English Class X Maths Sslllla

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    Statistics is the science of collection, classification, tabulation, analysis and

    interpretation of numerical data. Measures of central tendency give a birds eye view

    of the huge mass of statistical data. Measures of dispersion give an idea about thehomogeneity or hetrogeneity of the distribution.

    You are already familiar with

    * Collection of data and preparing frequency distribution table for a given set of scores.

    * Computation of the measures of central tendency namely arithmetic mean, median

    and mode for both ungrouped and grouped data.

    * Determination of the measures of dispersion such as the range, quartile deviation

    and mean deviation for given distribution.

    This chapter deals with the measures of dispersion namely the standard deviation

    and co-efficient of variation.

    After learning this chapter you will be able to :

    * Compute the measure standard deviation for both ungrouped and grouped data.

    * Determine the relative measure co-efficient of variation and understand the variability.

    * Interpret the data based on these statistical measures.Dispersion refers to the variability in the size of items of the distribution. The degree

    to which numerical data tend to spread about an average value, is the variation or

    dispersion of the data.

    The measures of dispersion, which are in common use are : (i) Range (ii) Quartile

    Deviation (iii) Mean Deviation and (iv) Standard Deviation.

    We have already discussed about the measures of dispersion such as range, quartile

    deviation and mean deviation. Besides these measures of dispersion, Standard Deviation

    is an ideal measure of dispersion.

    1. Standard Deviation :

    Statistical data is the set of observations of characteristics of individuals as age,

    height, weight, income, marks scored etc., These characteristics are generally called

    variables. The values of variables may be close to the arithmetical average or scattered

    away from the average. The values vary from the mean and the measure of such variation

    is called the Variance of the distribution.

    The square root of Variance is the Standard Deviation of the distribution.

    3 STATISTICS

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    2. Standard Deviation for an Ungrouped Data

    Model 1 : Calculation of variance and standard Deviation in a series of individual

    observations.

    X be the variable given by X : x1, x

    2 x

    3 .... x

    n

    N : The number of items

    X : The Arithmetic Mean

    D : Deviation from the Arithmetic Mean = (X X )

    The Variance of the distribution is given by Variance =2

    DN

    1

    Symbolically Variance is denoted by22

    D

    N

    1 = . As the Standard Deviation

    is the square root of Variance of the distribution.

    S.D = Variance =

    S.D =N

    D2

    The Standard Deviation is conventionally represented by the Greekletter Sigma :

    Standard Deviation is the square root of the arithmetic average of

    the squares of the deviations from the mean.

    Working Rule :

    When individual observations are given.1) The arithmetic mean is computed.

    2) The deviations of the individual scores from the arithmetic mean are obtained.

    3) The squares of the deviations are calculated.

    4) The sum of the squared deviations is divided by the number of observations.

    5) This value is the Variance.

    6) The positive square root of Variance is the standard deviation.

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    Worked Examples :

    1) Calculate the Variance and Standard Deviation of the following 10 scores.

    14, 16, 21, 9, 16, 17, 14, 12, 11 and 20. Also interpret the results.

    Table of calculation

    Scores Deviation Square of

    X (D = X X ) the deviation D2

    14 -1 1

    16 +1 1

    21 +6 36

    9 -6 36

    16 +1 1

    17 +2 4

    14 -1 112 -3 9

    11 -4 16

    20 +5 25

    X =150 2D = 130

    Solution : The number of observations = N = 10

    (i) Find the arithmetic mean by using

    Mean =N

    XX =

    =10

    150Arithmetic Mean = 15

    The average score = 15

    (ii) Find the variance by using

    Variance = 2DN

    1

    2 =

    10

    1 (130) = 13 Variance = 13

    (iii) Find the standard deviation by using

    S.D = Variance

    = 13 = 3.6 S.D. of the scores = 3.6

    On an average, the individual scores deviate from the arithmetic mean by 3.6

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    2) The runs scored by a batsman in eight innings are given as 35, 42, 23, 34, 39,

    36, 32 and 31 Find (i) the average score and (ii) the standard deviation of scores.

    How do you interpret the results?

