Transcript Normal Distribution

Random Variables and Probability
Distributions
• Random Variables - Random responses corresponding
to subjects randomly selected from a population.
• Probability Distributions - A listing of the possible
outcomes and their probabilities (discrete r.v.s) or their
densities (continuous r.v.s)
• Normal Distribution - Bell-shaped continuous
distribution widely used in statistical inference
• Sampling Distributions - Distributions corresponding
to sample statistics (such as mean and proportion)
computed from random samples
Normal Distribution
• Bell-shaped, symmetric family of distributions
• Classified by 2 parameters: Mean (m) and standard
deviation (s). These represent location and spread
• Random variables that are approximately normal have
the following properties wrt individual measurements:
–
–
–
–
Approximately half (50%) fall above (and below) mean
Approximately 68% fall within 1 standard deviation of mean
Approximately 95% fall within 2 standard deviations of mean
Virtually all fall within 3 standard deviations of mean
• Notation when Y is normally distributed with mean m
and standard deviation s :
Y ~ N (m ,s )
Normal Distribution
P(Y  m )  0.50 P( m  s  Y  m  s )  0.68 P( m  2s  Y  m  2s )  0.95
Example - Heights of U.S. Adults
• Female and Male adult heights are well approximated by
normal distributions: YF~N(63.7,2.5) YM~N(69.1,2.6)
20
20
18
16
14
12
10
10
8
6
4
Std. Dev = 2.48
Std. Dev = 2.61
2
Mean = 63.7
Mean = 69.1
0
N = 99.68
55.5
57.5
56.5
59.5
58.5
61.5
60.5
63.5
62.5
65.5
64.5
67.5
66.5
INCHESF
69.5
68.5
70.5
N = 99.23
0
59.5 61.5 63.5 65.5 67.5 69.5 71.5 73.5 75.5
60.5 62.5 64.5 66.5 68.5 70.5 72.5 74.5 76.5
INCHESM
Cases weighted by PCTM
Cases weighted by PCTF
Source: Statistical Abstract of the U.S. (1992)
Standard Normal (Z) Distribution
• Problem: Unlimited number of possible normal
distributions (- < m <  , s > 0)
• Solution: Standardize the random variable to have
mean 0 and standard deviation 1
Y ~ N (m ,s )  Z 
Y m
s
~ N (0,1)
• Probabilities of certain ranges of values and specific
percentiles of interest can be obtained through the
standard normal (Z) distribution
Standard Normal (Z) Distribution
• Standard Normal Distribution Characteristics:
–
–
–
–
a
za
P(Z  0) = P(Y  m ) = 0.5000
P(-1  Z  1) = P(m-s  Y  m+s ) = 0.6826
P(-2  Z  2) = P(m-2s  Y  m+2s ) = 0.9544
P(Z  za) = P(Z  -za) = a (using Z-table)
0.500
0.000
0.100
1.282
0.050
1.645
0.025
1.960
0.010
2.326
0.005
2.576
Finding Probabilities of Specific Ranges
• Step 1 - Identify the normal distribution of interest (e.g.
its mean (m) and standard deviation (s) )
• Step 2 - Identify the range of values that you wish to
determine the probability of observing (YL , YU), where
often the upper or lower bounds are  or -
• Step 3 - Transform YL and YU into Z-values:
ZL 
YL  m
s
ZU 
YU  m
s
• Step 4 - Obtain P(ZL Z  ZU) from Z-table
Example - Adult Female Heights
• What is the probability a randomly selected female is
5’10” or taller (70 inches)?
• Step 1 - Y ~ N(63.7 , 2.5)
• Step 2 - YL = 70.0 YU = 
• Step 3 70.0  63.7
ZL 
 2.52
ZU  
2.5
• Step 4 - P(Y  70) = P(Z  2.52) = .0059 (  1/170)
z
2.4
2.5
2.6
.00
.0082
.0062
.0047
.01
.0080
.0060
.0045
.02
.0078
.0059
.0044
.03
.0075
.0057
.0043
Finding Percentiles of a Distribution
• Step 1 - Identify the normal distribution of interest
(e.g. its mean (m) and standard deviation (s) )
• Step 2 - Determine the percentile of interest
100p% (e.g. the 90th percentile is the cut-off where only
90% of scores are below and 10% are above)
• Step 3 - Turn the percentile of interest into a tail
probability a and corresponding z-value (zp):
– If 100p  50 then a = 1-p and zp = za
– If 100p < 50 then a = p and zp = -za
• Step 4 - Transform zp back to original units:
Yp  m  z s
p
Example - Adult Male Heights
•
•
•
•
Above what height do the tallest 5% of males lie above?
Step 1 - Y ~ N(69.1 , 2.6)
Step 2 - Want to determine 95th percentile (p = .95)
Step 3 - Since 100p > 50, a = 1-p = 0.05
zp = za = z.05 = 1.645
• Step 4 - Y.95 = 69.1 + (1.645)(2.6) = 73.4
z
1.5
1.6
1.7
.03
.0630
.0516
.0418
.04
.0618
.0505
.0409
.05
.0606
.0495
.0401
.06
.0594
.0485
.0392
Statistical Models
• When making statistical inference it is useful to
write random variables in terms of model
parameters and random errors
Y  m  (Y  m )  m  
 Y m
• Here m is a fixed constant and  is a random variable
• In practice m will be unknown, and we will use sample data to
estimate or make statements regarding its value
Sampling Distributions and the Central
Limit Theorem
• Sample statistics based on random samples are also
random variables and have sampling distributions that
are probability distributions for the statistic (outcomes
that would vary across samples)
• When samples are large and measurements independent
then many estimators have normal sampling
distributions (CLT):
 s 
Y ~ N  m,

n

– Sample Mean:
– Sample Proportion:


(1   ) 

 ~ N   ,

n


^
Example - Adult Female Heights
• Random samples of n = 100 females to be selected
• For each sample, the sample mean is computed
• Sampling distribution:
2.5 

Y ~ N  63.5,
  N (63.5,0.25)
100 

• Note that approximately 95% of all possible random
samples of 100 females will have sample means between
63.0 and 64.0 inches