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    NCERT Solutions for Class 11 Maths Chapter 15 Statistics

    Edited By Ramraj Saini | Updated on Sep 25, 2023 10:18 PM IST

    Statistics Class 11 Questions And Answers

    NCERT Solutions for Class 11 Maths Chapter 15 Statistics are provided here. You have learnt the basic statistics like mean, mode and median known as "measures of the central tendency" in the previous Class NCERT syllabus. In this NCERT book chapter, you will learn important topics like measures of dispersion, mean deviation, standard deviation, quartile deviation, range and many more. In class 11 maths chapter 15 question answer, you will get questions related to all the above topics. These NCERT solutions are prepared by expert team at Careers360 keeping in my the latest syllabus of CBSE 2023-24, easy and simple language, step by step comprehensive coverage of the concepts.

    This statistics class 11 solutions is important for the CBSE class 11 final examination as well as in the various competitive exams like JEE Main, BITSAT, etc . As statistics deal with the collected data for specific purposes and in the digital world, everything that we do comes in the form of data. This is a very useful technique to interpret, analyze the data, and make a conclusive decision based on the data. Class 11 maths chapter 15 NCERT solutions will build your fundamentals of statistics which will be useful to learn advanced statistical techniques. There are 27 questions in 3 exercises in the NCERT textbook. Interested students can find all NCERT solutions for class 11 at one place which will help to understand the concepts in an easy way.

    Also Read :

    Statistics Class 11 Questions And Answers PDF Free Download

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    Statistics Class 11 Solutions - Important Formulae

    Measure of Dispersion:

    Dispersion measures the degree of variation in the values of a variable. It quantifies how scattered observations are around the central value in a distribution.

    Range:

    Range is the simplest measure of dispersion. It is defined as the difference between the largest and smallest observations in a distribution.

    Range of distribution = Largest observation – Smallest observation.

    Mean Deviation:

    Mean Deviation for Ungrouped Data:

    • For n observations x1, x2, x3, ..., xn, the mean deviation about their mean x¯ is calculated as:

      • MD(\bar x) = (Σ |xᵢ - x¯|) / n

    • The mean deviation about its median M is calculated as:

      • MD(M) = (Σ |xᵢ - M|) / n

    Mean Deviation for Discrete Frequency Distribution:

    • For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the mean deviation is calculated as:

      • MD(\bar x) = (Σ fᵢ |xᵢ - x¯|) / (Σ fᵢ) = (Σ fᵢ |xᵢ - x¯|) / N

    Variance:

    • Variance is the average of the squared deviations from the mean x¯. If x, x, …, xₙ are n observations with mean x¯, the variance denoted by σ² is calculated as:

      • σ² = (Σ(xᵢ - x¯)²) / n

    Standard Deviation:

    • Standard deviation, denoted as σ, is the square root of the variance σ². If σ² is the variance, then the standard deviation is given by:

      • σ = √(σ²)

    • For a discrete frequency distribution with values xᵢ, frequencies fᵢ, mean x¯, and total frequency N, the standard deviation is calculated as:

      • σ = √((Σ fᵢ(xᵢ - x¯)²) / N)

    Coefficient of Variation:

    • The coefficient of variation (CV) is used to compare two or more frequency distributions. It is defined as:

      • Coefficient of Variation = (Standard Deviation / Mean) × 100

      • CV = (σ / x¯) × 100

    Free download NCERT Solutions for Class 11 Maths Chapter 15 Statistics for CBSE Exam.

    Statistics Class 11 NCERT Solutions (Intext Questions and Exercise)

    Class 11 maths chapter 15 question answer - Exercise: 15.1

    Question:1 . Find the mean deviation about the mean for the data.

    \small 4,7,8,9,10,12,13,17

    Answer:

    Mean ( \overline{x} ) of the given data:

    \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{4+ 7+ 8+ 9+ 10+ 12+ 13+ 17}{8} = 10

    The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

    6, 3, 2, 1, 0, 2, 3, 7

    \therefore \sum_{i=1}^{8}|x_i - 10| = 24

    \therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

    = \frac{24}{8} = 3

    Hence, the mean deviation about the mean is 3.

    Question:2. Find the mean deviation about the mean for the data.

    \small 38,70,48,40,42,55,63,46,54,44

    Answer:

    Mean ( \overline{x} ) of the given data:

    \\ \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{38+70+48+40+42+55+63+46+54+44}{10} \\ = \frac{500}{10} = 50

    The respective absolute values of the deviations from mean, |x_i - \overline{x}| are

    12, 20, 2, 10, 8, 5, 13, 4, 4, 6

    \therefore \sum_{i=1}^{8}|x_i - 50| = 84

    \therefore M.D.(\overline{x}) = \frac{1}{n}\sum_{i=1}^{n}|x_i - \overline{x}|

    = \frac{84}{10} = 8.4

    Hence, the mean deviation about the mean is 8.4.

    Question:3. Find the mean deviation about the median.

    \small 13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17

    Answer:

    Number of observations, n = 12, which is even.

    Arranging the values in ascending order:

    10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18.

    Now, Median (M)

    \\ = \frac{(\frac{12}{2})^{th} observation + (\frac{12}{2}+ 1)^{th} observation}{2} \\ = \frac{13 + 14}{2} = \frac{27}{2}= 13.5

    The respective absolute values of the deviations from median, |x_i - M| are

    3.5, 2.5, 2.5, 1.5, 0.5, 0.5, 0.5, 2.5, 2.5, 3.5, 3.5, 4.5

    \therefore \sum_{i=1}^{8}|x_i - 13.5| = 28

    \therefore M.D.(M) = \frac{1}{12}\sum_{i=1}^{n}|x_i - M|

    = \frac{28}{12} = 2.33

    Hence, the mean deviation about the median is 2.33.

    Question:4. Find the mean deviation about the median.

    \small 36, 72, 46, 42, 60, 45, 53, 46, 51, 49

    Answer:

    Number of observations, n = 10, which is even.

    Arranging the values in ascending order:

    36, 42, 45, 46, 46, 49, 51, 53, 60, 72

    Now, Median (M)

    \\ = \frac{(\frac{10}{2})^{th} observation + (\frac{10}{2}+ 1)^{th} observation}{2} \\ = \frac{46 + 49}{2} = \frac{95}{2}= 47.5

    The respective absolute values of the deviations from median, |x_i - M| are

    11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5

    \therefore \sum_{i=1}^{8}|x_i - 47.5| = 70

    \therefore M.D.(M) = \frac{1}{10}\sum_{i=1}^{n}|x_i - M|

    = \frac{70}{10} = 7

    Hence, the mean deviation about the median is 7.

    Question:5 Find the mean deviation about the mean.

    \small \\x_i\hspace{1cm}5\hspace{1cm}10\hspace{1cm}15\hspace{1cm}20\hspace{1cm}25\\\small f_i\hspace{1cm}7\hspace{1cm}4\hspace{1.15cm}6\hspace{1.22cm}3\hspace{1.3cm}5

    Answer:

    x_i
    f_i
    f_ix_i
    |x_i - \overline{x}|
    f_i|x_i - \overline{x}|
    5
    7
    35
    9
    63
    10
    4
    40
    4
    16
    15
    6
    90
    1
    6
    20
    3
    60
    6
    18
    25
    5
    125
    11
    55

    \sum{f_i}
    = 25
    \sum f_ix_i
    = 350

    \sum f_i|x_i - \overline{x}|
    =158

    N = \sum_{i=1}^{5}{f_i} = 25 ; \sum_{i=1}^{5}{f_ix_i} = 350

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{350}{12} = 14

    Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

    \sum f_i|x_i - \overline{x}| = 158

    \therefore M.D.(\overline{x}) = \frac{1}{25}\sum_{i=1}^{n}f_i|x_i - \overline{x}|

    = \frac{158}{25} = 6.32

    Hence, the mean deviation about the mean is 6.32

    Question:6. Find the mean deviation about the mean.

