Statistics I HSSC I FBISE, Solved Paper 2009, MCQS, Short Questions, Extensive Questions

Statistics I HSSC I FBISE, Solved Paper 2009, MCQS, Short Questions, Extensive Questions
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In this post, we are going to discuss, Statistics I HSSC I FBISE, Solved Paper 2009, MCQS, Short Questions, Extensive Questions. Solved Paper of 2008 is already published on the website. If you want to read the solved papers of Business Statistics ICOM II of FBISE, BISELHR, BISERWP and other boards, just explore the resource of the site. All the papers and reading resource including solved papers from intermediate, O, A level, graduation and masters level is given the website.

Table of Contents

Statistics I HSSC I FBISE, Solved Paper 2009, MCQS, Short Questions, Extensive Questions

Solved by Iftikhar Ali M.Sc. Economics, MCOM Finance Lecturer Statistics, Finance & Accounting

MCQS

Q.1 Circle the Correct Option i.e. A/B/C/D. Each Part Carries 1 Mark.
(i)The branch of statistics concerned with the procedure & methodology for obtaining valid conclusion is:
 A. Descriptive StatisticsB. Inferential StatisticsC. Sample StatisticsD. Deductive Statistics 
      
(ii)Statistics are collected in a:
 A. Simple MannerB. Systematic MannerC. Haphazard MannerD. None of these 
      
(iii)Issue a new ID Card is an example of:
 A. CensusB. RegistrationC. SamplingD. Survey 
      
(iv)A Statistical Table has at least:
 A. Three PartsB. Four PartsC. Five PartsD. Two Parts 
      
(v)Graph of a Time Series is called:
 A. HistogramB. OgiveC. HistorigramD. Polygon 
      
(vi)A curve whose tail is longer to the right is called:
 A. Negatively SkewedB. Positively SkewedC. SymmetricalD. None of these 
      
(vii)If mean is less than mode, the distribution is:
 A. Skewed to rightB. Skewed to leftC. SymmetricalD. None of these 
      
(viii)If arithmetic mean is 82 and median is 78 then appropriate value for mode is:
 A. 60B. 70C. 50D. 80 
      
(ix)In a certain distribution the first and second moment about the value 2 are -1 and 16 respectively. Variance will be equal to:
 A. 17B. 15C. 16D. None of these 
      
(x)If the price of a commodity remains at Rs.20 per kg for the whole month then measure of dispersion will be:
 A. 20 Rs.B. 2 Rs.C. 0D. None of these 
      
(xi)Var (2X + 3) is equal to:
 A. 5 var(x)B. 4 Var(x) + 3C. 4 Var(x)D. 2 Var(x) 
      
(xii)If all the values are of equal importance then the index numbers are called:
 A. Weighted Index NumbersB. Un-weighted Index NumbersC. Composite Index NumbersD. Value Index Numbers 
      
(xiii)The Index Number for base period is always equal to:
 A. OneB. ZeroC. 100D. None of these 
      
(xiv)If Y = 2 + 0.6X then value of Y intercept is equal to:
 A. 0.6B. 2C. 0D. None of these 
      
(xv)If r = 0.6 and byx = 1.8 then bxy will be:
 A. 3B. 5C. 0.2D. 0 
      
(xvi)If byx = -1.36 and bxy = -0.34 then rxy =:
 A. 0.46B. 0.21C. 0.68D. -0.68 
      
(xvii)If a straight line is fitted to the time series, then:
 A. \[\sum y\ > \ \sum y\hat{}\ \]B. \[ \sum y\ < \ \sum\widehat{y}\ \] C. \[ \mathbf{\sum(y\ – \ }\widehat{\mathbf{y}}\mathbf{)\ = \ 0}\ \]D. Both A & C 

Short Questions

SECTION-B

Q.2 Attempt any fourteen parts.

(i) Write any three characteristics of statistics.

Answer:

  1. It consists of aggregates of facts
  2. It should be numerically expressed
  3. It is effected by many causes
  4. It must be enumerated or estimated accurately
  5. It should be collected in a systematic manner
  6. It should be collected for a predetermined purpose

(ii) Differentiate between qualitative data and quantitative data with examples.

Answer:

A data expressed in terms of quantity, numeric & digits is called quantitative data for example income, saving, expenditures, GDP, NNP etc.

Whereas a data expressed in terms of attribute or quality is qualitative data for example color of hairs, color of eyes, habits etc.

