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Chapter 6: Probability Distributions. Section 6.1: How Can We Summarize Possible Outcomes and Their Probabilities?. Learning Objectives. Random variable Probability distributions for discrete random variables Mean of a probability distribution

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chapter 6 probability distributions

Chapter 6: Probability Distributions

Section 6.1: How Can We Summarize Possible Outcomes and Their Probabilities?

learning objectives
Learning Objectives
  • Random variable
  • Probability distributions for discrete random variables
  • Mean of a probability distribution
  • Summarizing the spread of a probability distribution
  • Probability distribution for continuous random variables
learning objective 1 randomness
Learning Objective 1:Randomness
  • The numerical values that a variable assumes are the result of some random phenomenon:
    • Selecting a random sample for a population

or

    • Performing a randomized experiment
learning objective 1 random variable
Learning Objective 1:Random Variable
  • A random variable is a numerical measurement of the outcome of a random phenomenon.
learning objective 1 random variable1
Learning Objective 1:Random Variable
  • Use letters near the end of the alphabet, such as x, to symbolize
    • Variables
    • A particular value of the random variable
  • Use a capital letter, such as X, to refer to the random variable itself.

Example: Flip a coin three times

    • X=number of heads in the 3 flips; defines the random variable
    • x=2; represents a possible value of the random variable
learning objective 2 probability distribution
Learning Objective 2:Probability Distribution
  • The probability distribution of a random variable specifies its possible values and their probabilities.

Note: It is the randomness of the variable that allows us to specify probabilities for the outcomes

learning objective 2 probability distribution of a discrete random variable
Learning Objective 2:Probability Distribution of a Discrete Random Variable
  • A discrete random variableX has separate values (such as 0,1,2,…) as its possible outcomes
  • Its probability distribution assigns a probability P(x) to each possible value x:
    • For each x, the probability P(x) falls between 0 and 1
    • The sum of the probabilities for all the possible x values equals 1
learning objective 2 example
Learning Objective 2:Example
  • What is the estimated probability of at least three home runs?

P(3)+P(4)+P(5)=0.13+0.03+0.01=0.17

learning objective 3 the mean of a discrete probability distribution
Learning Objective 3:The Mean of a Discrete Probability Distribution
  • The mean of a probability distribution for a discrete random variable is

where the sum is taken over all possible values of x.

  • The mean of a probability distribution is denoted by the parameter, µ.
  • The mean is a weighted average; values of x that are more likely receive greater weight P(x)
learning objective 3 expected value of x
Learning Objective 3:Expected Value of X
  • The mean of a probability distribution of a random variable X is also called the expected value of X.
  • The expected value reflects not what we’ll observe in a single observation, but rather that we expect for the average in a long run of observations.
  • It is not unusual for the expected value of a random variable to equal a number that is NOT a possible outcome.
learning objective 3 example
Learning Objective 3:Example
  • Find the mean of this probability distribution.

The mean:

= 0(0.23) + 1(0.38) + 2(0.22) + 3(0.13) + 4(0.03) + 5(0.01) = 1.38

learning objective 4 the standard deviation of a probability distribution
Learning Objective 4:The Standard Deviation of a Probability Distribution

The standard deviation of a probability distribution, denoted by the parameter, σ, measures its spread.

  • Larger values of σ correspond to greater spread.
  • Roughly, σ describes how far the random variable falls, on the average, from the mean of its distribution
learning objective 5 continuous random variable
Learning Objective 5:Continuous Random Variable
  • A continuous random variable has an infinite continuum of possible values in an interval.
  • Examples are: time, age and size measures such as height and weight.
  • Continuous variables are measured in a discrete manner because of rounding.
learning objective 5 probability distribution of a continuous random variable
Learning Objective 5:Probability Distribution of a Continuous Random Variable
  • A continuous random variable has possible values that form an interval.
  • Its probability distribution is specified by a curve.
  • Each interval has probability between 0 and 1.
  • The interval containing all possible values has probability equal to 1.
chapter 6 probability distributions1

Chapter 6: Probability Distributions

Section 6.2: How Can We Find Probabilities for Bell-Shaped Distributions?

