Solved Paper Statistics I FBISE 2016

Solved Paper Statistics I FBISE 2016
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Solved Paper Statistics I FBISE 2016 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 FBISE, BISERWP, BISELHR, MU, DU, PU, NCERT, CBSE & others & across all the statistics, business & finance disciplines.

Table of Contents

Solved Paper Statistics I FBISE 2016

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MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)A specific character of population is called:
 A. StatisticB. ParameterC. VariableD. Sample 
      
(ii)Listing of the data in order of Numerical Magnitude is called:
 A. Raw DataB. Arrayed DataC. Discrete DataD. Continuous Data 
      
(iii)The word ogive is also used for:
 A. Frequency PolygonB. Cumulative FrequencyC. Frequency CurveD. Histogram 
      
(iv)A variable that assumes any value within a range is called:
 A. Discrete VariableB. Continuous VariableC. Independent VariableD. Dependent Variable 
      
(v)The average of upper and lower class limit is called:
 A. Class BoundaryB. Class FrequencyC. Class MarkD. Class Limit 
      
(vi)A pie diagram is represented by a:
 A. RectangleB. TriangleC. CircleD. Square 
      
(vii)Step-Deviation method or coding method is used for computation of the:
 A. Geometric MeanB. Harmonic MeanC. Arithmetic MeanD. Weighted Mean 
      
(viii)A Curve that tails off to the right end is called:
 A. SymmetricalB. Negatively SkewedC. Positively SkewedD. Both A & B 
      
(ix)The Sample mean X̅ is a:
 A. ParameterB. ConstantC. VariableD. Statistic 
      
(x)The variance is zero only if all observations are:
 A. DifferentB. SquareC. Square rootD. Same 
      
(xi)The range of the values -5, -8, -10, 0, 6, 10 is:
 A. 0B. 10C. -10D. 20 
      
(xii)Bowley’s Coefficient of skewness lies between:
 A. 0 and 1B. -2 and +2C. -1 and 0D. -1 and +1 
      
(xiii)Index for base period is always taken as:
 A. 100B. OneC. 200D. Zero 
      
(xiv)Price relative computed by Chain Base method is called:
 A. Link RelativeB. Chain IndicesC. Price RelativesD. None of these 
      
(xv)Base year quantities as weights are used in:
 A. Laspeyre’s MethodB. Paasche’s MethodC. Fisher’s Ideal MethodD. None of these 
      
(xvi)In simple regression equation, the number of variables involved is:
 A. 2B. 1C. 0D. 3 
      
(xvii)Depression in Business is:
 A. CyclicalB. SecularC. SeasonalD. Irregular 

Short Questions

(i) Differentiate between parameter & statistic.

Answer:

A measure computed from a population data is called parameter. For example, a population mean is a parameter. For example, below given formula is population average and it is parameter.

µ= ∑X/N

A measure computed from a sample data is called statistic. For example, sample mean is a statistic. For example, below given formula is sample average and it is statistic.

    \[ \ \left( \overline{X} = \frac{\sum x}{n} \right)\ \ \]

(ii) Name the branches of statistics.

Answer:

There are two branches of statistics. Descriptive & Inferential.

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.

(iii) Differentiate between grouped and ungrouped data.

Answer:

Ungrouped Data

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

Grouped Data

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

(iv) The mean of 5 observations is 60. Another item is included in the observations and now the mean becomes 62. Find the included item.

Solution:

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

    \[ \mathbf{60 =}\frac{\mathbf{\sum}\mathbf{X}}{\mathbf{5}}\ \]

    \[ \mathbf{\sum}\mathbf{X = 60\ }\mathbf{\times}\mathbf{\ 5 = 300}\ \]

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

    \[ \mathbf{62 =}\frac{\mathbf{\sum}\mathbf{X}}{\mathbf{6}}\ \]

    \[ \mathbf{\sum}\mathbf{X = 62\ }\mathbf{\times}\mathbf{\ 6 = 372}\ \]

372 – 300 = 72 new included item is 72

(v)∑(x – 10) = 2.8 and n = 5. Find the sample mean.

Solution:

Data

∑D = 2.8, A = 10, n = 5

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

    \[ \overline{\mathbf{X}}\mathbf{=}\mathbf{10 +}\frac{\mathbf{2.8}}{\mathbf{5}}\ \]

    \[ \overline{\mathbf{X}}\mathbf{= 10 + 0.56}\ \]

    \[ \overline{\mathbf{X}}\mathbf{= 10.56}\ \]

(vi) What is meant by measures of central tendency?