 (iii) Write four important bases of classification of data.

Answer:

(1) Qualitative Base:

When the data are classified according to some quality or attributes such as sex, religion, literacy, intelligence etc.

(2) Quantitative Base:

When the data are classified by quantitative characteristics like heights, weights, ages, income etc.

(3) Geographical Base:

When the data are classified by geographical regions or location, like states, provinces, cities, countries etc.

(4) Chronological or Temporal Base:

When the data are classified or arranged by their time of occurrence, such as years, months, weeks, days etc…

For Example: Time series data.

(iv)Write any six rules for drawing graphs.

Answer:

  1. Give your graph a descriptive title.
  2. Determine the range of data.
  3. Draw axis with proper names on x-axis and y-axis.
  4. Write out a KEY/LEGEND.
  5. Write dots according to relative data.
  6. Draw the line and joint dots.

(v)A variable y is determined from a variable x by the equation y = 10 – 4x. Find y for x = -3, -2, -1, 0, 1, 2, 3, 4, 5. Show that y̅ = 10 – 4x̅

Solution

xy = 10 – 4xy
-3y = 10 – 4(-3)22
-2y = 10 – 4(-2)18
-1y = 10 – 4(-1)14
0y = 10 – 4(0)10
1y = 10 – 4(1)6
2y = 10 – 4(2)2
3y = 10 – 4(3)-2
4y = 10 – 4(4)-6
5y = 10 – 4(5)-10
∑X=9 ∑Y=54

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

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

y̅ = 10 – 4x̅

\[ \textbf{6 = 10 — 4(1)} \]

\[ \textbf{6 = 6} \]

(vi) Three teachers of statistics reported mean examination grades of 75, 82 and 84 for their classes which consist of 30, 25 and 17 students respectively. What is mean grade of all the classes?

Solution:

\[ \overline{\mathbf{X}}\mathbf{1\ = \ 75,\ }\overline{\mathbf{X}}\mathbf{2\ = \ 82,\ }\overline{\mathbf{X}}\mathbf{3\ = \ 84,\ n}\mathbf{1\ = \ 30,\ n}\mathbf{2\ = \ 25,\ n}\mathbf{3\ = 17}\ \]

\[ \overline{\mathbf{X}}\mathbf{c =}\frac{\overline{\mathbf{X}}\mathbf{1}\mathbf{n}\mathbf{1 +}\overline{\mathbf{X}}\mathbf{2}\mathbf{n}\mathbf{2 + \ }\overline{\mathbf{X}}\mathbf{3}\mathbf{n}\mathbf{3}}{\mathbf{n}\mathbf{1 + n}\mathbf{2 + n}\mathbf{3}}\mathbf{= \ }\frac{\left( \mathbf{75} \right)\mathbf{(30) +}\left( \mathbf{82} \right)\mathbf{(25) + \ }\left( \mathbf{84} \right)\mathbf{(17)}}{\mathbf{30 + 25 + 17}}\mathbf{=}\frac{\mathbf{2250 + 2050 + 1428}}{\mathbf{72}}\mathbf{=}\frac{\mathbf{5728}}{\mathbf{72}}\mathbf{= 79.56}\ \]

(vii) The mean of 200 items is 48 and their standard deviation is 3. Find the sum of squares of items.

Solution

Mean of 200 items (x̄) = 48

Standard deviation (σ) = 3 here we have to answer for two questions that is sum of all items (Σ x) and sum of squares all items (Σ x²).

\[ \overline{X} = \frac{\Sigma\ x}{n}\ \ \]

\[ 48 = \frac{\Sigma\ x}{200}\ \ \]

\[ \Sigma\ x = 48\ \times \ 200 = 9600 = sum\ of\ all\ items\ \]

To find sum of squares of all items, we have to find variance (σ²).

\[ \sigma^{2} = \frac{\Sigma\ x²}{n} – \ {(\frac{\Sigma x}{n})}^{2}\ \]

\[ ({(3)}^{2} = \frac{\Sigma\ x²}{200} – \ {(48)}^{2}\ \]

\[ 9 = \frac{\Sigma\ x²}{200} – \ 2304\ \]

\[ \frac{\Sigma\ x²}{200} = 9 + \ 2304\ \]

\[ \frac{\Sigma\ x²}{200} = 2313\ \]

\[ \Sigma\ x² = 2313\ \times \ 200 = 462600\ \] 

Sum of squares of all items = 462600

(viii) (a) In a moderately skewed distribution the value of mean and median are 120 and 110 respectively. Find the value of mode. (b) The geometric mean of a series of 4 items is 10.2. Find the product of all the values.

