Mean Deviation

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  • Q1

    Compute the mean deviation about the median for the following frequency distribution.

    Variable(x)

    8

    10

    15

    20

    25

    32

    35

    Frequency(f)

    3

    2

    4

    7

    4

    3

    7

    Marks:3
    Answer:

    We construct the following table. Firstly, we compute the median.

    x

    f

    (c.f.)

    xi - Median

    |xi - Median|

    f|xi - Median|

    8

    3

    3

    -12

    12

    36

    10

    2

    5

    -10

    10

    20

    15

    4

    9

    -5

    5

    20

    20

    7

    16

    0

    0

    0

    25

    4

    20

    5

    5

    20

    32

    3

    23

    12

    12

    36

    35

    7

    30

    15

    15

    105

    f=N= 30

    f|xi - Median| = 237

    Here N = 30, which is even.
    The median will be an average of the (N/2)th and (N+1)/2 th observations. i.e., 15th and 16th observations.
    So, Median = (20+20)/ 2 = 20.

    View Answer
  • Q2

    Find the mean deviation about the median for the following data:

    Class

    0-10

    10-20

    20-30

    30-40

    40-50

    50-60

    Frequency

    7

    15

    6

    16

    2

    4

    Marks:5
    Answer:

    We construct the following table:

    The class-interval containing (N/2)th or 25th item is 20-30. Therefore, it is the median class.

    Here, l = 20, N = 50, C = 22, f= 6 and h = 10

    = 20+{(50/2 - 22)}/6 10=25

    M.D.( ) = = 610/50= 12.2

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  • Q3

    Find the mean deviation about the mean for the following data:

    Class Interval

    10-20

    20-30

    30-40

    40-50

    50-60

    60-70

    70-80

    Frequency

    2

    3

    8

    14

    8

    3

    2

    Marks:5
    Answer:

    We construct the following table:

    Class Interval

    f

    Midpoint of class, x

    fx


    = x – 45

    f

    10-20

    2

    15

    30

    -30

    30

    60

    20-30

    3

    25

    75

    -20

    20

    60

    30-40

    8

    35

    280

    -10

    10

    80

    40-50

    14

    45

    630

    0

    0

    0

    50-60

    8

    55

    440

    10

    10

    80

    60-70

    3

    65

    195

    20

    20

    60

    70-80

    2

    75

    150

    30

    30

    60

    N = 40

    fx=1800

    =400

    = (fx) / N = 1800/40 = 45

    M.D.() =( )/N = 400/40=10

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  • Q4

    Find the mean deviation about the mean for the following data :

    x

    3

    5

    7

    9

    11

    13

    f

    2

    7

    10

    9

    5

    2

    Marks:3
    Answer:

    We construct the following table:

    x

    f

    fx


    = x – 7.8

    f

    3

    2

    6

    - 4.8

    4.8

    9.6

    5

    7

    35

    - 2.8

    2.8

    19.6

    7

    10

    70

    - 0.8

    0.8

    8.0

    9

    9

    81

    1.2

    1.2

    10.8

    11

    5

    55

    3.2

    3.2

    16.0

    13

    2

    26

    5.2

    5.2

    10.4

    N = 35

    fx=273

    = 74.4

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  • Q5

    Find the mean deviation about the mean for the following data :
    6, 7, 11, 15, 16, 22, 18, 12, 10.

    Marks:2
    Answer:

    View Answer