    \small \\x_i\hspace{1cm}10\hspace{1cm}30\hspace{1cm}50\hspace{1cm}70\hspace{1cm}90\\\small f_i\hspace{1cm}4\hspace{1.1cm}24\hspace{1.1cm}28\hspace{1.1cm}16\hspace{1.2cm}8

    Answer:

    x_i
    f_i
    f_ix_i
    |x_i - \overline{x}|
    f_i|x_i - \overline{x}|
    10
    4
    40
    40
    160
    30
    24
    720
    20
    480
    50
    28
    1400
    0
    0
    70
    16
    1120
    20
    320
    90
    8
    720
    40
    320

    \sum{f_i}
    = 80
    \sum f_ix_i
    = 4000

    \sum f_i|x_i - \overline{x}|
    =1280

    N = \sum_{i=1}^{5}{f_i} = 80 ; \sum_{i=1}^{5}{f_ix_i} = 4000

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{4000}{80} = 50

    Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

    \sum f_i|x_i - \overline{x}| = 1280

    \therefore M.D.(\overline{x}) = \frac{1}{80}\sum_{i=1}^{5}f_i|x_i - \overline{x}|

    = \frac{1280}{80} = 16

    Hence, the mean deviation about the mean is 16

    Question:7. Find the mean deviation about the median.

    \small x_i \small 5 \small 7 \small 9 \small 10 \small 12 \small 15

    \small f_i \small 8 \small 6 \small 2 \small 2 \small 2 \small 6

    Answer:

    x_i
    f_i
    c.f.
    |x_i - M|
    f_i|x_i - M|
    5
    8
    8
    2
    16
    7
    6
    14
    0
    0
    9
    2
    16
    2
    4
    10
    2
    18
    3
    6
    12
    2
    20
    5
    10
    15
    6
    26
    8
    48

    Now, N = 26 which is even.

    Median is the mean of 13^{th} and 14^{th} observations.

    Both these observations lie in the cumulative frequency 14, for which the corresponding observation is 7.

    Therefore, Median, M = \frac{13^{th} observation + 14^{th} observation}{2} = \frac{7 + 7}{2} = \frac{14 }{2} = 7

    Now, we calculate the absolute values of the deviations from median, |x_i - M| and

    \sum f_i|x_i - M| = 84

    \therefore M.D.(M) = \frac{1}{26}\sum_{i=1}^{6}|x_i - M|

    = \frac{84}{26} = 3.23

    Hence, the mean deviation about the median is 3.23

    Question:8 Find the mean deviation about the median.

    \small x_i \small 15 \small 21 \small 27 \small 30 \small 35

    \small f_i \small 3 \small 5 \small 6 \small 7 \small 8

    Answer:

    x_i
    f_i
    c.f.
    |x_i - M|
    f_i|x_i - M|
    15
    3
    3
    13.5
    40.5
    21
    5
    8
    7.5
    37.5
    27
    6
    14
    1.5
    9
    30
    7
    21
    1.5
    10.5
    35
    8
    29
    6.5
    52

    Now, N = 30, which is even.

    Median is the mean of 15^{th} and 16^{th} observations.

    Both these observations lie in the cumulative frequency 21, for which the corresponding observation is 30.

    Therefore, Median, M = \frac{15^{th} observation + 16^{th} observation}{2} = \frac{30 + 30}{2} = 30

    Now, we calculate the absolute values of the deviations from median, |x_i - M| and

    \sum f_i|x_i - M| = 149.5

    \therefore M.D.(M) = \frac{1}{29}\sum_{i=1}^{5}|x_i - M|

    = \frac{149.5}{29} = 5.1

    Hence, the mean deviation about the median is 5.1

    Question:9. Find the mean deviation about the mean.

    Income per day in Rs
    \small 0-100
    \small 100-200
    \small 200-300
    \small 300-400
    \small 400-500
    \small 500-600
    \small 600-700
    \small 700-800
    Number of persons
    \small 4
    \small 8
    \small 9
    \small 10
    \small 7
    \small 5
    \small 4
    \small 3






    Answer:

    Income
    per day
    Number of
    Persons f_i
    Mid
    Points x_i
    f_ix_i
    |x_i - \overline{x}|
    f_i|x_i - \overline{x}|
    0 -100
    4
    50
    200
    308
    1232
    100 -200
    8
    150
    1200
    208
    1664
    200-300
    9
    250
    2250
    108
    972
    300-400
    10
    350
    3500
    8
    80
    400-500
    7
    450
    3150
    92
    644
    500-600
    5
    550
    2750
    192
    960
    600-700
    4
    650
    2600
    292
    1168
    700-800
    3
    750
    2250
    392
    1176

    \sum{f_i}
    =50

    \sum f_ix_i
    =17900

    \sum f_i|x_i - \overline{x}|
    =7896


    N = \sum_{i=1}^{8}{f_i} = 50 ; \sum_{i=1}^{8}{f_ix_i} = 17900

    \overline{x} = \frac{1}{N}\sum_{i=1}^{8}f_ix_i = \frac{17900}{50} = 358

    Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

    \sum f_i|x_i - \overline{x}| = 7896

    \therefore M.D.(\overline{x}) = \frac{1}{50}\sum_{i=1}^{8}f_i|x_i - \overline{x}|

    = \frac{7896}{50} = 157.92

    Hence, the mean deviation about the mean is 157.92

    Question:10. Find the mean deviation about the mean.

    Height in cms
    \small 95-105
    \small 105-115
    \small 115-125
    \small 125-135
    \small 135-145
    \small 145-155
    Number of persond
    \small 9
    \small 13
    \small 26
    \small 30
    \small 12
    \small 10

    Answer:

    Height
    in cms
    Number of
    Persons f_i
    Mid
    Points x_i
    f_ix_i
    |x_i - \overline{x}|
    f_i|x_i - \overline{x}|
    95 -105
    9
    100
    900
    25.3
    227.7
    105 -115
    13
    110
    1430
    15.3
    198.9
    115-125
    26
    120
    3120
    5.3
    137.8
    125-135
    30
    130
    3900
    4.7
    141
    135-145
    12
    140
    1680
    14.7
    176.4
    145-155
    10
    150
    1500
    24.7
    247

    \sum{f_i}
    =100

    \sum f_ix_i
    =12530

    \sum f_i|x_i - \overline{x}|
    =1128.8


    N = \sum_{i=1}^{6}{f_i} = 100 ; \sum_{i=1}^{6}{f_ix_i} = 12530

    \overline{x} = \frac{1}{N}\sum_{i=1}^{6}f_ix_i = \frac{12530}{100} = 125.3

    Now, we calculate the absolute values of the deviations from mean, |x_i - \overline{x}| and

    \sum f_i|x_i - \overline{x}| = 1128.8

    \therefore M.D.(\overline{x}) = \frac{1}{100}\sum_{i=1}^{6}f_i|x_i - \overline{x}|

    = \frac{1128.8}{100} = 11.29

    Hence, the mean deviation about the mean is 11.29

    Question:11. Find the mean deviation about median for the following data :

    Marks
    \small 0-10
    \small 10-20
    \small 20-30
    \small 30-40
    \small 40-50
    \small 50-60
    Number of girls
    \small 6
    \small 8
    \small 14
    \small 16
    \small 4
    \small 2

    Answer:

    Marks
    Number of
    Girls f_i
    Cumulative
    Frequency c.f.
    Mid
    Points x_i
    |x_i - M|
    f_i|x_i - M|
    0-10
    6
    6
    5
    22.85
    137.1
    10-20
    8
    14
    15
    12.85
    102.8
    20-30
    14
    28
    25
    2.85
    39.9
    30-40
    16
    44
    35
    7.15
    114.4
    40-50
    4
    48
    45
    17.15
    68.6
    50-60
    2
    50
    55
    27.15
    54.3





    \sum f_i|x_i - M|
    =517.1

    Now, N = 50, which is even.

