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Normal Distribution
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Finding the Area
Under the Normal Curve
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Finding the z-value
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Describing Normal Distribution
A normal distribution is a continuous,
symmetric, bell-shaped distribution of a variable.
The known characteristics of the normal curve
make it possible to estimate the probability of
occurrence of any value of a normally distributed
variable.
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Most scientific and business data and natural relationships, T
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such as weight, height, etc., when displayed using a histogram
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frequency curve are bell-shaped, and symmetrical, known as
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NORMAL DISTRIBUTION.
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Many things closely follow a Normal
Distribution:
heights of people
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size of things produced by machines
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errors in measurements
blood pressure
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scores on a test
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Describing Normal Distribution
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Properties of a Normal Distribution
The following are the properties of a normal
distribution:
1.) The distribution is bell-shaped.
2.) The mean, median, and mode are equal and are located at the center of the
distribution.
3.) The normal distribution is unimodal.
4.) The normal distribution curve is symmetric about the mean (the shape are same on
both sides).
5.) The normal distribution is continuous.
6.) The normal curve is asymptotic (it never touches the x-axis).
7.) The total area under the normal distribution curve is 1.00 or 100%.
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Properties of a Normal Distribution
8. The area under the part of a normal curve that lies within 1 standard deviation of the mean
68%; within 2 standard deviation, about 95%; and with 3 standard deviation, about 99.7%.
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Finding the Area Under
the Normal Curve
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Finding the Area Under the Normal Curve
Example 1: Find the area under the standard normal distribution
curve between z = 0 and z = 1.25.
Step 1: Draw the figure and represent
the area.
Step 2: Look up the z value in the table.
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Hence, the area is 0.3944 or 39.44%.
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Finding the Area Under the Normal Curve
Example 2: Find the area under the standard normal distribution
curve between z = 0 and z = -2.13.
Step 1: Draw the figure and represent
the area.
Step 2: Look up the z value in the table.
The z-table can
also be used even when the
given z value is negative.
They make no difference
with the positive ones since
we are looking for the area.
Therefore, the area between
the given values of z is
0.4834 or 48.34%.
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Finding the Area Under the Normal Curve
Example 3: Find the area under the standard normal distribution
curve to the right of z = 0.96.
Step 1: Draw the figure and represent
0.5 or 50%
the area.
Step 2: Look up the z value in the table.
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0 0.96
The area required is at the right tail of the normal curve. The
area obtained in Step 2 is the area from 0 to 0.96. To find the
reqjuired area, subtract 0.3315 from 0.5 (area of half of the normal
curve). Thus, 0.5 – 0.3315 is 0.1685 or 16.85%.
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Finding the Area Under the Normal Curve
Example 4: Find the area under the standard normal
distribution curve to the left of z = 1.08.
Step 1: Draw the figure and represent
the area.0.5 or 50%
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0 1.08
Step 2: Look up the z value in the table.
The area required is at the left tail of the normal curve. The
area obtained in Step 2 is the area from 0 to 1.08. To find the
reqjuired area, add 0.3599 to 0.5 (area of half of the normal
curve). Therefore, 0.5 + 0.3599 is 0.8599 or 85.99%.
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Finding the Area Under the Normal Curve
Example 5: Find the area under the standard normal
distribution curve between z = 0.34 and z = 1.05.
Step 1: Draw the figure and represent
the area.
Step 2: Look up the z value in the table.
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0 0.34 1.05
The area required is the difference between the area between z=0 and z=1.05
and the area between z=0 and z=0.34. The areas obtained in Step 2 are 0.3531
and 0.1331 respectively. To find the reqjuired area, subtract the two areas
obtained. Thus, 0.3531 – 0.1331 is 0.22 or 22%.
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Finding the Area Under the Normal Curve
Example 6: Find the area under the standard normal
distribution curve between z = -1.11 and z = 2.75.
Step 1: Draw the figure and represent
the area.
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-1.11
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2.75
Step 2: Look up the z value in the table.
The area required is the
sum of the areas between z=0
and z=-1.11 and between z=0
and z=2.75. The areas
obtained in Step 2 are 0.3665
and 0.4970 respectively. To
find the required area, add
the two areas obtained.
Therefore, the area is 0.3665 +
0.4970 which is 0.8635 or
86.35%.
