Spring 2009: Section 4 – Lecture 4 Reading: Chapter 4 Chapter 7.
CHAPTER 4
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Transcript of CHAPTER 4
CHAPTER 4
• 4.1 - Discrete Models General distributions Classical: Binomial, Poisson, etc.
• 4.2 - Continuous Models General distributions Classical: Normal, etc.
X
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~ The Normal Distribution ~(a.k.a. “The Bell Curve”)
Johann Carl Friedrich Gauss 1777-1855
μ
σ
mean
standard deviationX ~ N(μ, σ)
• Symmetric, unimodal• Models many (but not
all) natural systems• Mathematical
properties make it useful to work with
Standard Normal DistributionZ ~ N(0, 1)
Z
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21( )2
z
z e
density function
The cumulative distribution function (cdf) is denoted by (z). It is tabulated, and computable in R via the command pnorm.
SPECIA
L CASE
Total Area = 11
Z
1
Standard Normal DistributionZ ~ N(0, 1)
Example
Find P(Z 1.2).
1.2
“z-score”
Total Area = 1
Z
Standard Normal DistributionZ ~ N(0, 1)
Example
Find P(Z 1.2).
1
1.2
Use the included table.
“z-score”
Total Area = 1
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Lecture Notes Appendix…
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Z
Standard Normal DistributionZ ~ N(0, 1)
Example
Find P(Z 1.2).
1
1.2
Use the included table.
0.88493
Use R:
> pnorm(1.2) [1] 0.8849303
“z-score”
P(Z > 1.2)
0.11507
Total Area = 1
Note: Because this is a continuous distribution, P(Z = 1.2) = 0, so there is no difference between P(Z > 1.2) and P(Z 1.2), etc.
Standard Normal DistributionZ ~ N(0, 1)
Z
XZ
μσ
X ~ N(μ, σ)
1
Any normal distribution can be transformed to the standard normal distribution via a simple change of variable.
Why be concerned about this, when most “bell curves” don’t have mean = 0, and standard deviation = 1?
Year 2010X ~ N(25.4, 1.5)
μ = 25.4
σ = 1.5
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Example
Random Variable X = Age at first birth
POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?
P(X < 27.2) = ?
27.2
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Example
Random Variable X = Age at first birth
POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?
P(X < 27.2) = ?
Year 2010X ~ N(25.4, 1.5) σ = 1.5
μ = 33
The x-score = 27.2 must first be transformed to a corresponding z-score.
μ = 25.4μ = 25.4 27.2
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Example
Random Variable X = Age at first birth
POPULATION Question: What proportion of the population had their first child before the age of 27.2 years old?
P(X < 27.2) = ?
XZ
27.2 25.4
1.5Z
1.2Z
σ = 1.5
μ = 33
P(Z < 1.2) = 0.88493
Using R:
> pnorm(27.2, 25.4, 1.5) [1] 0.8849303
Year 2010X ~ N(25.4, 1.5)
μ = 25.4 27.2
Z
What symmetric interval about the mean 0 contains 95% of the population values?
That is…
1
Standard Normal DistributionZ ~ N(0, 1)
Z
0.95
0.025 0.025
+z.025 = ?-z.025 = ?
What symmetric interval about the mean 0 contains 95% of the population values?
That is…
Standard Normal DistributionZ ~ N(0, 1)
Use the included table.
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Lecture Notes Appendix…
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Use the included table.
+z.025 = ?+z.025 = +1.96-z.025 = ?
Standard Normal DistributionZ ~ N(0, 1)
Z
0.95
0.025 0.025
What symmetric interval about the mean 0 contains 95% of the population values?
-z.025 = -1.96 “.025 critical values”
Use R:
> qnorm(.025) [1] -1.959964
> qnorm(.975) [1] 1.959964
+z.025 = ?+z.025 = +1.96-z.025 = ?
Standard Normal DistributionZ ~ N(0, 1)
Z
0.95
0.025 0.025
What symmetric interval about the mean 0 contains 95% of the population values?
-z.025 = -1.96 “.025 critical values”
25.41.961.5
X
XZ
25.4 (1.96)(1.5)X
What symmetric interval about the mean age of 25.4 contains 95% of the population values?
25.4 2.94X
22.46 X 28.34 yrs
X ~ N(μ, σ) X ~ N(25.4, 1.5)
> areas = c(.025, .975)> qnorm(areas, 25.4, 1.5) [1] 22.46005 28.33995
Use the included table.
Standard Normal DistributionZ ~ N(0, 1)
Z
0.90
0.05 0.05
+z.05 = ?-z.05 = ?
What symmetric interval about the mean 0 contains 90% of the population values?
Similarly…
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…so average 1.64 and 1.65 0.95 average of 0.94950 and 0.95053…
Use the included table.
-z.05 = ?-z.05 = -1.645
Standard Normal DistributionZ ~ N(0, 1)
Z
0.90
0.05 0.05
+z.05 = ?
What symmetric interval about the mean 0 contains 90% of the population values?
Similarly…+z.05 = +1.645“.05 critical values”
Use R:
> qnorm(.05) [1] -1.644854
> qnorm(.95) [1] 1.644854
-z.05 = ?-z.05 = -1.645
Standard Normal DistributionZ ~ N(0, 1)
Z
0.90
0.05 0.05
+z.05 = ?
What symmetric interval about the mean 0 contains 100(1 – )% of the population values?
Similarly…+z.05 = +1.645“.05 critical values”
In general….
1 –
/ 2 / 2
-z / 2 +z / 2“ / 2 critical values”
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Normal Approximation to the Binomial Distributioncontinuous discrete
Suppose a certain outcome exists in a population, with constant probability .
P(Success) = P(Failure) = 1 –
We will randomly select a random sample of n individuals, so that the binary “Success vs. Failure” outcome of any individual is independent of the binary outcome of any other individual, i.e., n Bernoulli trials (e.g., coin tosses).
Discrete random variableX = # Successes in sample(0, 1, 2, 3, …,, n)
Discrete random variableX = # Successes in sample(0, 1, 2, 3, …,, n)
Then X is said to follow a Binomial distribution, written X ~ Bin(n, ), with “probability function”
f(x) = , x = 0, 1, 2, …, n.
x n xnx (1 )
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> dbinom(10, 100, .2) [1] 0.00336282
Area
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> pbinom(10, 100, .2) [1] 0.005696381
Area
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Therefore, if…
X ~ Bin(n, ) with n 15 and n (1 – ) 15,
then… , (1 .X N n n
That is…
(1 )ˆ ,X Nn n
“Sampling Distribution” of ̂
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Classical Continuous Probability Distributions
● Normal distribution
● Log-Normal ~ X is not normally distributed (e.g., skewed), but Y = “logarithm of X” is normally distributed
● Student’s t-distribution ~ Similar to normal distr, more flexible
● F-distribution ~ Used when comparing multiple group means
● Chi-squared distribution ~ Used extensively in categorical data analysis
● Others for specialized applications ~ Gamma, Beta, Weibull…