The mean of the following frequency table is 53. But the frequen
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    The following table represents the expenditure on the usage of water by 70 families in a locality. Find the average expenditure per family.

    Exp. water (in Rs.)

    15-20

    20-25

    25-30

    30-35

    35-40

    40-45

    45-50

    50-55

    No. of families

    7

    8

    7

    8

    10

    15

    7

    8

     
    Here,  we have                             A (assumed

    Here,  we have
                                 A (assumed mean) = 37.5,   sum from blank to blank of f subscript i equals 70 comma space space space sum from blank to blank of f subscript i u subscript i equals negative 21 space a n d space h space equals space 5

    Now,                     top enclose straight x equals A plus fraction numerator begin display style sum from blank to blank of end style f subscript i u subscript i over denominator begin display style sum from blank to blank of end style f subscript i end fraction cross times h equals 37.5 plus open parentheses fraction numerator negative 21 over denominator 70 end fraction close parentheses cross times 5 equals 37.5 minus 21 over 14 equals 27.5 minus 1.5 equals 3.6

    Hence, the average expenditure per family is 36.
    101 Views
    Answer
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    The mean of the following frequency table is 53. But the frequencies f1 and f2 in the classes 20-40 and 60-80 arc missing. Find the missing frequencies.   

    Age (in years)

    0-20

    20-40

    40-60

    60-80

    80-100

    Total

    Number of people

    15

    f1

    21

    f2

    17

    100

    Age (in yrs) class interval

    Mid values xi

    Frequency fi

    fi.xi

    0-20

    10

    15

    150

    20-40

    30

    f1

    30f1

    40-60

    50

    21

    1050

    60-80

    70

    f2

    70f2

    80-100

    90

    17

    1530
       

    Σfi = f1 + f2 + 53

    Σfixi = 2730 + 30f1 + 70f2

    It is given that :
                                                      

                            15 + fi + 21 + f2 + 17 = 100
                                          f1 + f2 + 53 = 100
                                                 f1 + f2 = 47

    Now,                                 

                                 53 = 

                                 53 x 100 = 2730 + 30f1 + 70f2 
                                 30f1 + 70f2  = 5300 - 2730
                                 30f1 + 70f2 = 2570
                                 3f1 + 7f2 = 257
    Thus, we have             f1 + f2 = 47
                                     3f1 + 7f2 = 257

    Multiplying equation (i) by 3, we get 

                                         3f1 + 7f2 = 141 
     and                               3f1 + 7f2 = 257 

    Subtracting (iv) from (iii), we get
                                  
                                     (3f1 + 3f2) - (3f1 + 7f2) = 141 - 257
                                 3f1 + 3f2 - 3f1-7f2 = -116
                                 -4f2 = -116
                                    f2 = 29

    Putting the value of f2 in (i), we get,

                              f1 + f2 = 47 
                         f1 + 29 = 47
                        f1= 1= 47 - 29
                        f1 = 18
    Henc, f1 = 18 and f2= 29                     
     
                    
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    The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequency f1 and f2.

    C.I.

    0-20

    20-40

    40-60

    60-80

    80-100

    100-120

    Total

    f

    5

    f1

    10

    f2

    7

    8

    50
    Here, we have                                    

    Here, we have               sum from blank to blank of straight f subscript straight i straight u subscript straight i space equals space 28 space plus space straight f subscript 2 minus straight f subscript 1
                                     
                                       top enclose straight x space equals space 62.8 space a n d space h space equals space 20 space a n d space sum from blank to blank of f subscript i equals 50

    It is given that,             sum from blank to blank of f subscript i equals 50

    rightwards double arrow                              30 + f1 + f2   = 50
    rightwards double arrow                              f1 + f2    = 20

    Now,                           top enclose straight x space equals space A plus fraction numerator begin display style sum from blank to blank of f subscript i u subscript i end style over denominator begin display style sum from blank to blank of f subscript i end style end fraction cross times h

    rightwards double arrow                           62.8 space equals space 50 space plus space fraction numerator 28 minus straight f subscript 1 plus straight f subscript 2 over denominator 50 end fraction cross times 20

    rightwards double arrow                           62.8 space equals fraction numerator 1 left parenthesis 28 minus f subscript 1 plus f subscript 2 right parenthesis over denominator 5 end fraction plus 50 

    rightwards double arrow                           62.8 space equals fraction numerator 56 minus 2 straight f subscript straight i plus 2 straight f subscript 2 plus 250 over denominator 5 end fraction
    rightwards double arrow                            314 = -2f1 + 2f2 + 306 rightwards double arrow  2f1- 2f2 = -8
    Adding (i) and (ii), we get            f1  - f2  = -4
    Putting this value in (i), we get    2f1 = 16 rightwards double arrow f1  = 8
    Hence,                                       f1 = 8,   f2 = 12


