Statistics Questions and Answers - Form 4 Topical Mathematics

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Questions

  1. The table below shows the masses to the nearest kg of a number of people.
    Mass (kg)
    Frequency
    50 – 54
    19
    55 – 59
    23
    60 – 64
    40
    65 – 69
    28
    70 – 74
    17
    75 – 79
    9
    80 – 84
    4
    1. Using an assumed mean of 67.0, calculate to one decimal place the mean mass.
    2. Calculate to one decimal place the standard deviation of the distribution.
  2. Use only a ruler and pair of compasses in this question;
    1. construct triangle ABC in which AB = 7cm, BC = 6cm and AC = 5cm
    2. On the same diagram construct the circumcircle of triangle ABC and measure its radius
    3. Construct the tangent to the circle at C and the internal bisector of angle BAC. If these lines meet at D, measure the length of AD
  3. Below is a histogram drawn by a student of Got Osimbo Girls Secondary School.
    1. Develop a frequency distribution table from the histogram above.
    2. Use the frequency distribution table above to calculate;
      1. The inter-quartile range.
      2. The sixth decile.
  4. ABC is a triangle drawn to scale. A point x moves inside the triangle such that
    1. AX 4 cm
    2. BX >CX
    3. Angle BCX Angle XCA.
      Show the locus of X.
  5. The following able shows the distribution of marks of 80 students
    Marks 1-10  11-20  21-30  31-40  41-50  51-60  61-70  71-80  81-90  91-100
    Frequency  1 6 10 20 15 5 14 5 3 1
    1. Calculate the mean mark
    2. Calculate the semi-interquartile range
    3. Workout the standard deviation for the distribution
  6. The table below shows the marks of 90 students in a mathematical test
    Marks 5-9  10-14  15-19  20-24  25-29  30-34  35-39
    No. of students  2 13 31 23 14 X 1
    1. Find X
    2. State the modal class
    3. Using a working mean of 22, calculate the;
      1. Mean mark
      2. Standard deviation
  7.  
    1. Using a ruler and a pair of compasses only construct triangle PQR in which PQ = 5cm, PR = 4cm and ∠PQR = 30o
    2. Measure
      1. RQ
      2. ∠PQR
    3. Construct a circle, centre O such that the circle passes through vertices P, Q, and R
    4. Calculate the area of the circle
  8. The ages of 100 people who attended a wedding were recorded in the distribution table below
    Age 0-19  20-39  40-59  60-79  80-99
    Frequency  7 21 38 27 7
    1. Draw the cumulative frequency
    2. From the curve determine:
      1. Median
      2. Inter quartile range
      3. 7th Decile
      4. 60th Percentile
  9. The marks obtained by 10 students in a maths test were:-
    25, 24, 22, 23, x, 26, 21, 23, 22 and 27
    The sum of the squares of the marks, Σx2 = 5154
    1. Calculate the
      1. value of x
      2. Standard deviation
    2. If each mark is increased by 3, write down the:-
      1. New mean
      2. New standard deviation
  10. 40 form four students sat for a mathematics test and their marks were distributed as follows:-
    Marks 1 – 10  11-20 21- 30  31 – 40  41 – 50  51 – 60  61 – 70  71 – 80  81 – 90  91 - 100
    No. of
    students 
    1 3 4 7 12 9 2 1 0 1
    1. Using 45.6 as the working mean, calculate;
      1. The actual mean.
      2. The standard deviation.
    2. When ranked from first to last, what mark was scored by the 30th student? (Give your answer correct to 3 s.f.)
  11. The table below shows the distribution of marks scored by pupils in a maths test at Nyabisawa Girls.
    Marks 11 – 20  21 – 30  31 – 40  41 – 50  51 – 60  61 – 70  71 – 80  81 – 90
    Frequency  2 5 6 10 14 11 9 3
    1. Using an Assumed mean 45.5, calculate the mean score.
    2. Calculate the median mark.
    3. Calculate the standard deviation.
    4. State the modal class.
  12. The table below shows the marks scored in a mathematics test by a form four class;
    Marks 20-29  30-39  40-49  50-59  60-69  70-79  80-89
    No. of students  4 26 72 53 25 9 11
    1. Using an assumed mean of 54.5, calculate:-
      1. The mean
      2. The standard deviation
    2. Calculate the inter quartile range