Solution

(a)

Mode = 3 median – 2 mean

Mode = 3 (110) – 2 (120)

Mode = 330 – 240

Mode = 90

(b)

\[ \sqrt{\mathbf{a.b.c.d}}\mathbf{= 10.2}\ \]

Taking square root on both sides

\[ {\mathbf{(}\sqrt{\mathbf{a.b.c.d}}\mathbf{)}}^{\mathbf{2}}\mathbf{= \ }\mathbf{(10.2)}^{\mathbf{2}}\ \]

\[ \mathbf{abcd = 10824.3216}\ \]

\[ \mathbf{So\ product\ of\ 4\ numbers\ is = 10824.3216}\ \]

(ix) In a certain distribution Mean = 50, Median = 48 and coefficient of skewness = -1. Find the value of variance.

Solution

\[ \mathbf{Skewness =}\frac{\mathbf{3(Mean – Median)}}{\mathbf{S.D}}\ \] 

\[ \mathbf{- 1 =}\frac{\mathbf{3(50 – 48)}}{\mathbf{S.D}}\mathbf{\ }\ \]

\[ \mathbf{- 1 =}\frac{\mathbf{6}}{\mathbf{S.D}}\mathbf{\ }\ \]

\[ \mathbf{- 1(S.D) = 6\ }\ \]

\[ \mathbf{S.D =}\frac{\mathbf{6}}{\mathbf{1}}\mathbf{= 6}\ \]

\[ \mathbf{\sigma}^{\mathbf{2}}\mathbf{=}\mathbf{(\sigma)}^{\mathbf{2}}\mathbf{= \ }\mathbf{(6)}^{\mathbf{2}}\mathbf{= 36\ }\ \]

(x) The first two moments of a distribution about the value 5 of a variable are 2 and 32. Find mean, variance and co-efficient of variation.

Solution

\[ Here\ A\ = \ 5,\ m´1 = 2,\ m´2 = 32\ \] 

\[ Mean\ \overline{X} = A + m´1 = 5 + 2 = 7\ \]

\[ m2 = m´2 – (m´1)^{2} = 32 – (2)^{2} = 28 = Variance\ S^{2}\ \]

\[ Standard\ deviation\ S = \sqrt{S^{2}} = \sqrt{28} = 5.2915\ \]

\[ C.V = \frac{S}{\overline{X}} \times 100 = \frac{5.2915}{7} \times 100 = 75.59\ \]

So Mean = 7, Variance = 28 and C.V = 75.59

(xi) For each of the following tell whether the distribution is symmetrical, positively skewed or negatively skewed:

(a) Mean = 25, Mode = 60

(b) Q1 = 136, Median = 160, Q3 = 184

(c) Mean = 78, Median = 61

Answer:

(a) Negative Skewness

(b) Positive Skewness

(c) Positive Skewness

(xii) (a) Define an index number. (b) Compute Fisher’s Price Index number if base year weighted price index is 110 and current year weighted price index is 115.

Solution:

(a) Index Number is a statistical measure to calculate the percentage change in price or quantity of a product or products with respect to time. Index number related to one product is called simple index number whereas related to more than one product is called composite index number.

(b) \[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ Number =}\sqrt{\mathbf{L\ \times \ P}}\ \]

\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ Number =}\sqrt{\mathbf{110\ \times \ 115}}\ \]

\[ \mathbf{Fishe}\mathbf{r}^{\mathbf{‘}}\mathbf{s\ Ideal\ Index\ Number = 112.47}\ \]

(xiii) Differentiate between simple & composite index number.

Answer:

An index number in which the price or quantity is taken related to one product is called simple index number which can be calculated through two methods (i) Fixed base (ii) chain base. Whereas an index number in which price or quantity is taken related to more than one product is called composite index number which has further two types (i) Un-weighted (ii) Weighted.

(xiv) If n = 8, x̅ = 45, y̅ = 72.125, σx = 22.91, σy = 13.96, ∑xy = 28480. Find the value of r.

