The Normal Distribution

Continuous Probability Distributions

■ A continuous variable is a variable that can

assume any value on a continuum (can assume
an uncountable number of values)
■ thickness of an item
■ time required to complete a task
■ temperature of a solution
■ height, in inches

■ These can potentially take on any value

depending only on the ability to precisely and
accurately measure
The Normal Distribution

■ Bell Shaped
■ Symmetrical f(X)
■ Mean, Median and Mode
are Equal
Location is determined by the σ
mean, μ X
Spread is determined by the -∞ μ +∞
standard deviation, σ
Mean
The random variable has an = Median
infinite theoretical range: = Mode
+ ∞ to − ∞
By varying the parameters μ and σ, we
obtain different normal distributions
A
B
C

A and B have the same mean but different standard deviations.

B and C have different means and different standard deviations.
The Normal Distribution Shape

f(X) Changing μ shifts the

distribution left or right.

Changing σ increases
or decreases the
σ spread.

μ X
The Standardized Normal

■ Any normal distribution (with any mean and

standard deviation combination) can be
transformed into the standardized normal
distribution (Z)

■ To compute normal probabilities need to

transform X units into Z units

■ The standardized normal distribution (Z) has a

mean of 0 and a standard deviation of 1
Translation to the Standardized
Normal Distribution

■ Translate from X to the standardized normal

(the “Z” distribution) by subtracting the mean
of X and dividing by its standard deviation:

The Z distribution always has mean = 0 and

standard deviation = 1
The Standardized
Normal Distribution

■ Also known as the “Z” distribution

■ Mean is 0
■ Standard Deviation is 1
f(Z)

1
Z
0
Values above the mean have positive Z-values.
Values below the mean have negative Z-values.
Example

■ If X is distributed normally with mean of $100

and standard deviation of $50, the Z value for
X = $200 is

■ This says that X = $200 is two standard

deviations (2 increments of $50 units) above
the mean of $100.
Comparing X and Z units

$100 $200 $X (μ = $100, σ = $50)

0 2.0 Z (μ = 0, σ = 1)
Note that the shape of the distribution is the same,
only the scale has changed. We can express the
problem in the original units (X in dollars) or in
standardized units (Z)
Finding Normal Probabilities

Probability is measured by the area under

the curve
f(X) P (a ≤ X ≤ b)
= P (a < X < b)
(Note that the probability
of any individual value is
zero)

a b X
Probability as
Area Under the Curve
The total area under the curve is 1.0, and the curve is
symmetric, so half is above the mean, half is below

f(X)

0.5 0.5

μ X
The Standardized Normal Table

■ The Cumulative Standardized Normal table in

\gives the probability less than a desired value
of Z (i.e., from negative infinity to Z)

0.9772
Example:
P(Z < 2.00) = 0.9772

0 2.00 Z
https://www.z-table.com/
The Standardized Normal Table
(continued)

The column gives the value of

Z to the second decimal point
Z 0.00 0.01 0.02 …

0.0
The row shows
the value of Z 0.1
to the first . The value within the
.
decimal point . table gives the
2.0 .9772 probability from Z = − ∞
up to the desired Z
2.0 value
P(Z < 2.00) = 0.9772
General Procedure for Finding
Normal Probabilities

To find P(a < X < b) when X is

distributed normally:
■ Draw the normal curve for the problem in
terms of X

■ Translate X-values to Z-values

■ Use the Standardized Normal Table

Finding Normal Probabilities
■ Let X represent the time it takes (in seconds) to
download an image file from the internet.
■ Suppose X is normal with a mean of18.0
seconds and a standard deviation of 5.0
seconds. Find P(X < 18.6)

X
18.0
18.6
Finding Normal Probabilities
(continued)
■ Let X represent the time it takes, in seconds to download an image file from
the internet.
■ Suppose X is normal with a mean of 18.0 seconds and a standard deviation
of 5.0 seconds. Find P(X < 18.6)