Answers

  1.  
    Mass
    kg
    Mid
    term
    x
    F d=xA  fd d2 fd2
    50 - 54
    55 - 59
    60 - 64
    65 - 69
    70 - 74
    75 - 79
    80 - 84
    52
    57
    62
    67
    72
    77
    82
    19
    23
    40
    28
    17
    9 4
    -15
    -10
    -5
    0 5
    10
    15
    -285
    -230
    -200
    0
    85
    90
    60
    225
    100
    25
    0
    25
    100
    225
    4275
    2300
    1000
    0
    425
    900
    900
       Σf =140     Σfs= -480   Σfd2 = 9800
    Marks awarded for table as follows:-
    Σf = 140 B1
    Column for d B1
    Column for fd B1
    Σfd = - 480 B1
    Column for d2 = 9800 B1
    Σfd = 9800B1
    x =A + Σfd
                Σf
    = 67.0 + - 480
                    140
    = 67.0 – 3.43 = 63.57 ………… M1
    = 63.6 kg …………… A1
    Standard deviation = Σfd2 - Σfd
                                    Σf       Σf
    statistics ans1
  2. = 8/150 + 6/150 + 9/300+ 3/300
    =40/300 = 2/15
    1. Construction of AB B1
      Construction of BC B1
      Construction of AC B1
    2. Construction of bisect of AC B1
      Construction of bisect BC B1
      Radius 3.6 cm B1
    3. Construction of bisect ∠ CAB B1 OC B1
      Construction of AD B1 AD = 12.8cm B1
  3.  
    1.  
      Class f x d = A – x  fd d2 fd2
      41 – 50
      51 – 55
      56 – 65
      66 – 70
      71 – 85
      20
      60
      60
      50
      15
      45.5
      53
      60.5
      68
      73
      15
      7.5

      -7.5
      -12.5
      300
      450

      -375
      187.5
      225
      56.25
      0
      56.25
      156.25
      4500
      3375
      0
      2812.50
      2343.75
               Σfd = 562.5     Σfd2 13031.25
    2.  
      statistics ans 3b
  4. 15 (ax)4 (-2/x2)2 = 4860
    60a4 = 4860
    a4= 81
    a = 3
  5.  
    Marks(x)  Freq.(f)  fx d=x-x  d2 Fd2
    5.5
    15.5
    25.5
    35.5
    45.5
    55.5
    65.6
    75.5
    85.5
    95.5
    1
    6
    10
    20
    15
    5
    14
    5
    3
    1
    5.5
    99
    255
    710
    682.5
    277.5
    917
    377.5
    256.5
    95.5
    -40.45
    -30.45
    -20.45
    -10.45
    -0.45
    9.55
    19.55
    29.55
    39.55
    49.55
    1636
    927.2
    418.2
    109.2
    0.2025
    91.20
    382.2
    873.2
    1564
    2455
    1636
    5563
    4182
    2184
    3038
    456
    535
    4366
    4692
    2455
       Σf = 80 Σfx = 3676     Σfx2= 33,923
    1. Mean = Σfx = 3676
                   Σf      80
      = 45.95
    2. Q1 = 30.5 + x 10
                        14
      = 62.64
      S.I.R = ½ (62.64 -32)
      = 15.32
    3. Standard deviation
      statistics ans 5c
  6.  
    1. x = 90 – (2 +13 + 51 + 27 + 14 + 1)
      = 90 – 84 = 6
    2. 15 – 19
    3.  
      1.  
        Class x f D= x-A  fd D2 Fd2
        5-9 7 2 -15 -30 225  450
        10-14  12  13  -10 -130  100  1300
        15-19  17  31  -5 -155  25 775
        20-24  22  23  0 0 0 0
        25-29  27  14  5 70 350  4900
        30-34  32  6 10 60 600  3600
        35-39  37  1 15 15 225  225
        Ef = 90 Efd = 170 Efd2 = 11250
        Mean = E + d + A
                       Ef
        = -170 + 22
             90
        = 22 – 1.888 = 20.11
      2. S.d = Efd - [Efd]2
                    Ef       Ef
        = 122 – (-1.888)2
        = 125 – 3.566 = 121.4
        = 11.02
  7.  
    1.  
      statistics ans 7
    2.  
      1. RQ = 7.5 ±0.1
      2. PRQ 40° ±1
    3. B1 circle through P, Q and R
    4. r = 4.1 cm
      A =πr2
      22/7 x 4.1 x 4.1 = 52.83cm2
  8.  
    1.  
      Class limits f cf
      -0.5 – 19.5 7 7
      19.5- 39.5 21 28
      39.5 – 59.5 38 66
      59.5 – 79.5 27 93
      79.5 -= 99.5 7 100
    2.  from the curve
      1. median = 52. M1 A1
      2. Inter quartile range = 66-38 = 28.
      3. 7th 7/10 = 62.46marks
      4. 60th percentile – 56.34
  9.  
      1. 252 + 242 + 222+ 232 + x2 + 262 + 212 + 232 + 222 + 272 = 5154
        5.625 +576 + 2(484) + 2(529) + 676 + 441 + 729 + x2 = 5154
        X2 = 81
        X =9
      2. X = 222 = 22.2
               10
        Σ(X – x)2 = 2.82 + 1.82 + 0.22 + 0.82
        13.22 + 3.82 + 1.22 + 0.82 + 0.22 + 4.82
        (x-x)2= 7.84 + 3.24 2(0.04) + 2(0.64) +174.24 + 14.44 + 1.44 + 23.04
        = 225.6
              10
        s.d 22.56
        = 4.75
      1. New mean = 22.2 + 3
        = 25.2
      2. s.d = 4.75
  10.  
      1. x = A + ∑fd
                     ∑f
        = 45.6 + (-74)
                       40
        = 43.75


