chapter 11 section 1

Sullivan – Fundamentals of Statistics – 2nd Edition – Chapter 11 Section 1 – Slide 1 of 34

Chapter 11Section 1

Random Variables


Chapter 11 – Section 1

● Learning objectives Distinguish between discrete and continuous random

variables Identify discrete probability distributions Construct probability histograms Compute and interpret the mean of a discrete random

variable Interpret the mean of a discrete random variable as

an expected value Compute the variance and standard deviation of a

discrete random variable

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● A random variable is a numeric measure of the outcome of a probability experiment Random variables reflect measurements that can

change as the experiment is repeated Random variables are denoted with capital letters,

typically using X (and Y and Z …) Values are usually written with lower case letters,

typically using x (and y and z ...)



● Examples● Tossing four coins and counting the number of

heads The number could be 0, 1, 2, 3, or 4 The number could change when we toss another four

coins

● Examples● Tossing four coins and counting the number of

heads The number could be 0, 1, 2, 3, or 4 The number could change when we toss another four

coins● Measuring the heights of students

The heights could change from student to student



● A discrete random variable is a random variable that has either a finite or a countable number of values A finite number of values such as {0, 1, 2, 3, and 4} A countable number of values such as {1, 2, 3, …}

● A discrete random variable is a random variable that has either a finite or a countable number of values A finite number of values such as {0, 1, 2, 3, and 4} A countable number of values such as {1, 2, 3, …}

● Discrete random variables are designed to model discrete variables

● Discrete random variables are often “counts of …”



● An example of a discrete random variable● The number of heads in tossing 3 coins (a finite

number of possible values)


number of possible values) There are four possible values – 0 heads, 1 head, 2

heads, and 3 heads



heads, and 3 heads A finite number of possible values – a discrete

random variable



heads, and 3 heads A finite number of possible values – a discrete

random variable This fits our general concept that discrete random

variables are often “counts of …”



● Other examples of discrete random variables● Other examples of discrete random variables● The possible rolls when rolling a pair of dice

A finite number of possible pairs, ranging from (1,1) to (6,6)

● Other examples of discrete random variables● The possible rolls when rolling a pair of dice


● The number of pages in statistics textbooks A countable number of possible values

● Other examples of discrete random variables● The possible rolls when rolling a pair of dice


● The number of pages in statistics textbooks A countable number of possible values

● The number of visitors to the White House in a day A countable number of possible values



● A continuous random variable is a random variable that has an infinite, and more than countable, number of values The values are any number in an interval

● A continuous random variable is a random variable that has an infinite, and more than countable, number of values The values are any number in an interval

● Continuous random variables are designed to model continuous variables (see section 1.1)

● Continuous random variables are often “measurements of …”



● An example of a continuous random variable● The possible temperature in Chicago at noon

tomorrow, measured in degrees Fahrenheit


tomorrow, measured in degrees Fahrenheit The possible values (assuming that we can measure

temperature to great accuracy) are in an interval


tomorrow, measured in degrees Fahrenheit The possible values (assuming that we can measure

temperature to great accuracy) are in an interval The interval may be something like (–20,110)


tomorrow, measured in degrees The possible values (assuming that we can measure

temperature to great accuracy) are in an interval The interval may be something like (–20,110) This fits our general concept that continuous random

variables are often “measurements of …”



● Other examples of continuous random variables● Other examples of continuous random variables● The height of a college student

A value in an interval between 3 and 8 feet

● Other examples of continuous random variables● The height of a college student

A value in an interval between 3 and 8 feet● The length of a country and western song

A value in an interval between 1 and 15 minutes

● Other examples of continuous random variables● The height of a college student

A value in an interval between 3 and 8 feet● The length of a country and western song

A value in an interval between 1 and 15 minutes








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● The probability distribution of a discrete random variable X relates the values of X with their corresponding probabilities

● The probability distribution of a discrete random variable X relates the values of X with their corresponding probabilities

● A distribution could be In the form of a table In the form of a graph In the form of a mathematical formula



● If X is a discrete random variable and x is a possible value for X, then we write P(x) as the probability that X is equal to x

