Normal Distribution.pptx

Statistics and Probability
Quarter 3: Normal Distribution
Ma’am Neomy Angela L. Tolentino

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“
What is a normal distribution?
A normal distribution (Gaussian distribution) is a
probability distribution that is symmetric about the
mean, showing that data near the mean are more
frequent in occurrence than data far from the mean.

“
𝑓 𝑥 =
1
𝜎 2𝜋
𝑒
−
1
2
𝑥−𝜇
𝜎
2
f(x) = the height of the curve
particular values of x
X = any score in the
distribution
𝜎 = standard deviation of the
population
𝜇 = mean of the population
𝜋 = 3.1416
𝑒 = 2.7183

“
In graph form, normal
distribution will appear as a bell
curve.

Properties of the normal distribution:
1
The graph is a
continuous curve
and has a domain
− ∞ < 𝑋 < ∞.

As the x gets larger in either positive direction, the tail of the curve approaches but will never
touch the horizontal axis. The same thing happens when x gets smaller in the negative
direction.
2
The graph is asymptotic
to the x-axis. The value
of the variable gets closer
and closer but will never
be equal to 0.

▹ The mean (𝜇) indicates the highest peak of the
curve and is found at the center.
▹ Note:
3The highest point on
the curve occurs at
𝑥 = 𝜇 (mean)
𝜇 = 𝑚𝑒𝑎𝑛
𝜎 = 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛

4
The curve is
symmetrical about
the mean.

5
The total area in the
normal distribution
under the curve is
equal to 1.
100% or 1

6
In general, the graph of
a normal distribution is
a bell-shaped curve
with two inflection
points, one on the left
and another on the
right. Inflection points
are the points that
mark the change in the
curve’s concavity

7
Every normal
curve
corresponds
to the
“empirical
rule” (also
called the 68 –
95 – 99.7%
rule):
0.3413
or
34.13%
0.4772
or
47.72%
0.4987
or
49.87%
0.4987
or
49.87%
0.4772
or
47.72%
0.3413
or
34.13%
68.26%
95.44%
99.74%

Example #1:
Suppose the
mean is 60 and
the standard
deviation is 5.
Sketch a normal
curve for the
distribution.
60
𝝁 =
5
𝝈 =
65 70
55
50
0 1 2
-1
-2
45
-3
75
3
5 5
Z scores

How to
compute for
the z-score
given the raw
score and
standard
deviation?
1. Use the formula 𝑧 =
𝑥−𝜇
𝜎
2. Substitute the values of the mean, raw
score and standard deviation to the
formula.
3. Leave your final answer in decimal form.

Example #1:
Q1:What is the z score if
x = 66? x = 57? x = 46?
𝑧 =
𝑥 − 𝜇
𝜎
𝜎 = 5, 𝜇 = 60
x = 66
𝑧 − 𝑧 𝑠𝑐𝑜𝑟𝑒
𝑥 – raw score/ observed
value
𝜇 − 𝑚𝑒𝑎𝑛
𝜎 − 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛
𝑧 =
66 − 60
5
𝑧 =
6
5
𝑧 = 1.2
x = 57
𝜎 = 5, 𝜇 = 60
𝑧 =
57 − 60
5
𝑧 =
−3
5
𝑧 = −0.6
x = 46
𝜎 = 5, 𝜇 = 60
𝑧 =
46 − 60
5
𝑧 =
−4
5
𝑧 = −0.8

Example #2:
Find for the z score of the following:
𝝈 = 𝟓
𝑧 =
50 − 40
5
𝑧 =
10
5
𝑧 = 2
𝒙 𝝈 𝝁
50 5 40
40 8 52
36 6 28
60 10 74
75 15 82
x = 50 𝝁 = 𝟒𝟎
𝒛 =
𝒙 − 𝝁
𝝈
𝝈 = 𝟖
x = 40 𝝁 = 𝟓𝟐
𝑧 =
40 − 52
8
𝑧 =
−12
8
= −
3
2
𝑧 = 1.5
𝑧 = 1.33
𝑧 = −1.4
𝑧 = −0.47

How to sketch
a normal
distribution
and its area or
compute its
probability? Table of Areas under the Normal
Curve is also known as the z-table.

Four steps in Finding the Areas Under
the Normal Curve Given a z-value
Step 1: Express the given z-
value into a three-digit form
Step 2: Using the z-table, find
the first two digits on the left
column
Step 3: Match the third digit
with the appropriate column
on the right
Step 4: Read the area (or
probability) at the intersection
of the row and the column
Example: 0.78
Area: 0.2823

𝑧 = 1.2
Sketch the normal distribution of each
scores and indicate its area.
Area: 0.3849
60 65 70
55
50
45 75
1.2
65
38.49%

𝑧 = −0.6
Area: 0.2257
60 65 70
55
50
45 75
-0.6
57
22.57%

𝑧 = −0.8
Area: 0.2881
60 65 70
55
50
45 75
-0.8
46
28.81%

Place your answer in a 1 whole sheet of paper to be submitted on the day of
retrieval
Sketch the normal distribution of each z-scores and indicate its area.
𝑧 = 2 𝑧 = 1.5 𝑧 = 1.33 𝑧 = −1.4 𝑧 = −0.47
Assignment #1

Compute probability of
P(z>1.27),
P(z<1.31) ,
P(-1.4<z<2.71) &
P(1.5<z<2).

Compute
probability of
P(z>1.27)
1.27
1. Get the probability of
the given z score
2. Subtract the
probability from
0.5000

Compute
probability of
P(z>1.27)
1.27
the given z score
2. Subtract the
probability from
0.5000
0.3980
0.5000 Solution:
= 0.5000 – 0.3980
= 0.102 or 10.2%

Compute
probability of
P(z<1.31)
1.31
the given z score
2. Add the probability
of the z score to
0.5000
0.4049
0.5000
Solution:
= 0.5000 + 0.4049
= 0.9049 or 90.49%

Compute
probability of
P(-1.4<z<2.71)
2.71
the given z score
2. Add the probability
of the two z scores
0.4966
0.4192
-1.4
Solution:
= 0.4192 + 0.4966 = 0.9158 or 91.58

Compute
probability of
P(1.5<z<2)
2
the given z score
2. Subtract the
probability of the two
z scores
0.4772
0.4332
1.5
Solution:
= 0.4772 - 0.4332
= 0.044 or 4.4%

Summary in computing the probability given:
a. Area less than +z
ex. P(z<1) = ADD
the area of the z
score to 0.5000
c. Area greater than
+z ex. P(z>1) =
SUBTRACT the
area of the z score
from 0.5000
b. Area greater than
-z ex. P(z>-1) = ADD
the area of the z
score to 0.5000
d. Area less than -z ex.
P(z<-1) = SUBTRACT
the area of the z score
from 0.5000
e. Area between –z and
+z ex. P(-1<z<-1) = ADD
the area of the two z
scores
e. Area between –z and –z / +z
and +z
ex. P(1<z<2) / P(-2<z<-1) =
SUBTRACT the area of the
two z scores

Place your answer in a 1 whole sheet of paper to be submitted on the day of
retrieval.
Compute for the probability of the following and sketch the graph.
P(z>2.13)
Assignment #2
P(-2.19<z<0.16) P(-1.74<z<-0.25) P(z<0.15)

Thank you for listening!
Have a good day ahead
everyone!

Normal Distribution.pptx

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