Answer:

Measures of central tendency is a statistical concept and related to descriptive branch of statistics in which average or central point of the data is calculated. There are various forms of averages such as arithmetic mean, median, mode, harmonic mean and geometric mean can be calculated in order to know the central point or average of the data.

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Solved Paper Statistics I FBISE 2016

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

Solution

XX²
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) If var(X) = 25 then find Var (2X + 4) .

Solution

Var (2X + 4)

4 Var(X) + Var(4)

4(25) + 0 = 100

(ix)The first two moments of a distribution about the value 5 of a variable are 2 and 32. Find Variance.

Solution:

Second moment is always equal to variance so we are going to calculate second moment below:

    \[ m_{2} = m_{2}^{'} - \left( m_{1}^{'} \right)^{2}\ \]

    \[ m_{2} = 32 - (2)^{2}\ \]

    \[ m_{2} = 32 - 4 = 28\ \]

So variance is 28

(x) Define the standard deviation.

Answer:

Standard deviation is a measure of dispersion. It is an absolute measure of dispersion. It is a square root of variance. It calculates the spread or dispersion of the data. There are various formulas to calculate it as one of the is given below:

Ungrouped Data

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

Grouped Data

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

(xi) ∑poqo = 1500 and ∑pnqo = 2040. Find base year weighted index.

Solution:

    \[ \mathbf{Base}\mathbf{\ }\mathbf{Year\ Weighted\ Index =}\frac{\mathbf{\sum pnqo}}{\mathbf{\sum poqo}}\mathbf{\ }\mathbf{\times}\mathbf{\ 100}\ \]

    \[ \mathbf{Base\ Year\ Weighted\ Index}\mathbf{= \ }\frac{\mathbf{2040}}{\mathbf{1500}}\mathbf{\ }\mathbf{\times}\mathbf{\ 100}\ \]

    \[ \mathbf{Base\ Year\ Weighted\ Index}\mathbf{= 136\ }\ \]

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

Answer:

Economic Measurement: Measures changes in economic variables like prices, wages, and production.

Cost of Living: Assesses changes in the cost of living for households.

Policy Formulation: Aids policymakers in planning and decision-making.

Business Trends: Tracks business performance and market trends.

Inflation Tracking: Monitors inflation rates and price stability.

International Comparison: Facilitates comparison of economic conditions across countries.

Base for Contracts: Used in wage agreements, rent contracts, and price adjustments.

Investment Analysis: Helps investors analyze market conditions and trends.

(xiii) X̅=50, Y̅ = 110 and a = 10. Find the value of slope b.

Solution

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

    \[ \mathbf{10 = 110 - b}\left( \mathbf{50} \right)\ \]

    \[ \mathbf{- b\ }\left( \mathbf{50} \right)\mathbf{= 10 - 110}\ \]

    \[ \mathbf{- b\ }\left( \mathbf{50} \right)\mathbf{= - 100}\ \]

    \[ \mathbf{b\ =}\frac{\mathbf{100}}{\mathbf{50}}\mathbf{= 2}\ \]

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

Solution

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

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

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

    \[ \mathbf{rxy = \ 0.8}\ \]

(xv) Write down the properties of correlation coefficient.

Answer:

(i) The correlation coefficient is symmetrical with respect to X and Y i.e.

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

(ii) The correlation coefficient is the geometric mean of the two regression coefficients.

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

(iii) The correlation coefficient is independent of origin and unit of measurement i.e.

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

(iv) The correlation coefficient lies between -1 and +1 i.e. -1 ≤ r ≤ +1

(xvi) ∑x = 0, ∑y=41172 and n = 10. Find the value of X intercept a.

Solution:

    \[ \mathbf{a =}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{41172}}{\mathbf{10}}\mathbf{= 4117.2}\ \]

    \[ \mathbf{a = 4117.2}\ \]

(xvii) ∑x = 0, ∑y=245, ∑x² = 28, ∑xy=66 and n = 7. Fit a linear trend.

Solution:

    \[ \mathbf{b = \ }\frac{\mathbf{\sum XY}}{\mathbf{\sum}\mathbf{X}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{66}}{\mathbf{28}}\mathbf{= 2.35}\ \]

    \[ \mathbf{a =}\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{=}\frac{\mathbf{245}}{\mathbf{7}}\mathbf{= 35}\ \]

    \[ \mathbf{a = 35}\ \]

So Ŷ = a + bx = 35 + 2.35X

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

(xix) Differentiate between signal & noise.

