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Measures of Central Tendency Test - 28

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Measures of Central Tendency Test - 28
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  • Question 1
    1 / -0
    Below is given frequency distribution of marks (out of 100) obtained by the students.
    Calculate mean marks scored by a student.
    Marks0 - 1010 - 2020 - 3030 - 4040 - 50 50 - 6060 - 7070 - 8080 - 9090 - 100
    No. of Students35710121512628
    Solution
     cicifif_i xix_i fixif_ix_i 
     0-10 3 5 15
     10-20 5 15 75
     20-30 7 25 175
     30-40 10 35 350
     40-50 12 45 540
     50-60 15 55 825
     60-70 12 65 780
     70-80 6 75 450
     80-90 2 85 170
     90-100 8 95 760
    Consider the following table, to calculate mean:
    N=Σfi=80N=\Sigma f_i=80          
     Σfixi=4140\Sigma f_ix_i=4140

    Mean x=ΣfixiN\overline x=\dfrac {\Sigma f_ix_i}{N}
    x=414080=51.75\therefore \overline x=\dfrac{4140}{80}=51.75
    Mean marks scored by a student is 51.7551.75 
    Hence, option BB is correct.
  • Question 2
    1 / -0
    Find the mode of the following:
    18,14,22,23,14,18,17,28,28,14,25,1418,14,22,23,14,18,17,28,28,14,25,14
    Solution
    The numbers are: 18,14,22,23,14,18,17,28,28,14,25,1418,14,22,23,14,18,17,28,28,14,25,14
    Arranging in ascending order: 14,14,14,14,17,18,18,  22,23,25,28,2814, 14, 14, 14, 17, 18, 18,  22, 23,25, 28, 28
    1414 occurs 44 times and maximum number of times. 
    Hence, the mode of the above data is 1414
  • Question 3
    1 / -0
    If 2525 is the arithmetic mean of X X and 46,46, then find X.X.
    Solution
    Since 2525 is the arithmetic mean of XX and 4646

    25=(X+46)2\therefore 25=\displaystyle \frac{(X+46)}{2}
     X+46=50\therefore  X+46=50
     X=4\therefore  X=4
  • Question 4
    1 / -0
    Following table gives age groupwise distribution of people suffering from 'Asthama' due to air pollution in certain city. Find mean age of person suffering from 'Asthama' .
    Age (in years)7 - 1111 - 1515 - 1919 - 2323 - 2727 - 3131 - 3535 - 39
    No. of People 5 9 13 21 16 15 12  9
    Solution
    Consider the following table, to calculate mean:
     cicifi f_i xix_i fixif_ix_i
     7-11 5 9 45
     11-15 9 13 117
     15-19 13 17 221
     19-23 21 21 441
     23-27 16 25 400
     27-31 15 29 435
     31-35 1233  396
     35-39 9 37 333
    N=Σfi=100N=\Sigma f_i=100          
     Σfixi=2388\Sigma f_ix_i=2388

    Mean x=ΣfixiN\overline x=\dfrac {\Sigma f_ix_i}{N}
    x=2388100=23.88\therefore \overline x=\dfrac{2388}{100}=23.88
    Mean age of people suffering from Asthama is 23.8823.88years
    Hence, option AA is correct.
  • Question 5
    1 / -0
    Following table gives frequency distribution of milk (in litres) given per week by 50 cows.
    Find average (mean) amount of milk given by a cow by 'shift of origin method'.
    Milk (in litres)24 - 3030 - 3636 - 4242 - 4848 - 5454 - 6060 - 6666 - 72 
    72 - 7878 - 8484 - 90
    No. of cows13855584623
    Solution
    Consider the following table, to calculate mean by "shift of origin method":
    xix_i=mid value of class interval
    Assumed mean a=57a=57

     cici fif_i xix_idi=xiad_i=x_i-a fidif_id_i
    24-30 127 27-57= -30 -30
    30-36 333 33-57= -24 -72
    36-42 839 39-57= -18 -144
    42-48  545  45-57= -12 -60
    48-54  551 51-57= -6 -30
    54-60  557 57-57=0 0
    60-66  863  63-57=6  48
    66-72  469 69-57=12 48
    72-78  675  75-57=18 108
    78-84 281  81-57=24 48
    84-9087  87-57=30 90
     N=Σfi=50N=\Sigma f_i=50          
     Σfidi=6\Sigma f_id_i=6

