Measures of Central Tendency Questions and Answers - Form 2 Topical Mathematics

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Questions

  1. The results of a mathematics test that a hundred students took are as shown below:-
    Marks No. of students
     30-34
     35-39
     40-44
     45-49
     50-54
     55-59
     60-64
     65-69
     4
     6
     10
     14
     X
     24
     14
     6
    1. Determine
      1. the value of X
      2. The modal class
    2. Calculate the mean
    3. The median
  2. Without using logarithms or calculator evaluate:
    2log105 – 3log102 + log1032
  3. The table below shows heights of 50 students :-
    Height (cm) Frequency
     140-144
     145-149
     150-154
     155-159
     160-164
    3
    15
    19
    11
    2
    1. State the modal class
    2. Calculate the median height
  4. In an experiment, the height of 100 seedlings were measured to the nearest centimeter and the results were recorded as shown below;
     Height(cm)  20-24  25-29  30-34  35-39  40-44  45-49
     Frequency  3  19  25  20  18  15
    Calculate the median height
  5. Given that x = -4 is a root of the equation 2x2 + 6x – 2k = 0; Find;
    1. the value of k
    2. the second root
  6. The table below shows the distribution of marks obtained by some candidates in a mathematics test
     Marks  30-39  40-49  50-59  60-69  70-79  80-89  90-99
     No. of candidates  2  3  10  12  8  3  2
     C.f              
    1. state the total number of candidates who sat the test
    2. state the modal class
    3. calculate the mean mark using an assumed mean of 64.5 marks
    4. calculate the median mark
  7. Find these statistics of the following data 4, 2, 2, 6, 1, 3, 4, 1, 4
    1. Mode
    2. Median
    3. Mean
  8. The marks scored by a group of form two students in a mathematical test were as recorded in the table below:-
      1. State the modal class
      2. Determine the class in which the median mark lies
      3. Using an assumed mean of 54.5, calculate the mean mark
  9. Six weeks after planting, the height of maize plants were measured correct to the nearest centimeter. The frequency distribution is given in the table below:
     Height (x)  ≤ < 4  ≤ < 8  ≤ < 12  12 ≤ < 16  16 ≤ < 20
     Frequency  3  8  19  14  6
    Estimate the median height of the plants
  10. Below are marks scored by student in maths talk in science congress.
     Marks  1 - 5  6 – 15  16 – 20  21 – 35  36 – 40  41 – 50
     No. of students  1  3  6  12  5  3
    Draw a histogram from the table above.

Answers

      1. 4 + 6 + 10 + 14 + x + 24 + 14 + 6 = 100
        78 + x = 100
        x = 22
      2. Modal class = 55 -59
         Marks  x  f  fx  cf
         30-34
        35-39
        40-44
        45-49
        50-54
        55-59
        60-64
        65-69
         32
        37
        42
        47
        52
        57
        62
        67
         4 
        6
        10
        14
        22
        24
        14
        6
         128
        222
        420
        659
        1144
        1368
        868
        462
         4
        10
        20
        34
        56
        80
        94
        100
                 B1    Σf=100
         B1
         Σfx=5210  B1

        Σfx=521
    1. Mean = 5210/100
      = 52.10
    2. Median = 49.5 + [50-34/22] x 5
      = 53.14
  1. Log10 52 – log10 23 + log 25
    Log10 [25 x 32/8]
    Log10 100 = log1010
    = 2 log 1010
    But log1010 = 1
    = 2
  2. Modal class 150-154
     Height  Frequency  c.f 
     140- 144
     145 – 149
     150 – 154
     155 – 159
     160 - 164
     3
     15
     19
     11
     2
     3
     18
     37
     48
     50
    Height Frequency c.f
    = 149.5 + (25-18)/19 x 5
    = 149.5 + 7/19 x 5
    = 149.5 + 1.842
    = 15.34
  3.  
     H  20-24  25-29  30-34  35-39  40-44  45-49
     C  3  19  25  20  18  15
     CF  3  22  47  67  85  100 
    Md = 34.5 + (50 – 47)/20 x 4
    = 34.5 + 12/20 = 35.1
    1. 2x2 + 6x – 2x = 0
      32 – 24 – 2x = 0
      -2x = -8
      x = 4
    2. 2x2 + 6x – 8 = 0
      x2 + 3x – 4= 0
      x2+ 4x – x - 4 = 0
      x(x – 4) – (x + 4) = 0
      (x – 1) (x + 4) = 0
      ∴ the other root is 1
  4. xf = 61 x 10 + 65.5 x 20 + 71 x 40 + 77 x 15
    = 610 + 1310 + 2840 + 1155
    = 5915
    xf = 5915
    f       85
    X Mean = 69.59
  5.  
     Marks  30-39  40-49  50-59  60-69  70-79  80-89  90-99
     No. of candidates  2  3  10  12  8  3  2
     C.F.  2  5  15  27 35   38  40
    1. Number who sat = 40








    2. The modal class = 60 – 69
    3.  Marks  x  f  X – 64.5= d  fd
       30-39
       40-49
       50-59
       60-69
       70-79
       80-89
       90-99
        34.5
       44.5
       54.5
       64.5
       74.5
      84.5
       94.5
       2 
       3
       10
       12
       8 
       3 
       2
       £f = 40
       -30
       -20
       -10
       0
       10
       20
       30
       -60
      -60
      -100
       0
       80
       60
       60
       £ fd = -20
      Mean = 64.5 + -20/40
      = 64.0
    4. The median mark
      = ½ (20th and 21st ) marks
      = ½ (59.5 + 5/12 x 10 + 59.5 + 6/12 x 10)
      = ½ (59.5 + 4.16666 + 59.5 + 5)
      = ½ (128.16666667) = 64.083
  6. 1, 1, 2, 2, 3, 4, 4, 6
    1. Mode = 4
    2. Median = 3
    3. Mean = 1 x2 + 2 x 2 + 3 x 1 + 4 x 3 + 6 x 1/9
      = 3
      1. Modal class = 60 – 69
      2. class where medium lies 
        median class 50- 59
         Class
         0 - 9
         10 – 19
         20 – 29
         30 – 39
         40 – 49
         50 – 59
         60 – 69
         70 – 79
         80 – 89
         90 – 99

        Centre X
         4.5
         14.5
         24.5
         34.5
         44.5
         54.5
         64.5
         74.5
         84.5
         94.5 

        Fd
         -50
         -80
         -120
         -140
         -100
         0
         200
         120
         90
         40
         εfd - 40 
         D= x – A
         -50
         -40
         -30
         -20
         -10
         0
         10
         20
         30
         40

        Mean = 54.5 – 40/70
        = 53.93
  7. Cumulative frequency
    3,11, 30, 44, 50
    Median = L1t (n/2 – cfa)/Fn
    = 8 + (25 – 11)/19 X 4
    = 10.947
  8.  
    measures of central tendency ans 11
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