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Chapter 6:Probability Distributions

Lecture Powerpoint

Discovering Statistics

2nd Edition Daniel T. Larose

+ Chapter 6 Overview

6.1 Discrete Random Variables

6.2 Binomial Probability Distribution

6.3 Poisson Probability Distribution

6.4 Continuous Random Variables and the Normal Probability Distribution

6.5 Standard Normal Distribution

6.6 Applications of the Normal Distribution

6.7 Normal Approximation to the Binomial Probability Distribution

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+ The Big Picture

Where we are coming from and where we are headed…

In Chapter 5, we learned about probability, which allows us to quantify the uncertainty involved in performing statistical inference.

However, we need a new set of tools in our probability toolbox: random variables and probability distributions. Here in Chapter 6, we learn these new tools, including the binomial distribution and the normal distribution.

Chapter 7, “Sampling Distributions,” is a pivotal chapter where we learn that statistics have predictable behavior, which allows us to perform the statistical inference we learn in the remainder of the book.

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+ 6.1: Discrete Random Variables

Objectives:

Identify random variables.

Explain what a discrete probability distribution is and construct probability distribution tables and graphs.

Calculate the mean, variance, and standard deviation of a discrete random variable.

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Random VariablesIn this chapter, we develop an approach that analyzes probability problems more efficiently than we did in Chapter 5. Recall from Chapter 1 that a variable is a characteristic that can assume different values.

A random variable is a variable that takes on quantitative values representing the results of a probability experiment, and thus its values are determined by chance. We denote random variables using capital letters such as X, Y, or Z.

Example

Let X = outcome of a single roll of a die

X could be 1, 2, 3, 4, 5, 6

P(X = 5) = 1/6

6Discrete and Continuous Random VariablesThere are two main types of random variables. The difference between the two types relates to the possible values that each type of random variable can assume.

Discrete and Continuous Random Variables

• A discrete random variable can take either a finite or a countable number of values. The values can be graphed as separate points on a number line.

• A continuous random variable can take on infinitely many values. The values form an interval on the number line.

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Discrete Probability Distributions

For every random variable, there is a probability distribution that allows us to view all possible values of the random variable at a glance.

A probability distribution of a discrete random variable provides all the possible values that the random variable can assume, together with the probability associated with each value. The probability distribution can take the form of a table, graph, or formula. Probability distributions describe populations, not samples.

When constructing a probability distribution of a discrete random variable,

•Note each possible value of X

•Note the probability associated with each value of X

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Mean of a Discrete Random Variable

We can calculate the mean and standard deviation of a discrete random variable X, just as we can calculate the mean and standard deviation of quantitative data.

Finding the Mean of a Discrete Random Variable XThe mean µ of a discrete random variable X is found as follows:1. Multiply each possible value of X by its probability2. Add the resulting products

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μ = X⋅ P(X)[ ]∑

X = age P(X)

15 0.07

16 0.17

17 0.29

18 0.47€

μ =15(0.07) +16(0.17) +17(0.29) +18(0.47)=17.16

The mean µ of a discrete random variable is also called the expected value or the expectation of the random variable X, denoted as E(X).

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Variability of a Discrete Random Variable

Since a discrete random variable takes on quantitative values, we use the variance or standard deviation of a random variable X to help us determine whether a particular value is unusual.

Formulas for the Variance and Standard Deviation of a Discrete Random Variable

Definition Formulas Computational Formulas

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σ2 = (X − μ)2⋅ P(X)[ ]∑σ = (X − μ)2⋅ P(X)[ ]∑

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σ2 = X 2⋅ P(X)[ ]∑ − μ 2

σ = X 2⋅ P(X)[ ]∑ − μ 2

X = age P(X)

15 0.07

16 0.17

17 0.29

18 0.47

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μ =17.16σ 2 = (15 −17.16)2(0.07) + ...+ (18 −17.16)2(0.47)σ 2 = 0.8944σ = 0.9457

+ 6.2: Binomial Probability Distribution

Objectives:

Explain what constitutes a binomial experiment.

Compute probabilities using the binomial probability formula.

Find probabilities using the binomial tables.

Calculate the mean, variance, and standard deviation of the binomial random variable and find the mode of the distribution.

