4.1 probability distributions

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4.1 Probability

Distributions

Statistics

Mrs. Spitz

Fall 2010

Random Variables

• The outcome of a probability experiment is often a

count or a measure. When this occurs, the

outcome is called a random variable.

• A random variable, x, represents a numerical

value assigned to an outcome of a probability

experiment.• There are two types: discrete and continuous

Discrete v. continuous

• A random variable is discrete if it has a

finite or countable number of possibleoutcomes that can be listed.

• A random variable is continuous if it has an

infinite number of possible outcomes

represented by an interval on the number

So . . .

• Suppose you conduct an experiment of thenumber of calls a salesperson makes in oneday. The possible values of the random variable

are 0, 1, 2, 3, 4, and so on. Because the set ofpossible outcomes {0, 1, 2, 3 . . . } can be listed,x is a discrete random variable. You canrepresent its values as points on a number line.

0 1 2 3 4 5 6 7 8 x can only have wholenumber values 0, 1, 2, 3 . . .

So . . .

• A different way to conduct the study would be tomeasure the time (in hours) a salespersonspends making calls in one day. Because thetime spent making sales calls can be any

number from 0 to 24 (including fractions anddecimals), x is a continuous random variable.You can represent its values with an interval ona number line, but you cannot list all the possiblevalues.

0 3 6 9 12 15 18 21 24 x can have any valuebetween 0 and 24

Ex. 1: Discrete variables andcontinuous variables.

• Decide whether the random variable, x, isdiscrete or continuous. Explain your reasoning.

1. x represents the number of stocks in the Dow

Jones Industrial Average that have share priceincreases on a given day.

The number of stocks whose share valueincreases can be counted {0, 1, 2, 3 . . . }. So xis a discrete random variable.

Ex. 1: Discrete variables andcontinuous variables.

• Decide whether the random variable, x, isdiscrete or continuous. Explain yourreasoning.

2. x represents the volume of bottled waterin a 32-ounce container.

The amount of water in the container can beany volume between 0 and 32, so x is acontinuous random variable.

• It is important that you can distinguish between discrete and continuous variables

because different statistical techniques areused to analyze each. The remainder of thischapter focuses on discrete random

variables and their probability distributions.You will study continuous distributionslater.

Discrete Probability Distributions

• Each value of a discrete random variable can beassigned a probability. By listing each value of therandom variable with its corresponding probability, youare forming a probability distribution.

1. The probability of each value of the discrete randomvariable is between 0 and 1 inclusive. That is,

0 P(x) 1

2. The sum of all the probabilities is 1. That is,

P(x) = 1

Graphing

• Because probabilities represent relative

frequencies, a discrete probability

distribution can be graphed with a relative

frequency histogram.

Guidelines of Constructing

a Discrete Probability

DistributionLet x be a discrete random variable with possible

outcomes x1, x2, . . . xn.

1. Make a frequency distribution for the possibleoutcomes.

2. Find the sum of the frequencies.

3. Find the probability of each possible outcome by

dividing the frequency by the sum of thefrequencies.

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

Ex. 2: Constructing a DiscreteProbability Distribution

• An industrial psychologist has administered apersonality inventory test for passive-aggressivetraits to 150 employees. Individuals were rated

on a score from 1 to 5 where 1 was extremelypassive and 5 extremely aggressive. A score of3 indicated neither trait. The results are shownon the next slide. Construct a probability

distribution for the random variable, x. Thengraph the distribution.

Solution

Divide the frequency ofeach score by the totalnumber of scores (150) tofind the probability for

each value of the randomvariable.

P(1) = 24/150 = 0.16

P(2) = 33/150 = 0.22

P(3) = 42/150 = 0.28

P(4) = 30/150 = 0.2

P(5) = 21/150 = 0.14

Score (x) Frequency (f)

• The discreteprobabilitydistribution is shown

in the followingtable. Note thateach probability isbetween 0 and 1and the sum of the

probabilities is 1.

• The relativefrequencydistribution is also

shown at the right.The area of eachbar represents theprobability of aparticular outcome.

1 2 3 4 5

x 1 2 3 4 5

P(x) 0.16 0.22 0.28 0.2 0.14

Ex. 3: Verifying ProbabilityDistributions

• Verify that thedistribution is aprobability

distribution.

Days of Rain Probability

0 0.216

1 0.432

2 0.288

3 0.064

Solution: If the distributionis a probability

distribution, the (1) eachof probability is between0 and 1, inclusive and (2)the sum of theprobabilities equals 1.

