1. Normal Distribution 9.1 Continuous distribution
9.2 The normal distribution
9.3 A check for normality
9.4 Application of the normal distribution
9.5 Normal approximation to Binomial

2. 9.1 Continuous Distribution
For a discrete distribution, for example
Binomial distribution with n=5, and p=0.4, the probability distribution is x
f(x)

3. A probability histogramx
P(x)

4. How to describe the distribution of a continuous random variable?
For continuous random variable, we also represent probabilities by areas—not by areas of rectangles, but by areas under continuous curves.
For continuous random variables, the place of histograms will be taken by continuous curves.
Imagine a histogram with narrower and narrower classes. Then we can get a curve by joining the top of the rectangles. This continuous curve is called a probability density (or probability distribution).

5. Continuous distributions
For any x, P(X=x)=0. (For a continuous distribution, the area under a point is 0.)
Can’t use P(X=x) to describe the probability distribution of X
Instead, consider P(a≤X≤b)

6. Density function A curve f(x): f(x) ≥ 0 The area under the curve is 1
P(a≤X≤b) is the area between a and b

7. P(2≤X≤4)= P(2≤X<4)= P(2<X<4)

8. 9.2 The normal distribution
A normal curve: Bell shaped
Density is given by
μand σ2 are two parameters: mean and standard variance of a normal population
(σ is the standard deviation)

9. The normal—Bell shaped curve: μ=100, σ2=10

10. Normal curves: (μ=0, σ2=1) and (μ=5, σ 2=1)

11. Normal curves: (μ=0, σ2=1) and (μ=0, σ2=2)

12. Normal curves: (μ=0, σ2=1) and (μ=2, σ2=0.25)

13. The standard normal curve: μ=0, and σ2=1

14. How to calculate the probability of a normal random variable?
Each normal random variable, X, has a density function, say f(x) (it is a normal curve).
Probability P(a<X<b) is the area between a and b, under the normal curve f(x)
Table I in the back of the book gives areas for a standard normal curve with =0 and =1.
Probabilities for any normal curve (any  and ) can be rewritten in terms of a standard normal curve.

15. Table I: Normal-curve Areas
Table I on page
We need it for tests
Areas under standard normal curve
Areas between 0 and z (z>0)
How to get an area between a and b? when a<b, and a, b positive
area[0,b]–area[0,a]

16. Get the probability from standard normal table
z denotes a standard normal random variable
Standard normal curve is symmetric about the origin 0
Draw a graph

17. Table I: P(0<Z<z)
… … … … … … … …

18. Examples Example 9.1 P(0<Z<1) = 0.3413 Example 9.2
=P(0<Z<2)–P(0<Z<1)
=0.4772–0.3413
=0.1359

19. Examples
Example 9.3
P(Z≥1)
=0.5–P(0<Z<1)
=0.5–0.3413
=0.1587

20. Examples
Example 9.4
P(Z ≥ -1) =
=0.8413

21. Examples
Example 9.5
P(-2<Z<1) =
=0.8185

22. Examples Example 9.6 P(Z ≤ 1.87) =0.5+P(0<Z ≤ 1.87) =0.5+0.4693
=0.9693

23. Examples Example 9.7 P(Z<-1.87) = P(Z>1.87) = 0.5–0.4693=

24. From non-standard normal to standard normal
X is a normal random variable with mean μ, and standard deviation σ
Set Z=(X–μ)/σ
Z=standard unit or z-score of X
Then Z has a standard normal distribution and

25. Example 9.8 X is a normal random variable with μ=120, and σ=15
Find the probability P(X≤135)
Solution:

26. XZ x z-score of x Example 9.8 (continued) P(X≤150)
x=150  z-score z=( )/15=2
P(X≤150)=P(Z≤2)
= =

27. 9.3 Checking Normality
Most of the statistical tools we will use in this class assume normal distributions.
In order to know if these are the right tools for a particular job, we need to be able to assess if the data appear to have come from a normal population.
A normal plot gives a good visual check for normality.

