section 4.1 probability distributions larson/farber 4th ed

Section 4.1

Probability Distributions

Larson/Farber 4th ed

Section 4.1 Objectives

• Distinguish between discrete random variables and continuous random variables

• Construct a discrete probability distribution and its graph

• Determine if a distribution is a probability distribution

• Find the mean, variance, and standard deviation of a discrete probability distribution

• Find the expected value of a discrete probability distribution

Random Variables

Random Variable• Represents a numerical value associated with each

outcome of a probability distribution.• Denoted by x• Examples

x = The number of single occupancy vehicles on I5 during rush hour.

x = The amount of CO2 levels in the atmosphere.

Random Variables

Discrete Random Variable• Has a finite or countable number of possible

outcomes that can be listed.• Example

x = The number of single occupancy vehicles on I-5 during rush hour.

x

1 530 2 4

Random Variables

Continuous Random Variable• Has an uncountable number of possible outcomes,

represented by an interval on the number line.• Example

x = The amount of CO2 levels in the atmosphere.

x

1 2430 2 …

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution: Continuous random variable

1. x = The amount of water the people of Seattle drink per day (Million Gallons per day).

x

1 3230 2 …

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution: Discrete random variable

2. x = The number of umbrellas sold in Seattle in the Month of February (in thousands).

x

1 3030 2 …

Discrete Probability Distributions

Discrete probability distribution• Lists each possible value the random variable can

assume, together with its probability. • Must satisfy the following conditions:

In Words In Symbols1. The probability of each value of

the discrete random variable is between 0 and 1, inclusive.

2. The sum of all the probabilities is 1.

0 ≤ P (x) ≤ 1

ΣP (x) = 1

Constructing a Discrete Probability Distribution

1. Make a frequency distribution for the possible outcomes.

2. Find the sum of the frequencies.

3. Find the probability of each possible outcome by dividing its frequency by the sum of the frequencies.

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

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

Example: Constructing a Discrete Probability Distribution

Australia's unique wildlife apparently risks being hunted to extinction unless the cat population is controlled. Native fauna is ill-equipped to deal with this naturalized predator. Three types of cat are recognized: domestic cats which are wholly dependent on humans, unowned stray cats which rely on humans to some extent and feral cats whose reliance on humans is minimal. They can breed 3 or 4 times a year, averaging 4 kittens per litter and can rapidly establish colonies wherever there is a good food source.


Example: Constructing a Discrete Probability Distribution

The number of house holds with cats in Darwin Australia from a survey of 2668 house holds.

Cats, x Frequency, f

0 1941

1 349

2 203

3 78

4 57

5 40

∑ f = 2668

Solution: Constructing a Discrete Probability Distribution

• Divide the frequency of each score by the total number of individuals in the study to find the probability for each value of the random variable.

x 0 1 2 3 4 5

P(x) 0.73 0.13 0.08 0.03 0.02 0.01

• Discrete probability distribution:



This is a valid discrete probability distribution since

1. Each probability is between 0 and 1, inclusive,0 ≤ P(x) ≤ 1.

2. The sum of the probabilities equals 1, ΣP(x) = 0.73 + 0.13 + 0.08 + 0.03 + 0.02 + 0.01 = 1.


x 0 1 2 3 4 5

P(x) 0.73 0.13 0.08 0.03 0.02 0.01


• Histogram

Because the width of each bar is one, the area of each bar is equal to the probability of a particular outcome.


Mean

Mean of a discrete probability distribution• μ = ΣxP(x)• Each value of x is multiplied by its corresponding

probability and the products are added.


x P(x)

0 0.73

1 0.13

2 0.08

3 0.03

4 0.02

5 0.01

Example: Finding the MeanThe probability distribution for the cats in Darwin. Find the mean.

Solution:

xP(x)

0*0.73 0

1*0.13 0.13

2*0.08 0.15

3*0.03 0.09

4*0.02 0.09

5*0.01 0.07

Variance and Standard Deviation

Variance of a discrete probability distribution• σ2 = Σ(x – μ)2P(x)

Standard deviation of a discrete probability distribution

•


Example: Finding the Variance and Standard Deviation

The probability distribution for cats per household in Darwin. Find the variance and standard deviation. ( μ = 0.53)

x P(x)

0 0.73

1 0.13

2 0.08

3 0.03

4 0.02

5 0.01


x P(x) x - μ (x - μ)2 (x - μ)2P(x)

0 0.73 -0.53 0.28 0.21

1 0.13 0.47 0.22 0.03

2 0.08 1.47 2.16 0.16

3 0.03 2.47 6.10 0.18

4 0.02 3.47 12.03 0.26

5 0.01 4.47 19.97 0.30

1.14

Solution: Finding the Variance and Standard Deviation

Standard Deviation:

Variance:


(x - μ)2P(x)

0.21

0.03

0.16

0.18

0.26

0.30

1.14

Expected Value

Expected value of a discrete random variable • Equal to the mean of the random variable.• E(x) = μ = ΣxP(x)


Example: Finding an Expected Value

#46 - A charity organization is selling $4 raffle tickets as part of a fund-raising program. The first prize is a boat valued at $3150 and the second prize is a camping tent valued at $450. The remaining 15 prizes are $25 gift certificates. The number of tickets sold is 5000.

What is the expected value of your gain?


Solution: Finding an Expected Value

• To find the gain for each prize, subtractthe price of the ticket from the prize: Your gain for the $3150 prize is $3150 – $4 =

$3146 Your gain for the $450 prize is $450 – $4 = $446 Your gain for the $25 prizes is $25 – $4 = $21

• If you do not win a prize, your gain is $0 – $4 = –$4


Solution: Finding an Expected Value

• Probability distribution for the possible gains (outcomes)

Gain, x 3150 - 4 = 3146 450 - 4 = 446 25 - 4 = 21 –$4

P(x)

You can expect to lose an average of $3.21 for each ticket you buy.


Section 4.1 Summary

• Distinguished between discrete random variables and continuous random variables

• Constructed a discrete probability distribution and its graph

• Determined if a distribution is a probability distribution

• Found the mean, variance, and standard deviation of a discrete probability distribution

• Found the expected value of a discrete probability distribution


section 4.1 probability distributions larson/farber 4th ed

Documents