    The class interval containing \left (\frac{N}{2} \right)^{th} or 25^{th} item is 20-30. Therefore, 20-30 is the median class.

    We know,

    Median = l + \frac{\frac{N}{2}- C}{f}\times h

    Here, l = 20, C = 14, f = 14, h = 10 and N = 50

    Therefore, Median = 20 + \frac{25 - 14}{14}\times 10 = 20 + 7.85 = 27.85

    Now, we calculate the absolute values of the deviations from median, |x_i - M| and

    \sum f_i|x_i - M| = 517.1

    \therefore M.D.(M) = \frac{1}{50}\sum_{i=1}^{6}f_i|x_i - M|

    = \frac{517.1}{50} = 10.34

    Hence, the mean deviation about the median is 10.34

    Question:12 Calculate the mean deviation about median age for the age distribution of \small 100 persons given below:

    Age (in years)
    \small 16-20
    \small 21-25
    \small 26-30
    \small 31-35
    \small 36-40
    \small 41-45
    \small 46-50
    \small 51-55
    \small 5
    \small 6
    \small 12
    \small 14
    \small 26
    \small 12
    \small 16
    \small 9

    [ Hint Convert the given data into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval]

    Answer:

    Age
    (in years)
    Number
    f_i
    Cumulative
    Frequency c.f.
    Mid
    Points x_i
    |x_i - M|
    f_i|x_i - M|
    15.5-20.5
    5
    5
    18
    20
    100
    20.5-25.5
    6
    11
    23
    15
    90
    25.5-30.5
    12
    23
    28
    10
    120
    30.5-35.5
    14
    37
    33
    5
    70
    35.5-40.5
    26
    63
    38
    0
    0
    40.5-45.5
    12
    75
    43
    5
    60
    45.5-50.5
    16
    91
    48
    10
    160
    50.5-55.5
    9
    100
    53
    15
    135





    \sum f_i|x_i - M|
    =735

    Now, N = 100, which is even.

    The class interval containing \left (\frac{N}{2} \right)^{th} or 50^{th} item is 35.5-40.5. Therefore, 35.5-40.5 is the median class.

    We know,

    Median = l + \frac{\frac{N}{2}- C}{f}\times h

    Here, l = 35.5, C = 37, f = 26, h = 5 and N = 100

    Therefore, Median = 35.5 + \frac{50 - 37}{26}\times 5 = 35.5 + 2.5 = 38

    Now, we calculate the absolute values of the deviations from median, |x_i - M| and

    \sum f_i|x_i - M| = 735

    \therefore M.D.(M) = \frac{1}{100}\sum_{i=1}^{8}f_i|x_i - M|

    = \frac{735}{100} = 7.35

    Hence, the mean deviation about the median is 7.35

    Class 11 maths chapter 15 ncert solutions - Exercise: 15.2

    Question:1. Find the mean and variance for each of the data.

    \small 6, 7, 10, 12, 13, 4, 8, 12

    Answer:

    Mean ( \overline{x} ) of the given data:

    \overline{x} = \frac{1}{8}\sum_{i=1}^{8}x_i = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9

    The respective values of the deviations from mean, (x_i - \overline{x}) are

    -3, -2, 1 3 4 -5 -1 3

    \therefore \sum_{i=1}^{8}(x_i - 10)^2 = 74

    \therefore \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \frac{1}{8}\sum_{i=1}^{8}(x_i - \overline{x})^2= \frac{74}{8} = 9.25

    Hence, Mean = 9 and Variance = 9.25

    Question:2. Find the mean and variance for each of the data.

    First n natural numbers.

    Answer:

    Mean ( \overline{x} ) of first n natural numbers:

    \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{\frac{n(n+1)}{2}}{n} = \frac{n+1}{2}

    We know, Variance \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}\left (x_i - \frac{n+1}{2} \right )^2

    We know that (a-b)^2 = a^2 - 2ab + b^2

    \\ \therefore n\sigma^2 = \sum_{i=1}^{n}x_i^2 + \sum_{i=1}^{n}(\frac{n+1}{2})^2 - 2\sum_{i=1}^{n}x_i\frac{n+1}{2} \\ = \frac{n(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4}\times n - 2.\frac{(n+1)}{2}.\frac{n(n+1)}{2}

    \\ \implies \sigma^2 = \frac{(n+1)(2n+1)}{6} + \frac{(n+1)^2}{4} - \frac{(n+1)^2}{2}

    \\ = \frac{(n+1)(2n+1)}{6} - \frac{(n+1)^2}{4} \\ = (n+1)\left [\frac{4n+2 - 3n -3}{12} \right ] \\ = (n+1).\frac{(n-1)}{12} \\ = \frac{n^2-1}{12}

    Hence, Mean = \frac{n+1}{2} and Variance = \frac{n^2-1}{12}

    Question:3. Find the mean and variance for each of the data

    First 10 multiples of 3

    Answer:

    First 10 multiples of 3 are:

    3, 6, 9, 12, 15, 18, 21, 24, 27, 30

    Mean ( \overline{x} ) of the above values:

    \\ \overline{x} = \frac{1}{10}\sum_{i=1}^{10}x_i = \frac{3+ 6+ 9+ 12+ 15+ 18+ 21+ 24+ 27+ 30}{10} \\ = 3.\frac{\frac{10(10+1)}{2}}{10} = 16.5

    The respective values of the deviations from mean, (x_i - \overline{x}) are

    -13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

    \therefore \sum_{i=1}^{10}(x_i - 16.5)^2 = 742.5

    \therefore \sigma^2 = \frac{1}{n}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \frac{1}{10}\sum_{i=1}^{10}(x_i - \overline{x})^2= \frac{742.5}{10} = 74.25

    Hence, Mean = 16.5 and Variance = 74.25

    Question:4. Find the mean and variance for each of the data.

    \small x_i
    6
    10
    14
    18
    24
    28
    30
    \small f_i
    2
    4
    7
    12
    8
    4
    3


    Answer:

    x_i
    f_i
    f_ix_i
    (x_i - \overline{x})
    (x_i - \overline{x})^2
    f_i(x_i - \overline{x})^2
    6
    2
    12
    -13
    169
    338
    10
    4
    40
    -9
    81
    324
    14
    7
    98
    -5
    25
    175
    18
    12
    216
    -1
    1
    12
    24
    8
    192
    5
    25
    200
    28
    4
    112
    9
    81
    324
    30
    3
    90
    13
    169
    363

    \sum{f_i}
    = 40
    \sum f_ix_i
    = 760


    \sum f_i(x_i - \overline{x})^2
    =1736

    N = \sum_{i=1}^{7}{f_i} = 40 ; \sum_{i=1}^{7}{f_ix_i} = 760

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{760}{40} = 19

    We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \implies \sigma^2 = \frac{1736}{40} = 43.4

    Hence, Mean = 19 and Variance = 43.4

    Question:5. Find the mean and variance for each of the data.