28. Simulation: 100 observations, normal with mean=5, st dev=1
x<-rnorm(100, mean=5, sd=1)
qqnorm(x)

29. The plot below shows results on alpha-fetoprotein (AFP) levels in maternal blood for normal and Down’s syndrome fetuses.
Estimating a woman’s risk of having a preganancy
associated with Down’s syndrome using her age
and serum alpha-fetoprotein level
H.S.Cuckle, N.J.Wald, S.O.Thompson

30. Normal Plot The way these normal plots work is
Straight means that the data appear normal
Parallel means that the groups have similar variances.

31. Normal plot
In order to plot the data and check for normality, we compare
our observed data to
what we would expect from a sample of normal data.

32. To begin with, imagine taking n=5 random values from a standard normal population (=0, =1)
Let Z(1) Z(2) Z(3) Z(4) Z(5) be the ordered values. Suppose we do this over and over.
Sample Z(1) Z(2) Z(3) Z(4) Z(5)
… … … … … …
Forever ___ ___ ___ ___ ___
Mean
E(Z(1)) E(Z(2)) E(Z(3)) E(Z(4)) E(Z(5))
On average
the smallest of n=5 standard normal values is standard deviations below average
the second smallest of n=5 standard normal values is standard deviations below average
the middle of n=5 standard normal values is at the average, 0 standard deviations from average

33. The table of “rankits” from the Statistics in Biology table gives these expected values.
For larger n, space is saved by just giving the positive values. The negative values are a mirror image of the positive values, since a standard normal distribution is symmetric about its mean of zero.

34. Check for normality
If X is normal, how do ordered values of X, X(i) , relate to expected ordered Z values, E( Z(i) ) ?
For normal with mean  and standard deviation , the expected values of the data, X(i), will be a linear rescaling of standard normal expected values
E(X(i)) ≈  +  E( Z(i) )
The observed data X(i) will be approximately a linearly related to E( Z(i) ).
X(i) ≈  +  E( Z(i) )

35. If we plot the ordered X values versus E( Z(i) ), we should see roughly a straight line with
intercept 
slope 

36. Example Example: Lifetimes of springs under 900 N/mm2 stress
i E( Z(i) ) X(i)

37. The plot is fairly linear indicating that the data are
pretty similar to what we would expect from
normal data.

38. To compare results from different treatments, we can put more than one normal plot on the same graph.
The intercept for the 900 stress level is above the intercept for the 950 stress group, indicating that the mean lifetime of the 900 stress group is greater than the mean of the 950 stress group.
The slopes are similar, indicating that the variances or standard deviations are similar.

39. These plots were done in Excel
These plots were done in Excel. In Excel you can either enter values from the table of E(Z) values or generate approximations to these tables values.
One way to generate approximate E(Z) values is to generate evenly spaced percentiles of a standard normal, Z, distribution.
The ordered X values correspond roughly to particular percentiles of a normal distribution.
For example if we had n=5 values, the 3rd ordered values would be roughly the median or 50th percentile.
A common method is to use percentiles corresponding to

40. For n=5 this would give us
1 0.1
2 0.3
the 50th percentile
4 0.7
5 0.9
For E(Z) we would use corresponding percentiles of a standard normal Z distribution.
Percentiles expressed as fractions are called quantiles. The 0.5 quantile is the 50th percentile.
Normal plots from this perspective are sometimes called Q-Q plots, since we are plotting standard normal quantiles versus the associated quantiles of the observed data.

41. For n = 10 values for the spring data, the corresponding normal percentiles would be
i Z quantile

42. If you are doing the plot by hand it’s easiest to use the E(Z) table.
For assessing whether a plotted line is fairly parallel, either the E(Z) values or the normal quantiles work fine.
If you are doing the plot by hand it’s easiest to use the E(Z) table.
If you are doing these in Excel it’s easiest to use the normal quantiles.
The function NORMINV(p,0,1) finds the Z values corresponding to a given quantile.
This is the inverse of the function that finds the cumulative probability for a given Z value.
Z  NORMDIST  probability = NORMDIST(1.645, 0, 1, TRUE)  0.95
Probability  NORMINV  probability = NORMINV(0.95, 0, 1)  1.645
(The TRUE in NORMDIST says to return the cumulative probability rather than density curve height.)

