Probability Density Function
The word "density" is similar to the term "relative frequency". The total area under the curve of a probability density function such as that shown in the figure below 1 is = 1. As in the figure, the shaded area represents the probability of having (X) between two values on the x-axis.
The Normal Distribution
-(x-µ )2
2r 2
f(x) = 1 e
r (2p )½
Properties
1. Approximation for Binomial distribution - when np > 5 and n(1-p) > 5, the normal distribution provides an good approximation of the binomial distribution.
2. Standardization
In situation where we usually hypothesize that the theoretical distribution of a certain variable is normal, whereas the measurement of such variable may not give a normal distribution, standardization is used.
e.g. In the introduction of Sociology there are 200 students. We assumed that the performance of the students in the examination should be normally distributed. Furthermore, to give a reasonable distribution of marks, the mean should be 55 and standard deviation should be 10. After the examination, a lecturer marked the papers and the mean and standard deviation of the raw scores given by the lecturers are 50 and 6 respectively. To convert the raw score to standardize scores, the following steps were taken.
1. The standard score is obtained by Z = (x - 50)/6.
2. Then the converted (standardized) mark = 10(z) + 55.
For example, a raw score of 56 will be converted into 65.
3. Composite scores
When more than one measure is used to measure a variable, the distribution of each measure usually differs from each other. In order to obtain an unbiased measure using several different measurements, each sub-measure is standardized before added together.
Example
| Students | Marker I |
Marker II |
Average |
| A | 80 |
50 |
65.0 |
| B | 70 |
55 |
62.5 |
| C | 60 |
60 |
60.0 |
| D | 50 |
65 |
57.5 |
| E | 40 |
70 |
55.0 |
| Mean | 60 |
60 |
|
| r | 14 |
7 |
If we award marks according to the average of the marks given by Marker I and Marker II, obviously the final grades are more heavily affected by Marker I (who gives marks with a higher standard deviation) than by Marker II.
In order to compute the composite score the standardized scores of Marker I and II are averaged. In this example, if we decided that the ideal average score (µ ) and standard deviation (r ) should be 60 and 10 respectively, we can convert the z scores into standard score for each Marker. The results turn out to an average standardized score 60 for every student.
Marker I |
Marker II |
|||||
Student |
Raw score |
z=(x-µ )/r |
Standard score |
Raw score |
z=(x-µ )/r |
Standard score |
A |
80 |
1.4 |
74 |
50 |
-1.4 |
46 |
B |
70 |
0.7 |
67 |
55 |
-0.7 |
53 |
C |
60 |
0 |
60 |
60 |
0 |
60 |
D |
50 |
-0.7 |
53 |
65 |
0.7 |
67 |
E |
40 |
-1.4 |
46 |
70 |
1.4 |
74 |