The normal distribution

Binomial distribution is discrete events, (infected, not infected)

The normal distribution is a probability density function for a continuous variable, and is represented by a continuous curve.

Density = relative frequency of varites on the Y (horizontal) axis.

freq

Some variable

Area under curve is equal to the sum of expected frequencies

freq

Some variable

Cannot evaluate the probability of the variable being exactly equal to some value (that area of the curve is soooo small)

Must estimate the frequency of observations falling between two limits

2 3

Won’t work

freq

Some variable

estimate the frequency of observations falling between 2 and 2.2

2 3

A normal curve is defined as:

Y is the height of the ordinateμ is the mean σ is the standard deviation π is the constant 3.14159e is the base of natural logarithms and is equal to 2.718282x can take on any value from -infinity to +infinity.

122 e-(x-)2/22 *Y=

How is a normal probability distribution like a lion?

They both have a MEAN MEW.

The shape of a given normal curve results from different values of and

The mean, , determines the midpoint

The standard deviation, , changes the shape, it affects the spread or the dispersion of scores

The larger the value of the more dispersed the scores; the smaller the value, the less dispersed.

The mean, , determines the midpoint

A smaller , means less dispersion

Bonus question. What’s wrong with this graph?

There is not one Normal Curve

freq

Some variable

2 3

Normal curves can differ in location or shape

How to determine what proportion of a normal population lies above/below a certain level

120 cm

The average Hobbit

If distribution of Hobbit heights is normal with mean = 120 cm, SD = 20

Half < 120 & half >120

What is probability of finding a Hobbit taller than 130 cm??

Calculate the normal deviate

Z = Xi -

- Normal deviate- Test statistic

- Any point on normal curve- Here, 130 cm Mean

SD

Z = (130-120)/20 = 0.5

P (probability) (Xi >130 cm) = P (Z>0.50) = 0.3085 or 30.85%

Table B.2; ZarTable A S & R

In any normal population:

68.27% of measurements lie w/in ( 1)

99.73% of measurements lie w/in ( 3)

50% lie w/in ( 0.67 )

95% lie w/in ( 1.96 )

- hence the 95% confidence interval of a sample = X 1.96 * s

- the range within one is 95% confident that the true population mean, , is to be found

The normal distribution in biology

Binomial distribution (p + q)k

Imagine a trait is controlled by many factors, ex skin pigmentation.

When a factor is present, it contributes 1 unit of pigmentation

If 3 factors were present, the animal would have skin that was 3 units dark

Assume 0.5 probability of each factor being present: p (hence 0.5 probability of each factor being absent): q

If only one factor existed, (p + q)1; k=1

Half the animals would have it, half would notexpected proportion w 0 pigmentation unit=0.5expected proportion w 1 pigmentation unit=0.5

If two factors existed, (p + q)2; k=2

There will now be 3 classes: pp, pq, qq

Frequency of pp = 0.25 or (0.5)2

Frequency of pq = 0.5 or 2[0.5*0.5]Frequency of qq = 0.25 or (0.5)2

If k (number of independent factors) becomes large, the distribution produced by binomial expansion would come very close to the normal distribution

Many biological variables behave like this

When samples are large, this occurs even when the factors are not strictly independent, or not all equal in magnitude of effect.

Assessing Normality

I'm not an outlier; I just haven't found my distribution yet

Skewness: asymmetry, one tail is drawn out

Mean not equal to medianfr

eq

Some variable

2 3

kurtosis: the proportion of a curve located in the center, shoulders and tails

How fat or thin the tails are

leptokurticno shoulders

platykurticwide shoulders

More later on assessing normality using SAS

Revisit variance and SD relative to normal curve

X

XiXii

Xi - X2

Xii -X2

Total sum of

squares

n-1=

Mean SS =variance

I side =SD

Distribution of means assuming normality

If you take repeated samples of size N from a normally distributed population, the distribution of the the means of those samples will be normal

If you take repeated samples of size N from a non-normally distributed population, the distribution of the the means of those samples will tend towards normality

Central Limit Theorem

Calculate the variance of the mean of the distribution of means

Variance of mean = 2

n

Square root of the variance of the mean is the SD of the mean, also called the standard error

x =2

n

But rarely know pop parameters, so…….

sx =s2

n

For a sample,

sx =s

n

We’ll come back to this with more on testing differences between a mean and a value or between 2 means

or

Par

amet

er u

nit

s

Sample size (n)

Changes in s2, SD, & SE with increasing N

Variance, s2

Become more accurate approximations of “true”

SD, s2

SE becomes smaller with increased sampling

SE, SD/ n

What SD and SE mean

SD is a parameter of a natural population (even though real populations are constantly changing). Its size reflects real, natural variance. Big is not good, small is not good. SD just is. Natural dispersion of population

SE becomes smaller with increasing sample size, therefore reflects sampling effort. Accuracy of mean.

Both frequently reported (graphed) in ecological / biological literature. SE smaller, so often favored– but this is wrong reasoning!

Practically either OK, if you state which is shown and report n!