μ = 18 μ=0
σ=5 σ=1

18 18.6 X 0 0.12 Z

P(X < 18.6) P(Z < 0.12)

Solution: Finding P(Z < 0.12)

Standardized Normal Probability P(X < 18.6)

Table (Portion) = P(Z < 0.12)
Z .00 .01 .02 0.5478
0.0 .5000 .5040 .5080

0.1 .5398 .5438 .5478

0.2 .5793 .5832 .5871
Z
0.3 .6179 .6217 .6255 0.00
0.12
Finding Normal
Upper Tail Probabilities

■ Suppose X is normal with mean 18.0

and standard deviation 5.0.
■ Now Find P(X > 18.6)

X
18.0
18.6
Finding Normal
Upper Tail Probabilities
(continued)
■ Now Find P(X > 18.6)…
P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z ≤ 0.12)
= 1.0 - 0.5478 = 0.4522

0.5478
1.000 1.0 - 0.5478
= 0.4522

Z Z
0 0
0.12 0.12
Finding a Normal Probability
Between Two Values

■ Suppose X is normal with mean 18.0 and

standard deviation 5.0. Find P(18 < X < 18.6)
Calculate Z-values:

18 18.6 X
0 0.12 Z
P(18 < X < 18.6)
= P(0 < Z < 0.12)
 Solution: Finding P(0 < Z < 0.12)

Standardized Normal Probability P(18 < X < 18.6)

Table (Portion) = P(0 < Z < 0.12)
= P(Z < 0.12) – P(Z ≤ 0)
Z .00 .01 .02 = 0.5478 - 0.5000 = 0.0478
0.0 .5000 .5040 .5080 0.0478
0.5000
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871

0.3 .6179 .6217 .6255 Z

0.00
0.12
Probabilities in the Lower Tail

■ Suppose X is normal with mean 18.0

and standard deviation 5.0.
■ Now Find P(17.4 < X < 18)

X
18.0
17.4
 Probabilities in the Lower Tail
(continued)

Now Find P(17.4 < X < 18)…

P(17.4 < X < 18)
= P(-0.12 < Z < 0)
0.0478
= P(Z < 0) – P(Z ≤ -0.12)
= 0.5000 - 0.4522 = 0.0478
0.4522

The Normal distribution is

symmetric, so this probability
17.4 18.0 X
is the same as P(0 < Z < 0.12) Z
-0.12 0
Empirical Rule
What can we say about the distribution of values
around the mean? For any normal distribution:

f(X)
μ ± 1σ encloses about
68.26% of X’s
σ σ

μ-1σ μ μ+1σ X
68.26%
The Empirical Rule
(continued)

■ μ ± 2σ covers about 95.44% of X’s

■ μ ± 3σ covers about 99.73% of X’s

2σ 2σ 3σ 3σ
μ x μ x
95.44% 99.73%
Given a Normal Probability
Find the X Value

■ Steps to find the X value for a known

probability:
1. Find the Z value for the known probability
2. Convert to X units using the formula:
Finding the X value for a Known
Probability (continued)

Example:
■ Let X represent the time it takes (in seconds) to
download an image file from the internet.
■ Suppose X is normal with mean 18.0 and standard
deviation 5.0
■ Find X such that 20% of download times are less than
X.
0.2000

? 18.0 X
? 0 Z
 Find the Z value for
20% in the Lower Tail
1. Find the Z value for the known probability
Standardized Normal Probability ■ 20% area in the lower
Table (Portion) tail is consistent with a
Z value of -0.84
Z … .03 .04 .05
-0.9 … .1762 .1736 .1711
0.2000
-0.8 … .2033 .2005 .1977
-0.7 … .2327 .2296 .2266
? 18.0 X
-0.84 0 Z
Finding the X value

2. Convert to X units using the formula:

So 20% of the values from a distribution

with mean 18.0 and standard deviation
5.0 are less than 13.80
Finding the X value
Find X such that 80% of download times
are greater than X.

