Lesson5 The Normal Curve

QUEENSEN MERA S.
CANTONJOS
The Normal Curve
Normal Curve (or Bell Curve)

a graph that represents the probability density
function of a normal probability distribution. It is
also called a Gaussian curve named after the
mathematician, Carl Friedrich Gauss.
Example:
The graph below represents the normal curve where 𝜇 is
the mean and 𝜎 is the standard deviation.
The Normal Curve
Empirical Rules
The area of the region between one standard
deviation away from the mean is 0.6826, two standard
deviations away from the mean is 0.9544, and three
standard deviations away from the mean is 0.9974.
A. Area of the Region between One Standard Deviation
away from the Mean
The Normal Curve
Empirical Rules
B. Area of the Region between Two Standard Deviations
away from the Mean
The Normal Curve
Empirical Rules
C. Area of the Region between Three Standard Deviations
away from the Mean
Example 1:
What is the area of the scores in a normally

distributed data that is more than one standard
deviation above the mean?
Solution:
1. Shade the region in the normal curve that is one

standard deviation above the mean.
Example 1:

Solution:
2. Use the empirical

rule to find the area
of the given region.
Find the area
of the region one
standard deviation
away from the mean.
By empirical rule,
the area of this
region is 0.6826.
Example 1:

Solution:
Since the normal curve is
symmetric, we can obtain
the area of the region
between the mean and one
standard deviation above
the mean by dividing
0.6826 by 2.
To get the area of this region, divide 0.6826 by 2.
0.6826 ÷ 2 = 0.3413
Example 1:

Solution:
To get the area of the region more than one standard
deviation above the mean, subtract 0.3413 from 0.5 since half
of the area of a normal curve is 0.5.
0.5 − 0.3413 = 0.1587
Thus, the area of the region more than one standard

deviation above the mean is 𝟎. 𝟏𝟓𝟖𝟕.
0.6826 ÷ 2 = 0.3413
Example 2:
The mean and standard deviation of a normally

distributed scores are 15.8 and 2.2, respectively.
Construct the normal curve of the data.
Solution:
1. Solve for the scores one, two, and three standard
deviations away from the mean.
𝜇 = 15.8, 𝜎 = 2.2
a. One standard deviation away from the mean
𝜇 + 𝜎 = 15.8 + 2.2 = 18
𝜇 − 𝜎 = 15.8 − 2.2 = 13.6

Example 2:

Solution:
b. Two standard deviations away from the mean
𝜇 + 2𝜎 = 15.8 + 2(2.2)
= 15.8 + 4.4
= 20.2
𝜇 − 2𝜎 = 15.8 − 2(2.2)
= 15.8 − 4.4
= 11.4
Example 2:

Solution:
c. Three standard deviations away from the mean
𝜇 + 3𝜎 = 15.8 + 3(2.2)
= 15.8 + 6.6
= 22.4
𝜇 − 3𝜎 = 15.8 − 3(2.2)
= 15.8 − 6.6
= 9.2
Example 2:

2. Construct the normal curve using the set of scores
obtained from the previous step.
Using the mean as the center and the set of scores, we can
construct the normal curve as follows:
QUEENSEN MERA S. CANTONJOS
Solving Problems Involving the Normal Random Variable

Mathematician Karl Friedrich Gauss.
Example:
The graph below represents the normal curve where 𝜇 is
the mean and 𝜎 is the standard deviation.
Normal Curve Areas under the Empirical Rule

To solve problems involving the normal random
variable, we can use the following areas in the
normal curve based from the empirical rule.
A. Areas between the mean and one standard deviation
above or below the mean

B. Area between the mean and two standard deviations

C. Areas between the mean and three standard
deviations above or below the mean
Example 1:
A teacher recorded the weight of his students

and got an average of 132.5 kg with a standard
deviation of 13.4. Assuming that the weights are
normally distributed, what percent of the
students are between 119.1 kg and 159.3 kg?
Solution:
1. Construct the
normal curve of the
distribution and
locate the given
weights.
Example 1:

Solution:
2. Shade the area of the normal
curve based on the problem.
The problem asks for the

percent of the students between
119.1 kg and 159.3 kg. Thus,
shade the area between these
two weights in the normal
curve.
Example 1:

Solution:
3. Find the area of the region.
To find the area of the region, add the given areas below.
0.3413 + 0.4772 = 0.8185

Example 1:

Solution:
4. Convert the decimal to percent.
0.8185 100 = 81.85%
Thus, there are 𝟖𝟏. 𝟖𝟓% of the students who are between 119.1
kg and 159.3 kg.
Example 2:
The average price of carrots was ₱60.5 per

kilogram with a standard deviation of 2.4 for the
last three months. If the price is normally
distributed for 90 days, what percent of 90 days
when the price was above ₱62.9?
Solution:
1. Construct the normal curve of the distribution and
locate the given price.
Example 2:

Solution:
2. Shade the area of the
normal curve based on
the problem.
The problem asks for the

percent of 90 days that
the price was above
₱62.9. Thus, shade the
area from 62.9 and
above.
Example 2:

Solution:
3. Find the area of the
region.
To find the area of the

region, subtract the
given area below from
0.5.
0.5 − 0.3413 = 0.1587

Example 2:

Solution:
4. Convert the decimal to percent.
0.1587 100 = 15.87%
Thus, there are 𝟏𝟓. 𝟖𝟕% of 90 days that the price was above
₱62.9.
Key Points:
mathematician, Carl Friedrich Gauss.
Empirical Rules
Key Points:
Empirical Rules
A. Area of the Region between One Standard
Deviation away from the Mean
Key Points:
Empirical Rules
B. Area of the Region between Two Standard
Deviations away from the Mean
Key Points:
Empirical Rules
C. Area of the Region between Three Standard
Deviations away from the Mean
Key Points:
A. Areas between the mean and one standard
deviation above or below the mean
B. Area between the mean and two standard
Key Points:
A. Areas between the mean and one standard deviation
Key Points:
B. Area between the mean and two standard deviations
Key Points:

Lesson5 The Normal Curve

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QUEENSEN MERA S.

Normal Curve (or Bell Curve)

What is the area of the scores in a normally

1. Shade the region in the normal curve that is one

What is the area of the scores in a normally

2. Use the empirical

What is the area of the scores in a normally

What is the area of the scores in a normally

Thus, the area of the region more than one standard

The mean and standard deviation of a normally

a. One standard deviation away from the mean

𝜇 − 𝜎 = 15.8 − 2.2 = 13.6

The mean and standard deviation of a normally

The mean and standard deviation of a normally

The mean and standard deviation of a normally

Normal Curve (or Bell Curve)

Normal Curve Areas under the Empirical Rule

Normal Curve Areas under the Empirical Rule

Normal Curve Areas under the Empirical Rule

A teacher recorded the weight of his students

A teacher recorded the weight of his students

The problem asks for the

A teacher recorded the weight of his students

0.3413 + 0.4772 = 0.8185

A teacher recorded the weight of his students

4. Convert the decimal to percent.

0.8185 100 = 81.85%

The average price of carrots was ₱60.5 per

The average price of carrots was ₱60.5 per

The problem asks for the

The average price of carrots was ₱60.5 per

To find the area of the

0.5 − 0.3413 = 0.1587

The average price of carrots was ₱60.5 per

0.1587 100 = 15.87%

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