = 2. Find the value of intercept a.

Solution

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

    \[ \mathbf{a = 8 -}\left( \mathbf{2} \right)\left( \mathbf{1} \right)\mathbf{= \ 6}\ \]

(xiv) If byx = -1.6 and bxy = -0.4. Find the value of rxy.

Solution:

    \[ \mathbf{rxy = \ }\sqrt{\mathbf{byx\ }\mathbf{\times}\mathbf{bxy}}\ \]

    \[ \mathbf{rxy = \ }\sqrt{\mathbf{- 1.6\ }\mathbf{\times}\mathbf{\ -0.4}}\ \]

    \[ \mathbf{rxy = \ }\sqrt{\mathbf{0.64}}\ \]

    \[ \mathbf{rxy = \ 0.8}\ \]

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(xv) What is meant by residual?

Answer:

In regression analysis, the difference between the observed value of the dependent variable (y) and the predicted value (ŷ) is called the residual (e). Each data point has one residual. Residual = Observed value – Predicted value. e = y – ŷ both the sum and the mean of the residuals are equal to zero.

(xvi) ∑x = 0, ∑y=27.1, ∑xy=29.5, ∑x² = 330, n = 10. Determine the value of b:

Solution:

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

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{29.5}\mathbf{-}\mathbf{\ 10}\left( \frac{\mathbf{0}}{\mathbf{10}} \right)\left( \frac{\mathbf{27.1}}{\mathbf{10}} \right)}{\mathbf{330}\mathbf{- \ 10}\left( \mathbf{0} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{29.5}}{\mathbf{330}}\ \]

    \[ \mathbf{b}\mathbf{= 0.08}\ \]

(xvii) If y = 16, 18, 20, 22, 24, x = -2, -1, 0, 1, 2 and ŷ = 20 + 2x. Compute the sum of squares of residual.

Solution:

Sum of Square of Residual = ∑(y – ŷ)²:

YXŷ = 20 + 2x(y – ŷ)(y – ŷ
16-2ŷ = 20 + 2(-2) = 1616 – 16 = 00
18-1ŷ = 20 + 2(-1) = 1818 – 18 = 00
200ŷ = 20 + 2(0) = 2020 – 20 = 00
221ŷ = 20 + 2(1) = 2222 – 22 = 00
242ŷ = 20 + 2(2) = 2424 – 24 = 00
    ∑(y – ŷ)² =0

(xviii) What is meant by seasonal variation?

Answer

These are short term movements occurring in a data due to seasonal factors. For example,  it is commonly observed that the consumption of ice-cream during summer us generally high and hence sales of an ice-cream dealer would be higher in some months of the year while relatively lower during winter months.

(xix) Define irregular variation.

Answer: These are sudden changes occurring in a time series. These are sudden in nature that is why they cannot be explained through trends or seasonal or cyclical movements. For example sudden change due to tsunami or earthquake etc.

Extensive Questions

Q.3 (a) The following frequency distribution shows the hourly income of 100 households in a locality: Calculate the arithmetic mean and show that sum of deviations of values from their mean is zero.

Income (Rs)Frequency
35—3913
40—4415
45—4928
50—5417
55—5912
60—6410
65—695
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Solution (a)

Income (Rs)Frequency (f)XfXX – X̅f(X – X̅)
35—391337481-12.5-162.5
40—441542630-7.5-112.5
45—4928471316-2.5-70
50—5417528842.542.5
55—5912576847.590
60—64106262012.5125
65—6956733517.587.5
      
 100 4950 0
 ∑f = ∑fX = f(X – X̅) =

    \[ \overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{=}\mathbf{\ }\frac{\mathbf{4950}}{\mathbf{100}}\mathbf{= 49.5}\ \]

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

(b) Compute the Bowley’s coefficient of skewness and interpret its value of the data given below:

GroupFrequency
40—5012
50—6015
60—7016
70—8015
80—9012

Solution (b)

GroupFrequency (f)c.f  X
40—50121245
50—60152755
60—70164365
70—80155875
80—90127085
    
 70  
 ∑f =  

    \[ \frac{\mathbf{n}}{\mathbf{2}}\mathbf{= \ }\frac{\mathbf{70}}{\mathbf{2}}\mathbf{= 35}\ \]

    \[ \frac{\mathbf{1}\mathbf{n}}{\mathbf{4}}\mathbf{= \ }\frac{\mathbf{70}}{\mathbf{4}}\mathbf{= 17.5}\ \]

    \[ \frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{= \ }\frac{\mathbf{3(70)}}{\mathbf{4}}\mathbf{= 52.5}\ \]

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

    \[ \mathbf{Median}\mathbf{= 60 +}\frac{\mathbf{10}}{\mathbf{16}}\left( \mathbf{35 - \ 27} \right)\ \]

    \[ \mathbf{Median}\mathbf{= 60 +}\frac{\mathbf{80}}{\mathbf{16}}\mathbf{= 60 + 5}\ \]

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    \[ \mathbf{Median}\mathbf{= 65\ }\ \]

