The document discusses the standard normal distribution. It defines the standard normal distribution as having a mean of 0, a standard deviation of 1, and a bell-shaped curve. It provides examples of how to find probabilities and z-scores using the standard normal distribution table or calculator. For example, it shows how to find the probability of an event being below or above a given z-score, or between two z-scores. It also shows how to find the z-score corresponding to a given cumulative probability.

1. Elementary Statistics
Chapter 6:
Normal Probability
Distributions
6.1 The Standard
Normal Distribution
1

2. Chapter 6: Normal Probability Distribution
6.1 The Standard Normal Distribution
6.2 Real Applications of Normal Distributions
6.3 Sampling Distributions and Estimators
6.4 The Central Limit Theorem
6.5 Assessing Normality
6.6 Normal as Approximation to Binomial
2
Objectives:
• Identify distributions as symmetric or skewed.
• Identify the properties of a normal distribution.
• Find the area under the standard normal distribution, given various z values.
• Find probabilities for a normally distributed variable by transforming it into a standard normal variable.
• Find specific data values for given percentages, using the standard normal distribution.
• Use the central limit theorem to solve problems involving sample means for large samples.
• Use the normal approximation to compute probabilities for a binomial variable.

3. Probability Distribution Requirements (3):
1. There is a numerical (not categorical) random variable (RV) x, and its values are
associated with corresponding probabilities.
2. ∑P(x) = 1 where x assumes all possible values. (The sum of all probabilities must be 1,
but sums such as 0.999 or 1.001 are acceptable because they result from rounding errors.)
3. 0 ≤ P(x) ≤ 1 for every individual value of the random variable x. (That is, each
probability value must be between 0 and 1 inclusive.)
3
Recall: 5.1 Probability Distributions
Mean: ( ) ( )E x x p x   
Parameters of a Probability Distribution
2 2 2
SD: ( ) ( ) ( )x p x x p x      
2 2 2 2
Variance: ( ) ( ) ( )x p x x p x         

4. p = probability of success, q = 1 − p = probability of failure
Make sure that the values of p and x refer to the same category called a success (x
and p are consistent).
x: A specific number of successes in n trials: 0 ≤ 𝑥(𝑎 𝑤ℎ𝑜𝑙𝑒 #) ≤ 𝑛 & P(x):
probability of getting exactly x successes among the n trials
The word success as used here is arbitrary and does not necessarily represent
something good. Either of the two possible categories may be called the success S as
long as its probability is identified as p.
Parameters: n & p
Binomial Experiment and P. D. (properties)
1. n identical ( fixed # of) trials (Each repetition of the experiment)
2. Each has only 2 categories of outcomes
3. Probability stays constant
4. Independent trials
Recall: 5.2 Binomial Probability Distributions
( )
(1 )
n x n x
xp x C p q
np
npq
np p






 
4
2
Variance:  npq

5. Key Concept: In this section, we introduce Poisson probability distributions, which are another category of
discrete probability distributions.
A Poisson probability distribution is a discrete probability distribution that applies to occurrences of some event
over a specified interval. The random variable x is the number of occurrences of the event in an interval. The
interval can be time, distance, area, volume, or some similar unit.
The probability of the event occurring x times over an interval is given by : 𝑃 𝑥 =
𝜇 𝑥∙𝑒−𝜇
𝑥!
where e  2.71828, µ =
mean number of occurrences of the event in the intervals
Parameters: 𝜇 𝑎𝑛𝑑 𝜎 = 𝜇
Properties of the Poisson Probability Distribution
• A particular Poisson distribution is determined only by the mean, µ.
• A Poisson distribution has possible x values of 0, 1, 2, . . . with no upper limit.
Requirements:
1. The random variable x is the number of occurrences of an event in some interval.
2. The occurrences must be random.
3. The occurrences must be independent of each other.
4. The occurrences must be uniformly distributed over the interval being used.
Recall: 5.3 Poisson Probability Distributions
5

