The Normal Models/Distributions
Normal Distribution Parameters
The first parameter specified here is shape:
symmetric, unimodal, bell shaped. (A formula
defines the exact shape.)
For a Normal variable the mean  and standard
deviation  are needed to proceed.
All else follows from knowing these three things.
Z-scores (Standardized Values)
X  Mean
Z
StDev
Z measures
“how many standard deviations from the mean”
whether above (+) or below (-) the mean
X  Mean  Z  StDev
This is the same formula, solved for X.
You are required to know these formulas.
Example
The gestation period (length of pregnancy) for
women is Normally distributed with mean
 = 270 days and standard deviation  = 11 days.
Draw a picture!
Example
Use the mean to get started.
Example
On the templates, consecutive tick marks are one
standard deviation apart.
Example
237
248
259
270
281
Length of Pregnancy (Days)
292
303
68 – 95 – 99.7 Rule
also known as the
Empirical Rule
A reference to getting started with Normal
distributions.
About 68% of the data fall within 1 standard
deviation of the mean:   
Example
0.6827
Empirical (68 – 95 – 99.7) Rule
A reference to getting started with Normal
distributions.
About 95% (all but roughly 1 out of 20) of the
data fall within 2 standard deviations of the
mean:   2
Example
0.9544
Empirical (68 – 95 – 99.7) Rule
A reference to getting started with Normal
distributions.
About 99.7% (all but roughly 1 in 370) of the
data fall within 3 standard deviations of the
mean:   3
Example
0.9973
Data more than 3 from 
0.0027 or about 1 in 370 data values (on
average) fall outside   3.
Technically the minimum and maximum are -
and + (infinite). Values far to the right and left
have very very very very small likelihood.
Beyond  + 3
0.00135 =1/741 values more than 3 above 
0.0000317 =1/31574 values
more than 4 above 
If the curve were drawn 1 mm
high at 6 from , then it would
be drawn 65.66 km high at .
Example
The Z scale “goes with” the data scale.
Z = # of standard deviations from the mean.
Example
Determine the probability a pregnancy lasts
at least 274 days (that’s 9 months).
Finding Probabilities
To solve a problem asking for a probability with
for a Normal distribution:
Step 1. Draw a picture identifying the relevant area.
Example
Determine the probability a pregnancy lasts
at least 274 days.
Example
Determine the probability a pregnancy lasts
at least 274 days.
Example
Determine the probability a pregnancy lasts
at least 274 days.
Area = Probability
Finding Probabilities
To solve a problem asking for a probability with
for a Normal distribution:
Step 2. Convert the data value to the Z scale.
Data = 274
 = 270
 = 11 days
274  270
Z
 0.36
11
274 is 0.36 standard deviations above the mean
Example
Determine the probability a pregnancy lasts
at least 274 days.
Finding Probabilities
To solve a problem asking for a probability with
for a Normal distribution:
Step 3. Use the Table (A-2 in the book) or computer
or calculator to determine the percentile for the
value.
A Z-score of 0.36 matches with a probability of
0.6406. That is: Z = 0.36 is the 64.06 percentile.
Example
Determine the probability a pregnancy lasts
at least 274 days.
0.6406
Finding Probabilities
To solve a problem asking for a probability with
for a Normal distribution:
Step 4. Problems asking for the probability of a
result above or between will require further
operations, because of the way percentiles are
oriented.
Example
0.6406
1 – 0.6406
= 0.3594
coincidence
Example - Solution
0.3594 is the probability that a pregnancy lasts at
least 274 days.
That is:
35.94% of all pregnancies last at least 274 days.
(Probability calculations always apply to
populations.)
274 days is the 64 percentile of gestation times.
Example
The gestation period (length of pregnancy) for
women is Normally distributed with mean
 = 270 days and standard deviation  = 11 days.
Example
A women gave birth to a child on October 2, 2007. She
identifies a man as the father. The case is in court.
The man testifies he is not the father, and documents
that he was out of the country from December 19, 2006
to January 30, 2007.
December 19 is 287 days before October 2.
January 30 is 245 days before October 2.
What is the probability a pregnancy lasts between 245
and 287 days?
(If it’s high, then we have evidence – in the data –
supporting the man’s claim that he is not the father.)
Example
What’s the probability the pregnancy lasted
between 245 and 287 days?
Example