Solution

\[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{\sigma x.\sigma y}}\ \]

\[ \mathbf{r =}\frac{\frac{\mathbf{28480}}{\mathbf{8}}\mathbf{- \ }\left( \mathbf{45} \right)\left( \mathbf{72.125} \right)}{\left( \mathbf{22.91} \right)\mathbf{(13.96)}}\mathbf{\ }\ \]

\[ \mathbf{r =}\frac{\mathbf{3560\ –\ 3245.625}}{\mathbf{319.8236}}\ \]

\[ \mathbf{r =}\frac{\mathbf{314.375}}{\mathbf{319.8236}}\ \]

\[ \mathbf{r = 0.98}\textbf{(Positive Relation)} \]

(xv) What are the properties of correlation coefficient?

Answer:

  • Its answer always ranges from -1 to +1
  • Geometric Mean of regression coefficients is equal to \[ r = \sqrt{byx.bxy}\ \]
  • It has five possible outcomes e.g. negative, perfectly negative, positive, and perfectly positive & Zero correlation.
  • It is denoted by “r” 

(xvi) If the regression line of y on x and x on y are respectively 2x – 3y = 0 and 4y – 5x = 8. Find the value of two regression coefficients of y on x and x on y.

Solution:

Regression Line Y on X 

\[ \widehat{Y}\ = \ a\ + \ bx\ = \ 2x\ –\ 3y\ = \ 0\ \ \]

\[ \ – 3y\ = \ 0\ –\ 2x\ \ \] 

\[ 3y\ = \ 2x\ \ \]

\[ y\ = \frac{2x}{3}\ = \ \mathbf{0.67}\mathbf{x}\ \]

Regression Line X on Y

\[ \widehat{X} = \ a\ + \ by\ = \ 4y\ –\ 5x\ = \ 8\ \] 

\[ – 5x\ = \ 8\ –\ 4y\ \ \] 

\[ \widehat{x} = \ \frac{8}{5} – \ \frac{4y}{5} = 1.6 – \mathbf{0.8}\mathbf{y}\ \]

So regression coefficient of y on x = byx = 0.67 & regression coefficient of x on y = bxy = 0.80

(xvii) (a) Determine the estimated regression equation Y = a + bx if n = 10, x̅ = 10, y̅ = 20, ∑xy = 1000, ∑x² = 2000

(b) If x̅ = 50, y̅ = 110 and a = 10 what is the value of b.

Solution:

(a)Regression Equation is: \[ \widehat{\mathbf{Y}}\mathbf{\ = \ a\ + \ bx}\ \]  where:

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

\[ \mathbf{b = \ }\frac{\mathbf{1000 – 10(10)(20)}}{\mathbf{2000 – \ 10(10)²}}\mathbf{= \ }\frac{\mathbf{- 1000}}{\mathbf{1000}}\mathbf{= \ – 1\ }\ \]

\[ \mathbf{a = 20 -}\left( \mathbf{- 1} \right)\mathbf{(10)}\ \]

\[ \mathbf{a = 30}\ \] 

\[ \mathbf{So\ }\widehat{\mathbf{Y}}\mathbf{\ = \ a\ + \ bx\ = \ 30\ –1}\mathbf{x}\ \]

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

\[ \mathbf{10 = (110) – \ b(50)}\ \]

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

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

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

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

(xviii) Fit a straight line y = a + bx for the year 1986 – 1992 (both inclusive), ∑x = 0, ∑y = 245, ∑x² = 28, ∑xy = 66. Find the trend values.

Solution:

Date:

∑x = 0, ∑y = 245, ∑x² = 28, ∑xy = 66, n = 7 if 1986 and 1992 both inclusive

\[ \mathbf{b = \ }\frac{\mathbf{66}}{\mathbf{28}}\mathbf{= \ 2.357\ }\ \]

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

\[ \mathbf{So\ }\widehat{\mathbf{Y}}\mathbf{\ = \ a\ + \ bx\ = \ 35\ + \ 2.357}\mathbf{X}\ \] 

YearXTrend Values \[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\mathbf{X}\ \]
1986-3\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{- 3} \right)\mathbf{= 27.929}\ \]
1987-2\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{- 2} \right)\mathbf{= 30.286}\ \]
1988-1\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{- 1} \right)\mathbf{= 32.643}\ \] 
19890\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{0} \right)\mathbf{= 35}\ \]
19901\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{1} \right)\mathbf{= 37.357}\ \]
19912\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{2} \right)\mathbf{= 39.714}\ \]
19923\[ \widehat{\mathbf{Y}}\mathbf{\ = \ 35\ + \ 2.357}\left( \mathbf{3} \right)\mathbf{= 42.071}\ \]

(xviv) What is meant by Time Series? What are the components of Time Series? Write multiplicative model of Time Series?