    \small x_i
    92
    93
    97
    98
    102
    104
    109
    \small f_i
    3
    2
    3
    2
    6
    3
    3


    Answer:

    x_i
    f_i
    f_ix_i
    (x_i - \overline{x})
    (x_i - \overline{x})^2
    f_i(x_i - \overline{x})^2
    92
    3
    276
    -8
    64
    192
    93
    2
    186
    -7
    49
    98
    97
    3
    291
    -3
    9
    27
    98
    2
    196
    -2
    4
    8
    102
    6
    612
    2
    4
    24
    104
    3
    312
    4
    16
    48
    109
    3
    327
    9
    81
    243

    \sum{f_i}
    = 22
    \sum f_ix_i
    = 2200


    \sum f_i(x_i - \overline{x})^2
    =640

    N = \sum_{i=1}^{7}{f_i} = 22 ; \sum_{i=1}^{7}{f_ix_i} = 2200

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{2200}{22} = 100

    We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \implies \sigma^2 = \frac{640}{22} = 29.09

    Hence, Mean = 100 and Variance = 29.09

    Question:6 Find the mean and standard deviation using short-cut method.

    \small x_i
    60
    61
    62
    63
    64
    65
    66
    67
    68
    \small f_i
    2
    1
    12
    29
    25
    12
    10
    4
    5


    Answer:

    Let the assumed mean, A = 64 and h = 1

    x_i
    f_i
    y_i = \frac{x_i-A}{h}
    y_i^2
    f_iy_i
    f_iy_i^2
    60
    2
    -4
    16
    -8
    32
    61
    1
    -3
    9
    -3
    9
    62
    12
    -2
    4
    -24
    48
    63
    29
    -1
    1
    -29
    29
    64
    25
    0
    0
    0
    0
    65
    12
    1
    1
    12
    12
    66
    10
    2
    4
    20
    40
    67
    4
    3
    9
    12
    36
    68
    5
    4
    16
    20
    80

    \sum{f_i}
    =100


    \sum f_iy_i
    = 0
    \sum f_iy_i ^2
    =286

    N = \sum_{i=1}^{9}{f_i} = 100 ; \sum_{i=1}^{9}{f_iy_i} = 0

    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =64 + \frac{0}{100} = 64

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(286) - (0)^2 \right ] \\ = \frac{28600}{10000} = 2.86

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{2.86} = 1.691

    Hence, Mean = 64 and Standard Deviation = 1.691

    Question:​​​​​​​7. Find the mean and variance for the following frequency distributions.

    Classes
    0-30
    30-60
    60-90
    90-120
    120-150
    150-180
    180-210
    Frequencies
    2
    3
    5
    10
    3
    5
    2


    Answer:

    Classes
    Frequency
    f_i
    Mid point
    x_i
    f_ix_i
    (x_i - \overline{x})
    (x_i - \overline{x})^2
    f_i(x_i - \overline{x})^2
    0-30
    2
    15
    30
    -92
    8464
    16928
    30-60
    3
    45
    135
    -62
    3844
    11532
    60-90
    5
    75
    375
    -32
    1024
    5120
    90-120
    10
    105
    1050
    2
    4
    40
    120-150
    3
    135
    405
    28
    784
    2352
    150-180
    5
    165
    825
    58
    3364
    16820
    180-210
    2
    195
    390
    88
    7744
    15488

    \sum{f_i} = N
    = 30

    \sum f_ix_i
    = 3210


    \sum f_i(x_i - \overline{x})^2
    =68280


    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{3210}{30} = 107

    We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \implies \sigma^2 = \frac{68280}{30} = 2276

    Hence, Mean = 107 and Variance = 2276

    Question:​​​​​​​8. Find the mean and variance for the following frequency distributions.

    Classes
    0-10
    10-20
    20-30
    30-40
    40-50
    Frequencies
    5
    8
    15
    16
    6


    Answer:

    Classes
    Frequency
    f_i
    Mid-point
    x_i
    f_ix_i
    (x_i - \overline{x})
    (x_i - \overline{x})^2
    f_i(x_i - \overline{x})^2
    0-10
    5
    5
    25
    -22
    484
    2420
    10-20
    8
    15
    120
    -12
    144
    1152
    20-30
    15
    25
    375
    -2
    4
    60
    30-40
    16
    35
    560
    8
    64
    1024
    40-50
    6
    45
    270
    18
    324
    1944

    \sum{f_i} = N
    = 50

    \sum f_ix_i
    = 1350


    \sum f_i(x_i - \overline{x})^2
    =6600


    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i = \frac{1350}{50} = 27

    We know, Variance, \sigma^2 = \frac{1}{N}\sum_{i=1}^{n}(x_i - \overline{x})^2

    \implies \sigma^2 = \frac{6600}{50} = 132

    Hence, Mean = 27 and Variance = 132

    Question:9. Find the mean, variance and standard deviation using short-cut method.

    Height in cms
    70-75
    75-80
    80-85
    85-90
    90-95
    95-100
    100-105
    105-110
    110-115
    No. of students
    3
    4
    7
    7
    15
    9
    6
    6
    3






    Answer:

    Let the assumed mean, A = 92.5 and h = 5

    Height
    in cms
    Frequency
    f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    y_i^2
    f_iy_i
    f_iy_i^2
    70-75
    3
    72.5
    -4
    16
    -12
    48
    75-80
    4
    77.5
    -3
    9
    -12
    36
    80-85
    7
    82.5
    -2
    4
    -14
    28
    85-90
    7
    87.5
    -1
    1
    -7
    7
    90-95
    15
    92.5
    0
    0
    0
    0
    95-100
    9
    97.5
    1
    1
    9
    9
    100-105
    6
    102.5
    2
    4
    12
    24
    105-110
    6
    107.5
    3
    9
    18
    54
    110-115
    3
    112.5
    4
    16
    12
    48

    \sum{f_i} =N = 60



    \sum f_iy_i
    = 6
    \sum f_iy_i ^2
    =254

    Mean,

    \overline{y} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =92.5 + \frac{6}{60}\times5 = 93

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(60)^2}\left [60(254) - (6)^2 \right ] \\ = \frac{1}{(60)^2}\left [15240 - 36 \right ] \\ = \frac{15204}{144} = 105.583

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{105.583} = 10.275

    Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

    Question:10. The diameters of circles (in mm) drawn in a design are given below:

    Diameters
    33-36
    37-40
    41-44
    45-48
    49-52
    No. of circles
    15
    17
    21
    22
    25

    Calculate the standard deviation and mean diameter of the circles.
    [ Hint First make the data continuous by making the classes as 32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5 and then proceed.]

    Answer:

    Let the assumed mean, A = 92.5 and h = 5

    Diameters
    No. of
    circles f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    y_i^2
    f_iy_i
    f_iy_i^2
    32.5-36.5
    15
    34.5
    -2
    4
    -30
    60
    36.5-40.5
    17
    38.5
    -1
    1
    -17
    17
    40.5-44.5
    21
    42.5
    0
    0
    0
    0
    44.5-48.5
    22
    46.5
    1
    1
    22
    22
    48.5-52.5
    25
    50.5
    2
    4
    50
    100

    \sum{f_i} =N = 100



    \sum f_iy_i
    = 25
    \sum f_iy_i ^2
    =199

    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =42.5 + \frac{25}{100}\times4 = 43.5

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(100)^2}\left [100(199) - (25)^2 \right ]\times4^2 \\ = \frac{1}{625}\left [19900 - 625 \right ] \\ = \frac{19275}{625} = 30.84

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{30.84} = 5.553

    Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553

    Statistics class 11 NCERT solutions - Exercise: 15.3

    Question:1. From the data given below state which group is more variable, A or B?

    Marks
    10-20
    20-30
    30-40
    40-50
    50-60
    60-70
    70-80
    Group A
    9
    17
    32
    33
    40
    10
    9
    Group B
    10
    20
    30
    25
    43
    15
    7


    Answer:

    The group having a higher coefficient of variation will be more variable.

