STAT 221

1
STAT 221

• Chapter 5 Part A
• Discrete Probability Distributions

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The Random Variable - X

• In statistics, X is a symbol for the unknown
  numerical value that expresses one possible
  outcome of an experiment/event.
• In an experiment, any one outcome value that is
  assigned to X can be considered a random variable
  because X can turn out to be (or assume) any
  value in the set of all possible outcomes (and
  that set is called the sample space).

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Depending on the experimental situation, X may be
Discrete or Continuous

• A random variable can be classified as being
  either discrete or continuous depending on the
  numerical values it can assume.
• A discrete random variable is usually an integer
  value and may assume either a finite number of
  values or an infinite sequence of values.
• A continuous random variable can be an integer or
  non-integer value and may assume any numerical
  value in an interval or collection of intervals.

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Examples of experimental outcome values

• Examples of discrete random variables
• Number of babies born on a particular day
• Number of customers arriving at a drive-thru in
  any particular hour
• Number of questions answered correctly on a test.
• Examples of continuous random variables
• Waiting time between customer arrivals
• Distance that a person can run
• Weight of a person

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This chapter discusses experiments whose outcomes
would be discrete random variables

• A discrete random variable can have a finite
  number of values
• Let x number of questions answered correctly
  on a test. If the test has 100 questions, x can
  be any of 101 possible values (0, 1, 2, 3, 4
  100) note the finite upper limit of 100.
• Or a discrete random variable can have an
  infinite sequence of values
• Let x number of customers arriving in one day
• where x can take on the values 0, 1, 2, . . .
  note that there is no upper limit.

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Probability distributions of random variables

• x f (x)
• 0 .125 (1/8)
• .375 (3/8)
• .375 (3/8)
• 3 .125 (1/8)

• All possible values for x and their probabilities
  of occurrence can be plotted on a chart and we
  call that a probability distribution.
• A probability distribution can actually be a
  chart, table, or function that expresses the
  probability of each possible outcome in an
  experiment.
• Here is a probability distribution presented as a
  function f(x)
• As you can see in this example, some x-values are
  more likely to occur than others.

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A probability distribution
1
0
2
3
X of girls out of 3 children
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Developing Probability Distributions

• Probabilities of each possible value of random
  variable x can be developed using any of the
  three methods
• Relative frequency approach look at past
  history of occurrences to project expected
  probabilities
• Classical frequency approach use reasoning to
  deduce the expected probability (Ex a die has 6
  sides, therefore there is a 1-in-6 chance of
  getting each outcome).
• Subjective frequency approach make an educated
  guess.

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Here is a probability distribution for rolling a
die developed using the classical approach
This is a uniform distribution the probability
of occurrence of each possible value of x (1, 2,
3, 4, 5, or 6) is the same (uniform).
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Here is a probability distribution developed
using the relative frequency approach

• In this example, DiCarlo Motors used past history
  to project future probabilities of daily car
  sales.
• They know that over the past 300 days, they sold
  no cars on 54 of those days, 117 days they sold 1
  car, 72 days they sold 2 cars, 42 days they sold
  3 cars, 12 days they sold 4, and 3 days they sold
  5.

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Basic requirements of a probability distribution
f (x) 1 where x assumes all possible
values
0 ? f (x) ? 1 for every individual value of x
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Descriptive statistics for frequency distributions

• It is possible to calculate a mean, a variance
  and standard deviation from a probability
  distribution.
• The mean is called the expected value and it is a
  single measure of central tendency.
• The variance and standard deviations are measures
  of dispersion.

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Expected Value mean of a random variable

• The expected value of a random variable is the
  same as the mean a single value that represents
  the average of all possible outcomes.
• In chapter 3, we calculated a mean value for x by
  adding up all the actual outcomes (x-values) and
  dividing by the number of outcomes. That approach
  was appropriate when you have the actual observed
  values for x.
• When you do not have actual (observed) outcomes
  but you only have a probability distribution for
  expected outcomes, you must calculate the mean
  value of x using this formula
• E(x) ? ?xf(x)

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Example Rolling a die what is the Expected
Value?