    Solution : The number of individual observations = N = 8

    Table of calculation

    Scores Deviation Square of

    X (D = X X ) the deviation D2

    35 +1 1

    42 +8 64

    23 11 121

    34 0 0

    39 +5 25

    36 +2 4

    32 -2 431 -3 9

    X = 272 2D = 228

    i) Arithmetic MeanN

    XX

    ==

    =8

    272= 34 Arithmetic Mean = 34

    ii) S.DN

    D

    2=

    =8

    228 = 5.28 = 5.34 S.D. = 5.34

    The average of runs scored by the batsman is 34 and on an average, the individual

    scores deviate from the average score by 5.34 runs.

    Alternate method

    This method is used when the arithmetic mean ( X ) is not an integer.

    Step 1 : Assume one of the scores as an assumed mean (A)

    Step 2 : Find out the deviations from the assumed mean, D = (X-A)

    Step 3 : Find out the sum of the deviations, DStep 4 : Square the deviations and find out the sum of the squares of the deviations,

    2D

    Step 5 : By using the formula, determine the standard deviation =

    22

    N

    D

    N

    D

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    3) The marks obtained by 10 students in an examination are given as 43, 48, 55,

    57, 42, 50, 47, 48, 58 and 50. Find (i) the Mean and (ii) the Standard deviation.

    Interpret the results

    Solution : The number of individual observations = N = 10. Let the assumed mean

    A = 50

    Table of calculationScores Deviation Square of

    X (D = X X ) the deviation D2

    43 -7 49

    48 -2 4

    55 +5 25

    57 +7 49

    42 -8 64

    50 0 047 -3 9

    48 -2 4

    58 +8 64

    50 0 0

    X = 498 D = -2 2D = 268

    i) Arithmetic MeanN

    XX

    ==

    =10

    498= 49.8 Arithmetic Mean = 49.8

    The average of marks scored by 10 students is 49.8

    ii) S.D

    22

    N

    D

    N

    D

    ==

    =

    2

    10

    2

    10

    268

    = 2)2.0(8.26 = 04.08.26

    = 76.26 = 5.173 S.D = 5.173

    The individual scores differ from the average score by 5.173 marks.

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    Model 2 : Calculation of Standard Deviation for a grouped data:

    X be the variable and f be the frequency, then the distribution is given by

    X x1

    x2

    x3

    ........... xn-1

    xn

    f f1

    f2

    f3

    ........... f n-1

    fn

    N = f = The sum of the frequencies

    X = The Arithmetic Mean =N

    fX

    D = Deviation from the arithmetic mean = (X X )

    Variance of the distribution is given byN

    fD2

    Variance =N

    fD2

    2 =

    Standard deviation of the distribution is given by

    S.D. =N

    fD

    2=

    Working Rule

    To calculate the Variance and Standard Deviation

    Step 1: Determine the arithmetic mean X , by using the formulaN

    XX

    = or

    N

    fXX

    =

    Step 2: Calculate the values of deviations, using D = (XX ) for all the values of X.

    Step 3: Square the values of D and obtain D2.

    Step 4: Multiply the corresponding values of f and D2 and then determine 2fd

    Step 5: CalculateN

    fD2

    2 = This gives the Variance of the distribution.

    Step 6: Take the positive square root of the VarianceN

    fD

    2=

    This gives the standard deviation of the distribution.

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    4) Calculate (i) the arithmetic mean (ii) the standard deviation of the given distribution.

    X 10 15 20 25 30 35

    f 3 8 5 9 4 1

    Table of calculation

    Score frequency Deviation Square of f.D2

    X f fX D = (X X ) deviation D2

    10 3 30 -11 121 363

    15 8 120 -6 36 288

    20 5 100 -1 1 5

    25 9 225 +4 16 144

    30 4 120 +9 81 324

    35 1 35 +14 196 196

    f =30 f =630 f D2=1320

    Solution : The sum of the frequencies = N = f = 30

    i) Arithmetic Mean =N

    fXX

    =

    =30

    630 = 21

    Arithmetic Mean = 21

    The average score = 21

    ii) S.DN

    fD

    2==

    =30

    1320

    = 44

    = 6.63

    S.D. = 6.63

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    5) Calculate (i) Arithmetic Mean (ii) Variance and (iii) Standard deviation of the following

