Normal Distribution
Z-scores put to use!
Section 2.2
Reference Text:
The Practice of Statistics, Fourth Edition.
Starnes, Yates, Moore
Today’s Objectives
• The 68-95-99.7 Rule
• State mean an standard deviation for The
Standard Normal Distribution
• Given a raw score from a normal distribution,
find the standardized “z-score”
• Use the Table of Standard Normal
Probabilities to find the area under a given
section of the Standard Normal curve.
Standard Deviation
• In statistics the standard deviation (SD) (represented by the
Greek letter sigma, σ) measures the amount of variation or
dispersion from the average.
• A low standard deviation indicates that the data points tend to
be very close to the mean
• A high standard deviation indicates that the data points are
spread out over a large range of values.
The 68-95-99.7 Rule
• How many standard deviations do you think it
would take for us to have the entire sample or
population accounted for and just have a .03%
uncertainty?
• In other words, how many standard deviations
away from the mean encompasses almost all
objects in the study?
3

The 68-95-99.7 Rule
Although there are many Normal curves, they all have properties
in common.
Definition:
The 68-95-99.7 Rule (“The Empirical Rule”)
•Approximately 68% of the observations fall within σ of µ.
•Approximately 95% of the observations fall within 2σ of µ.
•Approximately 99.7% of the observations fall within 3σ of µ.
Normal Distributions
In the Normal distribution with mean µ and standard deviation σ:
The 68-95-99.7 Rule
• If I have data within 2 standard deviations,
then I'm accounting for 95% of
observations
• Question: what percent is in the left tail?
2.5%
2.5%
You Try!
• The distribution of number of movies AP Statistic students
watch in two weeks is close to normal. Suppose the
distribution is exactly Normal with mean 𝜇 = 6.84 and
standard deviation σ = 1.55 (this is non fiction data)
• A) Sketch a normal density curve for this distribution of movies
watched. Label the points that are one, two, and three SD
away from the mean.
• B) What percent of the movies is less that 3.74? Show your
work!
• C) What percent of scores are between 5.29 and 9.94? Show
work!
• Remember: Always put your answers back into context!
Complete sentence!
Example, p. 113
The distribution of Iowa Test of Basic Skills (ITBS) vocabulary
scores for 7th grade students in Gary, Indiana, is close to
Normal. Suppose the distribution is N(6.84, 1.55).
b) What percent of ITBS vocabulary scores are less than 3.74?
c) What percent of the scores are between 5.29 and 9.94?
Normal Distributions
a) Sketch the Normal density curve for this distribution.
Standardizing Observations
• All normal distributions have fundamentally the same
shape.
• If we measure the x axis in units of size σ about a
center of 0, then they are all exactly the same curve.
• This is called the Standard Normal Curve
– We abbreviate the normal dist. As 𝑁(𝜇, 𝜎)
• To standardize observations, we change from x values
(the raw observations)  z values (the standardized
observations) by the formula:
z
x

