Transcript Normal distribution

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
Area under curve is equal to the sum of expected
frequencies
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
freq
Won’t work
2
3
freq
estimate the frequency of observations falling
between 2 and 2.2
2
3
A normal curve is defined as:
Y=
1
2
2
-(x-)
/2
22 * e
Y is the height of the ordinate
μ is the mean
σ is the standard deviation
π is the constant 3.14159
e is the base of natural logarithms and is equal to
2.718282
x can take on any value from -infinity to +infinity.
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.
There is not one Normal Curve
The mean, , determines the midpoint
A smaller , means less dispersion
Bonus question. What’s wrong with this graph?
freq
Normal curves can differ in location or shape
2
3
Some variable
How to determine what proportion of a normal
population lies above/below a certain level
If distribution of Hobbit
heights is normal with mean
= 120 cm, SD = 20
Half < 120 & half >120
120 cm
What is probability of finding
a Hobbit taller than 130 cm??
The average
Hobbit
Calculate the normal deviate
- Any point on normal curve
- Here, 130 cm
Mean
Xi - 
Z=

- Normal deviate
- Test statistic
SD
Z = (130-120)/20 = 0.5
Table B.2; Zar
Table A S & R
P (probability) (Xi >130 cm) = P (Z>0.50) = 0.3085 or
30.85%
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 not
expected proportion w 0 pigmentation unit=0.5
expected 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
freq
Mean not equal to median
2
3
Some variable
kurtosis: the proportion of a curve located in the
center, shoulders and tails
How fat or thin the tails are
leptokurtic
no shoulders
platykurtic
wide shoulders
More later on assessing normality using SAS
Revisit variance and SD relative to normal curve
X
Xi
X
i
i
2
Xi - X
2
Xii -X
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 nonnormally 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
2
Variance of mean =
n
Square root of the variance of the mean is the SD
of the mean, also called the standard error
2
x
=
n
But rarely know pop
parameters, so…….
For a sample,
s2
sx
=
n
or
s
sx
=
n
 We’ll come back to this with more on testing
differences between a mean and a value or between
2 means
Changes in s2, SD, & SE with increasing N
Parameter units
Variance, s2
SD, s2
SE, SD/ n
Sample size (n)
Become more
accurate
approximations
of “true”
SE becomes
smaller with
increased
sampling
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!