Republic of the Philippines

City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

DETAILED LEARNING MODULE

Module VI:
NORMAL DISTRIBUTION

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

Module No. V

“Be approximately right rather than exactly

wrong”
John Tukey
I. Introduction
An assessment of the normality of data is a prerequisite for many statistical tests as normal
data is an underlying assumption in parametric testing. The normal distribution is used in
analysis of data in determining parametric and non-parametric test. Its graph is called a
normal curve. The mathematical equation of the normal curve was first described in 1733 by
De Moivre.

A population investigated in a certain school regarding the academic performance of

students and has a characteristic that follows a normal distribution. If we are to study the
grade point average (GPA) of the students with a population (N= 2, 500), we may find that
the majority of the students’ population will yield excellent, very good, good, satisfactory,
passed and failed.

If the grades of the students are plotted on a graph with the frequency of students in
the ordinate or y axis and their grades in the abscissa or x axis, we probably approximate a
bell shaped curve like the figure 8 below.

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

Figure 8: Normal Curve

II. Learning Objectives

At the end of Module 6, the students are expected to:
1. Identify the properties of a normal curve
2. Solve problems involving normal distributions

3. Learn and appreciate how to use the table of are under normal curve
III. Topics and Key Concepts
Properties of a Normal Curve
1. The mean, median and mode coincide at
one point at the center of the distribution
2. The curve is symmetrical and bell- shaped
3. The tail of the curve is asymptotic to the
horizontal line
4. Three standard deviation to the left & right
of the curve
Figure 9: Standard
5. The total area under normal curve is 100%.
Normal Curve

Standard Normal Curve

Formula for finding the standard score (z)
x− x̄
z=
s

where:
z = standard score
x̄ = mean
s = standard deviation
x = a given value of a particular variable

Consider the following procedures in determining the areas under the standard normal
curve:

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

1. If the areas above the mean or right of a positive z-score, subtract the value in the
table of the normal curve areas from 0.5000.
2. If the areas below the mean or left of a positive z-score, add the value in the table of
the normal curve areas to 0.5000.
3. If the areas above the mean or right of a negative z-score, add the value in the table of
the normal curve areas to 0.5000
4. If the areas below the mean or left of a negative z-score, subtract the value in the
table of the normal curve areas from 0.5000

Area under Normal curve

Example 1. Find the area under the normal curve from z ¿ 1.18.

Solution:

Step 1: Sketch the curve. Locate the

measurement of horizontal axis
on the indicated range of values.

Step 2: The area under the curve over the

description of the event. The shaded
part is the area under the normal
curve of the event z ¿ 1.18.

Step 3: Use the standard normal table (see Appendix A) to find this area. Now, a complete
copy of the table is not here. But, here's an abridged version to locate the area under
the normal curve.

z .00 .01 ... .08 .09

0.0 .0000 .0040 ... .0319 .0359
0.1 .0398 .0438 ... .0714 .0753
... ... ... ... ... ...

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

1.0 .3413 .3438 ... .3599 .3621

1.1 .3643 .3665 ... .3810 .3830
1.2 .3849 .3869 ... .3997 .4015
... ... ... ... ... ...

Corresponding to a measurement value of z = 1.18 is an area of 0.3810.

0.3810
0.5000
0.8810

¿
For the area under standard normal curve from z 1.18 is 0.8810.

Answer: 0.8810 or 88.10%.

Example 2.  What is the area under the normal curve from z ¿ -0.63?

Solution:

1. Identify the range of values described by " ¿ -0.63”.

2. Identify the area you need to find (shaded part of the figure below).
3. Look-up the appropriate area in your table. That area is 0.2357. Since the condition is
¿
z -0.63, we subtract 0.5 to 0.2357.

0.5000
0.2357
0.2643

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

Answer: 0.2643 or 26.43%.

Example 3. Find the area of the normal curve from z = -1.57 to z = 3.99.

Solution:

1. Identify the range of values described from z = -1.57 to z = 3.99.

2. Identify the area you need to find (shaded part of the figure below).
3. Look-up the appropriate area in your table. That area of 1.57 is 0.4418 and 3.99 is
0.5000. Since the condition is z = -1.57 to z = 3.99, we add the area 0.4418 to 0.5000.

0.4418
0.5000
0.9418

Answer: 0.9418 or 94.18%.

