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Business Mathematics And Statistics Question Papers And Answers

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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

    1. Using a ruler and a pair of compasses only construct triangle PQR in which PQ = 5cm, PR = 4cm and ∠ PQR = 30 o
    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. 7 th Decile
      4. 60 th 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, Σ x 2 = 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 30 th 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 d 2 = 9800 B 1
     Σfd = 9800B 1
     x =A + Σfd
                Σf
     = 67.0 + - 480
                    140
     = 67.0 – 3.43 = 63.57 ………… M1
     = 63.6 kg …………… A1
    Standard deviation = Σfd 2 - Σf d
                                    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


    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
      0
      -7.5
      -12.5      		 300
      450
      0
      -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
     60a 4 = 4860
     a 4 = 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 = Σf x = 3676
                   Σf      80
      = 45.95
    2. Q1 = 30.5 + 3 x 10
                         14
       = 62.64
       S.I.R = ½ (62.64 -32)
       = 15.32
    3. Standard deviation
      statistics ans 5c

    1. x = 90 – (2 +13 + 51 + 27 + 14 + 1)
       = 90 – 84 = 6
    2. 15 – 19

      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 Efd 2 = 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


    1. statistics ans 7
      1. RQ = 7.5 ± 0.1
      2. ∠PRQ 40° ± 1
    3. B1 circle through P, Q and R
    4. r = 4.1 cm
       A =π r 2
      22 /7 x 4.1 x 4.1 = 52.83cm2

    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

      1. 25 2 + 24 2 + 22 2 + 23 2 + x 2 + 262 + 21 2 + 23 2 + 22 2 + 27 2 = 5154
         5.625 +576 + 2(484) + 2(529) + 676 + 441 + 729 + x2 = 5154
         X 2 = 81
         X =9
      2. X = 222 = 22.2
                10
         Σ(X – x) 2 = 2.8 2 + 1.8 2 + 0.22 + 0.8 2
                 13.2 2 + 3.8 2 + 1.22 + 0.8 2 + 0.2 2 + 4.8 2
                 (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

      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
    2. 30 th student = 10 th from bottom
       30.5 + (10 – 8)10
                       7
       = 30.5 + 2.9 = 33.4 marks.

    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
    2. Q 1 = 39.5 + (50 – 30) x 10
                               72
       = 42.28
       Q 3 = 49.5 + (150-102) x 10
                               53
       = 58.56
       Q 3 – Q1 = 58.56 – 42.28
       = 16.28


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Business Mathematics And Statistics Question Papers And Answers

Source: https://www.easyelimu.com/high-school-notes/maths/form-4-topical/item/1174-statistics

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