● If X is a discrete random variable and x is a possible value for X, then we write P(x) as the probability that X is equal to x

● Examples In tossing one coin, if X is the number of heads, then

P(0) = 0.5 and P(1) = 0.5 In rolling one die, if X is the number rolled, then

P(1) = 1/6



● Properties of P(x)● Since P(x) form a probability distribution, they

must satisfy the rules of probability 0 ≤ P(x) ≤ 1 Σ P(x) = 1

● In the second rule, the Σ sign means to add up the P(x)’s for all the possible x’s



● An example of a discrete probability distribution

● All of the P(x) values are positive and they add up to 1

x P(x)1 .22 .65 .16 .1



● An example that is not a probability distribution

● Two things are wrong

x P(x)1 .22 .65 -.36 .1


● Two things are wrong P(5) is negative

x P(x)1 .22 .65 -.36 .1


● Two things are wrong P(5) is negative The P(x)’s do not add up to 1

x P(x)1 .22 .65 -.36 .1








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● A probability histogram is a histogram where The horizontal axis corresponds to the possible

values of X (i.e. the x’s) The vertical axis corresponds to the probabilities for

those values (i.e. the P(x)’s)● A probability histogram is very similar to a

relative frequency histogram



● An example of a probability histogram

● The histogram is drawn so that the height of the bar is the probability of that value








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● The mean of a probability distribution can be thought of in this way: There are various possible values of a discrete

random variable


random variable The values that have the higher probabilities are the

ones that occur more often



ones that occur more often The values that occur more often should have a larger

role in calculating the mean



ones that occur more often The values that occur more often should have a larger

role in calculating the mean The mean is the weighted average of the values,

weighted by the probabilities



● The mean of a discrete random variable isμX = Σ [ x • P(x) ]

● The mean of a discrete random variable isμX = Σ [ x • P(x) ]

● In this formula x are the possible values of X P(x) is the probability that x occurs Σ means to add up these terms for all the possible

values x



● Example of a calculation for the mean

x P(x)

1 0.2

2 0.6

5 0.1

6 0.1


x P(x) x • P(x)

1 0.2 0.2

2 0.6

5 0.1

6 0.1

Multiply


x P(x) x • P(x)

1 0.2 0.2

2 0.6 1.2

5 0.1 0.5

6 0.1 0.6

MultiplyMultiply again

Multiply again

Multiply again


● Add: 0.2 + 1.2 + 0.5 + 0.6 = 2.5● The mean of this discrete random variable is 2.5

Multiply again

Multiply again

Multiply again



● The calculation for this problem written out μX = Σ [ x • P(x) ]

= [1• 0.2] + [2• 0.6] + [5• 0.1] + [6• 0.1]

= 0.2 + 1.2 + 0.5 + 0.6= 2.5

● The mean of this discrete random variable is 2.5



● The mean can also be thought of this way (as in the Law of Large Numbers)

● The mean can also be thought of this way (as in the Law of Large Numbers) If we repeat the experiment many times

● The mean can also be thought of this way (as in the Law of Large Numbers) If we repeat the experiment many times If we record the result each time

● The mean can also be thought of this way (as in the Law of Large Numbers) If we repeat the experiment many times If we record the result each time If we calculate the mean of the results (this is just a

mean of a group of numbers)

● The mean can also be thought of this way (as in the Law of Large Numbers) If we repeat the experiment many times If we record the result each time If we calculate the mean of the results (this is just a

mean of a group of numbers) Then this mean of the results gets closer and closer

to the mean of the random variable








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● The expected value of a random variable is another term for its mean

● The expected value of a random variable is another term for its mean

● The term “expected value” illustrates the long term nature of the experiments – as we perform more and more experiments, the mean of the results of those experiments gets closer to the “expected value” of the random variable








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● The variance of a discrete random variable is computed similarly as for the mean


● The mean is the weighted sum of the valuesμX = Σ [ x • P(x) ]



● The variance is the weighted sum of the squared differences from the mean

σX2 = Σ [ (x – μX)2 • P(x) ]



● The variance is the weighted sum of the squared differences from the mean

σX2 = Σ [ (x – μX)2 • P(x) ]

