Transcript Week 6, Friday - gozips.uakron.edu

Continuous Random Variables
Week 6, Friday
Continuous Random Variable
The Discrete Random Variables we’ve seen so far are easy to represent
in table form; however, some random variables are impossible to
represent in this form.
Example: Consider the number of seconds it takes to run one mile. For
very fast runners, this number could be around 300. For very slow
runners, this number could be as high as 1,200. Furthermore, it is possible
to get more precise values by rounding to some decimal, for example
400.25 seconds. To put all of these values into a table would be
impractical. For the purpose of this class, we will define such random
variables as continuous random variables.
But if we don’t define X by a table then how do we define it?
ANSWER: with a function
Examples of a Continuous R.V.
Suppose I give a 10 point quiz. There are 100 questions, and partial
credit is possible for each one. Then the score for students would be a
continuous random variable because decimal scores such as 8.38495/10
may be possible, and these values are tough to put into a table.
The way we describe such a random variable is through a function.
1
Consider the following function as a probability model: f ( x)  10
where x takes on values between 0 and 10.
1
10
0
P[0 ≤ X ≤ 10] = 100%
What is the Area Under This Curve?
10 * (1/10) = 1.00 = 100%
How does this link to Probability?
For continuous random variables,
P[a ≤ X ≤ b] can be found by calculating
the area under the curve, between x=a and x=b
10
Examples of a Continuous R.V.
Suppose I give a 10 point quiz. There are 100 questions, and partial
credit is possible for each one. Then the score for students would be a
continuous random variable because decimal scores such as 8.38495/10
may be possible, and these values are tough to put into a table.
The way we describe such a random variable is through a function.
1
Consider the following function as a probability model: f ( x)  10
where x takes on values between 0 and 10.
P[passing] = ?
1
10
0
P[6 ≤ X ≤ 10] = 4/10 = 40%
What is the Area Under This Curve?
10 * (1/10) = 1.00 = 100%
How does this link to Probability?
For continuous random variables,
P[a ≤ X ≤ b] can be found by calculating
the area under the curve, between x=a and x=b
6
10
Examples of a Continuous R.V.
Suppose I give a 10 point quiz. There are 100 questions, and partial
credit is possible for each one. Then the score for students would be a
continuous random variable because decimal scores such as 8.38495/10
may be possible, and these values are tough to put into a table.
Alternative model: Define the function: f ( x) 
x
50
What is the Area Under This Curve?
10 * (.2) * (1/2) = 1.00 = 100%
Now, what is the probability of Passing the quiz?
P[passing] = 1 – area of the small triangle
= 1 – ( 4 * .2 / 2 )
= .6 = 60%
.2
0
6
10
The Most Important Continuous R.V.
Consider the function that is defined for every number from (-∞,∞)
f ( x) 

1
 2
What is the Area Under This Curve?
1 = 100%
What is the Symmetry/Skewness?
Symmetric
e
( x )2
2 2
How many Normal Distributions exist?
Infinite Amount
For each combination
of mu and sigma,
there is a unique
normal distribution
The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA


The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA


The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA


The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA


The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA


The Most Important Continuous R.V.
1
e
Don’t memorize the complicated formula: f ( x) 
 2

( x )2
2 2
If mu and sigma (mean and standard deviation) are given to you
then you should know exactly what the distribution looks like:
The distribution is CENTERED at MU
The distribution is more SPREAD OUT for larger values of SIGMA
N ( , )


The Most Important Continuous R.V.
Example: Brown University gives a standardized exam to all students
interested in joining the MBA program. The exam is graded in such a way
that the scores are normally distributed with a mean of 100 and standard
deviation of 10.
“Empirical Rule”: For normal distributions,
68% of the values are within 1 standard deviation of the mean,
95% of the values are within 2 standard deviation of the mean,
99.8% of the values are within 3 standard deviation of the mean
  10
  100
90
110
The Most Important Continuous R.V.
Example: Brown University gives a standardized exam to all students
interested in joining the MBA program. The exam is graded in such a way
that the scores are normally distributed with a mean of 100 and standard
deviation of 10.
“Empirical Rule”: For normal distributions,
68% of the values are within 1 standard deviation of the mean,
95% of the values are within 2 standard deviation of the mean,
99.8% of the values are within 3 standard deviation of the mean
  10
  100
80
90
110
120
The Most Important Continuous R.V.
Example: Brown University gives a standardized exam to all students
interested in joining the MBA program. The exam is graded in such a way
that the scores are normally distributed with a mean of 100 and standard
deviation of 10.
“Empirical Rule”: For normal distributions,
68% of the values are within 1 standard deviation of the mean,
95% of the values are within 2 standard deviation of the mean,
99.8% of the values are within 3 standard deviation of the mean
  10
  100
70
80
90
110
120
130
Calculations with the Normal Distribution
Recall the last example: Average test score (mu) is 100 with 10 as sigma.
What’s the probability of scoring over 105?
P[X ≥ 105] cannot
be found by simple geometry,
so we have to do something else
– a “Z calculation” – and look up
the probability on a table
  10
  100
105
Calculations with the Normal Distribution
Recall the last example: Average test score (mu) is 100 with 10 as sigma.
What’s the probability of scoring over 105?
Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)
Z
X 


105  100
 .5
10
 1
0
.5
Calculations with the Normal Distribution
Recall the last example: Average test score (mu) is 100 with 10 as sigma.
What’s the probability of scoring over 105?
Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)
Step 2: Look up the values in a “z table” (page A58, A59)
 1
0
.5
For any value k of Z,
you can look up the
probabity P[Z≤k]
Z
X 


105  100
 .5
10
We are interested in
the 6th row (0.5) and
the first column,
since the second
decimal point is 0
(0.50)
Therefore
P[Z ≤ 0.50] = .6915
Calculations with the Normal Distribution
Recall the last example: Average test score (mu) is 100 with 10 as sigma.
What’s the probability of scoring over 105?
Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)
Step 2: Look up the values in the a “z table” (page A58, A59)
P[Z ≤ 0.5] = .6915
 1
0
.5
P[Z > 0.5] = .3085