     
    2780 Views
    Answer

    Find the value of P, if the mean of the following distribution is 18.

    x:

    13

    15

    17

    19

    20 + p

    23

    f:

    8

    2

    3

    4

    5p

    6 

    xi

    fi

    fixi

    13

    8

    104

    15

    2

    30

    17

    3

    51

    19

    4

    76

    20 + p

    5p

    5p (20 + p)

    23

    6

    138
     

    Σfi = 23 + 5p

    Σfixi = 399 + 5p2 + 100p

     

    Here, we have                    top enclose straight x equals 18 comma space space sum from blank to blank of straight f space equals space 23 space plus space 5 straight p space and space sum from blank to blank of straight f subscript straight i straight x subscript straight i space equals 399 plus 5 straight p squared plus 100 straight p

    Now,                                 top enclose straight x space equals space fraction numerator begin display style sum from blank to blank of end style f subscript i x subscript i over denominator begin display style sum from blank to blank of end style f subscript i end fraction equals fraction numerator 5 p squared plus 100 p plus 399 over denominator 5 p plus 23 end fraction

    rightwards double arrow                                     18 equals fraction numerator 5 straight p squared plus 10 straight p plus 399 over denominator 5 straight p plus 23 end fraction
    rightwards double arrow                      18(5p + 23) = 5p2 + 100p + 399
    rightwards double arrow                      90p + 414 = 5p2 + 100p + 399
    rightwards double arrow                      5p2 + 10p - 15 = 0
    rightwards double arrow                      p2 + 2p - 3 = 0
    rightwards double arrow                      p2 + 3p - p - 3 = 0
    rightwards double arrow                     p(p + 3) - 1 (p + 3) = 0
    rightwards double arrow                     (p - 1) (p + 3) = 0
    rightwards double arrow                      p - 1 = 0 or p + 3 = 0
    Since                   p =  1 or  p = - 3 neglected
    Therefore,            p = 1 


    1602 Views
    Answer

    The mean of the following frequency distribution is 132 and the sum of observations is 50. Find the missing frequency f1 and f2.

    Class

    0-40

    40-80

    80-120

    120-160

    160-200

    200-240

    Freq.

    4

    7

    f1

    12

    f2

    9
    Here, we have                      A (assumed mean) = 100

    Here, we have                     sum from blank to blank of f subscript i space equals space 50 comma space sum from blank to blank of f subscript i u subscript i space equals space 2 f subscript 2 space plus space 24 comma space A (assumed mean) = 100
                                              
                                              top enclose straight x space equals space 132 space a n d space h space equals space 40

    It is given that                      sum from blank to blank of f subscript i space equals space 50
    rightwards double arrow                           f1 + f2 + 32 = 50
    rightwards double arrow                              f1 + f3  = 18

    Now,                       top enclose straight x space equals space A plus fraction numerator begin display style sum from blank to blank of end style f subscript i u subscript i over denominator begin display style sum from blank to blank of end style f subscript i end fraction cross times h

    rightwards double arrow                       132 = 100 + fraction numerator 24 plus 2 straight f subscript 2 over denominator 50 end fraction cross times 40

    rightwards double arrow                        132 = 100 + fraction numerator 4 left parenthesis 2 straight f subscript 2 plus 24 right parenthesis over denominator 5 end fraction

    rightwards double arrow                       132 = fraction numerator 500 plus 8 straight f subscript 2 plus 96 over denominator 5 end fraction
    rightwards double arrow                    132 x 5 = 8f2 + 596
    rightwards double arrow                     660 = 8f2 + 596
    rightwards double arrow                      8f2  = 64
    rightwards double arrow                       f2 = 8
    Putting this value in (i), we get

                         straight f subscript 1 plus 8 space equals space 18
    rightwards double arrow                f1  = 10
    Hence.           f1 = 10,  f2 = 8


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