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Measures of central tendency 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 |
(a) Determine (i) the value of X
(ii) The modal class
(b) Calculate the mean
(c) 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 |
(a) State the modal class
(b) 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;
(a) the value of k
(b) the second root
Marks | 60 – 62 | 63 – 68 | 69 – 73 | 74 – 80 |
Frequency | 10 | 20 | 40 | 15 |
7. 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 |
a) state the total number of candidates who sat the test
b) state the modal class
c) calculate the mean mark using an assumed mean of 64.5 marks
d) calculate the median mark
8. Find these statistics of the following data 4, 2, 2, 6, 1, 3, 4, 1, 4
a) Mode
b) Median
c) Mean
9. (a) The marks scored by a group of form two students in a mathematical test were as recorded
in the table below:-
Marks | 0-9 | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 | 80-89 | 90-99 |
Frequency | 1 | 2 | 4 | 7 | 10 | 16 | 20 | 6 | 3 | 1 |
(a) (i) State the modal class
(ii) Determine the class in which the median mark lies
(iii) Using an assumed mean of 54.5, calculate the mean mark
10. 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
11. 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.
Measures of central tendency Answers
1. 4 + 6 + 10 + 14 + x + 24 + 14 + 6 = 100
78 + x = 100
(i) x = 22
(ii) Modal class = 55 -59
Marks | x | | x | c |
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 | = 100 B1 | x = 5210 | B1 | |
x = 5210
- Mean = 5210
100
= 52.10
(ii) Median = 49.5 + 50-34 x 5
22
= 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) x 5
19
= 149.5 + 7 x 5
19
= 149.5 + 1.842
= 15.34
4.
H | 20-24 | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 |
F | 3 | 19 | 25 | 20 | 18 | 15 |
CF | 3 | 22 | 47 | 67 | 85 | 100 |
Md = 34.5 + (50 – 47) x 4
20
= 34.5 + 12/20 = 35.1
5. a) 2x2 + 6x – 2x = 0
32 – 24 – 2x = 0
-2x = -8
x = 4
b) 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 |
- Number who sat = 40
- The modal class = 60 – 69
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
d) The median mark
= ½ (20th and 21st ) marks
= ½ (59.5 + 5 x 10 + 59.5 + 6 x 10)
12 12
= ½ (59.5 + 4.16666 + 59.5 + 5)
= ½ (128.16666667) = 64.083
8. 1, 1, 2, 2, 3, 4, 4, 6
a) Mode = 4
b) Median = 3
c) Mean = 1 x2 + 2 x 2 + 3 x 1 + 4 x 3 + 6 x 1
9
= 3
9. a) i) Modal class = 60 – 69
ii) 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) X 4
19
= 10.947
11.
