STATISTICS AND PROBABILITY
Module 5: Week 5: Third Quarter
Title of the Activity:
Normal Curve Distribution
ENTRY BEHAVIOUR
Brief Introduction
This module will introduce the concept of a normal random variable. This will also
enlighten your mind regarding normal curve; answer the question what is
probability/percentile; and how to identify the value of a probability using standard normal
table.
Learning Competencies
At the end of this module, the learner should be able to;
1. Illustrates a normal random variable and its characteristics;
2. Identifies region under the normal curve corresponding to different
standard normal values;
3. Converts a normal random variable to a standard normal variable and
vice versa; and
4. Computes probabilities and percentiles using the standard normal
table.
Objectives
The objective of this module is for the students to define what normal curve
distribution is, identify the area under the normal curve distribution given the value of z,
and, use the standard normal table in finding probabilities.
Let’s Recall
(Measures of Central Tendency, histogram and frequency polygon)
Let’s have a simple review of the measures of central tendencies, the mean,
median, and mode.
̅) – refers to the average of the elements in the distribution.
Mean (𝒙
Ʃ𝒙
𝒙̅ = , where x = the values in the distribution
𝒏
n = the number of values
Median (𝒙 ̃) – refers to the middlemost value or elements in the
distribution after arranging the values either in ascending or
descending. If the number of elements is ODD, get the average of the
two middlemost values.
Mode (𝒙 ̂) – refers to the frequently occurs value or elements in the
distribution.
Let’s consider the distribution below in order for us to figure out what are the
characteristics of a normal curve distribution.
1. Given the table below, study the values of mean, median, and mode.
X f Since, there is an odd
Ʃ𝑓𝑥 number of elements in a
1 1 𝑥̅ = 𝑛 ̃= 2.
distribution, therefore, 𝒙
2 3 10
= 5
3 1 Take a look on column f, it
̅=2
𝒙 shows that element 2 repeated
n=5
̂= 2.
thrice, therefore, 𝒙
What is histogram? Histogram is a vertical bar graph with no space between the
bars. The histogram of the given distribution was shown below:
3 3
f 2 f 2
1 1
1 2 3 1 2 3
X X
Histogram Frequency Polygon
2
2. Another table shown below, study the values of mean, median, and mode.
X f
1 1 Since, there is an odd
2 3
number of elements in a
Ʃ𝑓𝑥 ̃= 4.
distribution, therefore, 𝒙
3 4 𝑥̅ = 𝑛
88
4 6 = 22
5 4 Take a look on column f, it
̅=4
𝒙 shows that element 2 repeated
6 3 ̂= 4.
thrice, therefore, 𝒙
7 1
n = 22
Line of symmetry
6 6
5 5
f 4 f 4
3 3
2 2
1 1
1 2 3 4 5 6 7 1 2 3 4 5 6 7
X X
Histogram Frequency Polygon
Conclusion: The mean, median, and mode coincide at the center of the distribution.
If you will draw an imaginary vertical line passing the center of the distribution (on
the histogram), you will notice that the parts are mirror image of one another, or simply they
are symmetrical.
Let’s Understand
Going back to the frequency polygon (in the above example), can you connect the
points freehand? Is it similar to the curve at the right?
What is normal curve or normal curve
distribution?
Normal curve or normal curve
distribution is a continuous, symmetric, bell- Figure 1
shaped distribution of a variable.
3
Properties of the Normal Curve Distribution
1. The normal curve is a bell-shaped.
2. The mean, median, and mode coincide at the center of the distribution.
3. The normal distribution curve is unimodal (it has only one mode).
4. The curve is symmetric about the mean.
5. The curve never touches the x-axis or asymptotic to the horizontal line.
6. The total area under the normal curve distributions is equal to 1.00 or 100%.
7. The width of the distribution is determined by standard deviation.
Since each normally distributed variable has its own mean and standard deviation,
to address the uniformity, standard normal distribution was introduced.
