11. The Normal distributions

The Practice of Statistics in the Life SciencesSecond Edition

2012 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 curves are used to model many biological variables. They can describe a population distribution or a probability distribution .

Normalor Gaussiandistributions are a family of symmetrical, bell-shaped density curves defined by a mean (mu) and a standard deviation (sigma): N(,).

xx

2

21

21)(

=

x

exfInflection point

Inflection point

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

A family of density curves

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

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).

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 mor

e

Height (inches)

Perc

ent

Guinea pigs survival timesafter inoculation of a pathogen are clearly not a good candidate for a Normal model!

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.

The 689599.7 rule for any N(,)

Number of times from the center

All normal curves N(,) share the same properties:

To obtain any other area under a Normal curve, use either technology or Table B.

Population of young adultsN(0,1)

zStandardized bone density (no units)

What percent of young adults have osteoporosis or osteopenia?

World Health Organization definitions of osteoporosisbased 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 deviation below the young adult mean (z between 2.5 and 1).

Osteoporosis Bone density is 2.5 standard deviation or more below the young adult mean (z 2.5).

zStandardized bone density (no units)

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. -4 -2 0 2 4

Young adults N(0,1) Women 70-79 N(-2,1)

What is the probability that a randomly chosen woman in her 70s has

osteoporosis or osteopenia (< 1 on the standard scale)?

We can standardize data by computing a z-score:

If x has the N(,) distribution, then z has the N(0,1) distribution.

N(0,1)

=>

z

x

N(64.5, 2.5)

Standardized height (no units)

z =(x )

The standard Normal distribution

z =(x )

A z-score measures the number of standard deviations that a data value x is from the mean .

Standardizing: z-scores

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

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

1 ,for ==+=+=

zx

When x is 1 standard deviation larger than the mean, then z = 1.

222 ,2for ==+=+=

zx

When x is 2 standard deviations larger than the mean, then z = 2.

mean = 64.5"

standard deviation = 2.5"

height x = 67"

We calculate z, the standardized value of x:

mean from dev. stand. 1 15.25.2

5.2)5.6467( ,)( =>==== zxz

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(, ) = N(64.5, 2.5)

= 64.5 x = 67 z = 0 z = 1

Womens heights follow the N(64.5,2.5)

distribution. What percent of women are

shorter than 67 inches tall (thats 56)?

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

Area 0.84

Area 0.16

N(, ) = N(64.5, 2.5)

= 64.5 x = 67 z = 1

84.13% of women are shorter than 67.

Therefore, 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.

Tips on using Table B

Because of the curves symmetry,

there are two 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

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.

area between z1 and z2 =area left of z1 area left of z2

Dont 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

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)?

111 148 185 222 259 296 333

37

x zarea left

area right

240 0.49 69% 31%200 -0.59 28% 72%

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-valuefrom 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

zxxz

221 236 251 266 281 296 311

Gestation time (days)

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 , , 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

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 z-

scores. 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.

Normal quantile plots

Roughly normal

(~ straight-line pattern)

Right skewed

(most of the data points are short survival times, while a few are

longer survival times)

11. The Normal distributionsObjectives (PSLS Chapter 11)Slide Number 3A family of density curvesSlide Number 5The 689599.7 rule for any N(,)Slide Number 7Slide Number 8The standard Normal distributionStandardizing: z-scoresSlide Number 11Using Table BSlide Number 13Tips on using Table BUsing Table B to find a middle areaSlide Number 16Slide Number 17Slide Number 18Slide Number 19Slide Number 20Normal quantile plotsSlide Number 22