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

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. 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)	0 ≤ x < 4	4 ≤ x < 8	8 ≤ x < 12	12 ≤ x < 16	16 ≤ x < 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. 1. 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
   2. Mean = ⁵²¹⁰/₁₀₀
      = 52.10
   3. Median = 49.5 + [^50-34/₂₂] x 5
      = 53.14
2. Log₁₀ 5² – log₁₀ 2³ + log 2⁵
   Log₁₀ [^{25 x 32}/₈]
   Log₁₀ 100 = log1₀10
   = 2 log ₁₀10
   But log₁₀10 = 1
   ∴ = 2
3. 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
4.  
   

   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
5. 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
6. ∑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
7.  
   

   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 (20^th and 21^st ) 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
8. 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
9. 1. 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
10. Cumulative frequency
    3,11, 30, 44, 50
    Median = L1t (ⁿ/₂ – cfa)/_Fn
    = 8 + ^{(25 – 11)}/₁₉ X 4
    = 10.947
11.