Download CHAPTER 7: NORMAL DISTRIBUTION

Mr. Mark Anthony Garcia, M.S.
Mathematics Department
De La Salle University
Normal Distribution
A normal distribution is a probability
distribution that plots all of its values in a
symmetrical fashion and most of the
results are situated around the
probability's mean.
 Values are equally likely to plot either
above or below the mean. Grouping
takes place at values that are close to
the
mean
and
then
tails
off
symmetrically away from the mean.

Normal Distribution

The normal distribution is represented
by a bell-shaped curve called the normal
curve and the area under the curve
represents the probability of the normal
random variable X.
Situation: Normal Distribution
Let X be the height of a student from
DLSU and X is a normal random variable.
Suppose that the mean height of all DLSU
students is 𝜇 = 163𝑐𝑚 with standard
deviation 𝜎 = 20𝑐𝑚.
𝝁−𝝈
𝝁 − 𝟐𝝈
𝝁 − 𝟑𝝈
143
123
103
𝝁+𝝈
𝝁 + 𝟐𝝈
𝝁 + 𝟑𝝈
183
203
223
Situation: Normal Distribution
Properties of the Normal Curve
It is a bell-shaped curve.
 The mode, which is the point on the
horizontal axis where the curve is a
maximum, occurs at x = μ. This means
that the mean is equal to the mode.
 The curve is symmetric about a vertical
axis through the mean μ. The mean
divides the set of data into two equal
parts. This means that the mean is
equal to the median.

Properties of the Normal Curve
The normal curve approaches the
horizontal axis asymptotically as we
proceed in either direction away from
the mean. (The graph approaches the xaxis but the graph will never intersect
the x-axis).
 The total area under the curve and
above the horizontal axis is equal to 1.

Formula: Normal Distribution
The formula for the normal distribution is
given by
Comparing Normal Curves
Consider the figure below.
Comparing Normal Curves
Observe that the blue, red and yellow
normal curves have the same mean
because they are centered at 𝜇 = 0 but
with different heights because of different
frequencies. However, the green normal
curve has mean 𝜇 = −2.
Comparing normal curves
Moreover, all the normal curves shown
have different variances which measures
the dispersion or spread of the values in
the data set. This is illustrated by the width
of the curve. It can be seen from the
figure, that the yellow normal curve has
the largest width and with variance 𝜎 2 = 5.
Comparing Normal Curves
To avoid having normal distributions with
different
means
and
standard
deviations, we convert the normal
random variable X into the standard
normal random variable Z.
 The standard normal random variable Z
has mean equal to zero ( 𝜇 = 0 ) and
standard deviation equal to one (𝜎 = 1).

Standard Normal Distribution
To convert the values of the normal
random variable X to the standard normal
random variable Z, we use the formula
𝑋−𝜇
given by 𝑍 =
.
𝜎
Example: From X to Z
Suppose that X is the normal random
variable with mean 𝜇 = 550
and
standard deviation 𝜎 = 150. What is the
probability that X is greater than 600?
 In symbols, we have 𝑃(𝑋 > 600).


𝑋−𝜇
,
𝜎
Using the formula 𝑍 =
substitute
𝑋 = 600, 𝜇 = 550 and 𝜎 = 150.
Example 1: Normal Distribution
Hupper Corporation produces many types of
softdrinks, including Orange Cola. The filling
machines are adjusted to pour 12 ounces of
soda into each 12-ounce can of Orange cola.
However, the actual amount of soda poured
into each can is not exactly 12 ounces; it
varies from can to can. It has been observed
that the net amount of soda in such a can has
a normal distribution with a mean of 12
ounces and a standard deviation of 0.015
ounce.
Example 1: Normal Distribution
A.
What is the probability that a randomly
selected can of Orange Cola contains
11.97 to 11.99 ounces of soda?
Probability for a Range
From X Value
11.97
To X Value
11.99
Z Value for 11.97
-2
Z Value for 11.99
-0.666667
P(X<=11.97)
0.0228
P(X<=11.99)
0.2525
P(11.97<=X<=11.99)
0.2297
Example 1: Normal Distribution
The probability that a can of Orange
Cola will have between 11.97 to 11.99
ounces is 0.2297.
 If there are 1000 cans of Orange Cola,
how many of the 1000 cans will have
between 11.97 to 11.99 ounces?
 The
number
of
cans
is
0.2297 1000 = 229.7
or
approximately 230 cans.

