Transcript The Normal distributions

The Normal distributions
PSLS chapter 11
© 2009 W.H. Freeman and Company
Objectives (PSLS 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 distributions
Normal – or Gaussian – distributions are a family of symmetrical, bell
shaped 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
x
Normal curves are used to model many biological variables.
They can describe the population distribution or density curve.
x
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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
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70
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68
67
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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.

A family of density curves
Here means are the same (m = 15)
while standard deviations are
different (s = 2, 4, and 6).
0
2
4
6
8
10
12
14
16
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20
22
24
26
28
30
Here means are different
(m = 10, 15, and 20) while standard
deviations are the same (s = 3)
0
2
4
6
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10
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20
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The 68–95–99.7 rule for any N(μ,σ)

About 68% of all observations
Inflection point
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.
mean µ = 64.5
standard deviation s = 2.5
N(µ, s) = N(64.5, 2.5)
Reminder: µ (mu) is the mean of the idealized curve, while x is the mean of a sample.
σ (sigma) is the standard deviation of the idealized curve, while s is the s.d. of a sample.
The standard Normal distribution
Because all Normal distributions share the same properties, we can
standardize to transform any Normal curve N(m,s) into the standard
Normal curve N(0,1).
N(64.5, 2.5)
N(0,1)
=>
x
z
Standardized height, standard
deviation units
For each x we calculate a new value, z (called a z-score).
Standardizing: calculating 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.
X = Women heights follow the N(64.5”, 2.5”)
N(µ, s) =
N(64.5, 2.5)
distribution. What percent of women are
shorter than 67 inches tall (that’s 5’7”)?
Area= ???
Area = ???
mean µ = 64.5"
standard deviation s = 2.5"
height x = 67"
m = 64.5” x = 67”
z=0
z=1
X
Z
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, i.e., less
than, 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, i.e.,
P{Z < 1.00} = 0.8413
N(µ, s) = N(64.5, 2.5)
 84.13% of women are shorter than 67”.
P{X < 67}
= P{Z < 1}
= Area
≈ 0.84
P{X > 67}
= P{Z > 1}
= Area
≈ 0.16
 The complementary, or 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 2 ways of finding the
area under N(0,1) curve to the
Area = 0.9901
right of a z value.
Area = 0.0099
z = -2.33
area right of z = area left of -z
area right of z =
1
-
area left of z
More tips on using Table B
To calculate the area between 2 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 z2 – area left of z1
 The area under N(0,1) for a single value of z is zero
Inverse Normal calculations
You may also seek the range of values that correspond to a given
proportion/ area under the curve. For that, 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
Vitamins and better food: The lengths of pregnancies when malnourished
mothers are given vitamins and better food is approximately N(266, 15). How long
are the 75% longest pregnancies in this population?
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%.
upper 75%
We find z ≈ -0.67
z
(x  m)
s
206
 x  m  ( z *s )
221
236
?
251
266
281
296
Gestation time (days)
x  266  (0.67 *15)
 The 75% longest pregnancies in this
x  255.95  256
population are about 256 days or longer.
311
Checking your cholesterol
High levels of total serum cholesterol
increase the risk of cardiovascular disease.
Cholesterol levels above 240 mg/dl demand
medical attention because they place the
subject at high risk of CV disease.
In the hope of extending treatment benefits to patients with early disease,
various professional societies have recommended a lower threshold value
for diagnosis.
Levels above 200 mg/dl are considered elevated cholesterol and may place
the person at some risk of cardiovascular disease.
The cholesterol levels for women aged 20 to 34 follow an approximately Normal
distribution with mean 185 mg/dl and standard deviation 39 mg/dl.
What is the probability that a young woman has highNormal
cholesterol (> 240 mg/dl)?
What is the probability she has an elevated cholesterol (between 200 and 240)?
x
z
area
left
area
right
240
1.41
92%
8%
200
0.38
65%
35%
39
68
29
68
107 146
107
146
185
224
185
263
302 224
341
263
302
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
29
111 68
107
148 146
185
185
224
263
222
302 259
341
296
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