probability 4.1

Probability DistributionsProbability DistributionsChapter 4.1Chapter 4.1

Objectives• Distinguish between discrete random

variables and continuous random variables• Learn to construct a discrete probability

distribution and its graph• Learn to find the mean, variance, and

standard deviation of a discrete probability distribution

• Learn to find the expected value of a discrete probability distribution

Discrete & Continuous• Discrete data is countable

– Number of students– Number of sales calls

• Continuous data has an infinite number of possible outcomes – Amount of time– Temperature

Random Variables• A random variable x represents a numerical

value associated with each outcome of a probability experiment.

• The random variable is discrete if it has a finite or countable number of possible outcomes that can be listed.

• The variable is continuous if it has an uncountable number of possible outcomes, represented by an interval on the number line.

Discrete or Continuous?• The number of stocks in the Dow

Jones Industrial Average that have a share price increase on a given day.

Discrete or Continuous?• The length of time it takes to

complete a test.

Discrete or Continuous?• The number of home runs hit during a

baseball game

Discrete Probability Distributions

• A discrete probability distribution lists each possible value the random variable can assume, together with its probability.

• The Rules:– The probability of each value is between

0 and 1.– The sum of all the probabilities is 1.

Example – Roll 2 dice:Total Probability

2 0.02783 0.05564 0.08335 0.11116 0.13897 0.16678 0.13899 0.111110 0.083311 0.055612 0.0278

1.0000

Constructing a Discrete Probability Distribution

• Let x be a discrete random variable with the possible outcomes x1, x2, . . . xn

2. Make a frequency distribution for the possible outcomes.

3. Find the sum of the frequencies.4. Find the probability of each possible outcome by

dividing its frequency by the sum of the frequencies.

5. Check that each probability is between 0 and 1 and that the sum is 1.

Example• A sociologist surveyed 200 households

in a small town and asked how many dependents they had.

• The results were:Dependent Children Number of Households

0 14

1 40

2 76

3 44

4 26

Example• Find the sum of the frequencies:• 200• Find the probability of each possible

outcome:Dependent Children

Number of Households

Percentage

0 14 7%

1 40 20%

2 76 38%

3 44 22%

4 26 13%

Example• Check that each probability is

between 0 and 1 and that the sum is 1Dependent Children

Number of Households

Percentage

0 14 7%

1 40 20%

2 76 38%

3 44 22%

4 26 13%

Mean• How would you find the average?

Dependent Children

Number of Households

Percentage

0 14 7%

1 40 20%

2 76 38%

3 44 22%

4 26 13%

Mean• How would you find the average?

Dependent Children

Number of Households

Percentage

x(P(x)

0 14 7% 0*.07 0

1 40 20% 1*.20 .20

2 76 38% 2*.38 .76

3 44 22% 3*.22 .66

4 26 13% 4*.13 .52

Total 200 100% 2.14

Mean• The mean of a discrete random variable is

given by:• μ = ∑xP(x)• Each value of x is multiplied by its

corresponding probability and the products are added.

How would you find the variance & standard deviation?Dependent Children

Number of Households

Percentage

0 14 7%

1 40 20%

2 76 38%

3 44 22%

4 26 13%

How would you find the variance & standard deviation?

• standard deviation is √10.098 = 3.177

Dependent Children

Number of Households

Percentage x - μ (x – μ)2

0 14 7% -2.14 4.5796

1 40 20% -1.14 1.2996

2 76 38% -.14 .0196

3 44 22% .86 .7396

4 26 13% 1.86 3.4596

Total 200 100% 10.098

Mean, Variance, & Standard Deviation

• The variance of a discrete random variable is given by:

• σ2 = ∑(x-μ) 2 P(x)• A shortcut of this formula is:• σ2 = ∑x2 P(x) - μ2

• The standard deviation is:• σ = √σ2

Expected Value• The expected value of a discrete

random variable is equal to the mean of the random variable.

Expected Value Example• At a raffle, 1500 tickets are sold at $2 each for

4 prizes of $500, $250, $150, and $75. You buy 1 ticket. What is the expected value of your gain?

• What are your possible gains?• $500-$2 = $498• $248, $148, $73, and $-2• What is the possibility of each of these?• 1/1500 for the first 4, 1496/1500 for the -2

Expected Value Example• E(x) = $498 * 1/1500 + $248*1/1500 +

$148*1/1500 + $73* 1/1500 + -2*1496/1500

• = -$1.35• Because the expected value is negative,

you can expect to lose an average of $1.35 for each ticket you purchase.

Homework• P. 179 Do 1-10 together• P. 179 12-28, 34 evens