43. Excel File of Lifetime of Springs Data

44. For data that are not normal
Many types of data tend to follow a normal distribution, but many data sets aren’t particularly normal.
If the data aren’t fairly normal we have several options
Transform the data, meaning change the scale.
A log or ln scale is most common.
Weights of fish
Concentrations
Bilirubin levels in blood
pH is a log scale
RNA expression levels in a microarray experiment
A reciprocal (1/Y) change of times to rates
Other powers
Square root for Poisson variables

46. Non-normal data continued
Use a different distribution other than a normal distribution
Weibull distribution for lifetimes
Motors at General Electric
Patients in a clinical trial

47. Weibull Distributions (Time to Failure – Non-binomial & Non-normal)
Infant Mortality: Fail immediately or last a long time
Early Failure: These do not fail immediately, but many do fail early
Old-age Wearout: Very few of these fail until they were out

48. Non-normal data continued
Use a nonparametric methods which doesn’t assume any distribution
Finding a distribution that models the data well rather than nonparametric
Allows us to develop a more complete model
Allows us to generalize to other situations
Gives us more precise information for the same amount of effort

49. The methods in this class largely apply to normal data or data that we can transform to normal.
The EPA fish example is a good example of transforming data with a log transformation.
Geometric means and harmonic means arise when we are working with transformed data.
For example fish weights are usually analyzed in the log scale. Having a mean in the log scale we want to put this value back into the original scale, for example grams.
The back-transformed mean from the log scale is the geometric mean.
The back-transformed mean from a reciprocal scale (rates), is the harmonic mean.
Back-transformed differences between geometric means correspond to ratios in the original scale.

50. Suppose ln(X) = Y ~ N(, 2).
This means Y (or ln(x)) distributed as normal with mean  and variance 2.
The geometric mean is e, the back- transformed population mean in the ln scale.
If we have the difference between two means in the ln scale then back- transforming give us
= ratio of geometric means.

51. About geometric means
A fact is that if the variances of both populations are the same, then the ratio of the population geometric means is the same as the ratio of the population means.

52. Question: Why not just use the means in the original scale?
Answer: Means are best when populations are normal. Using the ratio of the geometric means will give us a more precise estimate of the true ratio than using the ratio of the means in the original scale.

53. A similar fact explains why we use means rather than medians.
For a normal population the mean is the same as the median. We could use either the sample mean or the sample median to estimate .
BUT, the mean will be a more precise guess (estimate of) the true value, .
It would take us roughly 50% more values (larger n) using the median as our guess at  to accomplish the same degree of precision as we get using the mean as our guess at .

54. 9.4 Application of the normal distribution
Public Health Service
Health Examination Survey
6,672 Americans years old
The woman’s heights were approximately
normal with 63 and standard deviation
2.5 .
What percentage of women were over 68
tall?

55. Solution: X=height P(X>68)=P(Z>(68-63)/2.5)) =P(Z>2)=
=0.0228

56. Continuity Correction for a Better Approximation
Sometimes only integer values are possible for x.
x=score of LSAT
x=# of heads in 10 tosses of a fair coin
A normal approximation is more accurate with a “continuity correction”

57. 1976 LSAT Approximately normal
mean 650, st. dev 60
P(X≥680)P(Z>( )/60)
=P(Z>0.49) =
=0.3121

58. 9.5 Normal Approximation to Binomial
A binomial distribution:
n=10, p=0.5
μ=np=5
σ2=np(1-p)=2.5  σ=1.58
P(X≥7)= from Binomial
P(X≥7)= P(Z>(6.5-5)/1.58)
=P(Z>0.95)
= =0.1711
from normal approximation

59. Dots: Binomial Probabilities Smoot Line: Normal Curve With Same Mean and Variance

60. Normal Approximation Is Good If
The normal curve has the same mean and standard deviation as binomial
np>5 and n(1-p)>5
Continuity correction is made

61. Example
Records show that 60% of the customers of a service station pay with a credit card. Use normal approximation to find the probabilities that among 100 customers
At most 65 will pay with a credit
At least 55 will pay with a credit
Between 55 and 65 will pay with a credit card
Exactly 65 will pay with a credit card

62. Solution: X=# of customers who pay with a credit card μ=np=60,
σ2=np(1-p)=24  σ=4.8990

63. Normal Approximation 3. 4.