Answer:

Time series is a statistical measure to compute the trend of a certain variable with respect to time. Multiplicative model is a common approach to modeling time–series data (Y) in which it is assumed that the four components of a time series; trend component (T), seasonal component (S), cyclical component (C) and irregular component (I), are multiplied to form the values of the time series at each time period.

Y = T x S x C x I

Extensive Questions

Section C

Q.3 Calculate the first four moments about:

(i) Origin (ii) Mean (iii) x = 12

GroupFrequency
0–42
5–95
10–1412
15–199
20–247

Solution (i) Moments about Origin X=0

GroupFrequency (f)Xfx \[ fx^2 \]\[ fx^3 \]\[ fx^4 \]
0–422481632
5–95735245171512005
10–141212144172820736248832
15–19917153260144217751689
20–247221543388745361639792
       
 35 49079701412202652350
 \[ \sum f \] \[ \sum fx \]\[ \sum fx^2 \]\[ \sum fx^3 \]\[ \sum fx^4 \]

\[ \mathbf{m´1 =}\frac{\mathbf{\sum fx}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{490}}{\mathbf{35}}\mathbf{= 14}\ \]

\[ \mathbf{m´2 =}\frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{2}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{7970}}{\mathbf{35}}\mathbf{= 227.71}\ \]

\[ \mathbf{m´3 =}\frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{3}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{141220}}{\mathbf{35}}\mathbf{= 4034.85}\ \]

\[ \mathbf{m´4 =}\frac{\mathbf{\sum f}\mathbf{x}^{\mathbf{4}}}{\mathbf{\sum f}}\mathbf{=}\frac{\mathbf{2652350}}{\mathbf{35}}\mathbf{= 75781.42}\ \]

Solution (ii) Moments about Mean

Moments about the Arithmetic mean are: Relation Method

\[ \mathbf{m}\mathbf{1 = 0}\ \]

\[ \mathbf{m}\mathbf{2 = m}\mathbf{2´ -}\left( \mathbf{m}\mathbf{1´} \right)^{\mathbf{2}}\ \]

\[ \mathbf{m}\mathbf{2 = 227.71 -}\left( \mathbf{14} \right)^{\mathbf{2}}\mathbf{= 31.71}\ \]

\[ \mathbf{m}\mathbf{3 = m}\mathbf{3´ – 3}\mathbf{m}\mathbf{1´m}\mathbf{2´ + 2}\left( \mathbf{m}\mathbf{1´} \right)^{\mathbf{3}}\ \]

\[ \mathbf{m}\mathbf{3}\mathbf{= 4034.85 – 3}\left( \mathbf{14} \right)\left( \mathbf{227.71} \right)\mathbf{+ 2(14)³}\ \]

\[ \mathbf{m}\mathbf{3 =}\mathbf{4034.85}\mathbf{- 9563.82 + 5488 = \ – 40.97}\ \]

\[ \mathbf{m}\mathbf{4 = m}\mathbf{4´ – 4}\mathbf{m}\mathbf{1´m}\mathbf{3´ + 6}\left( \mathbf{m}\mathbf{1´} \right)^{\mathbf{2}}\mathbf{m}\mathbf{2´ – 3}\left( \mathbf{m}\mathbf{1´} \right)^{\mathbf{4}}\ \]

\[ \mathbf{m}\mathbf{4 =}\mathbf{75781.42}\mathbf{- \ 4}{\left( \mathbf{14} \right)\left( \mathbf{4034.85} \right)\mathbf{+ 6}\left( \mathbf{14} \right)}^{\mathbf{2}}\mathbf{(227.71) – 3}\mathbf{(14)}^{\mathbf{4}}\ \]

\[ \mathbf{m}\mathbf{4 =}\mathbf{75781.42}\mathbf{- \ 225951.6 + 267786.96 – 115248 = \ 2368.78}\ \]

Solution (iii) Moments about Origin X=12

GroupFrequency (f)XD = X – 12fD\[ fD^2 \]\[ fD^3 \]\[ fD^4 \]
0–422-10-20200-200020000
5–957-5-25125-6253125
10–14121200000
15–1991754522511255625
20–247221070700700070000
        