    Let the assumed mean, A = 45 and h = 10

    For Group A

    Marks
    Group A
    f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    = \frac{x_i-45}{10}
    y_i^2
    f_iy_i
    f_iy_i^2
    10-20
    9
    15
    -3
    9
    -27
    81
    20-30
    17
    25
    -2
    4
    -34
    68
    30-40
    32
    35
    -1
    1
    -32
    32
    40-50
    33
    45
    0
    0
    0
    0
    50-60
    40
    55
    1
    1
    40
    40
    60-70
    10
    65
    2
    4
    20
    40
    70-80
    9
    75
    3
    9
    27
    81

    \sum{f_i} =N = 150



    \sum f_iy_i
    = -6
    \sum f_iy_i ^2
    =342

    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(342) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [51264 \right ] \\ =227.84

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{227.84} = 15.09

    Coefficient of variation = \frac{\sigma}{\overline x}\times100

    C.V.(A) = \frac{15.09}{44.6}\times100 = 33.83

    Similarly,

    For Group B

    Marks
    Group A
    f_i
    Midpoint
    x_i
    \dpi{100} y_i = \frac{x_i-A}{h}
    = \frac{x_i-45}{10}
    y_i^2
    f_iy_i
    f_iy_i^2
    10-20
    10
    15
    -3
    9
    -30
    90
    20-30
    20
    25
    -2
    4
    -40
    80
    30-40
    30
    35
    -1
    1
    -30
    30
    40-50
    25
    45
    0
    0
    0
    0
    50-60
    43
    55
    1
    1
    43
    43
    60-70
    15
    65
    2
    4
    30
    60
    70-80
    7
    75
    3
    9
    21
    72

    \sum{f_i} =N = 150



    \sum f_iy_i
    = -6
    \sum f_iy_i ^2
    =375

    Mean,

    \overline{x} = A + \frac{1}{N}\sum_{i=1}^{n}f_iy_i\times h =45 + \frac{-6}{150}\times10 = 44.6

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]\times h^2

    \\ \implies \sigma^2 = \frac{1}{(150)^2}\left [150(375) - (-6)^2 \right ]\times10^2 \\ = \frac{1}{15^2}\left [56214 \right ] \\ =249.84

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{249.84} = 15.80

    Coefficient of variation = \frac{\sigma}{\overline x}\times100

    C.V.(B) = \frac{15.80}{44.6}\times100 = 35.42

    Since C.V.(B) > C.V.(A)

    Therefore, Group B is more variable.

    Question:2 From the prices of shares X and Y below, find out which is more stable in value:

    X
    35
    54
    52
    53
    56
    58
    52
    50
    51
    49
    Y
    108
    107
    105
    105
    106
    107
    104
    103
    104
    101


    Answer:

    X( x_i )
    Y( y_i )
    x_i^2
    y_i^2
    35
    108
    1225
    11664
    54
    107
    2916
    11449
    52
    105
    2704
    11025
    53
    105
    2809
    11025
    56
    106
    8136
    11236
    58
    107
    3364
    11449
    52
    104
    2704
    10816
    50
    103
    2500
    10609
    51
    104
    2601
    10816
    49
    101
    2401
    10201
    =510
    = 1050
    =26360
    =110290

    For X,

    Mean , \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i = \frac{510}{10} = 51

    Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum x_i^2 - (\sum x_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(26360) - (510)^2 \right ] \\ = \frac{1}{100}.\left [263600 - 260100 \right ] \\ \\ = 35

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{35} = 5.91

    C.V.(X) = \frac{\sigma}{\overline x}\times100 = \frac{5.91}{51}\times100 = 11.58

    Similarly, For Y,

    Mean , \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i = \frac{1050}{10} = 105

    Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum y_i^2 - (\sum y_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(10)^2}\left [10(110290) - (1050)^2 \right ] \\ = \frac{1}{100}.\left [1102900 - 1102500 \right ] \\ \\ = 4

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{4} = 2

    C.V.(Y) = \frac{\sigma}{\overline y}\times100 = \frac{2}{105}\times100 = 1.904

    Since C.V.(Y) < C.V.(X)

    Hence Y is more stable.

    Question:3(i) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


    No. of wage earners
    586
    648
    Mean of monthly wages
    Rs\hspace {1mm} 5253
    Rs\hspace {1mm} 5253
    Variance of the distribution of wages
    100
    121

    Which firm A or B pays larger amount as monthly wages?

    Answer:

    Given, Mean of monthly wages of firm A = 5253

    Number of wage earners = 586

    Total amount paid = 586 x 5253 = 30,78,258

    Again, Mean of monthly wages of firm B = 5253

    Number of wage earners = 648

    Total amount paid = 648 x 5253 = 34,03,944

    Hence firm B pays larger amount as monthly wages.

    Question:3(ii) An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:


    No. of wage earners
    586
    648
    Mean of monthly wages
    Rs\hspace {1mm} 5253
    Rs\hspace {1mm} 5253
    Variance of the distribution of wages
    100
    121

    Which firm, A or B, shows greater variability in individual wages?

    Answer:

    Given, Variance of firm A = 100

    Standard Deviation = \sigma_A = \sqrt{Variance}= \sqrt{100} = 10

    Again, Variance of firm B = 121

    Standard Deviation = \sigma_B = \sqrt{Variance}= \sqrt{121} = 11

    Since \sigma_B>\sigma_A , firm B has greater variability in individual wages.

    Question:4 The following is the record of goals scored by team A in a football session:

    No. of goals scored
    0
    1
    2
    3
    4
    No. of matches
    1
    9
    7
    5
    3

    For the team B, mean number of goals scored per match was 2 with a standard deviation 1.25 goals. Find which team may be considered more consistent?

    Answer:

    No. of goals
    scored x_i
    Frequency
    f_i
    x_i^2
    f_ix_i
    f_ix_i^2
    0
    1
    0
    0
    0
    1
    9
    1
    9
    9
    2
    7
    4
    14
    28
    3
    5
    9
    15
    45
    4
    3
    16
    12
    48

    \sum{f_i} =N = 25

    \sum f_ix_i
    = 50
    \sum f_ix_i ^2
    =130

    For Team A,

    Mean,

    \overline{x} = \frac{1}{N}\sum_{i=1}^{n}f_ix_i =\frac{50}{25}= 2

    We know, Variance, \sigma^2 = \frac{1}{N^2}\left [N\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(25)^2}\left [25(130) - (50)^2 \right ] \\ \\ = \frac{750}{625} =1.2

    We know, Standard Deviation = \sigma = \sqrt{Variance}

    \therefore \sigma = \sqrt{1.2} = 1.09

    C.V.(A) = \frac{\sigma}{\overline x}\times100 = \frac{1.09}{2}\times100 = 54.5

    For Team B,

    Mean = 2

    Standard deviation, \sigma = 1.25

    C.V.(B) = \frac{\sigma}{\overline x}\times100 = \frac{1.25}{2}\times100 = 62.5

    Since C.V. of firm B is more than C.V. of A.

    Therefore, Team A is more consistent.

    Question:5 The sum and sum of squares corresponding to length x (in cm) and weight y (in gm) of 50 plant products are given below:

    \sum_{i=1}^{50}x_i=212, \sum _{i=1}^{50}x_i^2=902.8,\sum _{i=1}^{50}y_i=261,\sum _{i=1}^{50}y_i^2=1457.6
    Which is more varying, the length or weight?

