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    NCERT Solutions for Class 11 Maths Chapter 15 – Statistics

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    Exercise 15.1 : Solutions of Questions on Page Number : 360
    Q1 :Find the mean deviation about the mean for the data
    4, 7, 8, 9, 10, 12, 13, 17
    Answer :
    The given data is
    4, 7, 8, 9, 10, 12, 13, 17
    Mean of the data,

    The deviations of the respective observations from the mean are 
    -6,-3, -2, -1, 0, 2, 3, 7
    The absolute values of the deviations, i.e., are
    6, 3, 2, 1, 0, 2, 3, 7
    The required mean deviation about the mean is

    NCERT Solutions for Maths Class 11 Chapter 15 – Statistics

    Q2 :Find the mean deviation about the mean for the data
    38, 70, 48, 40, 42, 55, 63, 46, 54, 44
    Answer :
    The given data is
    38, 70, 48, 40, 42, 55, 63, 46, 54, 44
    Mean of the given data,

    The deviations of the respective observations from the mean are 
    -12, 20,-2, -10, -8, 5, 13, -4, 4,-6
    The absolute values of the deviations, i.e.  , are
    12, 20, 2, 10, 8, 5, 13, 4, 4, 6
    The required mean deviation about the mean is

    Q3 :Find the mean deviation about the median for the data.
    13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
    Answer :
    The given data is
    13, 17, 16, 14, 11, 13, 10, 16, 11, 18, 12, 17
    Here, the numbers of observations are 12, which is even.
    Arranging the data in ascending order, we obtain
    10, 11, 11, 12, 13, 13, 14, 16, 16, 17, 17, 18

    The deviations of the respective observations from the median, i.e.  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
    The absolute values of the deviations,,  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
    The required mean deviation about the median is

    Q4 :Find the mean deviation about the median for the data
    36, 72, 46, 42, 60, 45, 53, 46, 51, 49
    Answer :
    The given data is
    36, 72, 46, 42, 60, 45, 53, 46, 51, 49
    Here, the number of observations is 10, which is even.
    Arranging the data in ascending order, we obtain
    36, 42, 45, 46, 46, 49, 51, 53, 60, 72

    The deviations of the respective observations from the median, i.e. are
    -11.5, -5.5, -2.5, -1.5, -1.5, 1.5, 3.5, 5.5, 12.5, 24.5
    The absolute values of the deviations,  are
    11.5, 5.5, 2.5, 1.5, 1.5, 1.5, 3.5, 5.5, 12.5, 24.5
    Thus, the required mean deviation about the median is

    Q5 :Find the mean deviation about the mean for the data.

    xi 5 10 15 20 25
    fi 7 4 6 3 5

    Answer :

     

    xi fi xifi    
    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
    25 350 158

    Q6 :Find the mean deviation about the mean for the data

    Xi 10 30 50 70 90
    fi 4 24 28 16 8


    Answer :

     Xi  fi  Xi fi    
    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
    80 4000 1280

     

    Q7 :Find the mean deviation about the median for the data.

    Xi 5 7 9 10 12 15
    fi 8 6 2 2 2 6

    Answer :
    The given observations are already in ascending order.
    Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

    xi fi c.f
    5 8 8
    7 6 14
    9 2 16
    10 21 18
    12 2 20
    15 6 26

    Here, N = 26, which is even.
    Median is the mean of 13th and 14th observations. Both of these observations lie in the
    cumulative frequency 14, for which the corresponding observation is 7.

    The absolute values of the deviations from median, i.e.are
    |xi – M|

    / Xi – M /  2  0  2  3  5  8
     fi  8  6  2  2  2  6
     fi / Xi – M /  16  0  4  6  10  48

    Q8 : Find the mean deviation about the median for the data

    xi 15 21 27 30 35
    fi 3 5 6 7 8

    Answer :
    The given observations are already in ascending order.
    Adding a column corresponding to cumulative frequencies of the given data, we obtain the following table.

    Xi fi c.f
     15  3 3
     21  5 8
     27  6 14
     30  7 21
     35  8 29

    Here, N = 29, which is odd.

    observation = 15th observation
    This observation lies in the cumulative frequency 21, for which the corresponding observation is 30.
    ∴ Median = 30
    The absolute values of the deviations from median, i.e.|xi – M|  are

    / Xi – M / 15 9 3 0 5
    fi 3 5 16 7 8
    fi / Xi – M / 45 45 18 0 40

    Q9 :Find the mean deviation about the mean for the data.