Answer:

Statistical signal

In Time Series analysis, Signal is a pattern of regular variation. It provides the meaningful insight of the data that express the genuine change or variation over the period of time or

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

Statistical noise

In Time Series Analysis, it represents the irregular, random variations of the data over the period of time. These random fluctuations or noise may be caused by errors, external factors. It is simply the disturbance of the data which is unpredictable or

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

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Extensive Questions

Q.3 (a) The following table shows the distribution of the maximum loads in short tons supported by certain cables produced by a company: Determine mean, median & mode

Max: Load (Short tons)No of cables
9.3—9.72
9.8—10.25
10.3—10.712
10.8—11.217
11.3—11.714
11.8—12.26
12.3—12.73
12.8—13.21

Solution (a)

Max: Load (Short tons)No of cables (f)Class BoundariesXfXC.f
9.3—9.729.25—9.759.5192
9.8—10.259.75—10.2510507
10.3—10.71210.25—10.7510.512619
10.8—11.21710.75—11.251118736
11.3—11.71411.25—11.7511.516150
11.8—12.2611.75—12.25127256
12.3—12.7312.25—12.7512.537.559
12.8—13.2112.75—13.25131360
      
 60  665.5 
 ∑f=  ∑fX= 

    \[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{\ }\ \]

    \[ \mathbf{A.M\ }\overline{\mathbf{X}}\mathbf{=}\frac{\mathbf{665.5}}{\mathbf{60}}\mathbf{\ = 11.09}\ \]

n/2=60/2=30 so model class is 10.75-11.25

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

    \[ \widetilde{\mathbf{X}}\mathbf{= 10.75 +}\frac{\mathbf{0.5}}{\mathbf{17}}\left( \mathbf{30 - 19} \right)\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 10.75 +}\frac{\mathbf{5.5}}{\mathbf{17}}\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 11.07}\ \]

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

    \[ \widehat{\mathbf{X}}\mathbf{= 10.75 +}\frac{\mathbf{17 - 12}}{\left( \mathbf{17 - 12} \right)\mathbf{+}\left( \mathbf{17 - 14} \right)}\mathbf{\ \times \ 0.5}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 10.75 +}\frac{\mathbf{5}}{\mathbf{8}}\mathbf{\ \times \ 0.5}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 10.75 +}\frac{\mathbf{2.5}}{\mathbf{8}}\ \]

    \[ \widehat{\mathbf{X}}\mathbf{= 11.06}\ \]

(b) Compute the mean deviation from median and its coefficient from the following data:

Daily WagesFrequency
200—25010
250—30020
300—35040
350—40020
4000—45010

Solution (b)

Daily WagesFrequencyXC.F|X – Median|f |X – Median|
200—25010225101001000
250—3002027530501000
300—350403257000
350—4002037590501000
4000—450104251001001000
      
 100   4000
 ∑f=   ∑f |X – Median| =

    \[ \frac{\mathbf{n}}{\mathbf{2}}\mathbf{= \ }\frac{\mathbf{100}}{\mathbf{2}}\mathbf{= 50}\ \]

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

    \[ \widetilde{\mathbf{X}}\mathbf{= 300 +}\frac{\mathbf{50}}{\mathbf{40}}\mathbf{\ }\left( \mathbf{50 - \ 30} \right)\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 300 +}\frac{\mathbf{1000}}{\mathbf{40}}\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 300 + 25}\ \]

    \[ \widetilde{\mathbf{X}}\mathbf{= 325\ }\mathbf{\ }\mathbf{\ }\ \]

    \[ \mathbf{M.D\ from\ Median = \ }\frac{\mathbf{\sum f\ }\left| \mathbf{X\ - \ Median} \right|}{\mathbf{\sum f}}\ \]

    \[ \mathbf{M.D\ from\ Median =}\frac{\mathbf{4000}}{\mathbf{100}}\ \]

    \[ \mathbf{M.D\ from\ Median = 40}\ \]

    \[ \mathbf{Coef.of\ M.D\ from\ Median = \ }\frac{\mathbf{M.D\ from\ Median\ }}{\mathbf{Median}}\ \]

    \[ \mathbf{Coef.of\ M.D\ from\ Median =}\frac{\mathbf{40}}{\mathbf{325}}\ \]

    \[ \mathbf{Coef.of\ M.D\ from\ Median = 0.1230}\ \]