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Binomial ExperimentsMany situations involve only two possible outcomes to a process. Methods have been developed to make it more convenient to analyze them. These methods begin with the definition of a binomial experiment.

Binomial Experiment

A probability experiment that satisfies the following four requirements is said to be a binomial experiment:

1.Each trial of the experiment has only two possible outcomes. One outcome is denoted a success and the other a failure.2.There is a fixed number of trials, known in advance.3.The experimental outcomes are independent of each other.4.The probability of observing a success is the same from trial to trial.

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Binomial ExperimentsThe outcomes of a binomial experiment, together with their probabilities, generate a special discrete probability distribution called the binomial probability distribution.

Notation for Binomial Experiments and Binomial Distribution

Symbol Meaning

S The outcome denoted as a successF The outcome denoted as a failureP(Success)=P(S)=pThe probability of observing a successP(Failure)=P(F)=1-pThe probability of observing a failuren The number of trials

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Binomial Probability DistributionWe can use the binomial probability distribution formula to find probabilities for the number of successes for any binomial experiment.

Binomial Probability Distribution Formula

The probability of observing exactly X successes in n trials of a binomial experiment is

We often call this the binomial probability formula.

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P(X) =(nCX )pX (1 − p)n−X

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P(X) =(nCX )P(Success)number of successes P(Failure)number of failures

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Binomial Mean, Variance, and Standard Deviation, and ModeSince the binomial random variable X is discrete, it also has a mean, variance, and standard deviation.

Mean, Variance, and Standard Deviation of a Binomial Random Variable X

• Mean (or expected value):

• Variance:

• Standard deviation:

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μ =n⋅ pσ 2 = n⋅ p⋅ (1− p)

σ = np(1− p)

The mode of a binomial distribution is the most likely outcome of the binomial experiment for the given values of n, p, and X, that is, the outcome with the largest probability.

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Binomial Example

Suppose we know the population proportion p of left-handed students is 0.10, and we have a random sample of 100 students.

Calculate the expected number of left-handed students.

Calculate the variance and standard deviation of the number of left-handed students.

Would 22 left-handed students out of 100 be considered an outlier?

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μ =100(0.10) =10

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σ 2 =100(0.10)(0.90)

σ = 9 = 3

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P(X = 22)=100C22(0.1)22(0.9)78 = 0.00019

This is highly unlikely.

+ 6.3: Poisson Probability Distribution

Objectives:

Explain the requirements for the Poisson probability distribution.

Compute probabilities for a Poisson random variable.

Calculate the mean, variance, and standard deviation of a Poisson random variable.

Use the Poisson distribution to approximate the binomial distribution.

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Requirements for the Poisson Distribution

The Poisson distribution, like the binomial distribution, is a discrete probability distribution. The Poisson probability distribution is used when we wish to find the probability of observing a certain number of occurrences of an event within a fixed interval of space or time.

Poisson Probability Distribution

The random variable X represents the number of occurrences of an event in an interval.

1.The occurrences must be random.2.Each occurrence must be independent.3.The occurrences must be uniformly distributed over the given interval.

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Computing Probabilities for a Poisson Random VariablePoisson Probability Distribution Formula

If the requirements are met, the probability that a particular event occurs X times within a given interval is

where µ is the mean of the Poisson probability distribution, e is a constant approximately equal to 2.718281828, and X is the number of occurrences of the event within the interval.€

P(X) =μ X ⋅ e−μ

X!

ExampleA study of the number of cardiac arrests to occur per week in a hospital found a Poisson distribution with mean µ=1.09. Find the probability of 2 cardiac arrests in a given week.

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P(2) =1.092⋅ e−1.09

2!≈ 0.1997

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Mean, Variance, and Standard Deviation for a Poisson Random Variable

Parameters of the Poisson Distribution

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Mean = μVariance=σ 2 =μ

Standard Deviation=σ = μ

ExampleA study of the number of cardiac arrests to occur per week in a hospital found a Poisson distribution with mean µ=1.09. Find the mean, variance, and standard deviation of X=the number of cardiac arrests in a given week.

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Mean = μ =1.09Variance =σ 2 = μ =1.09

Standard Deviation =σ = 1.09 =1.044

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Using the Poisson Distribution to Approximate the Binomial Distribution

Requirements for Using the Poisson Distribution to Approximate the Binomial Distribution

n ≥ 100 and np ≤ 10

where n is the number of trials and p is the probability of success for the binomial distribution.