● The standard deviation, as we’ve seen before, is the square root of the variance … σX = √ σX

2



● The variance formulaσX

2 = Σ [ (x – μX)2 • P(x) ]can involve calculations with many decimals or fractions

● An equivalent formula isσX

2 = [ Σ x2 • P(x) ] – μX2

● This formula is often easier to compute



● For variables and samples, we had the concept of a population variance (for the entire population) and a sample variance (for a sample from that population)

● These probability distributions model the complete population These are population variance formulas There is no analogy for sample variance here


Chapter 11Section 2

The BinomialProbability Distribution



● Learning objectives Determine whether a probability experiment is a

binomial experiment Compute probabilities of binomial experiments Compute the mean and standard deviation of a

binomial random variable Construct binomial probability histograms

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● A binomial experiment has the following structure

● A binomial experiment has the following structure The first test is performed … the result is either a

success or a failure


success or a failure The second test is performed … the result is either a

success or a failure. This result is independent of the first and the chance of success is the same


success or a failure The second test is performed … the result is either a

success or a failure. This result is independent of the first and the chance of success is the same

A third test is performed … the result is either a success or a failure. The result is independent of the first two and the chance of success is the same



● Example A card is drawn from a deck. A “success” is for that

card to be a heart … a “failure” is for any other suit


card to be a heart … a “failure” is for any other suit The card is then put back into the deck


card to be a heart … a “failure” is for any other suit The card is then put back into the deck A second card is drawn from the deck with the same

definition of success.



definition of success. The second card is put back into the deck



definition of success. The second card is put back into the deck We continue for 10 cards



● A binomial experiment is an experiment with the following characteristics

● A binomial experiment is an experiment with the following characteristics The experiment is performed a fixed number of times,

each time called a trial


each time called a trial The trials are independent


each time called a trial The trials are independent Each trial has two possible outcomes, usually called a

success and a failure


each time called a trial The trials are independent Each trial has two possible outcomes, usually called a

success and a failure The probability of success is the same for every trial



● Notation used for binomial distributions The number of trials is represented by n The probability of a success is represented by p The total number of successes in n trials is

represented by X● Because there cannot be a negative number of

successes, and because there cannot be more than n successes (out of n attempts)

0 ≤ X ≤ n



● In our card drawing example Each trial is the experiment of drawing one card The experiment is performed 10 times, so n = 10

● In our card drawing example Each trial is the experiment of drawing one card The experiment is performed 10 times, so n = 10 The trials are independent because the drawn card is

put back into the deck


put back into the deck Each trial has two possible outcomes, a “success” of

drawing a heart and a “failure” of drawing anything else

The probability of success is 0.25, the same for every trial, so p = 0.25


put back into the deck Each trial has two possible outcomes, a “success” of

drawing a heart and a “failure” of drawing anything else

The probability of success is 0.25, the same for every trial, so p = 0.25

X, the number of successes, is between 0 and 10



● The word “success” does not mean that this is a good outcome or that we want this to be the outcome

● A “success” in our card drawing experiment is to draw a heart

● If we are counting hearts, then this is the outcome that we are measuring

● There is no good or bad meaning to “success”






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● We would like to calculate the probabilities of X, i.e. P(0), P(1), P(2), …, P(n)

● Do a simpler example first For n = 3 trials With p = .4 probability of success Calculate P(2), the probability of 2 successes



● For 3 trials, the possible ways of getting exactly 2 successes are S S F S F S F S S

● For 3 trials, the possible ways of getting exactly 2 successes are S S F S F S F S S

● The probabilities for each (using the multiplication rule) are 0.4 • 0.4 • 0.6 = 0.096 0.4 • 0.6 • 0.4 = 0.096 0.6 • 0.4 • 0.4 = 0.096



● The total probability isP(2) = 0.096 + 0.096 + 0.096 = 0.288

● But there is a pattern Each way had the same probability … the probability

of 2 success (0.4 times 0.4) times the probability of 1 failure (0.6)