    distribution

    X 2.5 3.5 4.5 5.5 6.5

    f 4 3 5 10 3

    Table of calculation

    Score frequency Deviation Square of f.D2

    X f fX D = (X X ) deviation D2

    2.5 4 10.0 -2.2 4.84 19.36

    3.5 3 10.5 -1.2 1.44 4.32

    4.5 5 22.5 -0.2 0.04 0.20

    5.5 10 55.0 +0.8 0.64 6.406.5 3 19.5 +1.8 3.24 9.72

    f =25 f = 117.5 f D2=40.0

    Solution : The sum of the frequencies = N = f = 25

    i) Arithmetic MeanN

    XX

    ==

    =25

    5.117 = 4.5

    Arithmetic Mean = 4.7

    ii) Variance =N

    fD2

    2 =

    =2540 = 1.6

    Variance = 1.6

    iii) S.D = Variance

    = 6.1 = 1.26

    S.D = 1.26

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    Alternate method :

    Step 1: Take any one of the values of X as an assumed average (A)

    Step 2: Find out the deviations from the assumed average D = (X-A)

    Step 3: Multiply the deviations by their respective frequencies and obtain the sum, f D

    Step 4: Square the deviations (D2)

    Step 5: Multiply the squared deviations by their respective frequencies and obtain fD2

    Step 6: Compute Arithmetic Mean and Standard Deviation by using the formula,

    Arithmetic Mean = X = A +N

    fD

    Standard Deviation

    22

    N

    fD

    N

    fD

    ==

    6) Compute : (i) Arithmetic Mean and (ii) Standard Deviation for the following

    distribution and interpret the results

    Marks 35 40 45 50 55

    Number of students 2 4 8 5 1

    Solution : Choose the assumed mean A = 45

    Table of calculation

    Score frequency Deviation Square of f.D f.D2

    X f D = (X X ) deviation D2

    35 2 -10 100 -20 20040 4 -5 25 -20 100

    45 8 0 0 0 0

    50 5 +5 25 25 125

    55 1 +10 100 10 100

    N = 20 fD = -5 fD2 = 525

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    i) MeanN

    fDAX +==

    = 45 +20

    )5(

    = 45 0.25 Arithmetic Mean = 44.75

    Average Marks is 44.75

    ii) S.D

    22

    N

    fD

    N

    fD

    =

    2

    20

    5

    20

    525

    = 0625.025.26 =

    1875.26= = 5.117 S.D. = 5.117

    The average marks of 20 students is 44.75. The scores of the students deviate

    from the Mean score by about 5 marks.

    3. Calculation of Standard Deviation for a grouped data

    Standard deviation for a grouped data can also be determined as similar to

    ungrouped data.

    Know this :

    1) The average of the class intervals are represented by their middle points.

    2) The Deviations from the arithmetic mean are obtained.

    3) The squares of these deviations are multiplied by the respective frequencies.

    4) The total of these products is divided by the total of the frequencies and

    the result is the Variance.

    5) The positive square root of the Variance is the Standard Deviation of thedistribution.

    7) Calculate the Arithmetic Mean and Standard Deviation for the given frequency

    distribution.

    Class Interval frequency

    1-5 2

    6-10 3

    11-15 4

    16-20 1

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    Solution :

    Step 1: Find the mid-points for each class intervals (X)

    Step 2: Find the product f.X for each class interval.

    Class Frequency Mid-point Deviation

    Interval f X fX from Mean D2 f.D2

    D = (X X )

    1-5 2 3 6 -7 49 98

    6-10 3 8 24 -2 4 12

    11-15 4 13 52 3 9 36

    16-20 1 18 18 8 64 64

    N = 10 fX = 100 fD2 = 210

    Step 3: Find arithmetic mean, by using the formula

    Arithmetic MeanN

    fxX

    ==

    =10

    100Arithmetic Mean = 10

    Step 4: Find the deviation D from the arithmetic mean X , for each class interval.

    i.e., D = X X

    Step 5: Square the deviations and obtain D2.