learning objectives1
Learning Objectives
  • Normal Distribution
  • 68-95-99.7 Rule for normal distributions
  • Z-Scores and the Standard Normal Distribution
  • The Standard Normal Table: Finding Probabilities
  • Using the TI-calculator: find probabilities
learning objectives2
Learning Objectives
  • Using the Standard Normal Table in Reverse
  • Using the TI-calculator: find z-scores
  • Probabilities for Normally Distributed Random Variables
  • Percentiles for Normally Distributed Random Variables
  • Using Z-scores to Compare Distributions
learning objective 1 normal distribution
Learning Objective 1:Normal Distribution

The normal distribution is symmetric, bell-shaped and characterized by its mean µ and standard deviation .

  • The normal distribution is the most important distribution in statistics
    • Many distributions have an approximate normal distribution
    • Approximates many discrete distributions well when there are a large number of possible outcomes
    • Many statistical methods use it even when the data are not bell shaped
learning objective 1 normal distribution1
Learning Objective 1:Normal Distribution
  • Normal distributions are
    • Bell shaped
    • Symmetric around the mean
  • The mean () and the standard deviation () completely describe the density curve
    • Increasing/decreasing  moves the curve along the horizontal axis
    • Increasing/decreasing  controls the spread of the curve
learning objective 1 normal distribution2
Learning Objective 1:Normal Distribution
  • Within what interval do almost all of the men’s heights fall? Women’s height?
learning objective 2 68 95 99 7 rule for any normal curve
Learning Objective 2:68-95-99.7 Rule for Any Normal Curve
  • 68% of the observations fall within one standard deviation of the mean
  • 95% of the observations fall within two standard deviations of the mean
  • 99.7% of the observations fall within three standard deviations of the mean
learning objective 2 example 68 95 99 7 rule
Learning Objective 2:Example : 68-95-99.7% Rule
  • Heights of adult women
    • can be approximated by a normal distribution
    • = 65 inches; =3.5 inches
  • 68-95-99.7 Rule for women’s heights
    • 68% are between 61.5 and 68.5 inches

[ µ = 65  3.5 ]

    • 95% are between 58 and 72 inches

[ µ 2 = 65  2(3.5) = 65  7 ]

    • 99.7% are between 54.5 and 75.5 inches

[ µ 3 = 65  3(3.5) = 65  10.5 ]

learning objective 2 example 68 95 99 7 rule1

68%

(by 68-95-99.7 Rule)

?

16%

-1

+1

65 68.5 (height values)

? = 84%

Learning Objective 2:Example : 68-95-99.7% Rule
  • What proportion of women are less than 69 inches tall?
learning objective 3 z scores and the standard normal distribution
Learning Objective 3:Z-Scores and the Standard Normal Distribution
  • The z-score for a value x of a random variable is the number of standard deviations that x falls from the mean
  • A negative (positive) z-score indicates that the value is below (above) the mean
  • z-scores can be used to calculate the probabilities of a normal random variable using the normal tables in the back of the book
learning objective 3 z scores and the standard normal distribution1
Learning Objective 3:Z-Scores and the Standard Normal Distribution
  • A standard normal distribution has mean µ=0 and standard deviation σ=1
  • When a random variable has a normal distribution and its values are converted to z-scores by subtracting the mean and dividing by the standard deviation, the z-scores have the standard normal distribution.
learning objective 4 table a standard normal probabilities
Learning Objective 4:Table A: Standard Normal Probabilities

Table A enables us to find normal probabilities

  • It tabulates the normal cumulative probabilities falling below the point +z

To use the table:

  • Find the corresponding z-score
  • Look up the closest standardized score (z) in the table.
    • First column gives z to the first decimal place
    • First row gives the second decimal place of z
  • The corresponding probability found in the body of the table gives the probability of falling below the z-score
learning objective 4 example using table a
Learning Objective 4:Example: Using Table A
  • Find the probability that a normal random variable takes a value less than 1.43 standard deviations above µ; P(z<1.43)=.9236

TI Calculator = Normcdf(-1e99,1.43,0,1)= .9236

learning objective 4 example using table a1
Learning Objective 4:Example: Using Table A
  • Find the probability that a normal random variable takes a value greater than 1.43 standard deviations above µ: P(z>1.43)=1-.9236=.0764