If 20% of the values have DL times less

than 13.8 seconds, then 80% must have
DL times greater than 13.8 seconds
 Area of Standard Normal Distribution

P(Z a) area P(Z -a ) area

within the yellow within the blue
shade is P(a) shade is P(a)

The AREA on the LEFT of the value of z is the

PROBABILITY
33
 The area within the blue shade is 1 -
P(a)

The area within the blue shade is P(b) –

P(a) 34
 More Examples:
1. For some computers, the time period between charges
of the battery is normally distributed with a mean of 50
hours and a standard deviation of 15 hours. John has one
of these computers and needs to know the probability
that the time period will be between 50 and 70 hours.

Let x = random variable that represents the time

period. μ= 50 σ = 15
Formula: z = (X – μ) / σ
For x = 50 , z = (50 – 50) / 15 = 0

For x = 70 , z = (70 – 50) / 15 = 1.33

35
 0 1.33
50 70
P(yellow area) = P(1.33) – P(0)
Using the z-table (https://www.z-table.com/)

P(yellow area) = 0.9082– 0.500 = 0.4082

The probability that John’s computer has a time
period between 50 and 70 hours is equal to 0.4082 or
40.82% 36
2. The annual salaries of employees in a large company
are approximately normally distributed with a mean of
50,000 and a standard deviation of 20,000. What percent
of people earn between 45,000 and 65,000?.

Let x = random variable that represents the time

period. μ= 50,000 σ = 20,000

Formula: z = (X – μ) / σ
For x = 45,000 , z = (45,000 – 50,000) / 20,000 =
-0.25
For x = 65,000 , z = (65,000 – 50,000) / 20,000 = 0.75

37
Examples:

-0.25 0.75
P(yellow area) = P(0.75) – P(-0.25)
Using the z-table (https://www.z-table.com/)
P(yellow area) = 0.7734– 0.4013 = 0.372
The percent of people earning between 45,000 and
65,000 is 37.2%
38
Examples:
3. The speeds of cars are measured using a radar unit, on
a motorway. The speeds are normally distributed with a
mean of 90 km/hr and a standard deviation of 10 km/hr.
What is the probability that a car selected at chance is
moving at more than 100 km/hr?

Let x = random variable that represents the time

period. μ= 90 σ = 10

Formula: z = (X – μ) / σ
For x = 100 , z = (100 – 90) / 10 =1

39
Examples:

1
P(yellow area) = 1 – P(1)
Using the z-table (https://www.z-table.com/)
P(yellow area) = 1 – 0.8413 = 0.1587
The probability that a car is moving more than 100
km/hr is 15.87%
40
4. The scores on standardized admissions test are normally
distributed with a mean of 500 and a standard deviation of
100. What is the probability that a randomly selected student
will score between 400 and 600 on the test?
Let x = random variable that represents the time
period. μ= 500 σ = 100
Formula: z = (X – μ) / σ
For x = 400 , z = (400 – 500) / 100 = -1

For x = 600 , z = (600 – 500) / 100 = 1

41
Examples:

-1 1
P(yellow area) = P(1) – P(-1)
Using the z-table (https://www.z-table.com/)
P(yellow area) = 0.8413 - 0.1587 = 0.6826
The probability that a randomly selected student
will score between 400 and 600 on the test is
68.26% 42
 MORE Examples:
Ans: 59.48%
4. Most graduate schools of business require applicants for
admission to take the Graduate Management Admission
Council’s GMAT examination. Scores on the GMAT are
roughly normally distributed with a mean of 527 and a standard
deviation of 112. What is the probability of an individual scoring
above 500 on the GMAT?

500 527
For x = 500 , z = (500 – 527) / 112 = -0.24 43

P(yellow area) = 1-0.4052 = 0.5948 or 59.48%

 Z-table Link

https://www.z-table.com/

More problems and exercises:

https://mathcenter.oxford.emory.edu/
site/math117/probSetNormalDistributi
on/