    Income per day Number of Persons
    0-100 4
    100-200 8
    200-300 9
    300-400 10
    400-500 7
    500-600 5
    600-700 4
    700–800 3


    Answer :
    The following table is formed.

    Income per day Number of persons fi Mid point Xi fixi
    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 26002 292 1168
    700-800 3 750 2250 392 1176
    50 17900 7896



     

    Q10 :Find the mean deviation about the mean for the data

    Heights in cm Number of Boys
    95-105 9
    105-115 13
    115-125 26
    125-135 30
    135-145 12
    145-155 10

    Answer :
    The following table is formed.

    Heights in cm Number of boys  Mid point Xi fixi  
    95-105  9  100 900  25.3  227.7
     105-115  13 110  1430  15.3  198.8
     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

     

    here,


    Q11 :Calculate the mean deviation about median age for the age distribution of 100 persons given below:

    Age Number
    16-20 5
    21-25 6
    26-30 12
    31-35 14
    36-40 26
    41-45 12
    46-50 16
    51-55 9


    Answer :
    The given data is not continuous. Therefore, it has to be converted into continuous frequency distribution by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class interval.
    The table is formed as follows.

    Age Number fi c.f Mid Point xi
    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 58 15 135
    100 735

    The class interval containing the  or 50th item is 35.5 – 40.5.
    Therefore, 35.5 – 40.5 is the median class.
    It is known that,

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

    Thus, mean deviation about the median is given by,

    Exercise 15.2 : Solutions of Questions on Page Number : 371

    Q1 :Find the mean and variance for the data 6, 7, 10, 12, 13, 4, 8, 12
    Answer :
    6, 7, 10, 12, 13, 4, 8, 12
    Mean,

    The following table is obtained.

    xi
    6 -3 9
    7 -2 4
    10 -1 1
    12 3 9
    13 4 16
    4 -5 25
    8 -1 1
    12 3 9
    74

     

    Q2 :Find the mean and variance for the first n natural numbers
    Answer :
    The mean of first n natural numbers is calculated as follows.

    Q3 :Find the mean and variance for the first 10 multiples of 3
    Answer :
    The first 10 multiples of 3 are
    3, 6, 9, 12, 15, 18, 21, 24, 27, 30
    Here, number of observations, n = 10

    The following table is obtained.

    xi
    3 -13.5 182.25
    6 -10.5 110.25
    9 -7.5 56.25
    12 -4.5 20.25
    15 -1.5 2.25
    18 1.5 2.25
    21 4.5 20.25
    24 7.5 56.25
    27 10.5 110.25
    30 13.5 182.25
    742.5

     

    Q4 :Find the mean and variance for the data

    xi 6 10 14 18 28 24 30
    fi 2 4 7 12 8 4 3

    Answer :
    The data is obtained in tabular form as follows.

    xi fi xifi
    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 11 121 363
    40 760 1736

     

    Here, N = 40,

    Q5 :Find the mean and variance for the data

    xi 92 93 97 983 102 104 109
    fi 3 2 3 2 6 3 3

    Answer :
    The data is obtained in tabular form as follow

    xi fi xifi
    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
    22 2200 640

     

    Here, N = 22,

    Q6 :Find the mean and standard deviation using short-cut method.

    xi 60 61 62 63 64 65 66 67 68
    fi 2 1 12 29 25 12 10 4 5


    Answer :
    The data is obtained in tabular form as follows.

    xi fi yi2 fiyi fiyi2
    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
    100 220 0 286

     

    Mean, 

    Q7 :Find the mean and variance for the following frequency distribution.

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


    Answer :

    Class Frequency fi Mid Point xi yi2 fiyi fiyi
    0-30 2 15 -3 9 -6 18
    30-60 3 45 -2 4 -6 12
    60-90 5 75 -1 1 -5 5
    90-120 10 105 0 0 0 0
    120-150 3 135 1 1 3 3
    150-180 5 165 2 4 10 20
    180-210 2 195 3 9 6 18
    30 2 76

     

    Mean,


    Q8 :Find the mean and variance for the following frequency distribution.

