Measure of Dispersion

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 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 76.8 5.2 12.4 10.3 32 36 32
5

1. 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 76.8 32 36 32 2457.6 2530.8 2457.6
100 7446
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.

7
nx 10
Hence, ?x 100
ny 10
Hence, ?y 40
?X ?x ?y 10040 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.

1. 150, 870
2. 6.2, 1.249