1. STA 291 Spring 2008
Lecture 8
Dustin Lueker

2. Continuous Probability Distribution
Can not list all possible values with probabilities
Probabilities are assigned to intervals of numbers
Probability of an individual number is 0
Probabilities have to be between 0 and 1
Probability of the interval containing all possible values equals 1
Mathematically, a continuous probability distribution corresponds to a (density) function whose integral equals 1
STA 291 Spring 2008 Lecture 8

3. Continuous Probability Distribution
Let X=Weekly use of gasoline by adults in North America (in gallons)
P(16<X<21)=0.34
The probability that a randomly chosen adult in North America uses between 16 and 21 gallons of gas per week is 0.34
STA 291 Spring 2008 Lecture 8

4. Graphs for Probability Distributions
Discrete Variables:
Histogram
Height of the bar represents the probability
Continuous Variables:
Smooth, continuous curve
Area under the curve for an interval represents the probability of that interval
STA 291 Spring 2008 Lecture 8

5. Continuous Distributions
STA 291 Spring 2008 Lecture 8

6. The Normal Probability Distribution
Gaussian Distribution
Carl Friedrich Gauss ( )
Perfectly symmetric and bell-shaped
Empirical rule applies
Probability concentrated within 1 standard deviation of the mean is always 0.68
Probability concentrated within 2 standard deviations of the mean is always 0.95
Probability concentrated within 3 standard deviations of the mean is always 0.997
Characterized by two parameters
Mean = μ
Standard Deviation = σ
STA 291 Spring 2008 Lecture 8

7. Different Normal Distributions
STA 291 Spring 2008 Lecture 8

8. Normal Distribution and Empirical Rule
Assume that adult female height has a normal distribution with mean μ=165 cm and standard deviation σ=9 cm
With probability 0.68, a randomly selected adult female has height between μ - σ = 156 cm and
μ + σ = 174 cm
This means that on the normal distribution graph of adult female heights the area under the curve between 156 and 174 is .68
With probability 0.95, a randomly selected adult female has height between μ - 2σ = 147 cm and μ + 2σ = 183 cm
This means that on the normal distribution graph of adult female heights the area under the curve between 147and 183 is .95
STA 291 Spring 2008 Lecture 8

9. Normal Distribution
So far, we have looked at the probabilities within one, two, or three standard deviations from the mean using the Empirical Rule
(μ + σ, μ + 2σ, μ + 3σ)
How much probability is concentrated within standard deviations of the mean?
More general, how much probability is concentrated within any number (say z) of standard deviations of the mean?
STA 291 Spring 2008 Lecture 8

10. Normal Distribution Table
Table 3 (page B-8) shows for different values of z the probability between 0 and μ + zσ
Probability that a normal random variable takes any value between the mean and z standard deviations above the mean
Example
z =1.43, the tabulated value is .4236
That is, the probability between 0 and 1.43 of the standard normal distribution equals .4236
Symmetry
z = -1.43, the tabulated value is .4236
That is, the probability between and 0 of the standard normal distribution equals .4236
So, within 1.43 standard deviations of the mean is how much probability?
STA 291 Spring 2008 Lecture 8

11. Verifying the Empirical Rule
P(-1<Z<1) should be about 68%
P(-1<Z<1) = P(-1<Z<0) + P(0<Z<1) = 2*P(0<Z<1) = ?
P(-2<Z<2) should be about 95%
P(-2<Z<2) = P(-2<Z<0) + P(0<Z<2) = 2*P(0<Z<2) = ?
P(-3<Z<3) should be about 99.7%
P(-3<Z<3) = P(-3<Z<0) + P(0<Z<3) = 2*P(0<Z<3) = ?
STA 291 Spring 2008 Lecture 8

12. Working backwards
We can also use the table to find z-values for given probabilities
Find the z-value corresponding to a right- hand tail probability of 0.025
This corresponds to a probability of between 0 and z standard deviations
Table: z = 1.96
P(Z>1.96) = .025
STA 291 Spring 2008 Lecture 8

13. Finding z-Values for Percentiles
For a normal distribution, how many standard deviations from the mean is the 90th percentile?
What is the value of z such that 0.90 probability is less than z?
P(Z<z) = .90
If 0.9 probability is less than z, then there is 0.4 probability between 0 and z
Because there is 0.5 probability less than 0
This is because the entire curve has an area under it of 1, thus the area under half the curve is 0.5
z=1.28
The 90th percentile of a normal distribution is 1.28 standard deviations above the mean
STA 291 Spring 2008 Lecture 8

14. Finding z-Values for Two-Tail Probabilities
What is the z-value such that the probability is 0.1 that a normally distributed random variable falls more than z standard deviations above or below the mean?
Symmetry
We need to find the z-value such that the right-tail probability is 0.05 (more than z standard deviations above the mean)
z=1.65
10% probability for a normally distributed random variable is outside 1.65 standard deviations from the mean, and 90% is within 1.65 standard deviations from the mean
Find the z-value such that the probability is 0.5 that a normally distributed random variable falls more than z standard deviations above or below the mean
STA 291 Spring 2008 Lecture 8