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


    Answer :

    Class Frequency fi Mid Point xi yi2 fiyi fiyi2
    0-10 5 5 -2 4 -10 20
    10-20 18 15 -1 1 -8 8
    20-30 15 25 0 0 0 0
    30-40 16 35 1 1 16 16
    40-50 6 45 2 4 12 24
    50 10 68


    Mean, 

    Q9 :Find the mean, variance and standard deviation using short-cut method

    Heights in cm No. of children
    70-75 3
    75-80 4
    80-85 7
    85-90 7
    90-95 15
    95-100 9
    100-105 6
    105-110 6
    110-115 3


    Answer :

    Class Interval Frequency Mid Point yi fiyi fiyi2
    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 41 16 12 48
    60 6 254


    Mean,

    Q10 :The diameters of circles (in mm) drawn in a design are given below:

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


    Answer :

    Class Interval Frequency fi Mid-point xi fi2 fiyi fiyi2
    32.5-36.5  15 34.5  -2  4  -30  60
    36.5-40.5  17 34.5  -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
     100  25  199


    Here, N = 100, h = 4
    Let the assumed mean, A, be 42.5.
    Mean, 

    Exercise 15.3 : Solutions of Questions on Page Number : 375

    Q1 :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 :
    Firstly, the standard deviation of group A is calculated as follows.

    Marks Group A fi Mid Point xi yi2 fiyi fiyi2
    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 0 27 81
    150 -6 342

    Here, h = 10, N = 150, A = 45

    The standard deviation of group B is calculated as follows.

    Marks Group B fi Mid Point xi yi2 fiyi fiyi2
    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 63
    150 -6 366


    Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
    Thus, group B has more variability in the marks.

    Q2 :From the data given below state which group is more variable, A or B?

    Answer :
    Firstly, the standard deviation of group A is calculated as follows.
    Here, h = 10, N = 150, A = 45
    The standard deviation of group B is calculated as follows.
    Since the mean of both the groups is same, the group with greater standard deviation will be more variable.
    Thus, group B has more variability in the marks.

    Q3 :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 :
    The prices of the shares X are
    35, 54, 52, 53, 56, 58, 52, 50, 51, 49
    Here, the number of observations, N = 10

    Xi
    35 -16 256
    54 3 9
    52 1 1
    53 2 4
    56 5 25
    58 7 49
    52 1 1
    50 -1 1
    51 0 0
    49 -2 4
    350


    The following table is obtained corresponding to shares X.
    The prices of share Y are
    108, 107, 105, 105, 106, 107, 104, 103, 104, 101

    yi    
    108 3  9
    107 2  4
    105 0  0
    105 0  0
    106 1  1
    107 2  4
    104 -1  1
    103 -2  4
    104 -1  1
    101 -4  16
     40

    The following table is obtained corresponding to shares Y.
    C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
    Thus, the prices of shares Y are more stable than the prices of shares X.

    Q4 :An analysis of monthly wages paid to workers in two firms A and B, belonging to the same industry, gives the following results:

    Firm A Firm B
    No. of wage earners 586 648
    Mean of monthly wages Rs 5253 Rs 5253
    Variance of the distribution of wages 100 121

     

    (i) Which firm A or B pays larger amount as monthly wages?
    (ii) Which firm, A or B, shows greater variability in individual wages?
    Answer :
    (i) Monthly wages of firm A = Rs 5253
    Number of wage earners in firm A = 586
    ∴Total amount paid = Rs 5253 x 586
    Monthly wages of firm B = Rs 5253
    Number of wage earners in firm B = 648
    ∴Total amount paid = Rs 5253 x 648
    Thus, firm B pays the larger amount as monthly wages as the number of wage earners in firm B are more than the number of wage earners in firm A.
    (ii) Variance of the distribution of wages in firm A = 100
    ∴ Standard deviation of the distribution of wages in firm
    A ((σ1) = 
    Variance of the distribution of wages in firm = 121
    ∴ Standard deviation of the distribution of wages in firm 
    The mean of monthly wages of both the firms is same i.e., 5253. Therefore, the firm with greater standard deviation will have more variability.
    Thus, firm B has greater variability in the individual wages.

    Q5 :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 :
    The mean and the standard deviation of goals scored by team A are calculated as follows.

    No. of goals scored No. of matches fixi xi2 fixi2
    0 1 0 0 0
    1 9 9 1 9
    2 7 14 4 28
    3 5 15 9 48
    4 3 12 16 45
    25 50 130

     


    Thus, the mean of both the teams is same.

    The standard deviation of team B is 1.25 goals.
    The average number of goals scored by both the teams is same i.e., 2. Therefore, the team with lower standard deviation will be more consistent.
    Thus, team A is more consistent than team B.