34.13%
13.59%
13.59%
34.13%
2.28%
2.28%
µ-3σnormal
The standard µ-2σ µ-1σ
distribution µ a normal
is µ+1σ distribution
µ+2σ µ+3σ with a mean (µ) = 0 and
a standard deviation (σ) = 1.
34.13%
13.59%
13.59%
34.13%
2.28%
2.28%
-3 -2 -1 0 1 2 3 z - score
µ-3σ µ-2σ µ-1σ µ µ+1σ µ+2σ µ+3σ Raw
score (X)
Formula in finding standard score:
𝒓𝒂𝒘 𝒔𝒄𝒐𝒓𝒆 − 𝒎𝒆𝒂𝒏
Z=
𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏
or
𝒙− µ
𝒛=
𝝈
4
Process in Finding the Area under the Normal Curve Distribution:
1. Between 0 and any z – value 2. In any tail
Find the corresponding area Find the corresponding area
of z-value in the z-table. of z-value in the z-table.
Subtract the area from 0.5000
-z 0 0 z -z 0 0 z
3. Between two z-scores on the 4. Between two z-scores on the
same side of the mean opposite side of the mean
Find the corresponding area Find the corresponding area
of z-value in the z-table. of z-value in the z-table.
Subtract the smaller area Add the areas.
from the larger one.
-z2 -z1 0 0 z1 z2 -z1 0 z2
5. To the left of any z-score, where z 6. To the right of any z-score, where
is greater than the mean z is less than the mean
Find the corresponding area Find the corresponding area
of z-value in the z-table. of z-value in the z-table.
Add 0.5000 to the area. Add 0.5000 to the area.
-z 0
0 z
5
7. In any two tails
Find the corresponding areas
of z-values in the z-table.
Subtract both areas from
0.5000
Add the answers
-z 0 z
How to find the area in the z-table
Using the z-table, you can easily find the area of the z-value.
First, locate the whole number and the tenths on the first column,
Second, locate also the third digits (the hundredths) on the first row,
Then, the intersection of the row and column is the area of the z-value.
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
For example:
0.0
1. What is the area of
0.1
z = 2.23?
0.2
Referring to the z-
0.3
table, we can conclude
0.4
that the area of z = 2.23
.
.
is .4871.
.
2. What is the area of 2.1
z = 3.1? 2.2 .4871
.
Base on the z-table, if .
.
the z-value is greater
2.9
than 3.09, its area is 3.0
approximately equal to
Note: Always remember that z-value must be a three-digit
.4999. number.
6
Let’s Apply
A. Direction: Find the area that corresponds to the given z-value.
1. z = 0.96 6. z = 3.00
2. z = 1.74 7. z = - 3. 01
3. z = 2.18 8. z = - 2.48
4. z = 2. 69 9. z = 3.11
5. z = - 0. 48 10. z = - 1. 01
Let’s Analyze
Direction: Identify the region under the normal curve given the following conditions.
1. between z1 = 0 and z2 = 2.96
2. to the left of z = 1.78
3. to the right z = 0.18
4. between z1 = - 2.5 and z2 = 1.5
5. to the left of z = - 0. 88 and to the right of z2 = 2.56
Let’s Evaluate
Direction: Answer the following questions below. Illustrate the area referring
to using the normal curve.
1. Find the area between z = 1.68 and z = -1.37
2. Find the area to the left of z = 1.98
3. Find the area to the right of z = -2.13
4. Find the area between z = -2.38 and z = -0.82
5. Find the area to the right of z = 2.45 and to the left of z = -3.0
Let’s Create
Direction: Answer the problem below.
1. Find z0 such that P(z > z0) = 0.1234.
2. Find z0 such that P(-1.2 < z < z0) = 0.8671.
3. Find z0 such that P(z0 < z < 2.5) = 0.7672.
4. Find z0 such that P( - z0 < z < z0) = 0.86
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5. Find z0 such that the area between z0 and z = -0.5 is 0.2345. Give the two
possible answers.