What’s the probability the pregnancy lasted
between 245 and 287 days?
Example
What’s the probability the pregnancy lasted
between 245 and 287 days?
Area = Probability
Convert both data values to the Z scale.
Data = 245
245  270
Z
 2.27
11
Data = 287
287  270
Z
 1.55
11
245 is 2.27 standard deviations below the mean
287 is 1.55 standard deviations above the mean
Example
What’s the probability the pregnancy lasted
between 145 and 287 days?
Area = Probability
Example
What’s the probability the pregnancy lasted
between 145 and 287 days?
0.9394 = Area below 287 (1.55)
Example
What’s the probability the pregnancy lasted
between 145 and 287 days?
0.9394 = Area below 287 (1.55)
0.0116 = Area below 245 (-2.27)
Example
What’s the probability the pregnancy lasted
between 145 and 287 days?
0.9394 = Area below 287 (1.55)
0.0116 = Area below 245 (-2.27)
0.9278 = Area between 245 and 287
Example - Solution
0.9278 is the probability that a pregnancy lasts
between 245 and 287 days.
We found 245 to be the 1.16 percentile of pregnancy
lengths.
We found 287 to be the 93.94 percentile.
Example - Solution
0.9266 is the probability that the woman’s
pregnancy had a length that proves the man’s
innocence.
Note: This is not the probability the man is not
the father. In our way of thinking about things,
there is no such thing as “the probability the man
is the father.” Either he is or is not the father.
(We just don’t know.)
Example
Determine the first quartile of the pregnancy
lengths.
First draw a picture.
Example
Determine the first quartile.
0.2500 = Area
Example
Determine the first quartile of the pregnancy
lengths.
Notice that the Z value will be negative.
Example
Determine the first quartile.
0.2500
Example
Determine the first quartile of the pregnancy
lengths.
In problems like these – where the probability is
given – the steps of the previous problems are
reversed.
Example
The steps of the previous problems are reversed.
From the given probability / percentile,
determine the Z-score.
Since 0.2500 is the probability of a result less
than Q1:
Z  -0.67
(Better tools give -0.674.)
Example
Determine the first quartile.
0.2500
Example
You have the Z score (–0.67). Use either of the
given equations to obtain the data value X.
X  Mean
Z
StDev
X  270
 0.67 
11
 0.6711  X  270
 7.37  X  270
270  7.37  X
262.63  X
Example
Determine the first quartile.
Q1 = 262.63
0.2500
Example
You have the Z score (–0.67). Use either of the
given equations to obtain the data value X.
X  Mean  Z  StDev
X  270   0.67 11
X  270  0.67 11
X  270  7.37
X  262.63
Example
The first quartile is 270 – 7.37 = 262.63.
(7.37 below the mean)
The median is 270.
The third quartile is 270 + 7.37 = 277.37.
(7.37 above the mean)
The IQR is 2(7.37) = 14.74.
SD / IQR = 11 / 14.74 = 0.746
For any Normal distribution, the SD is about
3/4ths the IQR.
Example
At one hospital, women whose pregnancy falls
into the top 2% of lengths are recommended for
inducing. At what time (in days) is a woman
recommended for inducing?
In other words…
Find the length that separates the longest 2%
from the shortest 98% of pregnancies.
Example
Find the length that separates the longest 2%
from the shortest 98% of pregnancies.
Example
Find the length that separates the longest 2%
from the shortest 98% of pregnancies.
Area = 0.02
Area = 0.98
Example
Find the length that separates the longest 2%
from the shortest 98% of pregnancies.
X=+Z
X = 270 + 2.054  11
Area = 0.98
Area = 0.02
Example
Find the length that separates the longest 2%
from the shortest 98% of pregnancies.
X=+Z
X = 292.594
Area = 0.02
Area = 0.98
Example
At one hospital, women whose pregnancy falls
into the top 2% of lengths are recommended for
inducing. At what time (in days) is a woman
recommended for inducing?
ANS: A woman should be induced at 292.6 days.
(This is the 98th percentile of the distribution.)
Percentiles / Z scores
Example
25th percentile Z = -0.67

X = 262.6

X = 292.6
Example
98th percentile Z = 2.05
Percentiles / Z scores
Example
25th percentile Z = -0.67

X = 262.6
In general for any variable with Normal distribution
The 25th percentile is 0.67 standard deviations below
the mean.
Example
98th percentile Z = 2.05

X = 292.6
In general for any variable with Normal distribution
The 98th percentile is 2.05 standard deviations above
the mean.
IQR Rule of Thumb
For a Normal distribution,   ¾ IQR. The
standard deviation is fairly close to 75% of the
interquartile range.
0.6745  2/3
IQR(Z)  (4/3)
IQR  (4/3) SD
SD  (3/4) IQR