Ex. 3: Verifying ProbabilityDistributions

1. Each probability isbetween 0 and 1. Days of Rain Probability

0 0.216

1 0.432

2 0.288

3 0.064

2. P(x) = 0.216 + 0.432 +

0.288 + 0.064 = 1

Because both conditions are

met, the distribution is aprobability distribution.

Ex. 4: Probability Distributions

• Decide whether each distribution is a probabilitydistribution.

x 5 6 7 8

P(x) 0.28 0.21 0.43 0.15

Each probability is

between 0 and 1.

However, the sum of

the probabilities is1.07, which is greaterthan 1. So, it is NOTa probabilitydistribution.

Ex. 4: Probability Distributions

• Decide whether each distribution is a probabilitydistribution.

x 1 2 3 4

P(x) ½ ¼ 5/4 -1

However, P(3) andP(4) are not between0 and 1. So, it is NOTa probabilitydistribution.

The sum of theprobabilities is equalto 1.

Mean, Variance and Standard

Deviation• You can measure the central tendency of a

probability distribution with its mean, and measurethe variability with its variance and standard

deviation.

The mean of a discrete random variable is given by:

= xP(x).

Each value of x is multiplied by its corresponding probability and the products are added.

• The mean of the random variable represents

the “theoretical average” of a probabilityexperiment and sometimes is not a possible

outcome. If the experiment were performed

thousands of times, the mean of all the

outcomes would be close to the mean of therandom variable.

Ex. 5: Finding the Mean of a ProbabilityDistribution

The probabilitydistribution for thepersonality inventory

test for passive-aggressive traitsdiscussed in Ex. 2 isgiven at the right.

Find the mean score.What can youconclude?

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14

Organize your tables carefully.

• Use the table to organizeyour work as shown atthe left. From the table,you can see that the

mean is 2.94. A score of3 represents an individualwho is neither extremelypassive nor aggressive,

but is slightly closer topassive.

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

P(x) = 1 xP(x) = 2.94MEAN

• While the mean of the random variable of a

probability distribution describes a typical

outcome, it gives no information about how

the outcomes vary. To study the variation

of the outcomes, you can use the variance

and standard deviation of the randomvariable of a probability distribution.

Standard Deviation of a Discrete RandomVariable

• The variance of a discrete randomvariable is:

2 = (x - )2P(x)

The standard deviat ion is:

= √2

Ex. 6: Finding the variance and StandardDeviation

• The probabilitydistribution for thepersonality inventorytest for passive-aggressive traitsdiscussed in Ex. 2 isgiven at the right.Find the variance and

standard deviation ofthe probabilitydistribution.

x P(x)

1 0.16

2 0.22

3 0.28

4 0.20

5 0.14

x P(x) x - (x -)2 P(x)(x - )2

1 0.16 1 – 2.94 =

(-1.94)2 =

(0.16)(3.764) = 0.602

2 0.22 2 – 2.94 =

- 0.94

(-0.94)2 =

(0.22)(0.884) = 0.194

3 0.28 3 – 2.94 =

(0.06)2 =

(0.28)(0.004) = 0.001

4 0.20 4 – 2.94 =

(1.06)2 =

(0.20)(1.124) = 0.225

5 0.14 5 – 2.94 =

(2.06)2 =

(0.14)(4.244) = 0.594

P(x) = 1 P(x)(x - )2 = 1.616 =

VARIANCE

You know from ex. 5 that the mean of the distribution is = 2.94. Use a tablelike the one below to organize your work!

• So, the variance of 2 = 1.616 and thestandard deviation is

= √1.62 ≈1.27

Lots of steps, so organize your workcarefully. Most of you got the problemscorrect, but some missed the details.

Expected Value

• The expected value of a discrete randomvariable is equal to the mean of therandom variable.

Expected value = E (x) = = xP (x)

Ex. 7: Finding an Expected Value

• At a raffle, 1500 tickets are sold at $2 eachfor four prizes of $500, $250, $150 and$75. You buy one ticket. What is the

expected value of your gain?

• Note: Expected value plays a role indecision theory. Although probability can

never be negative, the expected value canbe negative.

Gain, x $498 $248 $148 $73 - $2

Probability, P(x)

1/1500 1/1500 1/1500 1/1500 1496/1500

SOLUTION: To find the gain for each prize, subtract theprice of the ticket from the prize. For instance, your gain for

the $500 prize is $500 - $2 = $498. Then write a probabilitydistribution for the possible gains (or outcomes).

Then, using the probability distribution, you can find theexpected value. E(x) = xP(x)

1496)2(

)(E(x)

Because the expected value is negative, you can expectto lose an average of $1.35 for each ticket you buy.

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