We can use the Poisson distribution to approximate the binomial distribution when the number of trials n is large and the probability of successes p is small, as measured by the following requirements.

If the requirements are met, then the mean of the Poisson distribution used to approximate the binomial distribution is given as

µ = np

+ 6.4: Continuous Random Variables and the Normal Probability DistributionObjectives:

Identify a continuous probability distribution and state the requirements.

Calculate probabilities for the uniform probability distributions.

Explain the properties of the normal probability distribution.

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Continuous Probability DistributionsContinuous random variables assume infinitely many possible values, with no gap between the values. For a given continuous random variable X, we are not interested in whether X equals any particular value. Rather we are interested in whether X is greater or less than a particular value or between two values.

Continuous Probability Distribution

A continuous probability distribution is a graph that indicates the range of values that the continuous random variable X can take and, above which, is drawn a density curve.

1.The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)2.The vertical height of the density curve can never be negative. That is, the density curve never goes below the horizontal axis.

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Calculating Probabilities for the Uniform Probability DistributionThe uniform probability distribution is a continuous distribution that has constant probability from left endpoint a to right endpoint b. Its curve is a flat, straight line, so that the shape of the distribution is a rectangle.

Probability for a Continuous Distribution

The probability that a continuous random variable X takes a value in an interval is equal to the area under the density curve above that interval.

X = waiting time for the campus shuttle bus

Find the probability you will wait between 2 and 4 minutes for the bus.

P(2≤X≤4) = 2(0.1) = 0.2

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Normal Probability Distribution

We now turn to what is considered to be the most important probability distribution in statistics: the normal probability distribution.

Properties of the Normal Probability Distribution

1.It is symmetric about the mean µ.2.The highest point occurs at X=µ.3.The total area under the curve = 1.4.The area under the curve to the left of µ and to the right of µ are both equal to 0.5.5.The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions.6.Values of X are always found on the horizontal axis. Probabilities are represented by areas under the curve.

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The Empirical Rule

+ 6.5: Standard Normal Distribution

Objectives:

Find areas under the standard normal curve, given a Z-value.

Find the standard normal Z-value, given an area.

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Standard Normal Distribution

There is one very special normal distribution called the standard normal distribution. The mean and standard deviation of the standard normal distribution make it unique.

The standard normal distribution is a normal distribution with

• mean µ = 0 and

• standard deviation σ = 1.

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Finding Areas Under the Standard Normal Curve

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Finding Areas Under the Standard Normal Curve: Case 1Find the area to the left of Z = 0.57.

1. Draw the standard normal curve and label Z = 0.57.

2. Shade to the left of 0.57.

3. Look at the intersection of row 0.5 and column 0.07. This is the area to the left of Z = 0.57.

Area = 0.7157.

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Finding Areas Under the Standard Normal Curve: Case 2

Find the area to the right of Z = -1.25.

1. Draw the standard normal curve and label Z = -1.25.

2. Shade to the right of -1.25.

3. Look at the intersection of row -1.2 and column 0.05. This is the area to the left of Z = -1.25. The area to the right is then

Area = 1 – 0.1056 = 0.8944.

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Finding Areas Under the Standard Normal Curve: Case 3Find the area between Z = -1 and Z = 1.

1. Draw the standard normal curve and label Z = -1 and Z = 1.

2. Shade the area between -1 and 1.

3. Find the area to the left of Z = -1 and the area to the left of Z = 1. Subtract the smaller area from the larger area to find the area in between.

Area = 0.8413 – 0.1587 = 0.6826.

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Finding Z-Values for a Given Area

Find the Z-value with area 0.90 to its left.

1. Draw the standard normal curve and label Z1

2. Shade the area to the left of Z1 and label with the given area of 0.90.

3. Find the value closest to 0.90 in the body of the Z table. This should be 0.8997. Move to the left to find the value 1.2, then move up from 0.8997 to find the value 0.08. Putting these values together, we get

Z1 = 1.2 + 0.08 = 1.28

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Finding Z-Values for a Given Area

Find the Z-value with area 0.03 to its right.

