## Representation of data Questions

1.  The height of 36 students in a class was recorded to the nearest centimeters as follows.

148  159  163  158  166  155  155  179  158  155  171  172  156  161  160  165  157  165  175  173  172  178  159  168  160  167  147  168  172  157  165  154  170  157  162  173

(a) Make a grouped table with 145.5 as lower class limit and class width of 5. (4mks)

2.  Below is a histogram, draw.

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Use the histogram above to complete the frequency table below:

 Length Frequency 11.5 ≤ x ≤13.513.5 ≤ x ≤15.515.5 ≤ x ≤ 17.517.5 ≤ x ≤23.5

3.  Wambui spent her salary as follows:

 Food 40% Transport 10% Education 20% Clothing 20% Rent 10%

Draw a pie chart to represent the above information

4.  The examination marks in a mathematics test for 60 students were as follows;-

 60 54 34 83 52 74 61 27 65 22 70 71 47 60 63 59 58 46 39 35 69 42 53 74 92 27 39 41 49 54 25 51 71 59 68 73 90 88 93 85 46 82 58 85 61 69 24 40 88 34 30 26 17 15 80 90 65 55 69 89 Class Tally Frequency Upper class limit 10-2930-3940-6970-7475-8990-99

From the table;

(a) State the modal class

(b) On the grid provided , draw a histogram to represent the above information

5.  The marks scored by 200 from 4 students of a school were recorded as in the table below.

 Marks 41 – 50 51 – 55 56 – 65 66 – 70 71 – 85 Frequency 21 62 55 50 12

1. On the graph paper provided, draw a histogram to represent this information.
2. On the same diagram, construct a frequency polygon.
3. Use your histogram to estimate the modal mark.

6.  The diagram below shows a histogram representing the marks obtained in a certain test:-

(a) If the frequency of the first class is 20, prepare a frequency distribution table for the data

(b) State the modal class

(c) Estimate:   (i) The mean mark   (ii) The median mark

7.  The data below shows the number of sessions different subjects are taught in a week.

Draw a pie chart to show the data:

 Subject Eng Maths Chemistry C.R.E No. of sessions 8 7 4 3

8.  The height of 50 athletes in Moi University team were shown below:

 Height (cm) 150-159 160-169 170-179 180-189 190-199 200-209 Frequency 2 9 12 16 7 4

i) State the modal class

ii) Calculate the median height of the athletes

9.  The table below shows the length of 40 mango tree leaves;

 Length (mm) Frequency Cumulative frequency 118-126127-135136-144145-153154-162163-171172-180 341012542 371729343840

(a) Determine the;

(i) Modal class

(ii) Median class

(b) Calculate; (i) the mean of the leaves   (ii) the median of the leaves

1.

(a)

 Class Tally mark Freq Freq D 145-149 // 2 2 150-154 / 1 1 155-159 //// //// / 11 11 169-164 //// 5 5 165-169 //// // 7 7 170-174 //// // 7 7 175-179 /// 3 3

B1

B1

B1

B1

B1

B1

B1

B1

B1

B1

Classes

Tally mark column

Freq. column

Freq density column

(can be implied)

Freq. density (y axis

Height (x axis)

Correct spacing as per scale

Histogram drawn (bars)

Joining mid point of the bars

Joining mid point of first class to 144.5

Joining mid point of last class to 179.5

10

1.

 Length11.5 ≤ x≤ 13.513.5 ≤ x≤ 15.515.5 ≤ x≤ 17.517.5 ≤ x≤ 23.5 Frequency6963

2.  Food: 40/100 x 360 = 144 °

Transport: 10/100 x 360 = 36°

Education: 20/100 x 360 = 72 °

Clothing: 20/100 x 360 = 72 °

Rent: 10/100 x 360 = 36°

3.  Class Tally Frequency Upper Limit

10 – 29 ∣∣∣∣
∣∣∣ 8 29.5 B2 for

30 – 39 ∣∣∣∣
∣ 6 39.5 all tally

40 – 69 ∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣ 28 69.5 B2 all

70 – 74 ∣∣∣∣
∣ 6 74.5 – frequency

75 – 89 ∣∣∣∣
∣∣∣ 8 89.5 – B1

90 – 99 ∣∣∣∣ 4 99.5 B1

Modal class 40 – 69 B1

4.  See the graph paper.

For correct class boundaries

For correct class intervals.

All frequency densities

Correct scale

All the bars drawn.

Top mid pts. Of bars indicated.

For the mid pts. Joint to make a polygon.

For correctly identifying the modal mark point.

For reading correctly the modal mark ≡ 53.5  0.1

5. (a)

 Marks Frequency 5-910-1920-3040-49 20504030

(b) Modal class is 10-19

(c)(i)

 Class x f fx Cf 5-9 7 20 140 20 10-19 14.5 50 725 70 20-39 29.5 40 1180 110 40-49 44.5 30 1335 140 F = 140 Fx =3380

x = fx = 3380 = 24.14

f 140

(ii) Median mark is at 70 + 71 = 70.5th position

Median = 119.5 + (0.5) x 20

40

= 19.5 + 0.25

= 19.75

6.  Total No. of sessions

= 8 + 7 + 4 + 3 = 22

Angle for:

English = 8/22 x 360 = 130.9°

Maths = 7/22 x 360 = 114.5°

Chemistry = 4/22 x 360 = 65.5°

CRE = 3/22 x 360 = 49.01°

7.  180 – 189

Class limits

 class limits f cf 149.5 159.5 2 2 159.5 169.5 9 11 169.5 179.5 12 23 179.5 189.5 16 39 189.5 199.5 7 46 199.5 209.5 4 50

Median = 50/2 = 25

179.5 + 25 – 23 x 10

16

= 179.5 + 20 = 180.75

16

179.5 + 26 – 23 x 10

16

179.5 + 30 = 181.38

16

180.75 + 181.38

2

= 181.06

8.  a)  i) 145 – 153

ii) Median class

(40 + 1/2)th value  median class = 145 – 153

This is the 20.5th value

The value also in the 145 – 153 class

b)

 Class x f fx 118- 126127- 135136 – 144145 – 153154 – 162163 – 171172 – 180 122131140 B1149158167176 3410 B212542 36652414001788790668352 Ef = 40 Efx = 5888

B2 for all values of fx correct and B1 for 4 values of fx and above orrect

Mean = Efx = 5888 = 147.2mm

Ef 40

Median 20th = 144.5 + (11/12 x 9) = 152.75

21st = 144.5 + (12/12 x 9) = 153.5

Median = 152.75 + 153.5 = 153.125

2

(Alternatively one could work out the 20.5 value directly using median formula)

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