    Answer:

    For lenght x,

    Mean, \overline{x} = \frac{1}{n}\sum_{i=1}^{n}x_i =\frac{212}{50}= 4.24

    We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_ix_i^2 - (\sum f_ix_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(902.8) - (212)^2 \right ] \\ \\ = \frac{196}{2500} =0.0784

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{0.0784} = 0.28

    C.V.(x) = \frac{\sigma}{\overline x}\times100 = \frac{0.28}{4.24}\times100 = 6.603

    For weight y,

    Mean,

    Mean, \overline{y} = \frac{1}{n}\sum_{i=1}^{n}y_i =\frac{261}{50}= 5.22

    We know, Variance, \sigma^2 = \frac{1}{n^2}\left [n\sum f_iy_i^2 - (\sum f_iy_i)^2 \right ]

    \\ \implies \sigma^2 = \frac{1}{(50)^2}\left [50(1457.6) - (261)^2 \right ] \\ \\ = \frac{4759}{2500} =1.9036

    We know, Standard Deviation = \sigma = \sqrt{Variance}= \sqrt{1.9036} = 1.37

    C.V.(y) = \frac{\sigma}{\overline y}\times100 = \frac{1.37}{5.22}\times100 = 26.24

    Since C.V.(y) > C.V.(x)

    Therefore, weight is more varying.

    Statistics maths chapter 15 class 11 - Miscellaneous Exercise

    Question:1 The mean and variance of eight observations are 9 and 9.25 , respectively. If six of the observations are 6,7,10,12,12 and 13 , find the remaining two observations.

    Answer:

    Given,

    The mean and variance of 8 observations are 9 and 9.25, respectively

    Let the remaining two observations be x and y,

    Observations: 6, 7, 10, 12, 12, 13, x, y.

    ∴ Mean, \overline X = \frac{6+ 7+ 10+ 12+ 12+ 13+ x+ y}{8} = 9

    60 + x + y = 72

    x + y = 12 -(i)

    Now, Variance

    = \frac{1}{n}\sum_{i=1}^8(x_i - x)^2 = 9.25

    \implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2 -18(x+y)+ 2.9^2 \right ]

    \implies 9.25 = \frac{1}{8}\left[(-3)^2+ (-2)^2+ 1^2+ 3^2+ 4^2+ x^2+ y^2+ -18(12)+ 2.9^2 \right ] (Using (i))

    \implies 9.25 = \frac{1}{8}\left[48+x^2+ y^2 -216+ 162 \right ] = \frac{1}{8}\left[x^2+ y^2 - 6 \right ]

    \implies x^2+ y^2 = 80 -(ii)

    Squaring (i), we get

    x^2+ y^2 +2xy= 144 (iii)

    (iii) - (ii) :

    2xy = 64 (iv)

    Now, (ii) - (iv):

    \\ x^2+ y^2 -2xy= 80-64 \\ \implies (x-y)^2 = 16 \\ \implies x-y = \pm 4 -(v)

    Hence, From (i) and (v):

    x – y = 4 \implies x = 8 and y = 4

    x – y = -4 \implies x = 4 and y = 8

    Therefore, The remaining observations are 4 and 8. (in no order)

    Question:2 The mean and variance of 7 observations are 8 and \small 16 , respectively. If five of the observations are \small 2,4,10,12,14 . Find the remaining two observations.

    Answer:

    Given,

    The mean and variance of 7 observations are 8 and 16, respectively

    Let the remaining two observations be x and y,

    Observations: 2, 4, 10, 12, 14, x, y

    ∴ Mean, \overline X = \frac{2+ 4+ 10+ 12+ 14+ x+ y}{7} = 8

    42 + x + y = 56

    x + y = 14 -(i)

    Now, Variance

    = \frac{1}{n}\sum_{i=1}^8(x_i - \overline x)^2 = 16

    \implies 16 = \frac{1}{7}\left[(-6)^2+ (-4)^2+ 2^2+ 4^2+ 6^2+ x^2+ y^2 -16(x+y)+ 2.8^2 \right ]

    \implies 16 = \frac{1}{7}\left[36+16+4+16+36+ x^2+ y^2 -16(14)+ 2(64) \right ] (Using (i))

    \implies 16 = \frac{1}{7}\left[108+x^2+ y^2 -96 \right ] = \frac{1}{7}\left[x^2+ y^2 + 12\right ]

    \implies x^2+ y^2 = 112- 12 =100 -(ii)

    Squaring (i), we get

    x^2+ y^2 +2xy= 196 (iii)

    (iii) - (ii) :

    2xy = 96 (iv)

    Now, (ii) - (iv):

    \\ x^2+ y^2 -2xy= 100-96 \\ \implies (x-y)^2 = 4 \\ \implies x-y = \pm 2 -(v)

    Hence, From (i) and (v):

    x – y = 2 \implies x = 8 and y = 6

    x – y = -2 \implies x = 6 and y = 8

    Therefore, The remaining observations are 6 and 8. (in no order)

    Question:3 The mean and standard deviation of six observations are \small 8 and \small 4 , respectively. If each observation is multiplied by \small 3 , find the new mean and new standard deviation of the resulting observations.

    Answer:

    Given,

    Mean = 8 and Standard deviation = 4

    Let the observations be x_1, x_2, x_3, x_4, x_5\ and\ x_6

    Mean, \overline x = \frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} = 8

    Now, Let y_i be the the resulting observations if each observation is multiplied by 3:

    \\ \overline y_i = 3\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{3}

    New mean, \overline y = \frac{y_1+ y_2+ y_3+ y_4+ y_5+ y_6}{6}

    = 3\left [\frac{x_1+ x_2+ x_3+ x_4+ x_5+ x_6}{6} \right] =3\times 8

    = 24

    We know that,

    Standard Deviation = \sigma = \sqrt{Variance}

    \dpi{100} =\sqrt{ \frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2}

    \dpi{100}\\ \implies 4^2=\frac{1}{6}\sum_{i=1}^6(x_i - \overline x)^2 \\ \implies \sum_{i=1}^6(x_i - \overline x)^2 = 6\times16 = 96 -(i)

    Now, Substituting the values of x_i\ and\ \overline x in (i):

    \dpi{100} \\ \implies \sum_{i=1}^6(\frac{y_i}{3} - \frac{\overline y}{3})^2 = 96 \\ \implies \sum_{i=1}^6(y_i - \overline y)^2 = 96\times9 =864

    Hence, the variance of the new observations = \frac{1}{6}\times864 = 144

    Therefore, Standard Deviation = \sigma = \sqrt{Variance} = \sqrt{144} = 12

    Question:4 Given that \small \bar {x} is the mean and \small \sigma^2 is the variance of \small n observations .Prove that the mean and variance of the observations \small ax_1,ax_2,ax_3,....,ax_n , are \small a\bar{x} and \small a^2 \sigma^2 respectively, \small (a \neq 0) .

    Answer:

    Given, Mean = \small \bar {x} and variance = \small \sigma^2

    Now, Let y_i be the the resulting observations if each observation is multiplied by a:

    \\ \overline y_i = a\overline x_i \\ \implies \overline x_i = \frac{\overline y_i}{a}

    \overline y =\frac{1}{n}\sum_{i=1}^ny_i = \frac{1}{n}\sum_{i=1}^nax_i

    \overline y = a\left [\frac{1}{n}\sum_{i=1}^nx_i \right] = a\overline x

    Hence the mean of the new observations \small ax_1,ax_2,ax_3,....,ax_n is \small a\bar{x}

    We know,

    \dpi{100} \sigma^2=\frac{1}{n}\sum_{i=1}^n(x_i - \overline x)^2

    Now, Substituting the values of x_i\ and\ \overline x :

    \\ \implies \sigma^2= \frac{1}{n}\sum_{i=1}^n(\frac{y_i}{a} - \frac{\overline y}{a})^2 \\ \implies a^2\sigma^2= \frac{1}{n}\sum_{i=1}^n(y_i - \overline y)^2

    Hence the variance of the new observations \small ax_1,ax_2,ax_3,....,ax_n is a^2\sigma^2

    Hence proved.