• We dont have to roll a die hundreds of times and
  actually observe the number of times we get a 1,
  2, 3 etc. to know that the probability of rolling
  each possible outcome (x) is 1-out-of-6.
• But using the formula on the previous slide, we
  can calculate the average outcome / expected
  value
• E(x) ? ?xf(x)
• 1 1/6 2 1/6 3 1/6 4 1/6 5
  1/6 6 1/6
• 2.7

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Calculating a variance and standard deviation

• Recall that variance and standard deviations are
  measures of dispersion (spread-out-ness).
• Again, in Chapter 3 we learned formulas for
  calculating the variance and standard deviation
  based on actual/observed values for x. When all
  we have is an frequency distribution for expected
  x, we can still calculate a variance and standard
  deviation for x but we must use this formula
• Var(x) ? 2 ?(x - ?)2f(x)

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Example JSL Appliances the expected value or
mean

• x f(x) xf(x)
• 0 .40 .00
• 1 .25 .25
• 2 .20 .40
• 3 .05 .15
• 4 .10 .40
• E(x) 1.20

Here is the probability distribution for the
number of TVs they expect to sell each day.
The mean or the expected number of TV sets they
expect to sell in a day is 1.2
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Example JSL Appliances the variance and std
deviation

• x x - ? (x - ?)2 f(x) (x - ?)2f(x)
• 0 -1.2 1.44 .40 .576
• 1 -0.2 0.04 .25 .010
• 2 0.8 0.64 .20 .128
• 3 1.8 3.24 .05 .162
• 4 2.8 7.84 .10 .784
• 1.660 ? ?
• The variance of daily sales is 1.66 TV sets
  squared.
• The standard deviation of sales is 1.2884 TV
  sets.

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Practice question Calculate the mean, variance,
and SDExcel worksheet Computers

• According to a survey, 95 of subscribers to the
  Wall Street Journal Interactive Edition have a
  computer at home. For those households, the
  probability distribution for the number of
  computers is given (see spreadsheet).
• A. What is the expected value of the number of
  computers per household?
• B. What is the variance of the number of
  computers per household?
• C. Make three intelligent statements about the
  number of computers owned by the Journals
  subscribers.

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Open the file DataSetsForCh5 and select the
worksheet Computers
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To calculate the expected value for laptops,
first multiply the number of computers (x) by the
frequency f(x) for the first row C4 A4 B4
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Copy the formula in C4 down to C7 to multiply the
number of computers (x) by the frequency f(x) for
the other possible values of x.
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Sum the column C8 sum(C4C7)
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Copy the sum into cell C10 because that is the
expected value C10 C8
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To calculate the variance, start by squaring the
first value x D4 A4 A4
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Copy the formula in D4 down to D7 to square the
remaining x-values
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Multiply squared-x by the frequency of x for the
first x-value E4 D4 B4
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Copy the formula in E4 down to D7 to multiply the
other squared-xs by their frequencies.
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Sum the values in columns D and E D8
SUM(D4D7) E8 SUM(E4E8)
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To calculate the variance, subtract the square of
the mean from the sum of column E C11 E8
(C10 C10)
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To calculate the standard deviation, take the
square root of the variance C12 sqrt(C11)
Resave the file.
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Draw some conclusions about number of computers
owned by the Journals subscribers.

• The average Wall Street Journal reader has 1.42
  computers in their household.
• The majority of households have 1 computer.
• If we added and subtracted 2 standard deviations
  (.75) from the mean, wed find that about 95 of
  households have roughly 0 to 3 computers.

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Developing probability distributions for discrete
random variables

• Experiments are conducted when performing an
  experimental study.
• Experiments yield results - a dataset of x-values
  each one considered to be a random value.
• Some experiments have certain common
  characteristics that enable them to classified as
  a binomial experiment.

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The Binomial Experiment

• Here are some examples of binomial experiments
• Tossing a coin n times and seeing how many of
  those outcomes are heads.
• A salesman calling on potential clients and
  seeing how many of those clients purchase his
  product.
• Gathering a sample of newborn babies and seeing
  how many of them have a certain characteristic or
  gender.