        Class Mis-pt x d = (x – A) Frequency f fd Fd2
        1 – 10
        11 – 20
        21 – 30
        31 – 40
        41 – 50
        51 – 60
        61 – 70
        71 – 80
        81 – 90
        91 – 100
        5.5
        15.5
        25.5
        35.5
        45.5
        55.5
        65.5
        75.5
        85.5
        95.5
        -40.1
        -30.1
        -20.1
        -10.1
        -0.1
        9.9
        19.9
        29.9
        39.9
        49.9
        1
        3
        4
        7
        12
        9
        2
        1
        0
        1
        -40.1
        -90.3
        -80.4
        -70.7
        -1.2
        89.1
        39.8
        29.9
        0
        49.9
        1608.01
        8154.05
        6464.16
        4998.49
        1.44
        7938.81
        1584.04
        894.01
        0
        2410.01
      2. Standard Deviation
        statistics ans 10aii
    1. 30th student = 10th from bottom
      30.5 + (10 – 8)10
                      7
      = 30.5 + 2.9 = 33.4 marks.
  11.  
    1. Mean 45. 5 + 530
                           60
      = 54.33
    2. Median = 50.5 + (30.5 – 23)10
                                     14
      = 55.86
    3. Standard deviation =

      statistics ans11c
    4. Modal class 51 – 60
  12.  
     x  f  d d2 fd fd2
    24.5 4 -30 900 -120 3600
    34.5  26 -20  400  -520  10400
    44.5  72 -10  100  -720  7200
    54.5  53 0 0 0 0
    64.5  25 10 100  250 2500
    74.5  9 20 400  180 3600
    84.5  11 30 900  330 9900
      200     -600 37200
      1. Mean = A + ∑fd
                           ∑f
        = 54.5 – 600
                      200
        = 51.5
      2. Standard deviation
        statistics ans12aii
    1. Q1= 39.5 + (50 – 30) x 10
                              72
      = 42.28
      Q3 = 49.5 + (150-102) x 10
                              53
      = 58.56
      Q3 – Q1 = 58.56 – 42.28
      = 16.28

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