    Question:5(i) The mean and standard deviation of \small 20 observations are found to be \small 10 and \small 2 , respectively. On rechecking, it was found that an observation \small 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

    If wrong item is omitted.

    Answer:

    Given,

    Number of observations, n = 20

    Also, Incorrect mean = 10

    And, Incorrect standard deviation = 2

    \overline x =\frac{1}{n}\sum_{i=1}^nx_i

    \implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i \implies \sum_{i=1}^{20}x_i = 200

    Thus, incorrect sum = 200

    Hence, correct sum of observations = 200 – 8 = 192

    Therefore, Correct Mean = (Correct Sum)/19

    =\frac{192}{19}

    = 10.1

    Now, Standard Deviation,

    \sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

    \\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080 ,which is the incorrect sum.

    Thus, New sum = Old sum - (8x8)

    = 2080 – 64

    = 2016

    Hence, Correct Standard Deviation =

    \sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2016}{19} - (10.1)^2}

    \dpi{100} = \sqrt{106.1 - 102.01} = \sqrt{4.09}

    =2.02

    Question:5(ii) The mean and standard deviation of \small 20 observations are found to be \small 10 and \small 2 , respectively. On rechecking, it was found that an observation \small 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:

    If it is replaced by \small 12.

    Answer:

    Given,

    Number of observations, n = 20

    Also, Incorrect mean = 10

    And, Incorrect standard deviation = 2

    \overline x =\frac{1}{n}\sum_{i=1}^nx_i

    \implies 10 =\frac{1}{20}\sum_{i=1}^{20}x_i \implies \sum_{i=1}^{20}x_i = 200

    Thus, incorrect sum = 200

    Hence, correct sum of observations = 200 – 8 + 12 = 204

    Therefore, Correct Mean = (Correct Sum)/20

    \dpi{100} =\frac{204}{20}

    = 10.2

    Now, Standard Deviation,

    \sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

    \\ \implies 2^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 4 + (\overline x)^2 \\ \implies \frac{1}{20}\sum_{i=1}^nx_i ^2 = 4 + 100 = 104 \\ \implies \sum_{i=1}^nx_i ^2 = 2080 ,which is the incorrect sum.

    Thus, New sum = Old sum - (8x8) + (12x12)

    = 2080 – 64 + 144

    = 2160

    Hence, Correct Standard Deviation =

    \sigma' =\sqrt{\frac{1}{n}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{2160}{20} - (10.2)^2}

    = \sqrt{108 - 104.04} = \sqrt{3.96}

    = 1.98

    Question:6 The mean and standard deviation of marks obtained by \small 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

    Subject
    Mathematics
    Physics
    Chemistry
    Mean
    \small 42
    \small 32
    \small 40.9
    Standard deviation
    \small 12
    \small 15
    \small 20

    Which of the three subjects shows highest variability in marks and which shows the lowest?

    Answer:

    Given,

    Standard deviation of physics = 15

    Standard deviation of chemistry = 20

    Standard deviation of mathematics = 12

    We know ,

    C.V. = \frac{Standard\ Deviation}{Mean}\times 100

    The subject with greater C.V will be more variable than others.

    C.V. _P= \frac{15}{32}\times 100 = 46.87

    C.V. _C= \frac{20}{40.9}\times 100 = 48.89

    C.V. _M= \frac{12}{42}\times 100 = 28.57

    Hence, Mathematics has lowest variability and Chemistry has highest variability.

    Question:7 The mean and standard deviation of a group of \small 100 observations were found to be \small 20 and \small 3 , respectively. Later on it was found that three observations were incorrect, which were recorded as \small 21,21 and \small 18 . Find the mean and standard deviation if the incorrect observations are omitted.

    Answer:

    Given,

    Initial Number of observations, n = 100

    \overline x =\frac{1}{n}\sum_{i=1}^nx_i

    \dpi{100} \implies 20 =\frac{1}{100}\sum_{i=1}^{100}x_i \implies \sum_{i=1}^{100}x_i = 2000

    Thus, incorrect sum = 2000

    Hence, New sum of observations = 2000 - 21-21-18 = 1940

    New number of observation, n' = 100-3 =97

    Therefore, New Mean = New Sum)/100

    =\frac{1940}{97}

    = 20

    Now, Standard Deviation,

    \sigma =\sqrt{\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\frac{1}{n}\sum_{i=1}^n x)^2}

    \\ \implies 3^2 =\frac{1}{n}\sum_{i=1}^nx_i ^2 - (\overline x)^2 \\ \implies \frac{1}{n}\sum_{i=1}^nx_i ^2 = 9 + (\overline x)^2 \\ \implies \frac{1}{100}\sum_{i=1}^nx_i ^2 = 9 + 400 = 409 \\ \implies \sum_{i=1}^nx_i ^2 = 40900 ,which is the incorrect sum.

    Thus, New sum = Old (Incorrect) sum - (21x21) - (21x21) - (18x18)

    = 40900 - 441 - 441 - 324

    = 39694

    Hence, Correct Standard Deviation =

    \sigma' =\sqrt{\frac{1}{n'}\left (\sum_{i=1}^nx_i ^2 \right )' - (\overline x')^2} = \sqrt{\frac{39694}{97} - (20)^2}

    = \sqrt{108 - 104.04} = \sqrt{3.96}

    = 3.036

    Summary And Formulae Of Statistics Maths Chapter 15 class 11

    • Measures of dispersion- Range, mean deviation, Quartile deviation, standard deviation, variance are measures of dispersion.
    • Range- The difference between the maximum and the minimum values of each series is called the ‘Range’ of the data. Range = Maximum Value – Minimum Value.
    • Mean- The arithmetic average of a range of values or quantities, found by adding all values and dividing by the number of values. For example, the mean of 4, 10, and 1 is (4+10+1)/3=15/3=5.

    \bar{X}=\frac{\Sigma X}{N}
    X represents values or scores,
    N represents the number of values or scores.

    • Median- If the total number of values(n) is an odd number, then the formula is-

    Median= (\frac{n+1}{2})^{th}\:term

    • Median- If the total number of values(n) is an even number, then the formula is-

    Median= \frac{(\frac{n}{2})^{th}\:term+(\frac{n+1}{2})^{th}\:term}{2}

    • Mode- It is the most frequently occurring value or number.
    • Mean deviation for ungrouped data-

    M.D.(\bar{x})=\frac{\Sigma \:|x_{i}-\bar{x}|}{n}, \:\:\:\:\:\:\:\:\: M.D.(M)=\frac{\Sigma \:|x_{i}-M|}{n}

    • Mean deviation for grouped data-

    M.D.(\bar{x})=\frac{\Sigma f_i\:|x_{i}-\bar{x}|}{N}, \:\:\:\:\:\:\:\:\: M.D.(M)=\frac{f_i\Sigma \:|x_{i}-M|}{N}, \: where \:N=\Sigma f_i

    interested students can practice class 11 maths ch 15 question answer using the following exercises.

    NCERT Solutions For Class 11 Mathematics

    Key Features Of Statistics Class 11 NCERT Solutions

    Easy to Understand: The ch 15 maths class 11 solutions are written in simple language and are easy to understand, which makes them accessible for students who may not have a strong background in statistics.

    Comprehensive Coverage: The statistics NCERT solutions cover all the important concepts and topics in Statistics Class 11 NCERT curriculum. This ensures that students are well-prepared and have a thorough understanding of the subject.

    Structured Format: The ch 15 maths class 11 solutions are presented in a structured format that helps students to organize their thoughts and approach problems in a systematic manner.

    NCERT Solutions For Class 11 - Subject Wise

    This chapter seems very easy but it very useful in the field of science, research reliable predictions, or forecasts for future use, decision making, etc. There are 7 questions are given in the miscellaneous exercise. Try to solve them also to get command in this chapter. In NCERT solutions for class 11 maths chapter 15 statistics, you will get solutions to miscellaneous exercise too.