 35  701250550098750
 \[ \sum f \] \[ \sum fD \]\[ \sum f D^2 \]\[ \sum f D^3 \]\[ \sum f D^4 \]

\[ \mathbf{m1´ =}\frac{\mathbf{\sum fD}}{\mathbf{\sum f}}\mathbf{\ =}\frac{\mathbf{70}}{\mathbf{35}}\mathbf{\ = 2}\ \]

\[ \mathbf{m2´ =}\frac{\mathbf{\sum fD²}}{\mathbf{\sum f}}\mathbf{\ =}\frac{\mathbf{1250}}{\mathbf{35}}\mathbf{\ = 35.71}\ \] 

\[ \mathbf{m3´ =}\frac{\mathbf{\sum fD³}}{\mathbf{\sum f}}\mathbf{\ =}\frac{\mathbf{5500}}{\mathbf{35}}\mathbf{\ = 157.14}\ \]

\[ \mathbf{m4´ =}\frac{\mathbf{\sum fD}^{\mathbf{4}}}{\mathbf{\sum f}}\mathbf{\ =}\frac{\mathbf{98750}}{\mathbf{35}}\mathbf{\ = 2821.42}\ \]

Q.4 Construct Consumer’s Price Index number from the following data using:

(i) Aggregative Expenditure Method Using 1948 as base year.

(ii) Family Budget Method Using 1948 as base year.

CommodityQuantity ConsumedUnit of PricePrice
19481960
Wheat20 KgRs. Per Kg1013.50
Dal8 KgRs. Per Kg1520
Edible Oil1.5 KgRs. Per Kg90.25200.50
Fuel4 MaundRs. Per Maund2.252.50
Clothing22 YardRs. Per Yard1.502.25

Solution

CommodityQuantity Consumed(qo)Unit of PricePrice\[ \mathbf{p}\mathbf{1}\mathbf{qo}\ \]\[ \mathbf{poqo = w}\ \]\[ \mathbf{I =}\frac{\mathbf{p}\mathbf{1}}{\mathbf{po}}\mathbf{\times 100}\ \]WI
1948(po)1960(p1)    
Wheat20 KgRs. Per Kg1013.50270200135.027000
Dal8 KgRs. Per Kg1520160120133.316000
Edible Oil1.5 KgRs. Per Kg90.25200.50300.75135.375222.230075
Fuel4 MaundRs. Per Maund2.252.50109111.11000
Clothing22 YardRs. Per Yard1.502.2549.533150.04950
     790.25497.375751.679025
     \[ \mathbf{\sum p}\mathbf{1}\mathbf{qo =}\ \]\[ \mathbf{\sum poqo = \sum W}\ \] \mathbf{\sum WI =}\  

\[ \left( \mathbf{i} \right)\mathbf{CPI\ by\ Aggregative\ Method\ 1960 =}\frac{\mathbf{\sum P}\mathbf{1}\mathbf{qo}}{\mathbf{\sum Poqo}}\mathbf{\ \times \ 100 = \ }\frac{\mathbf{790.25}}{\mathbf{497.375}}\mathbf{\ \times \ 100 = 158\ Approx}\ \]

\[ \left( \mathbf{ii} \right)\mathbf{CPI\ by\ Family\ budget\ Method\ 1960 =}\frac{\mathbf{\sum WI}}{\mathbf{\sum W}}\mathbf{\ = \ }\frac{\mathbf{79025}}{\mathbf{497.375}}\mathbf{\ \times \ 100 = 158\ Approx}\ \] 

Q.5 The following table shows the marks obtained by students in Economics and Statistics:

S/NoEconomics (X)Statistics (Y)
014876
024056
033240
043450
053034
065070
072656
085068
092240
104357

Find the least square line of y on x. Also find the regression coefficients and show that:

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

Solution

S/NoEconomics (X)Statistics (Y)X²Y²XY
14876230457763648
24056160031362240
33240102416001280
43450115625001700
5303490011561020
65070250049003500
7265667631361456
85068250046243400
922404841600880
104357184932492451
      