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Characteristics of a binomial experiment

• The experiment must consists of a sequence
  repeating the same step n times (the step could
  be flipping a coin, making a sales call,
  determining if a baby has a certain
  characteristic or gender, etc.)
• For each step, there are two possible outcomes
  referred to as success or failure.
• The probability of each outcome is the same at
  each step (so it must use with replacement
  sampling).
• The steps are independent (the outcome at each
  step is not conditional on previous steps
  outcomes).

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Practice Determine whether each of the given
experiments can be classified as a binomial
experiment

• Surveying 1012 people and recording whether there
  is a should not response to this question Do
  you think the cloning of humans should or should
  not be allowed?
• Rolling a loaded die 50 times and finding the
  number of times that 5 occurs.
• Determining whether each of 3000 heart pacemakers
  is acceptable or defective.
• Spinning a roulette wheel 12 times and finding
  the number of times that the outcome is an odd
  number.

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Answers

• Yes
• Yes
• Yes
• Yes

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The random variable (x) in a binomial experiment

• A binomial experiment can produce results (a data
  set of values for x) where x is the number of
  successes after n trials.
• So if the experiment is tossing a coin 10
  times, and on one trial we get this outcome
  HHHTTHTHTH, then x 6 for 6 times we got heads
  (a success).
• We could just as easily have assigned tails to
  be the success.

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Binomial experiments yield binomial distributions

• A binomial experiment can produce results (a data
  set of values for x) that can be plotted into a
  frequency distribution that we refer to as a
  binomial frequency distribution.

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The binomial frequency distribution

• The binomial frequency distribution (using table
  format) would list all the possible outcomes of x
  and their probability of occurrence.
• So, if our experiment is to toss a coin 10 times,
  our frequency distribution might look like this
• of times Heads Probability
• 0 .003
• 1 .007
• 2 .024
• 3 .078
• Etc.
• 10 .003

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But we dont have to actually perform n number of
trials to get an actual data set

• In a binomial experiment, we dont actually have
  to have observed frequencies (that would be
  obtained if we engaged in n number of trials or
  have a sample size of n).
• All we need to know is the probability of 1
  success. In this case, all we need to know is the
  probability of getting heads on one trial which
  would be 50.
• We can use the binomial formula to calculate
  the probability of getting 0 heads, 1, 2, 3, etc.
  on 10 trials.

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The binomial formula
where n number of trials x number of
successes among n trials p probability of
success in any one trial q probability of
failure in any one trial (q 1 p)
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Lets start with what is the probability of
getting 3 heads (out of 10 tosses)?
10 9 8
.125 .007813
3 2
.1172 or 11.72
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A closer examination of this formula

• Notice the combinations formula within the
  binomial probability formula?
• Thats because this part .53 .510-3
• is the probability of getting one specific
  sequence of 3 heads and 7 tails (e. g.,
  HHHTTTTTTT)
• This part 10 9 8 / 3 2 1
• calculates the number of different combinations
  that result in 3 heads and 7 tails

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What we have so far

• X P(x)
• 0 ?
• 1 ?
• 2 ?
• 3 11.72
• 4 ?
• 5 ?
• 6 ?
• 7 ?
• 8 ?
• 9 ?
• 10 ?

The probability distribution for getting x heads
out of 10 trials.
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Using Excel

• The binomial probability formula is one of
  Excels built-in formulas.
• To calculate the probabilities of the other 10
  possible outcomes, lets use Excel.
• Open the worksheet binomial and follow the steps
  on the next slides.

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Open the file DataSetsForCH5 and click the
worksheet Coin Toss
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1. Enter the n, p, and q values into cells C3
10 C4 .5 C5 1-C4
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2. Position the cursor in B8. 3. To use Excels
binomdist( ) formula, click the fx button on the
formula bar to start the formula wizard.
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4. In the category box, select Statistical,
5. In the select a function box, select
binomdist and click ok.
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6. Enter the indicated value as function
arguments. Use absolute addressing where
appropriate. 7. Click ok.
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8. See the formula result appear in cell B8. That
is the probability of getting 0 heads out of 10
trials.
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9. Copy the formula in B6 down to B18 to
calculate the probabilities for the other 10
possible values for x.
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10. Use the button on the formatting toolbar
to format the values to percentage and use the
increase decimal places button to specify 2
decimal places. This table shows the probability
of each possible outcome of the experiment.
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11. Sum the probabilities to make sure they sum
to 100 B19 sum(B8B18)
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Obtaining a mean, variance, and standard
deviation from a binomial distribution

