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 curves are used to model many biological variables.

They can describe the population distribution or density curve.

Normal – or Gaussian – distributions are a family of symmetrical, bell

shaped density curves defined by a mean (mu) and a standard

deviation (sigma): N().

xx

2

2

1

2

1)(

x

exf

Human heights, by

gender, can be modeled

quite accurately by a

Normal distribution.

0

2

4

6

8

10

12

14

16

18

unde

r 56 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71

72 o

r m

ore

Height (inches)

Per

cen

t

Guinea pigs survival times

after inoculation of a pathogen

are clearly not a good candidate

for a Normal model!

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28 30

A family of density curves

Here means are different

( = 10, 15, and 20) while

standard deviations are the same

( = 3)

Here means are the same ( = 15)

while standard deviations are

different ( = 2, 4, and 6).

mean µ = 64.5 standard deviation = 2.5

N(µ, ) = N(64.5, 2.5)

The 68–95–99.7 rule for any N(μ,σ)

Reminder: µ (mu) is the mean of the idealized curve, while 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.

About 68% of all observations

are within 1 standard deviation

(of the mean ().

About 95% of all observations

are within 2 of the mean .

Almost all (99.7%) observations

are within 3 of the mean.

Inflection point

x

Because all Normal distributions share the same properties, we can

standardize to transform any Normal curve N() into the standard

Normal curve N(0,1).

The standard Normal distribution

For each x we calculate a new value, z (called a z-score).

N(0,1)

=>

z

x

N(64.5, 2.5)

Standardized height, standard deviation units

z (x )

A z-score measures the number of standard deviations that a data

value x is from the mean .

Standardizing: calculating z-scores

When x is larger than the mean, z is positive.

When x is smaller than the mean, z is negative.

1 ,

zxfor

When x is 1 standard deviation larger

than the mean, then z = 1.

222

,2

zxfor

When x is 2 standard deviations larger

than the mean, then z = 2.

mean µ = 64.5"

standard deviation = 2.5"

height x = 67"

We calculate z, the standardized value of x:

( ) (67 64.5) 2.5

, 1 1 stand. dev. from mean2.5 2.5

xz z

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%.

Area= ???

Area = ???

N(µ, ) = N(64.5, 2.5)

= 64.5” x = 67”

z = 0 z = 1

X = Women heights follow the N(64.5”, 2.5”)

distribution. What percent of women are

shorter than 67 inches tall (that’s 5’7”)?

X

Z

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

P{X < 67} = P{Z < 1} = Area ≈ 0.84

P{X > 67} = P{Z > 1} = Area ≈ 0.16

N(µ, ) = N(64.5, 2.5)

= 64.5 x = 67 z = 1

84.13% of women are shorter than 67”.

The complementary, or 15.87% of women

are taller than 67" (5'6").

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

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

right of a z value.

area right of z = 1 - area left of z

Area = 0.9901

Area = 0.0099

z = -2.33

area right of z = 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.

area between z1 and z2 =

area left of z2 – area left of z1

Don’t subtract the z values!!!

Normal curves are not square!

Then subtract the

smaller area from the

larger area.

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

25695.255

)15*67.0(266

)*()(

x

x

zxx

z

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?

?

upper 75%

The 75% longest pregnancies in this

population are about 256 days or longer.

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

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 high cholesterol (> 240 mg/dl)?

What is the probability she has an elevated cholesterol (between 200 and 240)?

68 107 146 185 224 263 30229 68 107 146 185 224 263 302 341

39

x zarea left

area right

240 1.41 92% 8%

200 0.38 65% 35%

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)?

What percent have elevated cholesterol (between 200 and 240 mg/dl)?

111 148 185 222 259 296 33329 68 107 146 185 224 263 302 341

37

x zarea left

area right

240 0.49 69% 31%

200 -0.59 28% 72%