Sum375547149933167721575
 ∑X =∑Y =∑X² =∑Y² =∑XY =

\[ \overline{\mathbf{X}}\mathbf{\ =}\frac{\mathbf{\sum X}}{\mathbf{n}}\mathbf{= \ }\frac{\mathbf{375}}{\mathbf{10}}\mathbf{= 37.5\ ,\ }\overline{\mathbf{Y}}\mathbf{\ = \ }\frac{\mathbf{\sum Y}}{\mathbf{n}}\mathbf{= \ }\frac{\mathbf{547}}{\mathbf{10}}\mathbf{= 54.7\ }\ \]

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

\[ \mathbf{Sx =}\sqrt{\left\lbrack \frac{\mathbf{14993}}{\mathbf{10}}\mathbf{- \ }\left( \frac{\mathbf{375}}{\mathbf{10}} \right)^{\mathbf{2}} \right\rbrack}\ \]

\[ \mathbf{Sx =}\sqrt{\mathbf{1499.3 – 1406.25}}\ \]

\[ \mathbf{Sx = \sqrt{}(93.05)}\ \]

\[ \mathbf{Sx = 9.64}\ \]

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

\[ \mathbf{Sy =}\sqrt{\left\lbrack \frac{\mathbf{31677}}{\mathbf{10}}\mathbf{- \ }\left( \frac{\mathbf{547}}{\mathbf{10}} \right)^{\mathbf{2}} \right\rbrack}\ \]

\[ \mathbf{Sy =}\sqrt{\mathbf{3167.7 – 2992.09}}\ \]

\[ \mathbf{Sy =}\sqrt{\mathbf{175.61}}\ \]

\[ \mathbf{Sy = 13.25}\ \] 

\[ \mathbf{r =}\frac{\frac{\mathbf{\sum xy}}{\mathbf{n}}\mathbf{- \ }\left( \frac{\mathbf{\sum x}}{\mathbf{n}} \right)\left( \frac{\mathbf{\sum y}}{\mathbf{n}} \right)}{\mathbf{Sx\ \times Sy}}\ \]

\[ \mathbf{r =}\frac{\frac{\mathbf{21575}}{\mathbf{10}}\mathbf{- \ }\left( \frac{\mathbf{375}}{\mathbf{10}} \right)\left( \frac{\mathbf{547}}{\mathbf{10}} \right)}{\left( \mathbf{9.64} \right)\mathbf{(13.25)}}\ \]

\[ \mathbf{r =}\frac{\mathbf{2157.5\ –\ 2051.25}}{\mathbf{127.73}}\ \]

\[ \mathbf{r =}\frac{\mathbf{106.25}}{\mathbf{127.73}}\ \]

\[ \mathbf{r = 0.83}) \textbf{(Positive Relation)} \] 

\[ \mathbf{bxy = r}\left( \frac{\mathbf{Sx}}{\mathbf{Sy}} \right)\mathbf{= 0.83}\left( \frac{\mathbf{9.64}}{\mathbf{13.25}} \right)\mathbf{\ = 0.83(0.727) = 0.60341}\ \]

\[ \mathbf{byx = r}\left( \frac{\mathbf{Sy}}{\mathbf{Sx}} \right)\mathbf{= 0.83}\left( \frac{\mathbf{13.25}}{\mathbf{9.64}} \right)\mathbf{\ = 0.83(1.374) = 1.140}\ \]

Regression Equation Y on X

\[ \mathbf{(Y\ – \ }\overline{\mathbf{Y}}\mathbf{) = \ byx\ (X\ – \ }\overline{\mathbf{X}}\mathbf{)}\ \]

\[ \mathbf{(Y\ –\ 54.7)\ = \ 1.140}\left( \mathbf{X\ –\ 37.5} \right)\ \]

\[ \mathbf{(Y\ –\ 54.7)\ = \ 1.140}\mathbf{X\ –\ 42.75}\ \] 

\[ \mathbf{Y\ \ = \ 1.140}\mathbf{X\ –\ 42.75\ + \ 54.7}\ \]

\[ \mathbf{Y\ = \ 1.140}\mathbf{X\ + \ 11.95\ }\ \]

\[ \mathbf{\ }\widehat{\mathbf{Y}}\mathbf{\ = \ 11.95\ + \ 1.140}\mathbf{X}\ \]

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

\[ \mathbf{r = \ }\sqrt{\left( \mathbf{0.60341} \right)\mathbf{.(1.140)}}\ \]

\[ \mathbf{r = \ }\sqrt{\mathbf{0.6878874}}\ \]

\[ \mathbf{r = \ 0.83}\ \] 

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