6. Key Concept: the standard normal distribution is a specific normal distribution having the
following three properties:
1. Bell-shaped: The graph of the standard normal distribution is bell-shaped.
2. µ = 0: The standard normal distribution has a mean equal to 0.
3. σ = 1: The standard normal distribution has a standard deviation equal to 1.
In this section we develop the skill to find areas (or probabilities or relative frequencies)
corresponding to various regions under the graph of the standard normal distribution. In
addition, we find z scores that correspond to areas under the graph.
For a continuous random variable, the probability of a single value of x is always zero.
6.1 The Standard Normal Distribution
Normal Distribution
If a continuous random variable has a distribution with a graph that is
symmetric and bell-shaped, we say that it has a normal distribution.
6

7. The mathematical equation for the normal distribution is:
The shape and position of the normal distribution curve depend on two parameters,
the mean and the standard deviation.
Each normally distributed variable has its own normal distribution curve, which
depends on the values of the variable’s mean and standard deviation.
7
6.1 The Standard Normal Distribution
2
1
2
2
x
e
y


 
 
  
 

2.718, 3.14
population mean
population standard deviation
e 


 


2 2
( ) (2 )
2
x
e
y
 
 
 


8. • The normal distribution curve is bell-shaped.
• The mean, median, and mode are equal and located at the center of the distribution.
• The normal distribution curve is Unimodal (i.e., it has only one mode).
• The curve is symmetrical about the mean, which is equivalent to saying that its shape
is the same on both sides of a vertical line passing through the center.
• The curve is continuous—i.e., there are no gaps or holes.
• The curve never touches the x-axis. Theoretically, no matter how far in either
direction the curve extends, it never meets the x-axis—but it gets increasingly
closer.
• The total area under the curve = 1.00 or 100%.
• The area under the normal curve that lies within
 one standard deviation of the mean is approximately 0.68 (68%).
 two standard deviations of the mean is approximately 0.95 (95%).
 three standard deviations of the mean is approximately 0.997 ( 99.7%).
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6.1 The Standard Normal Distribution
2 2
( ) (2 )
2
x
e
y
 
 
 


9. Properties of uniform distribution:
1. The area under the graph of a continuous probability distribution is equal to 1.
2. There is a correspondence between area and probability, so probabilities can be found by
identifying the corresponding areas in the graph using this formula for the area of a
rectangle:
Area = height × width
• Uniform Distribution
 A continuous random variable has a uniform distribution if its values are spread evenly
over the range of possibilities. The graph of a uniform distribution results in a rectangular
shape.
• Density Curve
 The graph of any continuous probability distribution is called a density curve, and any
density curve must satisfy the requirement that the total area under the curve is exactly 1.
Because the total area under any density curve is equal to 1, there is a correspondence
between area and probability.
9
Uniform Distribution6.1 The Standard Normal Distribution

10. Example 1
At JFK airport in New York City, passengers arriving at the security checkpoint have waiting
times that are uniformly distributed between 0 minutes and 5 minutes. All of the different
possible waiting times are equally likely. Waiting times can be any value between 0 min and 5
min (it’s a continuous value), so it is possible to have a waiting time of 1.234567 min.
Find the probability that a randomly selected passenger has a waiting time of at least 2
minutes.
10
Uniform Distribution
The total area under the curve = 1.00
Height (the probability) = 1/width→ ℎ =
1
5
= 0.2
Solution: P(wait time of at least 2 min) = height × width of shaded area in the
figure = 0.2 × 3 = 0.6

11. Standard Normal Distribution: Parameters of µ = 0 and σ = 1
Since each normally distributed variable has its own mean and standard deviation, the
shape and location of these curves will vary. In practical applications, one would have
to have a table of areas under the curve for each variable. To simplify this, statisticians
use the standard normal distribution.
We can find areas (probabilities) for different regions under a normal model using
technology or a Z-Table. Calculators and software generally give more accurate results
than a Z-Table.
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6.1 The Standard Normal Distribution
value mean
standard deviation
z


x
z





12. Formats Used for Finding Normal Distribution Areas
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1. To the left of any z
value: Look up the z value in the
table and use the area given.
2. To the right of any
z value: Look up the z value
and subtract the area from 1.
3. Between two
z values: Look up
both z values and
subtract the
corresponding areas.