TI Calculator = Normcdf(1.43,1e99,0,1)= 0.0764

learning objective 4 example
Learning Objective 4:Example:
  • Find the probability that a normal random variable assumes a value within 1.43 standard deviations of µ
    • Probability below 1.43σ = .9236
    • Probability below -1.43σ = .0764 (1-.9236)
    • P(-1.43<z<1.43) =.9236-.0764=.8472

TI Calculator = Normcdf(-1.43,1.43,0,1)= .8472

learning objective 5 using the ti calculator
Learning Objective 5:Using the TI Calculator

To calculate the cumulative probability

  • 2nd DISTR; 2:normalcdf(lower bound, upper bound,mean,sd)
  • Use –1E99 for negative infinity and 1E99 for positive infinity
learning objective 5 find probabilities using ti calculator
Learning Objective 5:Find Probabilities Using TI Calculator
  • Find probability to the left of -1.64
    • P(z<-1.64)=normcdf(-1e99,-1.64,0,1)=.0505
  • Find probability to the right of 1.56
    • P(z>1.56)=normcdf(1.56,1e99,0,1)=.0594
  • Find probability between -.50 and 2.25
    • P(-.5<z<2.25)=normcdf(-.5,2.25,0,1)=.6793
learning objective 6 how can we find the value of z for a certain cumulative probability
Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability?
  • To solve some of our problems, we will need to find the value of z that corresponds to a certain normal cumulative probability
  • To do so, we use Table A in reverse
    • Rather than finding z using the first column (value of z up to one decimal) and the first row (second decimal of z)
      • Find the probability in the body of the table
      • The z-score is given by the corresponding values in the first column and row
learning objective 6 how can we find the value of z for a certain cumulative probability1
Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability?
  • Example: Find the value of z for a cumulative probability of 0.025.
  • Look up the cumulative probability of 0.025 in the body of Table A.
  • A cumulative probability of 0.025 corresponds to z = -1.96.
  • Thus, the probability that a normal

random variable falls at least 1.96

standard deviations below the

mean is 0.025.

learning objective 6 how can we find the value of z for a certain cumulative probability2
Learning Objective 6:How Can We Find the Value of z for a Certain Cumulative Probability?
  • Example: Find the value of z for a cumulative probability of 0.975.
  • Look up the cumulative probability of 0.975 in the body of Table A.
  • A cumulative probability of 0.975 corresponds to z = 1.96.
  • Thus, the probability that a normal

random variable takes a value no more

than 1.96 standard deviations above

the mean is 0.975.

learning objective 7 using the ti calculator to find z scores for a given probability
Learning Objective 7:Using the TI Calculator to Find Z-Scores for a Given Probability
  • 2nd DISTR 3:invNorm; Enter
  • invNorm(percentile,mean,sd)
    • Percentile is the probability under the curve from negative infinity to the z-score
  • Enter
learning objective 7 examples
Learning Objective 7:Examples
  • The probability that a standard normal random variable assumes a value that is ≤ z is 0.975. What is z? Invnorm(.975,0,1)=1.96
  • The probability that a standard normal random variable assumes a value that is > z is 0.0275.

What is z? Invnorm(.975,0,1)=1.96

  • The probability that a standard normal random variable assumes a value that is ≥ z is 0.881.

What is z? Invnorm(1-.881,0,1)=-1.18

  • The probability that a standard normal random variable assumes a value that is < z is 0.119.

