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The Normal distribution
The Practice of Statistics in the Life Sciences
Third Edition
© 2014 W. H. Freeman and Company
Objectives (PSLS Chapter 11)
The Normal distributions

Normal distributions

The 68-95-99.7 rule

The standard Normal distribution

Using the standard Normal table (Table B)

Inverse Normal calculations

Normal quantile plots
Normal distributions
Normal—or Gaussian—distributions are a family of symmetrical, bellshaped density curves defined by a mean m (mu) and a standard
deviation s (sigma): N(m,s).
1
f ( x) 
e
2
1  xm 
 

2 s 
2
Inflection point
Inflection point
x
Normal curves are used to model many biological variables. They
can describe a population distribution or a probability distribution.
x
A family of density curves
Here means are the same (m = 15)
whereas standard deviations are
different (s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
Here means are different
(m = 10, 15, and 20) whereas
standard deviations are the same
(s = 3).
0
2
4
6
8
10
12
14
16
18
20
22
24
26
28
30
18
16
14
12
10
8
6
4
2
Height (inches)
 Guinea pigs survival times
after inoculation of a pathogen are
clearly not a good candidate for a
Normal model!
72 or more
71
70
69
68
67
66
65
64
63
62
61
60
59
58
57
56
0
under 56
Percent
Human heights, by
gender, can be modeled
quite accurately by a
Normal distribution.

The 68–95–99.7 rule for any N(μ,σ)
All normal curves N(µ,σ) share the same properties:

About 68% of all observations are
within 1 standard deviation
(s) of the mean (m).

About 95% of all observations are
within 2 s of the mean m.

Almost all (99.7%) observations
are within 3 s of the mean.
Number of times σ from the center µ
To obtain any other area under a Normal curve, use either technology
or Table B.
World Health Organization definitions of osteoporosis
based on standardized bone density levels
Normal
Bone density is within 1 standard deviation (z > –1)
of the young adult mean or above.
Low bone
mass
Bone density is 1 to 2.5 standard deviations below
the young adult mean (z between –2.5 and –1).
Osteoporosis
Bone density is 2.5 standard deviations or more
below the young adult mean (z ≤ –2.5).
What percent of young adults
have osteoporosis or osteopenia?
Population of
young adults
N(0,1)
These are bone densities of
1 or less, representing the
area to the left of the middle
68% between 1 and +1.
So 16%.
z
Standardized bone density (no units)
Young adults N(0,1)
Women 70-79 N(-2,1)
Women aged 70 to 79 are
NOT young adults. The mean
bone density in this age is
about −2 on the standard
scale for young adults.
-6
-4
-2
0
2
4
z
Standardized bone density (no units)
What is the probability that a randomly chosen woman
in her 70s has osteoporosis or osteopenia (< −1)?
A) 97.5%
B) 95%
C) 84%
D) 68%
E) 50%
The standard Normal distribution
We can standardize data by computing a z-score:
z
(x  m )
s
If x has the N(m,s) distribution, then z has the N(0,1) distribution.
N(64.5, 2.5)
N(0,1)
=>
x
z
Standardized height (no units)
Standardizing: z-scores
A z-score measures the number of standard deviations that a data
value x is from the mean m.
z
(x  m )
s
When x is 1 standard deviation larger
than the mean, then z = 1.
for x  m  s , z 
m s  m s
 1
s
s
When x is 2 standard deviations larger
than the mean, then z = 2.
for x  m  2s , z 
m  2s  m 2s

2
s
s
When x is larger than the mean, z is positive.
When x is smaller than the mean, z is negative.
Women’s heights follow the N(64.5”,2.5”)
distribution. What percent of women are
shorter than 67 inches tall (that’s 5’6”)?
N(µ, s) =
N(64.5, 2.5)
Area= ???
Area = ???
mean µ = 64.5"
standard deviation s = 2.5"
height x = 67"
m = 64.5” x = 67”
z=0
z=1
We calculate z, the standardized value of x:
z
(x  m)
s
, z
(67  64.5) 2.5