    Step 6: Find the product, fD2 for each class interval and find fD2

    Step 7: Compute Standard Deviation by using

    S.D.N

    D.f

    2==

    10

    210=

    21= = 4.6 S.D. = 4.6

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    8) The distribution of the ages (in years) of 20 persons in a locality, are recorded

    as follows :

    Age (in years) 30-34 35-39 40-44 45-49 50-54

    Number of Persons 2 5 6 5 2

    Calculate (i) the average age (ii) the variance and (iii) standard deviation of the distribution

    of ages.

    Solution : Table of calculation

    Age Frequency Mid-point Deviation

    (in yrs) f X fX from Mean D2 f.D2

    C.I. D = (X- X )

    30-34 2 32 64 -10 100 200

    35-39 5 37 185 -5 25 12540-44 6 42 252 0 0 0

    45-49 5 47 235 +5 25 125

    50-54 2 52 104 +10 100 200

    20 840 650

    Observations : N = f = Sum of the frequencies = 20

    fX = 840, f.D2 = 650

    i) Arithmetic Mean =N

    fxX

    =

    =20

    840 = 42 Arithmetic Mean = 42

    Average age (in years) is 42

    ii) Variance =NfD

    22 =

    =20

    650 = 32.5 Variance = 32.5

    iii) S.D. = Variance

    = 5.32 = 5.7 S.D = 5.7

    Standard deviation of the distribution of ages (in years) is 5.7

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    Step - Deviation Method

    1) Choose one of the middle values of the class intervals (X) as an assumed

    mean (A).

    2) Find the step-deviations from the assumed mean as d =i

    AX , where i is the

    size of the class interval.3) Multiply these step-deviations by their respective frequencies and obtain the sum

    of the products, f.d.

    4) Square the step-deviations, (d2)

    5) Multiply the squares of the step deviations by their respective frequencies and obtain

    the sum of the products, f.d2

    6) Compute the arithmetic mean and standard deviation, by using

    Arithmetic Mean ixNfd

    AX

    +==

    Standard Deviation ixN

    fd

    N

    fd

    22

    ==

    9) The marks obtained by 60 students in a test are given as follows:

    Marks 5-15 15-25 25-35 35-45 45-55 55-65

    Number of students 8 12 20 10 7 3

    Calculate (i) the arithmetic mean and (ii) standard deviation of the distribution.

    Solution :

    Marks Mid point Number Step d2 f.d f.d2

    X of students deviation

    f d=

    i

    AX

    5-15 10 8 -2 4 -16 32

    15-25 20 12 -1 1 -12 12

    25-35 30 20 0 0 0 0

    35-45 40 10 +1 1 10 10

    45-55 50 7 +2 4 14 28

    55-65 60 3 +3 9 9 27

    N = 6 0 fd = 5 2fd =109

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    Choose the assumed mean A = 30

    Observations : Total frequency N = 60Size of the class interval i = 10

    f.d = 5 and f.d2 = 109

    i) Arithmetic Mean ixNfdAX

    +==

    10x60

    530

    +=

    6

    530+= = 30 + 0.83

    = 30.83 Arithmetic Mean = 30.83

    ii) S.D ixN

    fd

    N

    fd

    22

    ==

    10x60

    5

    60

    1092

    =

    ( ) ( ) 10x08.0817.1 2=

    10x0064.0817.1 = 10x81.1=

    = 13.45 x 10 = 13.45 S.D. = 13.45

    The Average of marks scored by 60 students is 30.83

    The Standard Deviation of the distribution of marks is 13.45

    4. Co-efficient of variation

    The co-efficient of variation is a relative measure of dispersion. It is based on

    the arithmetic mean and standard deviation of a frequency distribution.

    The co-efficient of variation (CV) is given by C.V =

    M.A

    D.Sx 100

    In symbols, C.V =

    X

    x 100

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    Remember :

    * Co-efficient of variation is a relative measure of dispersion.

    * It is generally expressed as a percentage.

    * Mean ( X ) and S.D. ( ) determine the co-efficient of variation.

    * It is independent of units.

    * Consistency or variability is determined by the co-efficient of variation.10) The total runs scored by two cricket players Arun and Bharath in 15 matches are

    1050 and 900 with standard deviations 4.2 and 3.0 respectively. Who is better

    run getter? Who is more consistent?