    Mean x=a+ΣfidiN\overline x=a +\dfrac {\Sigma f_id_i}{N}
    x=57+650=57.12\therefore \overline x=57 + \dfrac{6}{50}=57.12
    Average amount of milk given by cow is 57.12 57.12 litres
    Hence, option CC is correct.
  • Question 6
    1 / -0
    The following observation are arranged in ascending order. If the median of the data is 1717, find xx if the observations are 6,8,9,15,x,x+2,21,22,25,29.6, 8, 9, 15, x,x + 2, 21, 22, 25, 29.
    Solution
    The observations are: 6,8,9,15,x,x+2,21,22,25,296, 8, 9, 15, x,x+2,21,22,25,29
    The number of observations are 1010. Since, the observations are even, the median will be the mean of n2\cfrac{n}{2} and n2+1\cfrac{n}{2}+1 observations
    median == mean of 102\cfrac{10}{2} and 10+22\cfrac{10+2}{2} observations
    median == mean of 5th5^{th} and 6th6^{th} observations
    median = x+x+22\cfrac{x + x+2}{2}
    17=x+117 = x + 1
    x=16x = 16
  • Question 7
    1 / -0
    Below is given frequency distribution of dividend in percentage declared by 120 companies.
    Dividend (in %)10 - 1920 - 2930 - 3940 - 4950 - 59 60 -
    69
    70 - 79 
    No. of Companies515284215123
    Obtain mean dividend declared by a company by step deviation method.
    Solution
    Consider the following table, to calculate mean by "step deviation method":
    xix_i=mid value of class interval
    Assumed mean a=44.5a=44.5
    class interval c=10c=10
     cici fif_i xix_i ui=xiacu_i=\dfrac{x_i-a}{c}  fiuif_iu_i
     9.519.59.5-19.555 14.514.514.544.510=3\dfrac{14.5-44.5}{10}=-3 15-15
     19.529.519.5-29.51515 24.524.5 2-2 30-30
     29.539.529.5-39.52828 34.534.5 1-1 28-28
     39.549.539.5-49.54242 44.544.5 00 00
     49.559.549.5-59.51515 54.554.5 11 1515
     59.569.559.5-69.51212 64.564.5 22 2424
     69.579.569.5-79.533 74.574.5 33 99
     N=Σfi=120N=\Sigma f_i=120          
     Σfiui=25\Sigma f_iu_i=-25

    Mean x=a+ΣfiuiN×c\overline x=a +\dfrac {\Sigma f_iu_i}{N}\times c

    x=44.5+25120×10=42.416666742.42\therefore \overline x=44.5 + \dfrac{-25}{120} \times 10=42.4166667 \Rightarrow 42.42
    Hence, option BB is correct.
  • Question 8
    1 / -0
    Find the median of:
    25,16,26,16,32,31,19,2825,16,26,16,32,31,19,28 and 3535
    Solution

    Step -1: Arrange the given data in ascending order.\textbf{Step -1: Arrange the given data in ascending order.}

                     Arranging data in ascending order:\text{Arranging data in ascending order:}

                        16,16,19,25,26,28,31,32,3516,16,19,25,26,28,31,32,35

    Step -2: Find median of the given data.\textbf{Step -2: Find median of the given data.}

                     Here, number of observations, n=9(odd number)\text{Here, number of observations, }n=9(\text{odd number})

                     Median=(n+12)thobservation.\mathbf{\because\text{Median}=\left(\dfrac{n+1}{2}\right)^\text{th}observation.}

                     Median=(9+12)thobservation\therefore \text{Median}=\left(\dfrac{9+1}{2}\right)^\text{th}\text{observation}

                                        =(102)thobservation=\bigg(\dfrac{10}{2}\bigg)^\text{th}\text{observation}

                                        =5th observation=5^\text{th}\text{ observation}

                                        =26=26

    Hence, the correct option is C.\textbf{Hence, the correct option is C.}

  • Question 9
    1 / -0
    Find the median of the following data:
    18,19,20,23,22,20,17,19,2518,19, 20, 23, 22, 20, 17, 19, 25 and 2121
    Solution
    The given data series is: 18,19,20,23,22,20,17,19,25.2118,19, 20, 23, 22, 20, 17, 19, 25. 21
    Arranging it in increasing order: 17,18,19,19,20,20,21,22,23,2517, 18, 19, 19, 20, 20, 21, 22, 23, 25
    The number of terms in the series are 1010, which is even. Hence, the median will be the mean of the two middle numbers.
    Thus, median == mean of 5th5^{th} term and 6th6^{th} term
    Median =20+202= \dfrac{20 + 20}{2}
    Median =20= 20
  • Question 10
    1 / -0
    What is the mean of first 8 whole numbers?
    Solution
    First 88 whole numbers are 0,1,2,3,4,5,6,7 0,1,2,3,4,5,6,7

    We know that, Mean=Sum of all observationsTotal number of observations\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Total number of observations}}

     Mean=0+1+2+3+4+5+6+78\therefore \ \text{Mean}=\dfrac {0+1+2+3+4+5+6+7}{8}
    288=3.5\Rightarrow \dfrac {28}{8}=3.5
    Hence, the mean of first 88 whole numbers is 3.53.5.
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