1. Measure of Dispersion
Problem solving

2. The following table shows the results of student A and student B in
a test. Which student shows a more
consistent performance? Why?
Student
Mean marks
Standard Deviation A B 80 10 1

3. Answer The performance of student B is more consistent because the
standard deviation of his marks is
smaller.

4. The table below shows the marks for a test of 3 classes in form 5.
Between 5A and 5C, which class show a more consistent achievement?
Find the combined mean mark for the three classes.
Class
Mean marks
Standard Deviation
Numbers of students 5A 5B 5C
76.8
70.3
5.2
12.4
10.3 32 36

5. Numbers of students (n)
Class 5A shows a more consistent achievement because the standard deviation of its marks is smaller.
Class
Mean
Numbers of students (n)
Sum of marks
x = n 5A 5B 5C
76.8
70.3 32 36
2457.6
2530.8
100
7446
Combined mean marks, =
= 74.46

6. A set of 10 numbers has a mean of 10 and a
standard deviation of 2.
Another set of 10 numbers has a mean
of 4 and a standard deviation of 3.
Calculate the mean and standard
deviation of the 20 numbers.
Tips: Use to represent mean of first set of
numbers and to represent mean of the
second set of numbers.

7. nx =10
Hence, x = 100
ny =10
Hence, y = 40
X = x + y
= = 140

8. N = 20 x2 = 1040 y2 = 250 X2 = x2 + y2 = 1040 + 250 = 1290 HENCE,
MEAN = 7
STANDARD DEVIATION =3.937

9. EXERCISE The mean and variance of five
numbers are 4 and 2 respectively.
When another five numbers whose
sum is 40 and the sum of squares is
400 are added to the set of previous
date, find the mean and variance of the
10 numbers.
Mean = 6, variance = 13

10. EXERCISE
Given that 30 numbers in set y has a mean of 5 and variance of 2
Calculate y and y2
Another set of numbers which has a sum of 160 and a mean of 8 and a sum of squares of 1130 is added to the set y. Calculate the mean and the standard deviation of the numbers in the new set.
150, 870
6.2, 1.249