    Solution : Number of matches played = 15

    Average score of Arun =15

    1050 = 70

    Average score of Bharath =15

    900 = 60

    Player Mean S.D. C.V. = 100xX

    Arun 70 4.2 0.6100x

    70

    2.4=

    Bharath 60 3.0 0.5100x60

    0.3=

    The average score of Arun is greater than the average score of Bharath. HenceArun is a better run getter.

    The co-efficient of variation of Bharath is less than the co-efficient of variationof Arun. Hence Bharath is more consistent.

    11) The performance of 20 students in an examination in English, Mathematics andScience is given below.

    Subject Mean Score S.D. Score

    X

    English 56 5.75

    Mathematics 73 6.25

    Science 62 6.0

    Using this data, determine in which subject their performance is more consistent?

    X

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    (i) English : Given X = 56 and = 5.75

    C.V 100xX

    =

    100x56

    75.5

    =

    C.V. in English = 10.27

    (ii) Mathematics : Given X = 73 and = 6.25

    C.V 100xX

    =

    100x7325.6

    =

    C.V. in Mathematics = 8.56

    (iii) Science : Given X = 62 and = 6.0

    C.V 100xX

    =

    100x62

    0.6

    =

    C.V. in science = 9.68

    On comparison, the co-efficient of variation in mathematics is the least.

    Hence the performance in Mathematics is more consistent.

    12) The marks obtained by two candidates Asha and Bhanu in 5 tests are given below.

    Test No. 1 2 3 4 5

    Asha 58 65 58 64 55

    Bhanu 66 60 60 76 68

    (i) Who is a better scorer? and (ii) Who is more consistent?

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    Solution :

    Table of calculation

    Asha Bhanu

    Scores Deviation Deviation

    X from Mean D2 X from Mean D2

    D = (X X ) D = (X X )

    58 -2 4 66 0 0

    65 +5 25 60 -6 36

    58 -2 4 60 -6 36

    64 +4 16 76 +10 100

    55 -5 25 68 +2 4

    300 74 330 176

    (i) Asha :

    The number of tests N = 5

    Arithmetic MeanN

    XX

    ==

    5

    300=

    = 60 Arithmetic Mean = 60

    Standard DeviationN

    D

    2==

    5

    74= 5.14=

    = 3.85 S.D = 3.85

    Co-efficient of variation 100xX

    CV

    ==

    100x60

    85.3

    =

    = 6.42 C.V. of Asha = 6.42

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    (ii) Bhanu : The number of observations N = 5

    Arithmetic MeanN

    XX

    ==

    5

    330= Arithmetic Mean = 66

    Standard DeviationN

    D

    2==

    5

    176= 2.35= S.D. = 5.93

    Co-efficient of variation 100xX

    CV

    =

    100x66

    93.5

    = C.V. of Bhanu = 8.98

    The average scores of Bhanu is higher than the average score of Asha. HenceBhanu is a better scorer.

    The co-efficient of variation of Asha is less than the co-efficient of variation ofBhanu. Hence Asha is more consistent.

    13) Calculate the Standard Deviation and Co-efficient of variation for the followingdistribution.

    Marks (X) 10 20 30 40 50

    Number of students (f) 4 3 6 5 2

    Solution : The sum of the frequencies = N = 20

    Table of calculation

    Scores frequency Deviation Square of

    X f f.X. from Mean the Deviation f.D2

    D = (X X ) D2

    10 4 40 -19 361 1444

    20 3 60 -9 81 243

    30 6 180 +1 1 6

    40 5 200 +11 121 605

    50 2 100 +21 441 882

    N = 20 fX = 580 fD2 = 3180

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    i) Arithmetic MeanN

    fXX

    ==

    20

    580= = 29 Arithmetic Mean = 29

    ii) S.D.N

    D.f

    2==

    20

    3180=

    159= = 12.61 S.D. = 12.61

    iii) C.V 100xM.AD.S

    =

    100x29

    61.12

    =

    29

    1261= = 43.48 C.V. = 43.48

    14) The following frequency distribution shows the daily wages earned by 15 workers.