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

    \[ \mathbf{Q}\mathbf{1}\mathbf{= 50 +}\frac{\mathbf{10}}{\mathbf{15}}\left( \mathbf{17.5 - \ 12} \right)\ \]

    \[ \mathbf{Q}\mathbf{1}\mathbf{= 50 +}\frac{\mathbf{55}}{\mathbf{15}}\ \]

    \[ \mathbf{Q}\mathbf{1}\mathbf{= 50 + 3.67}\ \]

    \[ \mathbf{Q}\mathbf{1}\mathbf{= 53.67}\ \]

    \[ \mathbf{Q3 = \ l +}\frac{\mathbf{h}}{\mathbf{f}}\left( \frac{\mathbf{3}\mathbf{n}}{\mathbf{4}}\mathbf{- \ c} \right)\mathbf{\ }\ \]

    \[ \mathbf{Q}\mathbf{3}\mathbf{= 70 +}\frac{\mathbf{10}}{\mathbf{15}}\left( \mathbf{52.5 - \ 43} \right)\ \]

    \[ \mathbf{Q}\mathbf{3}\mathbf{= 70 +}\frac{\mathbf{95}}{\mathbf{15}}\ \]

    \[ \mathbf{Q}\mathbf{3}\mathbf{= 70 + 6.34}\ \]

    \[ \mathbf{Q}\mathbf{3}\mathbf{= 76.34}\ \]

    \[ \mathbf{Bowle}\mathbf{y}^{\mathbf{'}}\mathbf{s\ }\mathbf{C.S}\mathbf{= \ }\frac{\mathbf{Q}\mathbf{3 + Q}\mathbf{1 - 2\ Median}}{\mathbf{Q}\mathbf{3 - Q}\mathbf{1}}\ \]

    \[ \mathbf{Bowle}\mathbf{y}^{\mathbf{'}}\mathbf{s\ C.S}\mathbf{= \ }\frac{\mathbf{76.34 + 53.67 - 2\ }\left( \mathbf{65} \right)}{\mathbf{76.34 - 53.67}}\ \]

    \[ \mathbf{Bowle}\mathbf{y}^{\mathbf{'}}\mathbf{s\ C.S}\mathbf{=}\frac{\mathbf{130.01 - 130}}{\mathbf{22.67}}\ \]

    \[ \mathbf{Bowle}\mathbf{y}^{\mathbf{'}}\mathbf{s\ C.S}\mathbf{=}\frac{\mathbf{130 - 130}}{\mathbf{22.67}}\mathbf{= 0}\ \]

Q.4 Compute the index numbers using simple aggregative method with 1952 as base year:

Commodity1952195319541955
Wheat25.221.325.430.2
Rice15.916.318.919.3
Barley15.914.016.318.5
Jawar11.314.311.513.6
Grams13.013.513.613.9

Solution

Commodity1952195319541955
Wheat25.221.325.430.2
Rice15.916.318.919.3
Barley15.914.016.318.5
Jawar11.314.311.513.6
Grams13.013.513.613.9
     
 81.379.485.795.5
 ∑Po=∑P1=∑P2=∑P3=
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    \[ \mathbf{Aggregative\ Index\ for\ 1953 = \ }\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{1}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{o}}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1953}\mathbf{=}\frac{\mathbf{79.4}}{\mathbf{81.3}}\mathbf{\times 1}\mathbf{00}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1953}\mathbf{= 97.66}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1954 = \ }\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{2}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{o}}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1954}\mathbf{=}\frac{\mathbf{85.7}}{\mathbf{81.3}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1954}\mathbf{= 105.41}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1955 = \ }\frac{\mathbf{\sum}\mathbf{p}_{\mathbf{3}}}{\mathbf{\sum}\mathbf{p}_{\mathbf{o}}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1955}\mathbf{=}\frac{\mathbf{95.5}}{\mathbf{81.3}}\mathbf{\times 100}\ \]

    \[ \mathbf{Aggregative\ Index\ for\ 1955}\mathbf{= 117.46}\ \]

Q.5 (a) The following sample observations were randomly selected:

X453612
Y46578

Determine the value ŷ when x is 7.

Solution (a)

XYXY
44161616
56253630
3592515
67364942
1281446496
     
3030230190199
∑X =∑Y =∑X² =∑Y² =∑XY =

    \[ \overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{= \  = \ }\frac{\mathbf{30}}{\mathbf{6}}\mathbf{= 5}\ \]

    \[ \overline{\mathbf{Y}}\mathbf{= \ }\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{= \  = \ }\frac{\mathbf{30}}{\mathbf{6}}\mathbf{= 5}\ \]

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

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{199}\mathbf{-}\mathbf{\ 5}\left( \frac{\mathbf{30}}{\mathbf{5}} \right)\left( \frac{\mathbf{30}}{\mathbf{5}} \right)}{\mathbf{230}\mathbf{- \ 5}\left( \mathbf{6} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{19}}{\mathbf{50}}\mathbf{= 0.38}\ \]

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

    \[ \mathbf{a = 6 - \ 0.38(6)}\ \]

    \[ \mathbf{a = 6 - \ 2.28 = 3.72}\ \]

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

    \[ \widehat{\mathbf{y}}\mathbf{= 3.72 + (0.38)x}\ \]

If x = 7 then ŷ:

    \[ \widehat{\mathbf{y}}\mathbf{= 3.72 +}\left( \mathbf{0.38} \right)\mathbf{(7)}\ \]

    \[ \widehat{\mathbf{y}}\mathbf{= 6.38}\ \]

b. Compute 3 year and 5 year moving averages from the following data:

YearFactory Sales in Millions
19826.2
19837.8
19848.3
19859.3
19868.6
19877.8
19888.1
19897.9

Solution (b)

YearFactory Sales in Millions3 Year Moving Total3 Year Moving Average5 Year Moving Total5 Year Moving Average
19826.2  
19837.822.37.433
19848.325.48.46740.28.04
19859.326.28.73341.88.36
19868.625.78.56742.18.42
19877.824.58.16741.78.34
19888.123.87.933 
19897.9 
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