• Recall that we could derive a mean, variance,
  and std. deviation from a (generic) probability
  distribution.
• When the probabilities distribution is
  binomial, the formulas can be simplified to the
  following
• Expected Value E(x) ? np
• Variance Var(x) ? 2 np(1
  - p)
• Standard Deviation

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Calculate the expected value or mean H3 C3 C4
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Calculate the variance H4 C3 C4 C5
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Calculate the standard deviation H5 sqrt(H4)
Resave the file.
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Here is the frequency distribution notice the
symmetric characteristic of a binomial
distribution But as p gets larger (making q
smaller), the distribution becomes skewed to the
left.
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Binomial probability practice 1

• A multiple-choice test has six questions and each
  has 4 possible answers, one of which is correct.
• A. If you completely guess at the answer to each
  question, what is the probability of getting
  (exactly) the first two wrong and the last four
  right (hint use the multiplication rule).
• B. Begin with WWCCCC and make a complete list of
  all the different possible arrangements of two
  wrong and four right. What is the probability of
  getting each of these arrangements?
• C. What is the probability of getting exactly two
  wrong and four right? (In other words, what is
  the probability of getting any of these
  arrangements?)

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a. If you completely guess at the answer to each
question, what is the probability of getting the
first two wrong and the last four right (hint
use the multiplication rule).

• P(C) .25
• P(W) .75
• P(WWCCCC) .75 .75 .25 .25 .25..25
• P(WWCCCC) .00220

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b. Begin with WWCCCC and make a complete list of
all the different possible arrangements of two
wrong and four right. Find the probability of
each arrangement.
CWWCCCCWCWCCCWCCWCCWCCCWCCWWCCCCWCWC
WWCCCCCWCWCCCCWCCWCCCWCCCWCCWCCCCWC WCCCCCW
CCWCCWCCCWWCCCCWCWCCCCWW
There are 15 ways to get 2 wrong and 4 right.
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b. What is the probability of each of these
arrangements?

• In each case, the probability is .752 .254
• Or .00220

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c. What is the probability of getting any of
these arrangements?
n is 6 p is .75 (the probability of
guessing wrong) x is 2 q is .25 (the
probability of guessing right)
The probability of getting any specific sequence
of 2 Ws and 4 Cs
Same as 15 (the number of ways of getting 2 Ws
and 4 Cs)
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Binomial probability practice 2

• If you run a red traffic light at an intersection
  equipped with a camera monitor, there is a .1
  probability that you will be given a ticket. If
  you run a red traffic light at this intersection
  five different times, what is the probability of
  getting at least one ticket?

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The possible outcomes

• The (6) possible outcomes are
• 0 tickets
• 1 ticket
• 2 tickets
• 3 tickets
• 4 tickets
• 5 tickets

None
At least 1
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Remember how we solved it by finding the
probability of the complement?

• If A is (getting at least one ticket)
• Then not A is (getting 0 tickets)
• Rather than find the P(A), find the P(not A) and
  subtract it from 1
• P (0 tickets)
• P(.95) (.9) (.9) (.9) (.9) (.9) ) .59
• 1 - .59 .41
• Or 41

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Now we know how to solve it this way

• P(at least one)
• P(1) P(2) P(3) P(4) P(5)
• Where p .1 and q .9
• Refer to worksheet NumberOfTickets
• Use Excels BINOMDIST( ) to find the
  probabilities and sum them.

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Binomial probability practice 3

• Assume several samples of 6 AA flights were
  selected at random and the average on-time
  arrival rate of these flights was .723 (72.3)
• Based on this success rate of 72.3, create a
  frequency distribution.
• Find the probability that at most two American
  Airlines flights arrive on time.
• Find the probability that at least one AA flight
  arrives on time.

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