    Wages (in Rs.) 30-40 40-50 50-60 60-70 70-80

    No. of workers 2 3 5 3 2

    Find i) Arithmetic Mean ii) Standard Deviation and iii) Co-efficient of variation

    Solution :

    Table of calculation :

    Wages frequency Mid point Deviation Square

    in Rs. f X fX from Mean of the f.D2

    C.I D = (X X ) deviation

    D2

    30-40 2 35 70 -20 400 800

    40-50 3 45 135 -10 100 300

    50-60 5 55 275 0 0 0

    60-70 3 65 195 +10 100 300

    70-80 2 75 150 +20 400 800

    15 fx = 8252

    fd =2200

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    Observations: The sum of the frequencies = N = 15

    f.X = 825 f.D2 = 2200

    i) Arithmetic MeanN

    XX

    ==

    =15

    825

    = 55 Arithmetic Mean = 55

    The average wages (in Rs) is 55

    ii) S.DN

    D.f

    2==

    15

    2200

    = 67.146= = 12.11 S.D = 12.11

    The standard deviation of the distribution of wages (in Rs) is 12.11

    iii) C.V 100xM.A

    D.S

    =

    100x55

    11.12

    =

    =

    55

    1211 = 22.02 C.V = 22.02

    Exercise : 3

    1) Calculate the Standard Deviation from the following set of 10 observations.

    8, 9, 15, 23, 5, 11, 19, 8, 10 and 122) The scores of a batsman in 8 innings are given as 48, 40, 36, 35, 46, 42, 36

    and 37

    Find (i) Average score (ii) Variance and (iii) Standard Deviation of scores.

    3) Calculate Mean and standard deviation for the following distribution.

    X 5 15 25 35 45

    f 5 8 15 16 6

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    4) The marks scored by 60 students in a mathematics test are given below.

    Marks (X) 10 20 30 40 50 60

    No. of students (f) 8 12 20 10 7 3

    Find the Variance and Standard Deviation of the marks.

    5) Calculate (i) Arithmetic Mean and (ii) Standard Deviation for the following frequency

    distribution.

    Class Frequency

    Interval

    20-25 8

    25-30 3

    30-35 15

    35-40 12

    40-45 8

    45-50 4

    N 50

    6) The daily wages of 40 workers of a factory are given in the following table.

    Wages in Rs. 30-34 34-38 38-42 42-46 46-50 50-54

    No. of workers 4 7 9 11 6 3

    Calculate (i) Mean (ii) Variance and (iii) Standard Deviation of wages and also

    interpret the findings.

    7) Batsman A gets an average of 64 runs per innings with standard deviation of 18

    runs, while batsman B gets an average score of 43 runs with standard deviation

    of 9 runs in an equal number of innings. Discuss the efficiency and consistency of

    both the batsmen.

    8) In two factories A and B, located in the same industrial area, the average weekly

    wages in rupees and the Standard Deviations are as follows :

    Factory Average of S.D. of

    wages in Rs. wages in Rs.

    A 34.5 6.21

    B 28.5 4.56

    Determine which factory has greater variability in individual wages?

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    9) The Mean and Standard Deviation of the heights and weights of 20 persons are

    given below :

    Characteristic X Height (in cms) 175 3.5

    Weight (in Kgs) 70 2.1

    In which characteristic are they more variable?

    10) Calculate (i) Arithmetic Mean (ii) Standard Deviation and (iii) Co-efficient of variation

    for the following frequency distribution.

    Class Interval Frequency

    30-35 5

    35-40 10

    40-45 1645-50 15

    50-55 4

    Total 50

    11) The runs scored by two batsmen A and B in six innings are given as follows :

    Batsman A 48 50 54 46 48 54

    Batsman B 46 44 43 46 45 46

    Determine (i) Who is a better run getter? (ii) Who is a consistent player?

    12) Marks obtained in a test by X standard students of two sections A and B are

    given below :

    Marks No. of students No. of students

    in in

    Section A Section B

    25-30 5 5

    30-35 10 12

    35-40 25 20

    40-45 8 8

    45-50 2 5

    Determine : (i) Which sections performance is better? and (ii) Which sections performance

    is more variable?