AP Statistics
Empirical Rule
What does a population that is
normally distributed look like?
 = 80 and  = 10
X
50
60
70
80
90
100
110
Normal Distribution Features
• Symmetric, single-peaked, and bell-shaped.
• Reported: N   ,  
• Changing µ changes the location of the curve on
the horizontal axis with no change in spread.
• Changing  changes the spread of the curve.
•  is located at the inflection point of the curve.
68-95-99.7 Rule
68%
95%
99.7%
68-95-99.7% RULE
68-95-99.7 Rule—restated
68% of the observations fall within 1
standard deviation of the mean in either
direction.
95% of the observations fall within 2
standard deviations of the mean in either
direction.
99.7% of the observations fall within 3
standard deviations of the mean in either
direction.
68-95-99.7 Rule
34% 34%
68%
47.5%
47.5%
95%
49.85%
49.85%
99.7%
Average American adult male
height is 69 inches (5’ 9”) tall with a
standard deviation of 2.5 inches.
Let H~N(69, 2.5)
What is the likelihood that a randomly selected
adult male would have a height less than 69 inches?
Answer: P(h < 69) = .50
P represents Probability
h represents one adult
male height
Using the 68-95-99.7 Rule
Let H~N(69, 2.5)
What is the likelihood that a randomly selected adult
male will have a height between 64 and 74 inches?
P(64 < h < 74) = .95
Let H~N(69, 2.5)
What is the likelihood that a randomly selected adult
male would have a height of less than 66.5 inches?
P(h < 66.5) = 1 – (.50 + .34) = .16
OR  .50 - .34 = .16
Let H~N(69, 2.5)
What is the likelihood that a randomly selected adult
male would have a height of greater than 74 inches?
P(h > 74) = 1 – P(h  74)
= 1 – (.50 + .47.5)
= 1 - .975
= .025
Let H~N(69, 2.5)
What is the probability that a randomly selected
adult male would have a height between 64 and
76.5 inches?
P(64 < h < 76.5) = .475 + .4985
= .9735
More about Normal Curves
• Percentiles divide the area of the density
curve into fractions out of 100.
• Median: 50th percentile.
• Q1: 25th percentile.
• Q3: 75th percentile.
Exploring Data: Distributions
The 68-95-99.7 Rule
• The distribution of scores on tests such as
the SAT college entrance exams is close to
normal. Scores on each of the three sections
(math, critical reading, writing) of the SAT are
adjusted so that the mean score is about 500
and the standard deviation is about 100.
Using this information answer the following
questions.
The 68-95-99.7 Rule
•
•
•
•
•
•
•
A) How high must a student score to fall in
the top 25%?
This question is asking for the third quartile.
Q3 = (.67)(standard deviation)
= (.67)(100)
= 67 points above the mean
= 500 + 67 = 567
Scores above 567 are in the top 25%.
The 68-95-99.7 Rule
•
•
•
•
•
B)
What percent of scores fall
between 200 and 800?
These scores are 3 standard
deviations from the mean.
200 = 500 – 3(100)
800 = 500 + 3(100)
The percent of scores are 99.7%.
Why are Normal Curves Important?
• They provide good descriptions of some
distributions of real data (SAT, ACT,
characteristics of biological populations).
• They are a good approximation of the
results of chance outcomes (sum the two
die rolls).
• Many statistical procedures are based on
normal distributions.
However!
• Many distributions are not normal
(income). Don’t assume every situation is
approximated by a normal distribution.