## 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 30-3435-3940-4445-4950-5455-5960-6465-69 4 61014 ecolebooks.com X2414 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 140-144145-149150-154155-159160-164 31519112

(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 30-3435-3940-4445-4950-5455-5960-6465-69 3237424752576267 4 61014222414 6 128 222 420 65911441368 868 462 4 10 20 34 56 80 94100 B1  = 100 B1 x = 5210 B1

x = 5210

1. 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- 144145 – 149150 – 154155 – 159160 -164 31519112 318374850

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

1. Number who sat = 40
2. The modal class = 60 – 69
 Marks x f X – 64.5= d fd 30-3940-4950-5960-6970-7980-8990-99 34.544.554.564.574.584.594.5 231012832£f = 40 -30-20-100102030 -60-60-1000806060£ 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

 Class0 – 910 – 1920 – 2930 – 3940 – 4950 – 5960 – 6970 – 7980 – 8990 – 99 Centre X4.514.524.534.544.554.564.574.584.594.5 Fd-50-80-120-140-10002001209040εfd -40 D= x – A-50-40-30-20-10010203040

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.

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