    Q6 :The sum and sum of squares corresponding to length x (in cm) and weight y
    (in gm) of 50 plant products are given below:

    Which is more varying, the length or weight?
    Answer :

    Here, N = 50
    ∴ Mean, 




    Mean, 

    Thus, C.V. of weights is greater than the C.V. of lengths. Therefore, weights vary more than the lengths.

    Exercise Miscellaneous : Solutions of Questions on Page Number : 380
    Q1 :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 :
    Let the remaining two observations be x and y.
    Therefore, the observations are 6, 7, 10, 12, 12, 13, x, y.

    From (1), we obtain
    x2 + y2 + 2xy = 144 — (3)
    From (2) and (3), we obtain
    2xy = 64 —-(4)
    Subtracting (4) from (2), we obtain
    x2 + y2 – 2xy = 80 – 64 = 16
    ⇒ x – y = ± 4 — (5)
    Therefore, from (1) and (5), we obtain
    x = 8 and y = 4, when x – y = 4
    x = 4 and y = 8, when x -y = -4
    Thus, the remaining observations are 4 and 8.

    Q2 :The mean and variance of 7 observations are 8 and 16, respectively. If five of the observations are 2, 4, 10, 12 and 14. Find the remaining two observations.
    Answer :
    Let the remaining two observations be x and y.
    The observations are 2, 4, 10, 12, 14, x, y.

    From (1), we obtain
    x2 + y2 + 2xy = 196 ……… (3)
    From (2) and (3), we obtain
    2xy = 196 – 100
    ⇒ 2xy = 96 ……… (4)
    Subtracting (4) from (2), we obtain
    x2 + y2 – 2xy = 100 – 96
    ⇒ (x – y)2 = 4
    ⇒ x – y = ± 2 ……… (5)
    Therefore, from (1) and (5), we obtain
    x = 8 and y = 6 when x – y = 2
    x = 6 and y = 8 when x- y = – 2
    Thus, the remaining observations are 6 and 8.

    Q3 :The mean and standard deviation of six observations are 8 and 4, respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.
    Answer :
    Let the observations be x1, x2, x3, x4, x5, and x6.
    It is given that mean is 8 and standard deviation is 4.

    If each observation is multiplied by 3 and the resulting observations are yi, then

    From (1) and (2), it can be observed that,

    Substituting the values of xi and in (2), we obtain

    Therefore, variance of new observations = 
    Hence, the standard deviation of new observations is 

    Q4 :Given that is the mean and σ2 is the variance of n observations x1, x2 … xn. Prove that the mean and variance of the observations ax1, ax2, ax3 …axn are and a2 σ2, respectively (a ≠ 0).
    Answer :
    The given n observations are x1, x2 … xn.
    Mean =
    Variance = σ2

    If each observation is multiplied by a and the new observations are yi, then

    Therefore, mean of the observations, ax1, ax2 … axn, is .
    Substituting the values of xiand in (1), we obtain

    Thus, the variance of the observations, ax1, ax2 … axn, is a2 σ2.

    Q5 :The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases:
    (i) If wrong item is omitted.
    (ii) If it is replaced by 12.
    Answer :
    (i)     Number of observations (n) = 20
    Incorrect mean = 10
    Incorrect standard deviation = 2

    That is, incorrect sum of observations = 200
    Correct sum of observations = 200 – 8 = 192
    ∴ Correct mean 

    (ii) When 8 is replaced by 12,
    Incorrect sum of observations = 200
    ∴ Correct sum of observations = 200 – 8 + 12 = 204

    Q6 :The mean and standard deviation of marks obtained by 50 students of a class in three subjects, Mathematics, Physics and Chemistry are given below:

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

    Which of the three subjects shows the highest variability in marks and which shows the lowest?
    Answer :
    Standard deviation of Mathematics = 12
    Standard deviation of Physics = 15
    Standard deviation of Chemistry = 20
    The coefficient of variation (C.V.) is given by 

    The subject with greater C.V. is more variable than others.
    Therefore, the highest variability in marks is in Chemistry and the lowest variability in marks is in Mathematics.

    Q7 :The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively. Later on it was found that three observations were incorrect, which were recorded as 21, 21 and 18. Find the mean and standard deviation if the incorrect observations are omitted.
    Answer :
    Number of observations (n) = 1
    Incorrect mean 
    Incorrect standard deviation 
    ∴ Incorrect sum of observations = 2000
    ⇒ Correct sum of observations = 2000 – 21 21 -18 = 2000 – 60 = 1940

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