    NCERT Books and NCERT Syllabus

    Frequently Asked Question (FAQs)

    1. Explain the term standard deviation, which is discussed in NCERT solutions for class 11 maths chapter 15.

    The concept of standard deviation involves measuring the amount of variation or deviation present in a given set of values, and the range is determined by considering the level of standard deviation. For a clearer understanding of this term, students are encouraged to download the PDF solutions provided by Careers360, which are available in both chapter-wise and exercise-wise formats to cater to their specific needs.

    2. What are the topics discussed in Chapter 15 of NCERT Solutions for Class 11 Maths?

    The topics discussed in class 11 maths statistics are
    1. Introduction
    2. Methods of Dispersion
    3. Range
    4. Mean Deviation
    5. Variance and Standard Deviation
    6. Analysis of Frequency Distributions

    3. Are the NCERT Solutions for statistics class 11 maths helpful for the students?

    The NCERT Solutions for class 11 chapter 15 maths offer precise and straightforward explanations that aid students in achieving good grades in their exams. The solutions' methodical approach to problem-solving provides students with a clear understanding of the marks allocation as per the latest CBSE Syllabus 2023. By using these class 11th statistics solutions , students can identify their weak areas and work towards improving them, resulting in better academic performance.

    4. Where can I find the complete solutions of class 11 statistics solutions ?

    Students can find a detailed NCERT solutions for class 11 maths  by clicking on the link. they can practice these solutions to get confidence in the concepts and in-depth understanding that ultimately lead to high score in the exam.

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    A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

    Option 1)

    0.34\; J

    Option 2)

    0.16\; J

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    Option 4)

    0.67\; J

    A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

    Option 1)

    2.45×10−3 kg

    Option 2)

     6.45×10−3 kg

    Option 3)

     9.89×10−3 kg

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    12.89×10−3 kg

     

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    Option 1)

    2,000 \; J - 5,000\; J

    Option 2)

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    Option 3)

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    Option 4)

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    Option 1)

    K/2\,

    Option 2)

    \; K\;

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    2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

    Option 1)

    11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

    Option 2)

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    Option 1)

    0.02

    Option 2)

    3.125 × 10-2

    Option 3)

    1.25 × 10-2

    Option 4)

    2.5 × 10-2

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    Option 1)

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    Option 2)

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    Option 3)

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    Option 1)

    twice that in 60 g carbon

    Option 2)

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    Option 3)

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    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Team Lead

    A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

    2 Jobs Available
    Structural Engineer

    A Structural Engineer designs buildings, bridges, and other related structures. He or she analyzes the structures and makes sure the structures are strong enough to be used by the people. A career as a Structural Engineer requires working in the construction process. It comes under the civil engineering discipline. A Structure Engineer creates structural models with the help of computer-aided design software. 

    2 Jobs Available
    Architect

    Individuals in the architecture career are the building designers who plan the whole construction keeping the safety and requirements of the people. Individuals in architect career in India provides professional services for new constructions, alterations, renovations and several other activities. Individuals in architectural careers in India visit site locations to visualize their projects and prepare scaled drawings to submit to a client or employer as a design. Individuals in architecture careers also estimate build costs, materials needed, and the projected time frame to complete a build.

    2 Jobs Available
    Landscape Architect

    Having a landscape architecture career, you are involved in site analysis, site inventory, land planning, planting design, grading, stormwater management, suitable design, and construction specification. Frederick Law Olmsted, the designer of Central Park in New York introduced the title “landscape architect”. The Australian Institute of Landscape Architects (AILA) proclaims that "Landscape Architects research, plan, design and advise on the stewardship, conservation and sustainability of development of the environment and spaces, both within and beyond the built environment". Therefore, individuals who opt for a career as a landscape architect are those who are educated and experienced in landscape architecture. Students need to pursue various landscape architecture degrees, such as M.Des, M.Plan to become landscape architects. If you have more questions regarding a career as a landscape architect or how to become a landscape architect then you can read the article to get your doubts cleared. 

    2 Jobs Available
    Plumber

    An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

    2 Jobs Available
    Orthotist and Prosthetist

    Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

    6 Jobs Available
    Veterinary Doctor

    A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.

    5 Jobs Available
    Pathologist

    A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

    5 Jobs Available
    Gynaecologist

    Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

    4 Jobs Available
    Surgical Technologist

    When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications. 

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    3 Jobs Available
    Oncologist

    An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

    3 Jobs Available
    Chemical Pathologist

    Are you searching for a chemical pathologist job description? A chemical pathologist is a skilled professional in healthcare who utilises biochemical laboratory tests to diagnose disease by analysing the levels of various components or constituents in the patient’s body fluid. 

    2 Jobs Available
    Biochemical Engineer

    A Biochemical Engineer is a professional involved in the study of proteins, viruses, cells and other biological substances. He or she utilises his or her scientific knowledge to develop products, medicines or ways to improve quality and refine processes. A Biochemical Engineer studies chemical functions occurring in a living organism’s body. He or she utilises the observed knowledge to alter the composition of products and develop new processes. A Biochemical Engineer may develop biofuels or environmentally friendly methods to dispose of waste generated by industries. 

    2 Jobs Available
    Actor

    For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

    4 Jobs Available
    Acrobat

    Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

    3 Jobs Available
    Video Game Designer

    Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

    3 Jobs Available
    Talent Agent

    The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.

    If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.

    3 Jobs Available
    Radio Jockey

    Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

    A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

    3 Jobs Available
    Producer

    An individual who is pursuing a career as a producer is responsible for managing the business aspects of production. They are involved in each aspect of production from its inception to deception. Famous movie producers review the script, recommend changes and visualise the story. 

    They are responsible for overseeing the finance involved in the project and distributing the film for broadcasting on various platforms. A career as a producer is quite fulfilling as well as exhaustive in terms of playing different roles in order for a production to be successful. Famous movie producers are responsible for hiring creative and technical personnel on contract basis.

    2 Jobs Available
    Fashion Blogger

    Fashion bloggers use multiple social media platforms to recommend or share ideas related to fashion. A fashion blogger is a person who writes about fashion, publishes pictures of outfits, jewellery, accessories. Fashion blogger works as a model, journalist, and a stylist in the fashion industry. In current fashion times, these bloggers have crossed into becoming a star in fashion magazines, commercials, or campaigns. 

    2 Jobs Available
    Photographer

    Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

    2 Jobs Available
    Copy Writer

    In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

    5 Jobs Available
    Editor

    Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

    3 Jobs Available
    Journalist

    Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

    3 Jobs Available
    Publisher

    For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

    3 Jobs Available
    Vlogger

    In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

    3 Jobs Available
    Travel Journalist

    The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

    2 Jobs Available
    Videographer

    Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production. 

    2 Jobs Available
    SEO Analyst

    An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns. 

    2 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    Quality Controller

    A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

    A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

    3 Jobs Available
    Production Manager

    Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers. 

    Resource Links for Online MBA 

    3 Jobs Available
    QA Manager

    Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes. 

    2 Jobs Available
    QA Lead

    A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

    2 Jobs Available
    Reliability Engineer

    Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment. 

    2 Jobs Available
    Safety Manager

    A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    Information Security Manager

    Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

    3 Jobs Available
    Computer Programmer

    Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

    3 Jobs Available
    Product Manager

    A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

    3 Jobs Available
    ITSM Manager

    ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks. 

    3 Jobs Available
    .NET Developer

    .NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#. 

    2 Jobs Available
    Corporate Executive

    Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

    2 Jobs Available
    DevOps Architect

    A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties. 

    2 Jobs Available
    Cloud Solution Architect

    Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

    2 Jobs Available
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