Example 1: Normal Distribution
Suppose we delete the last row of the
PhStat output. How do we get
𝑃(11.97 ≤ 𝑋 ≤ 11.99)?
Probability for a Range
From X Value
11.97
To X Value
11.99
Z Value for 11.97
-2
Z Value for 11.99
-0.666667
P(X<=11.97)
0.0228
P(X<=11.99)
0.2525
Example 1: Normal Distribution
To determine the probability, we have
 𝑃 𝑋 ≤ 11.99 − 𝑃 𝑋 ≤ 11.97
 0.2525 − 0.0228 = 0.2297

Example 1: Normal Distribution
What percentage of the cola contains at
least 12.025 ounces of soda?
 𝑃 𝑋 > 12.025 = 0.0478
B.
Probability for X <=
X Value
12.025
Z Value
1.6666667
P(X<=12.025) 0.9522096
Probability for X >
X Value
12.025
Z Value
1.6666667
P(X>12.025)
0.0478
Example 1: Normal Distribution
Suppose that the output below is the only
given output.
Probability for X <=
X Value
12.025
Z Value
1.6666667
P(X<=12.025) 0.9522096
Example 1: Normal Distribution
𝑃 𝑋 > 12.025 = 1 − 𝑃(𝑋 ≤ 12.025)
 𝑃 𝑋 > 12.025 = 1 − 0.9522 = 0.0478
 The probability that the cola will have at
least 12.025 ounces of soda is 0.0478.

Example 2: Normal Distribution
The price of diesel oil over the past 24
months is normally distributed with a mean
of 41 pesos per liter and standard
deviation of 5 pesos per liter.
Example 2: Normal Distribution
What is the probability that the price of
diesel is at most 34 pesos per liter?
 𝑃 𝑋 ≤ 34 = 0.0808
A.
Probability for X <=
X Value
34
Z Value
-1.4
P(X<=34)
0.0808
Example 2: Normal Distribution
B.
For which amount can we find the
highest 10% of the diesel prices?
Find X and Z Given Cum. Pctage.
Cumulative Percentage 10.00%
Z Value
-1.2816
X Value
34.592
Find X and Z Given Cum. Pctage.
Cumulative Percentage 90.00%
Z Value
1.2816
X Value
47.408
Example 2: Normal Distribution
Since the highest 10% of the diesel
prices occurs at the rightmost part of the
normal curve, we get the area at the left
which is 90%.
 Thus, we use the table with cumulative
percentage equal to 90%.

Example 2: Normal Distribution
Getting the X value, we have 47.408.
 This means that that the probability that
diesel prices is more than 47.408 pesos
per liter is 10% or 0.10.

Example 2: Normal Distribution
Suppose that we delete the last row of the
table of cumulative percentage. How do
we find the value of X?
Find X and Z Given Cum. Pctage.
Cumulative Percentage 90.00%
Z Value
1.2816
X Value
47.408
Example 2: Normal Distribution

Using the formula 𝑍 =
𝑋−𝜇
,
𝜎
we derive X.
Then 𝑋 = 𝑍𝜎 + 𝜇.
 𝑋 = 1.2816 5 + 41 = 47.408

Exercises: Normal Distribution
1.
The TV ratings of the show The Big
Bang
Theory
are
approximately
normally distributed with mean 22.7 and
standard deviation 7.4.
Probability for X <=
X Value
27.6
Z Value
0.7027027
P(X<=27.6)
0.7588795
Probability for X >
X Value
26.1
Z Value
0.5
P(X>26.1)
0.3085
Exercises: Normal Distribution
What is the probability that for a given day
the show The Big Bang Theory will obtain
a rating of at least 27.6?
B. Given 15 episodes of the show The Big
Bang Theory, how many episodes would
get a rating of less than 26.1?
A.
Exercises: Normal Distribution
2.
In the November 1990 issue of
Chemical Engineering Progress, a
study discussed the percent purity of
oxygen from a certain supplier. Assume
that the mean was 99.61 with a
standard deviation of 0.08. Assume that
the distribution of percent purity was
approximately normal.
Exercises: Normal Distribution
A.
What percentage of the purity values
would you expect to be between 99.5
and 99.7?
Probability for X <=
X Value
99.5
Z Value
-1.375
P(X<=99.5) 0.0845657
Probability for X >
X Value
99.7
Z Value
1.125
P(X>99.7)
0.1303
Probability for a Range
From X Value
99.5
To X Value
99.7
Z Value for 99.5 -1.375
Z Value for 99.7
1.125
P(X<=99.5)
0.0846
P(X<=99.7)
0.8697
Exercises: Normal Distribution
B.
What purity value would you expect to
exceed exactly 5% of the population?
Find X and Z Given Cum. Pctage.
Cumulative Percentage
5.00%
Z Value
-1.644854
Find X and Z Given Cum. Pctage.
Cumulative Percentage
95.00%
Z Value
1.644854