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The z-value
A normal can be converted into a standard normal distribution by
obtaining the z value. A z value is the signed distance between a selected value,
designated x, and the mean 𝑥, divided by the standard deviation. It is also
called as z scores, the z statistics, the standard normal deviates, or the
standard normal values. In terms of formula:
𝒙−𝒙
𝒛=
𝒔
where: z = z value
x = the value of any particular observation or measurement
𝑥 = the mean of the distribution
s = standard deviation of the distribution
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Complete the table below
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x
110
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𝒙
108
108
s
10
10
z
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1.30
15
87
?
108
1.25
?
-0.64
-2.10
Solution for z:
𝑧=
𝑥 −𝑥
𝑠
=
110 −108
10
= 0.2
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Complete the table below
x
110
?
𝒙
108
108
s
10
10
z
0.20
1.30
15
87
?
108
1.25
?
-0.64
-2.10
Solution for x: 𝒛 =
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𝒙−𝒙
𝒔
𝒙 − 𝟏𝟎𝟖
𝟏. 𝟑𝟎 =
𝟏𝟎
𝟏𝟑 = 𝒙 − 𝟏𝟎𝟖
𝟏𝟑 + 𝟏𝟎𝟖 = 𝒙
𝒙 = 𝟏𝟐𝟏
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Complete the table below
x
110
121
𝒙
108
108
s
10
10
z
0.20
1.30
15
87
?
108
1.25
?
-0.64
-2.10
Solution for 𝒙: 𝒛 =
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𝒙 −𝒙
𝒔
𝟏𝟓 − 𝒙
−𝟎. 𝟔𝟒 =
𝟏. 𝟐𝟓
– 𝟎. 𝟖 = 𝟏𝟓 − 𝒙
𝒙 = 𝟏𝟓 + 𝟎. 𝟖
𝒙 = 𝟏𝟓. 𝟖
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Complete the table below
x
110
121
𝒙
108
108
s
10
10
z
0.20
1.30
15
87
15.8
108
1.25
?
-0.64
-2.10
Solution for s: 𝒛 =
𝒙 −𝒙
𝒔
𝟖𝟕 − 𝟏𝟎𝟖
−𝟐. 𝟏𝟎 =
𝒔
– 𝟐. 𝟏𝟎𝒔 = −𝟐𝟏
−𝟐. 𝟏𝟎𝒔
−𝟐𝟏
=
−𝟐. 𝟏𝟎
−𝟐. 𝟏𝟎
𝒔 = 𝟏𝟎
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Applications
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Example 1: The scores of 25 grade 6 pupils have a mean of 5.35 and
standard deviation of 0.45. What percentage of all these scores are
lower than 4?
Solution for z:
𝑥 − 𝑥 4 − 5.35
𝑧=
=
= −3.0
𝑠
0.45
*The area between 4 and 5.35 is
49.87%. Therefore to find the
percentage of those who
obtained lower than 4, subtract
49.87% from 50% (area to the
left of the mean). Thus, 0.13%
obtained scores lower than 4.
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Applications
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Example 2: You took an entrance examination and scored 68. The mean score
for the exam is 54 and the standard deviation is 10. How well did you score on
the test compared to the average test taker?
Solution for z:
𝑥 − 𝑥 68 − 54
𝑧=
=
= 1.4
𝑠
10
*The area from the score of the
average taker 54 to 68 is 41.92%.
Therefore, you performed
41.92% better than the average
takers.
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Applications
Example 3: To qualify for a Master’s degree program in Our Lady of Fatima University,
candidates must score in the top 20% on a mental ability test. The test has a mean of 180
and a standard deviation of 25. Find the lowest possible score to qualify. Assume the test
scores are normally distributed.
Step 2: Find the z value that corresponds to
Step 1: Draw the figure.
an area of 0.3000 or the closest to it.
20% or 0.2000
Determine the area
between 180 and x.
0.5 – 0.2 = 0.3000
180
x
The closest is 0.2995 and the corresponding z
value is 0.84.
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Applications
Example 3: To qualify for a Master’s degree program in Our Lady of Fatima University,
candidates must score in the top 20% on a mental ability test. The test has a mean of 180
and a standard deviation of 25. Find the lowest possible score to qualify. Assume the test
scores are normally distributed.
Step 3: Solve for x.
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Solution for 𝑥:
𝑥 − 180
0.84 =
25
0.84(25) = 𝑥 − 180
21 + 180 = 𝑥
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𝑥 = 201
A score of 201 should be used as a cut
off. Anybody scoring 201 or higher
qualifies for Master’s degree program in
Our Lady of Fatima University.
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Are there any questions
pertaining to normal
distribution? If there’s none,
prepare now for the offline
activity.
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