1. Draw the standard normal curve and label Z1

2. Shade the area to the right of Z1 and label with the given area of 0.03. Label the area to the left as 0.97

3. Find the value closest to 0.97 in the body of the Z table. This should be 0.9699. Move to the left to find the value 1.8, then move up from 0.9699 to find the value 0.08. Putting these values together, we get

Z = 1.8 + 0.08 = 1.88

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Finding Z-Values for a Given Area

Find the Z-values that mark the boundaries of the middle 95% of the area under the standard normal curve.

1. Draw the standard normal curve and label Z1 and Z2

2. Shade the area in between and label the area as 0.95. By symmetry, there is area 0.05 ÷ 2 = 0.025 in each tail.

3. Find the value closest to 0.025 in the body of the Z table. This corresponds to a Z-value of -1.96. By symmetry, the other Z-value is 1.96.

+ 6.6: Applications of the Normal Distribution

Objectives:

Compute probabilities for a given value of any normal random variable.

Find the appropriate value of any normal random variable, given an area or probability.

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Standardizing a Normal Random Variable

To standardize a normal random variable X, we transform that normal random variable X into the standard normal random variable Z.

Standardizing a Normal Random Variable

Any normal random variable X can be transformed into the standard normal random variable Z by standardizing X using the formula

xZ

μσ

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Probabilities for Any Normal Distribution

Finding Probabilities for Any Normal Distribution

1.Determine the random variable X, the mean µ, and the standard deviation σ. Draw the normal curve for X and shade the desired area.

2.Standardize by using the formula Z = (X - µ)/σ to find the values of Z.

3.Draw the standard normal curve and shade the area corresponding to the shaded area in the graph of X.

4.Find the area under the standard normal curve using either the Z table or technology. This area is equal to the area under the normal curve for X drawn in step 1.

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Normal Data Values for a Given Area or Probability

Finding Normal Data Values for a Given Area or Probability

1.Determine X, µ, and σ, and draw the normal curve for X. Shade the desired area. Mark the position of the unknown value X1.

2.Find the Z-value corresponding to the desired area.

3.Transform this value of Z into a value of X using the formula

X1 = Zσ + µ

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Example

Edmunds.com reported that the average amount that people were paying for a 2007 Toyota Camry XLE was $23,400. Let X = price, and assume that price follows a normal distribution with μ = $23,400 and σ = $1000.

Find the prices that separate the middle 95% of 2007 Toyota Camry XLE prices from the bottom 2.5% and the top 2.5%.

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ExampleFind the Z-values corresponding to the desired area.

The area to the left of X1 equals 0.025, and the area to the left of X2 equals 0.975.

Looking up area 0.025 on the inside of the Z table gives us Z1 = –1.96. Z2 = 1.96.

X1 = Z1σ + μ =(–1.96)(1000) + 23,400 = 21,440

X2 = Z2σ + μ =(1.96)(1000) + 23,400 = 25,360

The prices that separate the middle 95% of 2007 Toyota Camry XLE prices from the bottom 2.5% of prices and the top 2.5% of prices are $21,440 and $25,360.

+ 6.7: Normal Approximation to the Binomial Probability Distribution

Objectives:

Use the normal distribution to approximate probabilities of the binomial distribution.

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Normal Approximation to the Binomial DistributionRecall the binomial random variable X represents the number of successes in n trials and depends on the sample size n and probability of success p. For a given probability of success p, if the sample size n gets large enough, the binomial distribution begins to resemble the normal distribution.

n=4, p=0.2 n=64, p=0.2

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Normal Approximation to the Binomial Distribution

The Normal Approximation to the Binomial Probability Distribution

Consider the binomial random variable X with probability of success p and number of trials n. If np≥5 and n(1-p)≥5, the binomial distribution may be approximated by a normal distribution with mean np and standard deviation √(np(1-p).

Example 6.44

The binomial distribution with n = 64 and p = 0.2 can be approximated by a normal distribution since np = 12.8 and n(1-p) = 51.2

The mean of the normal distribution is np = 12.8 and the standard deviation is √np(1-p) = 3.2.

+ Chapter 6 Overview

6.1 Discrete Random Variables

6.2 Binomial Probability Distribution

6.3 Poisson Probability Distribution

6.4 Continuous Random Variables and the Normal Probability Distribution

6.5 Standard Normal Distribution

6.6 Applications of the Normal Distribution

6.7 Normal Approximation to the Binomial Probability Distribution

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