used in statistics

1. NORMAL
DISTRIBUTION
PREPARED BY:
FERLIZA B. LOPEZ

2. • The normal distribution is a theoretical
concept whose objective is to be able to
explain for some variables the relation
between the intervals of its values and their
corresponding probabilities.
• It is bell-shaped and is symmetric around a
vertical line created at the center.
• The curve has two tails on both sides that
extend indefinitely in opposite direction.

3. PROPERTIES OF
NORMAL DISTRIBUTION

4. EMPIRICAL RULE
FIGURE 1.2
The subdivision of the
horizontal axis into equal
sub intervals with 1 unit
equal to 1 standard
devation
- 3s - 2s - 1s x̅ 1s 2s 3s

5. • FROM FIGURE 1.2, WE SHALL CONSIDER THREE
SPECIFIC INTERVALS:
68%
(x̅ - 1s) – (x̅ + 1s)
95%
(x̅ - 2s) – (x̅ + 2s)
99%
(x̅ - 3s) – (x̅ + 3s)
Figure 1.3. The Approximate Areas of the Intervals Representing the Empirical Rule.
- 3s - 2s - 1s x̅ 1s 2s 3s

6. To illustrate the significance of the empirical
rule, consider the NCEE scores of the students
in a certain college whose mean is 75 and a
standard deviation of 8. Assuming normality of
the data, then from the empirical rule, we can
say that
a.) Approximately, 68% of the students in that college
have NCEE scores between 75 plus or minus 8, that is,
75 – 1 (8) - 75 + 1 (8)
(75 – 8) – (75 + 8)
67 - 83

7. b.) Approximately, 95% of the students in that college have
NCEE scores between 75 plus minus 2 times the standard
deviation, 8. Thus,
75 – 2 (8) – 75 + 2 (8)
75 – 16 – 75 + 16
59 – 91
c.) Approximately, 99% of the students in that college have
NCEE scores of 75 plus or minus 3 times the standard
deviation, 8. Thus,
75 – 3 (8) – 75 + 3 (8)
75 – 24 – 75 + 24
51 - 99

8. Figure 1.4 The Graph of the NCEE Scores of the Students in a Certain
College. The Length of the Subinterval is equal to 1 Standard Deviation.
We can now present the data under the normal
curve area where the horizontal axis is divided into
equal sub intervals.
51 59 67 75 83 91 99

9. STANDARD
or the Z -
The standard score z represents a normal
distribution with mean x̅ = 0 and a standard
deviation s = 1. Such transformation can be
obtained by using the formula below.
z = x - x̅
s
SCORE

10. GET READY FOR AN
EXERCISE

11. The average efficiency rating of the faculty members of
a certain university is 85 with a standard deviation of 4.
ANSWER:
73 77 81 85 89 93 97
PRESENT THE DATA UNDER A NORMAL
CURVE AREA WHERE THE HORIZONTAL
AXIS IS DIVIDED INTO EQUAL
SUBINTERVALS.

12. By using the standard score, the values 73, 77, 81, 85 and so
on can be transformed into the standard scores by using this
equation: z = x - x̅
s
To illustrate, suppose
x = 85 and s = 4 x = 81 and s = 4
z = 85 – 85 z = 81 - 85
4 4
= 0 = -4
4
= - 1

13. Such conversions can be continued and we shall come up with
a normal curve in standard units as shown below.
Figure 1.5 The Normal Curve with x̅ = 0 and a Standard Deviation of 1

14. NORMAL CURVE AREAS
The total area under the normal curve is equal to 1. Since
a normally distributed set of data is symmetric, then the
total area from z = 0 to the right is equal to 0.5. the area
under the curve of any two points in a horizontal
scale can be obtained with Table 1.1. Such values
of z can be positive or negative but the area under
the curve will always be positive.

17. 1.25
.3944

18. Note that Table 1.1 has no negative values for z. It does
not mean, however, that the area under the curve which is to
the left of zero cannot be determined. Such area can still be
obtained due to the symmetric property of the normal
curve. Hence, we can say that the area under the curve from
z = 0 to z = -1.25 is also equal to .3944.
Also, the interval represented by z = 0 to z = -1.25 is
symbolized as -1. 25 < z < 0.
It was mentioned earlier that the total area under
the curve is equal to 1. Thus, we can associate these
areas under the normal curve as probabilities.

19. Generally, the capital letter Z is used to represent a certain
event that is greater than a standard score represented by
small letter z. Such event is denoted by Z > z. An event
that is less than a standard score shall be denoted b Z < z.
To determine the area under the curve or the probability of
the event, we shall use the notations P(Z > z), P(Z < z),
and P( z1 < Z < z2).

20. Solution: To be able to have a clear picture of the area
under the normal curve, it is important that the said
area be identified through the graph.
EXAMPLE 1:
FIND THE PROBABILTY VALUE OF P(Z>1.5)
1.5
P( Z > 1.5) = 0.5 – P(0 < Z < 1.5)
= 0.5 - .4332
P(Z > 1.5) = .0668

21. EXAMPLE 2:
FIND THE PROBABILITY VALUE OF P(Z < -0.5)
-0.5
Solution: P(Z < -0.5) = .5 – P(0 < Z < -.5)
= .5 - .1915
P(Z < -0.5) = .3085

22. EXAMPLE 3:
FIND THE PROBABILITY VALUE OF P( -.5 < Z < 1.5)
-0.5 1.5
P (-.5 < Z < 1.5) = P(0 < Z < 0.5) + P (0 < Z < 1.5)
= .1915 + .4332
P (-.5 < Z < 1.5) = .6247

23. EXAMPLE 4:
FIND THE PROBABILITY VALUE OF P(1.25 < Z < 2.31)
1.25 2.31
P(1.25 < Z < 2.31) = P (0 < Z < 2.31) – (0 < Z < 1.25)
= .4896 - .3944
P(1.25 < Z < 2.31) = .0952

24. EXAMPLE 5:
COMPUTE THE PROBABILITY OF P(Z < 1.25)
1.25
P (Z < 1.25) = 0.5 + P (0 < Z < 1.25)
= 0.5 + .3944
P (Z < 1.25) = .8944

25. GET READY FOR
A QUIZ !! 

26. The examination results of a large group of students in
statistics are approximately normally distributed with a
mean of 60 and a standard deviation of 9. If a student is
chosen at random , what is the probability that his score is?
1.) below 45?
2.) above 73?
3.) below 85?
4.) between 50 and 75?
LOOK FOR THE
STANDARD SCORE
AND PROBABILITY

27. Solution for #1:
Given x̅ = 60 , s = 9 x = 45
z = x - x̅
s
= 45 – 60
9
= -1.67
P ( x < 45) = P (Z < -1.67)
= 0.5 – P (0 < Z < 1.67)
= 0.5 - .4525
P ( x < 45) = .0475
- 1.67