# Representation of Data Questions and Answers - Form 2 Topical Mathematics

## Questions

1. Below is a histogram, draw.

Use the histogram above to complete the frequency table below:
2. Wambui spent her salary as follows:

Draw a pie chart to represent the above information
3. The examination marks in a mathematics test for 60 students were as follows;-

From the table;
1. State the modal class
2. On the grid provided, draw a histogram to represent the above information
4. The marks scored by 200 from 4 students of a school were recorded as in the table below.

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

1. If the frequency of the first class is 20, prepare a frequency distribution table for the data
2. State the modal class
3. Estimate:
1. The mean mark
2. The median mark
6. The data below shows the number of sessions different subjects are taught in a week. Draw a pie chart to show the data:
7. The heights of 50 athletes in Moi University team were shown below:

1. State the modal class
2. Calculate the median height of the athletes
8. The table below shows the length of 40 mango tree leaves;

1. Determine the;
1. Modal class
2. Median class
2. Calculate;
1. The mean of the leaves
2. The median of the leaves

1. 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°

2. 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

1. Modal class is 10-19
1. Class

x= ∑fx/∑f = 3380/140 = 24.14
2. Median mark is at 70 + 71 = 70.5th position
Median = 119.5 + (0.5)/40 x 20
= 19.5 + 0.25
= 19.75
3. 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°
4. 180 – 189
Class limits

Median = 50/2 = 25
179.5 + (25–23)/16 x 10
= 179.5 + 20/16 = 180.75
179.5 + (26–23)/16 x 10
179.5 + 30/16 = 181.38
(180.75 + 181.38)/2
= 181.06
1. 145 – 153
2. Median class
(40 + ½)th value  » median class = 145 – 153
This is the 20.5th value
The value also in the 145 – 153 class

1. B2 for all values of fx correct and B1 for 4 values of fx and above orrect
Mean = Efx/Ef = 5888/40 = 147.2mm
Median 20th = 144.5 + (11/12 x 9) = 152.75
21st = 144.5 + (12/12 x 9) = 153.5
Median = (152.75 + 153.5)/2 = 153.125
(Alternatively, one could work out the 20.5 value directly using median formula)

• ✔ To read offline at any time.
• ✔ To Print at your convenience
• ✔ Share Easily with Friends / Students

### Related items

.
Subscribe now

access all the content at an affordable rate
or
Buy any individual paper or notes as a pdf via MPESA
and get it sent to you via WhatsApp