Statistics

Probability Distributions – Part 1

Warm-up

Suppose a student is totally unprepared for a five question true or false test and has to guess for every question. Getting one question correct is independent of getting another question correct. What is the probability that she guessed all five of them correctly?

Warm-up

A shipment of 120 fasteners that contain 4 defective fasteners was sent to a manufacturing plant. The quality-control manager at the manufacturing plant selects 5 fasteners and inspects them. What is the probability that exactly 1 fasteners is defective?

Warm-up

The table shows a probability distribution of the number of students using the math lab each day and the probability.

What is the probability of 10 students using the lab? Find P(X≤ 10). Find P(8 < X ≤ 14). What is the mean of the probability distribution?

X 6 8 10 12 14

P(X) .15 .3 ? .1 .1

Agenda

Warm-up 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

Summary Homework

Random Variables

Random Variable Represents a numerical value associated

with each outcome of a probability distribution.

Denoted by x Examples

x = Number of sales calls a salesperson makes in one day.

x = Hours spent on sales calls in one day.

Random Variables

Discrete Random Variable Has a finite or countable number of possible

outcomes that can be listed. Example

x = Number of sales calls a salesperson makes in one day.

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 = Hours spent on sales calls in one day.

x

1 2430 2 …

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Discrete random variable (The number of stocks whose share price increases can be counted.) x

1 3030 2 …

1. x = The number of stocks in the Dow Jones Industrial Average that have share price increaseson a given day.

Example: Random Variables

Decide whether the random variable x is discrete or continuous.

Solution:Continuous random variable (The amount of water can be any volume between 0 ounces and 32 ounces) x

1 3230 2 …

2. x = The volume of water in a 32-ouncecontainer.

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

An industrial psychologist administered a personality inventory test for passive-aggressive traits to 150 employees. Individuals were given a score from 1 to 5, where 1 was extremely passive and 5 extremely aggressive. A score of 3 indicated neither trait. Construct a probability distribution for the random variable x. Then graph the distribution using a histogram.

Score, x

Frequency, f

1 24

2 33

3 42

4 30

5 21

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.

24(1) 0.16

150P

33(2) 0.22

150P 42

(3) 0.28150

P

30(4) 0.20

150P

21(5) 0.14

150P

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14

• Discrete probability distribution:

Solution: Constructing a 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) = 0.16 + 0.22 + 0.28 + 0.20 + 0.14 = 1.

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.20 0.14

Solution: Constructing a Discrete Probability Distribution

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) xP(x)

1 0.16 1(0.16) = 0.16

2 0.22 2(0.22) = 0.44

3 0.28 3(0.28) = 0.84

4 0.20 4(0.20) = 0.80

5 0.14 5(0.14) = 0.70

Example: Finding the Mean

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the mean.

μ = ΣxP(x) = 2.94

Solution:

Variance and Standard Deviation

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

Standard deviation of a discrete probability distribution

2 2( ) ( )x P x

Example: Finding the Variance and Standard Deviation

The probability distribution for the personality inventory test for passive-aggressive traits is given. Find the variance and standard deviation. ( μ = 2.94)

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14

Solution: Finding the Variance and Standard Deviation

Recall μ = 2.94x P(x) x – μ (x – μ)2 (x – μ)2P(x)1 0.16 1 – 2.94 = –1.94 (–1.94)2 = 3.764 3.764(0.16) = 0.602

2 0.22 2 – 2.94 = –0.94 (–0.94)2 = 0.884 0.884(0.22) = 0.194

3 0.28 3 – 2.94 = 0.06 (0.06)2 = 0.004 0.004(0.28) = 0.001

4 0.20 4 – 2.94 = 1.06 (1.06)2 = 1.124 1.124(0.20) = 0.225

5 0.14 5 – 2.94 = 2.06 (2.06)2 = 4.244 4.244(0.14) = 0.594

2 1.616 1.3 Standard Deviation:

Variance:σ2 = Σ(x – μ)2P(x) = 1.616

Probability Distribution

Construct a probability distribution for the likelihood of the number of girls born to a family of 3 children.

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

At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy one ticket. 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 $500 prize is $500 – $2 = $498 Your gain for the $250 prize is $250 – $2 = $248 Your gain for the $150 prize is $150 – $2 = $148 Your gain for the $75 prize is $75 – $2 = $73

If you do not win a prize, your gain is $0 – $2 = –$2

Solution: Finding an Expected Value

Probability distribution for the possible gains (outcomes)

Gain, x $498 $248 $148 $73 –$2

P(x)1

1500

1

1500

1

1500

1

1500

1496

1500

( ) ( )

1 1 1 1 1496$498 $248 $148 $73 ( $2)1500 1500 1500 1500 1500$1.35

E x xP x

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

Expectation -Expectation - Example

A ski resort loses $70,000 per season when it does not snow very much and makes $250,000 when it snows a lot. The probability of it snowing at least 75 inches (i.e., a good season) is 40%. Find the expected profit.

Expectation -Expectation - Example

The expected profit = ($250,000)(0.40) + (–$70,000)(0.60) = $58,000.

What else would concern you?

Profit, (X) 250,000 -70,000

P(X) 0.40 0.60

Expectation - Lottery

If you play the Daily Number and you choose to play 4 numbers straight and bet $1, then you could win $5000? What is the expected value of the game?

Expectation - Lottery

Expected Value Table

You would expect to lose $.50 every time you play the game.

R (Return) P(R) R P(R)

$4999 .0001 .4999

- $1 .9999 -.9999

1.0000 -$.50

A term life insurance policy will pay a beneficiary a certain sum of money upon the death of the policy holder. These policies have premiums that must be paid annually. Suppose a life insurance company sells a $250,000 one year term life insurance policy to a 49-year-old female for $530. According to the National Vital Statistics Report, Vol. 47, No. 28, the probability the female will survive the year is 0.99791. Compute the expected value of this policy to the insurance company.

x P(x)

530 0.99791

530 – 250,000 = -249,470

0.00209

Survives

Does not survive

E(X) = 530(0.99791) + (-249,470)(0.00209)

= $7.50 6-31

Example – Expected Value of a Discrete Random Variable

Expected Value

You are at a carnival and want to play the “rubber ducky” game. There are 100 ducks floating in the pond and it costs $1 to play. There is 1 duck that pays $20, 2 ducks that pay $10, and 3 ducks that pay $5 if selected. If you played the game a number of times, what is the expected value of the game?

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

Homework

Pg. 179-183, # 1-40 even.