# 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 31519112
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. 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.

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 30-3435-3940-4445-4950-5455-5960-6465-69 3237424752576267 4 610142224146 12822242065911441368868462 4102034568094100 B1 Σf=100 B1 Σfx=5210 B1

Σfx=521
1. Mean = 5210/100
= 52.10
2. Median = 49.5 + [50-34/22] x 5
= 53.14
1. Log10 52 – log10 23 + log 25
Log10 [25 x 32/8]
Log10 100 = log1010
= 2 log 1010
But log1010 = 1
= 2
2. 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
3.
 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
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
4. 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
5.
 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 74.584.5 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 (20th and 21st ) 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
6. 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
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
7. Cumulative frequency
3,11, 30, 44, 50
Median = L1t (n/2 – cfa)/Fn
= 8 + (25 – 11)/19 X 4
= 10.947
8.

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