Statistics Questions and Answers - Form 4 Topical Mathematics

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 = 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. 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
   
   
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 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.  
      
      
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 + 3  x 10
                        14
      = 62.64
      S.I.R = ½ (62.64 -32)
      = 15.32
   3. Standard deviation
      
      
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.  
      
      
   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 =πr^{2
      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. 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
   2. 1. New mean = 22.2 + 3
         = 25.2
      2. s.d = 4.75
10.  
    
    1. 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
          
          
    2. 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 =
       
       
    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. 1. Mean = A + ∑fd
                             ∑f
          = 54.5 – 600
                        200
          = 51.5
       2. Standard deviation
          
          
    2. 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