The Standard Normal Distribution
• Notice that the z-score formula always
subtracts μ from each observation.
– So the mean is always shifted to zero 𝑥 − 𝜇
• Also notice that the shifted values are
divided by σ, the standard deviation.
– So the units along the z-axis represent
numbers of standard deviations
𝜎
• Thus the Standard Normal Distribution is
always N(0,1).
Example!
• The heights of young women are:
N(64.5, 2.5)
• Use the formula to find the z-score of a
woman 68 inches tall.
68  64.5
z
 1.4
2.5
• A woman’s standardized height is the
number of standard deviations by which
her height differs from the mean height of
all young women.
Break!
- 5 Minutes
Normal Distribution Calculations
• What proportion of all young women are less
than 68 inches tall?
– Notice that this does not fall conveniently on one of the σ
borders
– We already found that 68 inches corresponds
to a z-score of 1.4
• So what proportion of all standardized
observations fall to the left of z = 1.4?
• Since the area under the Standard Normal
Curve is always 1, we can ask instead, what is
the area under the curve and to the left of z=1.4
– For that, we need a table!!
The Standard Normal Table
• Find Table A of the handout
– It is also in your textbook in the very back
• Z-scores (to the nearest tenth) are in the left
column
– The other 10 columns round z to the nearest hundredth
• Find z = 1.4 in the table and read the area
– You should find area to the left = .9192
• So the proportion of observations less than z =
1.4 is about 91.92%
– Now put the answer in context: “About 91.92% of all
young women are 68 inches tall or less.”
• The Standard Normal Table
Because all Normal distributions are the same when we
standardize, we can find areas under any Normal curve from
a single table.
The Standard Normal Table
Table A is a table of areas under the standard Normal curve. The table
entry for each value z is the area under the curve to the left of z.
Suppose we want to find the
proportion of observations from the
standard Normal distribution that are
less than 0.81.
We can use Table A:
Z
.00
.01
.02
0.7
.7580
.7611
.7642
0.8
.7881
.7910
.7939
0.9
.8159
.8186
.8212
Normal Distributions
Definition:
P(z < 0.81) = .7910
What about area above a value?
• Still using the N(64.5, 2.5) distribution, what
proportion of young women have a height of
61.5 inches or taller?
𝑥 − 𝜇 61.5 − 64.5
𝑧=
=
= −1.2
𝜎
2.5
• From Table A, area to the left of -1.2 =.1151
– So area to the right = 1 - .1151 = .8849
• So about 88.49% of young women are 61.5”
tall or taller.
What about area between two
values?
• What proportion of young women are between 61.5”
and 68” tall?
The facts:
• We already know 68” gives z = 1.4 and area to the
left of .9192
• We also know 61.5” gives z = -1.2 and area to the
left of .1151
• So just subtract the two, and find the proportion
between 61.5” and 68”: .9192 - .1151 = .8041
• So about 80.41% of young women are between
61.5” and 68” tall
– Remember to write your answer IN CONTEXT!!!
Example, p. 117
• Finding Areas Under the Standard Normal Curve
Find the proportion of observations from the standard Normal distribution that
are between -1.25 and 0.81.
Normal Distributions
Can you find the same proportion using a different approach?
1 - (0.1056+0.2090) = 1 – 0.3146
= 0.6854
Given a proportion, find the observation x
(Working Backwards!)
• SAT Verbal scores are N(505, 110). How high must you
score to be in the top 10%?
• If you are in the top 10%, there must be 90% below you
(to the left).
• Find .90 (or close to it) in the body of Table A. What is
the z-score?
– You should have found z = 1.28
• Now solve the z definition equation for x
𝑥−𝜇
𝜎
𝑥 − 505
1.28 =
110
𝑥 = 1.28 ∗ 110 + 505
𝑥 =645.8
𝑧=
• So you need a score of at least 646 to be in the top 10%.
How to Solve Problems
Involving Normal Distribution
• State: Express the problem in terms of the observed
variable x (list what x represents, mean? SD?)
– (get your ducks in order to work with)
• Plan: Draw a picture of the distribution and shade the
area of interest under the curve.
• Do: Preform the calculations
– Standardize x to restate the problem in terms of standard
normal variable z
– Use Table A and the fact that the total area under the curve is 1
to find the required area under the standard normal curve
• Conclude: Write your conclusion in context of the
problem.
• Lets look at TB pg 120 “Tiger on the Range”
• Normal Distribution Calculations
When Tiger Woods hits his driver, the distance the ball travels can be
described by N(304, 8). What percent of Tiger’s drives travel between 305
and 325 yards?

305 - 304
 0.13
8
When x = 325, z =
325 - 304
 2.63
8
Normal Distributions

When x = 305, z =
Using Table A, we can find the area to the left of z=2.63 and the area to the left of z=0.13.
0.9957 – 0.5517 = 0.4440. About 44% of Tiger’s drives travel between 305 and 325 yards.
Video explaination
• https://www.youtube.com/watch?v=85G_P
LBTX00
Today’s Objectives
• The 68-95-99.7 Rule
• State mean an standard deviation for The
Standard Normal Distribution
• Given a raw score from a normal
distribution, find the standardized “z-score”
• Use the Table of Standard Normal
Probabilities to find the area under a given
section of the Standard Normal curve.
Test Results!
•
•
•
•
•
•
Grade:
Amount: Marginal %
……A......…….....5……….26%
…….B…………...5……...26% 84% Passed
…….C…………...6……...32%
…….D…………...2……...11%
…….F…..............1………..5% 16% Failed
• Mean: 81% Max: 100% Min: 50% No Outliers
Tracking AP Stats
• 2014-2015 (WHS)
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Ch. 1 Test
A5
B5
C6
D2
F1
Ch. 2 Test
Ch. 3 Test
14/15 VS 15/16 AP Stats
14/15 15/16
14/15 15/16
14/15 15/16
14/15 15/16
14/15 15/16
Chapter 1
Test
Chapter 2
Test
Chapter 3
Test
Chapter 4
Test
Chapter 5
Test
A -5
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B-5
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C-6
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F-1
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Homework
Worksheet 1-28 (Multiples of 3)
EXTRA CREDIT: You MAY get ONLY 2 additional
points added to your test grade for completing
the entire assignment 1-28.
However, regardless of your score, the maximum
you may achieve is an 80%.
Sooo, if you have a 79% and you do the extra
credit…your test grade will max at 80%