What is z? Invnorm(.119,0,1)= -1.18

learning objective 7 example
Learning Objective 7:Example
  • Find the z-score z such that the probability within z standard deviations of the mean is 0.50.
    • Invnorm(.75,0,1)= .67
    • Invnorm(.25,0,1)= -.67
  • Probability = P(-.67<Z<.67)=.5
learning objective 8 finding probabilities for normally distributed random variables
Learning Objective 8:Finding Probabilities for Normally Distributed Random Variables
  • State the problem in terms of the observed random variable X, i.e., P(X<x)
  • Standardize X to restate the problem in terms of a standard normal variable Z
  • Draw a picture to show the desired probability under the standard normal curve
  • Find the area under the standard normal curve using Table A
learning objective 8 p x x
Learning Objective 8:P(X<x)
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure less than 100?
  • P(X<100) =
  • Normcdf(-1E99,100,120,20)=.1587
  • 15.9% of adults have systolic blood pressure less than 100
learning objective 8 p x x1
Learning Objective 8:P(X>x)
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 100?
  • P(X>100) = 1 – P(X<100)
  • P(X>100)= 1-.1587=.8413
  • Normcdf(100,1e99,120,20)=.8413
  • 84.1% of adults have systolic blood pressure greater than 100
learning objective 8 p x x2
Learning Objective 8:P(X>x)
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure greater than 133?
  • P(X>133) = 1 – P(X<133)
  • P(X>133)= 1-.7422=.2578
  • Normcdf(133,1E99,120,20)=.2578
  • 25.8% of adults have systolic blood pressure greater than 133
learning objective 8 p a x b
Learning Objective 8: P(a<X<b)
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What percentage of adults have systolic blood pressure between 100 and 133?
  • P(100<X<133) = P(X<133)-P(X<100)
  • Normcdf(100,133,120,20)=.5835
  • 58% of adults have systolic blood pressure between 100 and 133
learning objective 9 find x value given area to left
Learning Objective 9:Find X Value Given Area to Left
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. What is the 1st quartile?
  • P(X<x)=.25, find x:
    • Look up .25 in the body of Table A to find z= -0.67
    • Solve equation to find x:
  • Check:
    • P(X<106.6) P(Z<-0.67)=0.25
    • TI Calculator = Invnorm(.25,120,20)=106.6
learning objective 9 find x value given area to right
Learning Objective 9:Find X Value Given Area to Right
  • Adult systolic blood pressure is normally distributed with µ = 120 and σ = 20. 10% of adults have systolic blood pressure above what level?
  • P(X>x)=.10, find x.
    • P(X>x)=1-P(X<x)
    • Look up 1-0.1=0.9 in the body of Table A to find z=1.28
    • Solve equation to find x:
  • Check:
    • P(X>145.6) =P(Z>1.28)=0.10
    • TI Calculator = Invnorm(.9,120,20)=145.6
learning objective 10 using z scores to compare distributions
Learning Objective 10:Using Z-scores to Compare Distributions

Z-scores can be used to compare observations from different normal distributions

  • Example:
    • You score 650 on the SAT which has =500 and

=100 and 30 on the ACT which has =21.0 and

=4.7. On which test did you perform better?

    • Compare z-scores

SAT: ACT:

    • Since your z-score is greater for the ACT, you performed better on this exam
chapter 6 probability distributions2

Chapter 6: Probability Distributions

Section 6.3: How Can We Find Probabilities When Each Observation Has Two Possible Outcomes?

learning objectives3
Learning Objectives
  • The Binomial Distribution
  • Conditions for a Binomial Distribution
  • Probabilities for a Binomial Distribution
  • Factorials
  • Examples using Binomial Distribution
  • Do the Binomial Conditions Apply?
  • Mean and Standard Deviation of the Binomial Distribution
  • Normal Approximation to the Binomial
learning objective 1 the binomial distribution
Learning Objective 1:The Binomial Distribution
  • Each observation is binary: it has one of two possible outcomes.
  • Examples:
    • Accept, or decline an offer from a bank for a credit card.
    • Have, or do not have, health insurance.
    • Vote yes or no on a referendum.
learning objective 2 conditions for the binomial distribution
Learning Objective 2:Conditions for the Binomial Distribution
  • Each of n trials has two possible outcomes: “success” or “failure”.
  • Each trial has the same probability of success, denoted by p.
  • The ntrials are independent.
  • The binomial random variable X is the number of successes in the n trials.
learning objective 3 probabilities for a binomial distribution
Learning Objective 3:Probabilities for a Binomial Distribution
  • Denote the probability of success on a trial by p.
  • For n independent trials, the probability of x successes equals:
learning objective 4 factorials
Learning Objective 4:Factorials

Rules for factorials:

  • n!=n*(n-1)*(n-2)…2*1
  • 1!=1
  • 0!=1

For example,

  • 4!=4*3*2*1=24
learning objective 5 example finding binomial probabilities
Learning Objective 5:Example: Finding Binomial Probabilities
  • John Doe claims to possess ESP.
  • An experiment is conducted:
    • A person in one room picks one of the integers 1, 2, 3, 4, 5 at random.
    • In another room, John Doe identifies the number he believes was picked.
    • Three trials are performed for the experiment.
    • Doe got the correct answer twice.
learning objective 5 example 1
Learning Objective 5:Example 1

If John Doe does not actually have ESP and is actually guessing the number, what is the probability that he’d make a correct guess on two of the three trials?

  • The three ways John Doe could make two correct guesses in three trials are: SSF, SFS, and FSS.
  • Each of these has probability: (0.2)2(0.8)=0.032.
  • The total probability of two correct guesses is 3(0.032)=0.096.
learning objective 5 example 11
Learning Objective 5:Example 1
  • The probability of exactly 2 correct guesses is the binomial probability with n = 3 trials, x = 2 correct guesses and p = 0.2 probability of a correct guess.

2nd Vars

0:binampdf(n,p,x)

Binampdf(3,.2,2)=0.096

learning objective 5 binomial example 2
Learning Objective 5:Binomial Example 2
  • 1000 employees, 50% Female
  • None of the 10 employees chosen for management training were female.
  • The probability that no females are chosen is:
  • Binompdf(10,.5,0)=9.765625E-4
  • It is very unlikely (one chance in a thousand) that none of the 10 selected for management training would be female if the employees were chosen randomly
learning objective 6 do the binomial conditions apply
Learning Objective 6:Do the Binomial Conditions Apply?
  • Before using the binomial distribution, check that its three conditions apply:
    • Binary data (success or failure).
    • The same probability of success for each trial (denoted by p).
    • Independent trials.
learning objective 6 do the binomial conditions apply to example 2
Learning Objective 6:Do the Binomial Conditions Apply to Example 2?
  • The data are binary (male, female).
  • If employees are selected randomly, the probability of selecting a female on a given trial is 0.50.
  • With random sampling of 10 employees from a large population, outcomes for one trial does not depend on the outcome of another trial
learning objective 7 binomial mean and standard deviation
Learning Objective 7:Binomial Mean and Standard Deviation
  • The binomial probability distribution for n trials with probability p of success on each trial has mean µ and standard deviation σ given by:
learning objective 7 example racial profiling
Learning Objective 7: Example: Racial Profiling?
  • Data:
    • 262 police car stops in Philadelphia in 1997.
    • 207 of the drivers stopped were African-American.
    • In 1997, Philadelphia’s population was 42.2% African-American.
    • Does the number of African-Americans stopped suggest possible bias, being higher than we would expect (other things being equal, such as the rate of violating traffic laws)?
learning objective 7 example racial profiling1
Learning Objective 7:Example: Racial Profiling?
  • Assume:
    • 262 car stops represent n = 262 trials.
    • Successive police car stops are independent.
    • P(driver is African-American) is p = 0.422.
  • Calculate the mean and standard deviation of this binomial distribution:
learning objective 7 example racial profiling2
Learning Objective 7: Example: Racial Profiling?
  • Recall: Empirical Rule
    • When a distribution is bell-shaped, close to 100% of the observations fall within 3 standard deviations of the mean.
learning objective 7 example racial profiling3
Learning Objective 7:Example: Racial Profiling?
  • If there is no racial profiling, we would not be surprised if between about 87 and 135 of the 262 drivers stopped were African-American.
  • The actual number stopped (207) is well above these values.
  • The number of African-Americans stopped is too high, even taking into account random variation.
  • Limitation of the analysis:
    • Different people do different amounts of driving, so we don’t really know that 42.2% of the potential stops were African-American.
learning objective 8 approximating the binomial distribution with the normal distribution
Learning Objective 8:Approximating the Binomial Distribution with the Normal Distribution
  • The binomial distribution can be well approximated by the normal distribution when the expected number of successes, np, and the expected number of failures, n(1-p) are both at least 15.
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