● The probability for each case is0.42 • 0.61



● There are 3 cases S S F could represent choosing a combination of

2 out of 3 … choosing the first and the second S F S could represent choosing a second

combination of 2 out of 3 … choosing the first and the third

F S S could represent choosing a third combination of 2 out of 3

● There are 3 cases S S F could represent choosing a combination of

2 out of 3 … choosing the first and the second S F S could represent choosing a second

combination of 2 out of 3 … choosing the first and the third

F S S could represent choosing a third combination of 2 out of 3

● These are the 3 = 3C2 ways to choose 2 out of 3



● Thus the total probabilityP(2) = .096 + .096 + .096 = .288

can also be written asP(2) = 3C2 • .42 • .61

● Thus the total probabilityP(2) = .096 + .096 + .096 = .288

can also be written asP(2) = 3C2 • .42 • .61

● In other words, the probability is The number of ways of choosing 2 out of 3, times The probability of 2 successes, times The probability of 1 failure



● The general formula for the binomial probabilities is just this


● For P(x), the probability of x successes, the probability is The number of ways of choosing x out of n, times The probability of x successes, times The probability of n-x failures


● For P(x), the probability of x successes, the probability is The number of ways of choosing x out of n, times The probability of x successes, times The probability of n-x failures

● This formula isP(x) = nCx px (1 – p)n-x



● Example● A student guesses at random on a multiple

choice quiz There are n = 10 questions in total There are 5 choices per question so that the

probability of success p = 1/5 = .2● What is the probability that the student gets 6

questions correct?



● Example continued● This is a binomial experiment

There are a finite number n = 10 of trials Each trial has two outcomes (a correct guess and an

incorrect guess) The probability of success is independent from trial to

trial (every one is a random guess) The probability of success p = .2 is the same for each

trial



● Example continued● The probability of 6 correct guesses is

P(x) = nCx px (1 – p)n-x



= 6C10 .26 .84

= 210 • .000064 • .4096= .005505



= 6C10 .26 .84

= 210 • .000064 • .4096= .005505

● This is less than a 1% chance



= 6C10 .26 .84

= 210 • .000064 • .4096= .005505

● This is less than a 1% chance● In fact, the chance of getting 6 or more correct

(i.e. a passing score) is also less than 1%



● Binomial calculations can be difficult because of the large numbers (the nCx) times the small numbers (the px and (1-p)n-x)

● It is possible to use tables to look up these probabilities

● It is best to use a calculator routine or a software program to compute these probabilities






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● We would like to find the mean of a binomial distribution

● Example There are 10 questions The probability of success is .20 on each one Then the expected number of successes would be

10 • .20 = 2● The general formula

μX = n p



● We would like to find the standard deviation and variance of a binomial distribution

● This calculation is more difficult● The standard deviation is

σX = √ n p (1 – p)and the variance is

σX2 = n p (1 – p)



● For our random guessing on a quiz problem n = 10 p = .2 x = 6


● Therefore The mean is np = 10 • .2 = 2 The variance is np(1-p) = 10 • .2 • .8 = .16 The standard deviation is √.16 = .4


● Therefore The mean is np = 10 • .2 = 2 The variance is np(1-p) = 10 • .2 • .8 = .16 The standard deviation is √.16 = .4

● Remember the empirical rule? A passing grade of 6 is 10 standard deviations from the mean …





binomial random variable Construct binomial probability histograms4

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● With the formula for the binomial probabilities P(x), we can construct histograms for the binomial distribution

● With the formula for the binomial probabilities P(x), we can construct histograms for the binomial distribution

● There are three different shapes for these histograms When p < .5, the histogram is skewed right When p = .5, the histogram is symmetric When p > .5, the histogram is skewed left



● For n = 10 and p = .2 (skewed right) Mean = 2 Standard deviation = .4



● For n = 10 and p = .5 (symmetric) Mean = 5 Standard deviation = .5



● For n = 10 and p = .8 (skewed left) Mean = 8 Standard deviation = .4



● Despite binomial distributions being skewed, the histograms appear more and more bell shaped as n gets larger

● This will be important!


Summary: Chapter 11 – Section 2

● Binomial random variables model a series of independent trials, each of which can be a success or a failure, each of which has the same probability of success

● The binomial random variable has mean equal to np and variance equal to np(1-p)

chapter 11 section 1

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