Q.4 Find Chain index from the following price relatives of the three commodities using geometric mean:

YearCommodities
ABC
1999255216330
2000186162384
2001312261333
2002279225462
2003180129495

Solution

YearLink Relatives
ABC
1999255216330
2000(186/255)×100=72.94(162/216)×100=75(384/330)×100=116.36
2001(312/186)×100=167.74(261/162)×100=161.12(333/384) × 100=86.71
2002(279/312) × 100=89.42(225/261) × 100=86.20(462/333)×100=138.73
2003(180/279) × 100=64.51(129/225) × 100=57.33(495/462)×100=107.14
Year 

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

 

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

1999 

    \[ \mathbf{(255 \times 216 \times 330)}^{\mathbf{1/3}}\mathbf{= 262.92}\ \]

 

    \[ \mathbf{262.92}\ \]

2000   

    \[ \mathbf{(72.94 \times 75 \times 116.36)}^{\mathbf{1/3}}\mathbf{= 86.02}\ \]

   

    \[ \frac{\mathbf{262.92\ \times \ 86.02}}{\mathbf{100}}\mathbf{= 226.16}\ \]

2001 

    \[ \mathbf{(167.74 \times 161.12 \times 86.71)}^{\mathbf{1/3}}\mathbf{= 132.82}\ \]

 

    \[ \frac{\mathbf{226.16\ \times \ 132.82}}{\mathbf{100}}\mathbf{= 300.38}\ \]

2002 

    \[ \mathbf{(89.42 \times 86.20 \times 138.73)}^{\mathbf{1/3}}\mathbf{= 102.25}\ \]

 

    \[ \frac{\mathbf{300.38\ \times \ 102.25}}{\mathbf{100}}\mathbf{= 307.13}\ \]

2003 

    \[ \mathbf{(64.51 \times 57.33 \times 107.14)}^{\mathbf{1/3}}\mathbf{= 73.44}\ \]

 

    \[ \frac{\mathbf{307.13\ \times \ 73.44}}{\mathbf{100}}\mathbf{= 225.55}\ \]

Q.5 (a) Show that the sum of errors and sum of squares of errors are zero for the following data:

X12345
Y01234

Solution (a)

XYX²Y²XYŷ = -1+(1)x  y-ŷ(y-ŷ)²
10100ŷ =-1+(1)(1)= 00 – 0 = 00
21412ŷ =-1+(1)(2)= 1 1 – 1 = 00
32946 ŷ =-1+(1)(3)= 22 – 2 = 00
4316912ŷ =-1+(1)(4)= 3  3 – 3 = 00
54251620ŷ =-1+(1)(5)= 4 4 – 4 = 00
        
1510553040 00
∑X=∑Y=∑X²=∑Y²=∑XY= 
∑ y-ŷ=		  ∑(y-ŷ)²

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

    \[ \overline{\mathbf{X}}\mathbf{= \ }\frac{\mathbf{15}}{\mathbf{5}}\mathbf{= 3}\ \]

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

    \[ \overline{\mathbf{Y}}\mathbf{= \ }\frac{\mathbf{10}}{\mathbf{5}}\mathbf{= 2}\ \]

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

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{40}\mathbf{-}\mathbf{\ 5}\left( \mathbf{3} \right)\left( \mathbf{2} \right)}{\mathbf{55}\mathbf{- \ 5}\left( \mathbf{3} \right)^{\mathbf{2}}}\ \]

    \[ \mathbf{b}\mathbf{=}\frac{\mathbf{10}}{\mathbf{10}}\mathbf{= 1}\ \]

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

    \[ \mathbf{a = 2 - \ 1(3)}\ \]

    \[ \mathbf{a = 2 - \ 3 = - 1}\ \]

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

    \[ \widehat{\mathbf{y}}\mathbf{= - 1 + (1)x}\ \]

b. Compute 4 year centred moving average for the following time series:

YearProduction in Millions (K.g)
1993331
1994344
1995349
1996332
1997364
1998395
1999400
2000410

Solution (b)

YearProduction in Millions (K.g)4 Years Moving Total2 Values Moving Total4 Year Centered Moving Average
1993331   
1994344
  1356
1995349 2745343.125
  1389  
1996332 2829353.625
  1440  
1997364 2931366.375
  1491  
1998395 3060382.5
  1569  
1999400 
2000410
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Solved Paper Statistics I FBISE 2016

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