Continuous Random Variables

Week 6, Friday

Continuous Random VariableThe 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

The way we describe such a random variable is through a function. Consider the following function as a probability model:where x takes on values between 0 and 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.

101)( xf

100

101

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 calculatingthe area under the curve, between x=a and x=b

P[0 ≤ X ≤ 10] = 100%

The way we describe such a random variable is through a function. Consider the following function as a probability model:where x takes on values between 0 and 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.

101)( xf

100

101

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 calculatingthe area under the curve, between x=a and x=b

P[6 ≤ X ≤ 10]P[passing] = ?

6

= 4/10 = 40%

Alternative model: Define the 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.

50)( xxf

100

.2

What is the Area Under This Curve?10 * (.2) * (1/2) = 1.00 = 100%

Now, what is the probability of Passing the quiz?

6

P[passing] = 1 – area of the small triangle = 1 – ( 4 * .2 / 2 ) = .6 = 60%

Consider the function that is defined for every number from (-∞,∞)

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

What is the Area Under This Curve?1 = 100%

What is the Symmetry/Skewness?Symmetric

How many Normal Distributions exist?Infinite Amount

For each combination of mu and sigma,

there is a uniquenormal distribution

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

Don’t memorize the complicated formula:

If mu and sigma (mean and standard deviation) are given to youthen you should know exactly what the distribution looks like:

The Most Important Continuous R.V.

2

2

2)(

21)(

x

exf

The distribution is CENTERED at MUThe distribution is more SPREAD OUT for larger values of SIGMA

),( N

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.

The Most Important Continuous R.V.

“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

100

10

90 110

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.

The Most Important Continuous R.V.

“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

100

10

90 11080 120

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.

The Most Important Continuous R.V.

“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

100

10

90 11080 12070 130

Recall the last example: Average test score (mu) is 100 with 10 as sigma.What’s the probability of scoring over 105?

Calculations with the Normal Distribution

100

10

105

P[X ≥ 105] cannotbe found by simple geometry,

so we have to do something else – a “Z calculation” – and look up

the probability on a table

Recall the last example: Average test score (mu) is 100 with 10 as sigma.What’s the probability of scoring over 105?

Calculations with the Normal Distribution

Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)

5.10100105

XZ

0

1

.5

Recall the last example: Average test score (mu) is 100 with 10 as sigma.What’s the probability of scoring over 105?

Calculations with the Normal Distribution

Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)

0

1

.5

Step 2: Look up the values in a “z table” (page A58, A59)

For any value k of Z, you can look up the

probabity P[Z≤k]

5.10100105

XZ

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

Recall the last example: Average test score (mu) is 100 with 10 as sigma.What’s the probability of scoring over 105?

Calculations with the Normal Distribution

Step 1: Transform the variable X, N(mu, sigma) to Z, N(0,1)

0

1

.5

Step 2: Look up the values in the a “z table” (page A58, A59)

P[Z ≤ 0.5] = .6915

P[Z > 0.5] = .3085