13. A bone mineral density test can help identify the presence or likelihood of osteoporosis
which is measured as a z score: ND(µ = 0 & σ = 1).
Q: A randomly selected adult undergoes a bone density test. Find the probability that
this person has a bone density test score less than 1.27.
13
Example 2 Normal Distribution
Locate 1.2 in the left column and 0.07 in the top row: Read the value
Interpretation: The probability that a randomly
selected person has a bone density test result below
1.27 is 0.8980 (Or: Conclude that 89.80% of people
have bone density levels below 1.27.
( 1.27) 0.8980P z  

14. 14
a. Find the area to the left of z = 2.06.
b. Find the area to the right of z = –1.19.
Example 3 SND: Standard Normal Distribution
( 2.06) 0.9803P z  
( 1.19) 0.8830P z   
TI Calculator:
Normal Distribution Area
1. 2nd + VARS
2. normalcdf(
3. 4 entries required
4. Left bound, Right
bound, value of the
Mean, Standard
deviation
5. Enter
6. For −∞, 𝒖𝒔𝒆 − 𝟏𝟎𝟎𝟎
7. For ∞, 𝒖𝒔𝒆 𝟏𝟎𝟎𝟎

15. 15
a. Find the area between z = +1.68 and z = –1.37.
b. Find the probability: P(0 < z < 2.32)
Example 4 SND: Standard Normal Distribution
(0 2.32)P z  
( 1.37 1.68)P z   
0.9535 0.0853 0.8682 
0.9898 0.5 0.4898  

16. Using the same bone density test, find
a. the probability that a randomly selected person has a result above −1.00 (which is considered to be
in the “normal” range of bone density readings).
b. A bone density reading between −1.00 and −2.50 indicates the subject has osteopenia, which is
some bone loss. Find the probability that a randomly selected subject has a reading between −1.00
and −2.50.
16
Example 5 Normal Distribution
1 − 0.1587 = 0.8413. ( 1)a P z   
Interpretation: The probability of
randomly selecting someone with a bone
density reading above −1 is 0.8413. We
could also say that 84.13% of people have
bone density levels above −1.00.
( 2.5 1)P z    0.1525
Interpretation: There is a probability of
0.1525 that a randomly selected subject has a
bone density reading between −1.00 and −2.50.
Or: 15.25% of people have osteopenia, with
bone density readings between −1.00 and −2.50.

17. Notation & Finding z Scores from Known Areas
P(a < z < b) denotes the probability that the z score is between a and b.
P(z > a) denotes the probability that the z score is greater than a.
P(z < a) denotes the probability that the z score is less than a.
17
Draw a bell-shaped curve and identify the region under the curve that
corresponds to the given probability. If that region is not a cumulative region
from the left, work instead with a known region that is a cumulative region
from the left.
Use technology or the Z-Table find the z score. With Table A-2, use the
cumulative area from the left, locate the closest probability in the body of the
table, and identify the corresponding z score.
Critical Value: For the standard normal distribution, a critical value is a z
score on the borderline separating those z scores that are significantly low or
significantly high.
Notation: The expression zα denotes the z score with an area of α to its right.

18. Find the value of z0.025. (Let α = 0.025 in the expression zα.)
18
Example 6 Normal Distribution
Solution
The notation of z0.025 is used to represent the z score with an area of 0.025 to its right.
Note: that the value of z = 1.96 has an area of 0.025 to its right, so z0.025 = 1.96.
Note that z0.025 corresponds to a cumulative left area of 0.975.
CAUTION:
When finding a value of zα for a particular value of α, note that α is the area to the right of zα,
but the Table and some technologies give cumulative areas to the left of a given z score. To
find the value of zα, resolve that conflict by using the value of 1 − α. For example, to find z0.1,
refer to the z score with an area of 0.9 to its left.
TI Calculator:
Normal Distribution: find
the Z-score
1. 2nd + VARS
2. invNorm(
3. 3 entries required
4. Left Area, value of the
Mean, Standard
deviation
5. Enter

19. 19
Find the z value such that
a. the area under the standard normal distribution
curve between 0 and the z value is 0.2123.
b. Find 𝑃10, the 10th percentile.
Example 7 SND: Standard Normal Distribution
(0 ) 0.2123P z z value   
Add 0.5000 to 0.2123 to get the
cumulative area = 0.7123, why?
The z value is 0.56.
𝑃10 = −1.28, why?
Area = 𝟏𝟎% = 𝟎. 𝟏