Example 4. Find the z score corresponding to the given area to the right of + z = 0.2000

Solution:
The shaded portion shows that the area to the right of z is 0.2000. To obtain the area
we are looking for, we subtract 0.2000 from the total area of the right half of the normal
curve. Hence, 0.5000 - 0.2000 = 0.3000. Referring to the figure below.

0.5000
0.2000
0.3000 0.2995 0.84

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

Referring to the table of the area of the normal curve, Appendix A,

we find the entry nearest to 0.3000 is 0.2996 and this corresponds to a z-score of 0.84.

Answer: 0.84 (because 20% of the observations are above 0.84).

Example 5. 

Find the z score corresponding to the given area to the right of +z is 0.3520.

Solution:
The shaded portion shows that the area to the right of z is 0.3520. To obtain the area
we are looking for, we subtract 0.3520 from the total area of the right half of the normal
curve. Hence,
0.5000 - 0.3520 = 0.1480. Referring to the figure below.

0.5000
0.3520
0.1480 0.38

Referring to the table of the area of the normal curve, we find the entry 0.1480 and
this corresponds to a z-score of 0.38.

Answer: 0.38

Application of the Standard Normal Curve

Example:

The mean weight of college students is 70 kg and the standard deviation is 3 kg. Assuming
that the weight is normally distributed, what is the probability that the students weigh:

a. between 60 kg and 75 kg.

b. more than 72 kg.
c. less than 64 kg.

Solution:

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

a. Between 60 and 75 kg

1. Convert raw score of 60 and 75 to z score

60−70
z= =−3. 33
3

75−70
z= =1 . 67
3

2. Sketch the curve and identify the area you need to find (shaded part of the figure
below

3. The z value of 3.33 gives an area of 0.4996, while z value of 1.67 corresponds to an
area of 0.4525. The shaded area is the sum between these two given areas. Therefore,
the required area is 0.4996 + 0.4525 = 0.9521 or 95.21%

0.4996
0.4525
0.9521

Answer: 0.9521 or 95.21%

b. more than 72 kg.

1. Convert raw score of 72 to z score

72−70
z= =0 . 67
3

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

2. Sketch the curve and identify the area you need to find (shaded part of the figure
below).

3. The z value of 0.67 gives an area of 0.2486. From the shaded area, we subtract 0.2486
from the total area of the right half of the normal curve. Hence, 0.5000 - 0.2486 =
0.0.2514 or 25.14%

Answer: 0.2514 or 25.14%

c. less than 64 kg

1. Convert raw score of 64 to z score

64−70 -2
z= =2
3

2. Sketch the curve and identify the area you need to find (shaded part of the figure
below).

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

3. The z value of 2.00 gives an

area of 0.4772. From the
shaded area, we subtract 0.4772
from the total area of the left
half of the normal curve.
Hence, 0.5000 - 0.4772 =
0.0228 or 2.28%

0.5000
0.4772
0.0228

Answer: 0.0228 or 2.28%

IV. Teaching and Learning Materials and Resources

1. PowerPoint Presentation/Module
2. Module for Offline students
3. Textbook
4. Calculator

V. Learning Task

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

Solve the following problem completely. Show diagrams and shade the required
area.
1. Find the z-score corresponding to the given area of the normal curve with the
following given values:

a. area to the left of z is .8461

b. area to the left of –z is .2514
c. area to the right of z is .4207

2. Find the area of the normal curve given the following:

a. from z=0 to z=1.43

b. from z= -1.23 to z=0
c. from z=2.38 to z=3.09
3. An entrance examination is to be conducted to 1500 incoming freshmen
students at Gordon College which is known to be normally distributed and has a
mean of 85 and a standard deviation of 10.

a. How many students would be expected to score above 95?

b. How many students would be expected to score between 75 and 115?

STA
TIS Who is this French mathematician who was a pioneer in the development of
analytic trigonometry and in the theory of probability.

TIC ________________________________

VI. Reference
 Paguio, D. et al. 2012. Statistics With Computer Based Discussions. Jimczyville
Publication

Standard Normal Distribution Areas

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 Republic of the Philippines
City of Olongapo
GORDON COLLEGE
Olongapo City Sports Complex, East Tapinac, Olongapo City
Tel. No. (047) 224-2089 loc. 314

This table shows the normal area between 0 and z.

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