 1  1 stand. dev. from mean
2.5
2.5
Given the 68-95-99.7 rule, the percent of women shorter than 67” should be,
approximately, .68 + half of (1 – .68) = .84, or 84%. The probability of
randomly selecting a woman shorter than 67” is also ~84%.
Using Table B
Table B gives the area under the standard Normal curve to the left of any z-value.
.0062 is the
area under
N(0,1) left
of z = –2.50
.0060 is the area
under N(0,1) left
of z = –2.51
0.0052 is the area
under N(0,1) left
of z = –2.56
(…)
For z = 1.00, the area
under the curve to the
left of z is 0.8413.
N(µ, s) = N(64.5, 2.5)
 84.13% of women are shorter than 67”.
Area ≈ 0.84
Area ≈ 0.16
 Therefore, 15.87% of women are taller than
67" (5'6").
m = 64.5
x = 67
z=1
Tips on using Table B
Because of the curve’s symmetry,
there are two ways of finding the
area under N(0,1) curve to the
right of a z-value.
Area = 0.9901
Area = 0.0099
z = -2.33
area right of z = area left of –z
area right of z =
1
–
area left of z
Using Table B to find a middle area
To calculate the area between two z-values, first get the area under
N(0,1) to the left for each z-value from Table B.
Then subtract the
smaller area from the
larger area.
Don’t subtract the z-values!!!
Normal curves are not square!
area between z1 and z2 =
area left of z1 – area left of z2
 The area under N(0,1) for a single value of z is zero.
The blood cholesterol levels of men aged 55 to 64 are approximately Normal with
mean 222 mg/dl and standard deviation 37 mg/dl.
What percent of middle-age men have high cholesterol
(> 240 mg/dl)?
Normal
What percent have elevated cholesterol (between 200 and 240 mg/dl)?
z
area
left
area
right
0.49
69%
31%
200 -0.59
28%
72%
x
240
37
111 68
29
148 146
107
185
185
224
222
263
302 259
341
296
333
Inverse Normal calculations
You may also seek the range of values that correspond to a given
proportion/ area under the curve.
For that, use technology or use Table B backward.

First find the desired
area/ proportion in the
body of the table,

then read the
corresponding z-value
from the left column
and top row.
For a left area of 1.25% (0.0125),
the z-value is –2.24.
The hatching weights of commercial chickens can be modeled accurately using a
Normal distribution with mean μ = 45 g and standard deviation σ = 4 g.
What is the third quartile of the distribution of hatching weights?
We know μ, σ, and the area
under the curve; we want x.
Table B gives the area left of z
 look for the lower 25%
We find z ≈ 0.67
z
(x  m)
s
 x  m  ( z *s )
x  45  (0.67 * 4)
x  47.68
Q3 ≈ 47.7 g
The blood cholesterol levels of men aged 55 to 64 are approximately normal
with mean 222 mg/dl and standard deviation 37 mg/dl.
What range of values corresponds to the 10% highest cholesterol levels?
A) > 175
B) > 247
C) > 269
D) > 288
Normal
z
Area
left
Area
right
1.28
90%
10%
37
29
111
68
107
148
146
185
185
224
263
222
302
341
259
Cholesterol level (mg/dl)
296
333
Normal quantile plots
One way to assess if a data set has an approximately Normal
distribution is to plot the data on a Normal quantile plot.
The data points are ranked and the percentile ranks are converted to zscores. The z-scores are then used for the horizontal axis and the actual
data values are used for the vertical axis. Use technology to obtain Normal
quantile plots.
If the data have approximately a Normal distribution, the Normal
quantile plot will have roughly a straight-line pattern.
Roughly normal
Right skewed
(~ straight-line pattern)
(most of the data points are short
survival times, while a few are
longer survival times)