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M116 – NOTES – CH 5
Chapter 4 - Fundamentals of Probability
Experiment: Any process that allows researchers to obtain observations
Sample Space: All possible outcomes of an experiment
Simple Event: Consists of a single outcome of an experiment
Event: Consists of one or more outcomes of an experiment
Notation for Probabilities
Probability of Event A is denoted P(A)
Round-Off-Rule for Probability: Use 3 significant digits as decimals (or use fraction form).
Finding Probabilities with the Classical Approach (Requires Equally Likely Outcomes)
method
P( A) 
number.of .ways. A.can.occur
s

number.of .different.simple.events n
Probability Values
▪ For any event A, 0  P( A)  1
▪ The probability of an impossible event is zero
▪ The probability of a certain event is one
Complementary Events
C
The complement of event A, denoted by A (other books may use A’ or A ), consists of all
simple outcomes in the sample space not making up event A.
Rule of Complementary Events
Since P(A) + P( A ) = 1
then
P(A) = 1 – P( A )
and P( A ) = 1 – P(A)
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Section 5.1 – Probability Distributions
A random variable is a variable (typically represented by x) that has a numeric value, determined by
chance, for each possible outcome of an experiment
Examples:
The number of students passing a certain class
The average height of the students in a class
The number of girls in a family of 5 children
The sum on the faces of two rolled dice
The number of defective parts in a sample of 20
The average daily temperature
A word about randomness
The word randomness suggests unpredictability.
Randomness and uncertainty are vague concepts that deal with variation.
A simple example of randomness involves a coin toss. The outcome of the toss is uncertain. Since the
coin tossing experiment is unpredictable, the outcome is said to exhibit randomness.
Even though individual flips of a coin are unpredictable, if we flip the coin a large number of times, a
pattern will emerge. Roughly half of the flips will be heads and half will be tails.
This long-run regularity of a random event is described with probability. Our discussions of randomness
will be limited to phenomenon that in the short run are not exactly predictable but do exhibit long run
regularity.
A discrete random variable has either a finite or a countable number of values. This chapter deals
with discrete random variables.
A continuous random variable has infinitely many values, and those values can be associated with
measurements on a continuous scale in such a way that there are no gaps or interruptions.
A probability distribution is a graph, table, or formula that gives the probability for each possible value
of the random variable.
(Notice: similar to relative frequency tables, histograms)
A probability histogram is a way to graph a probability distribution.
The vertical scale shows probabilities instead of relative frequencies.
Note that the area of these rectangles is the same as the probabilities.
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M116 – NOTES – CH 5
Section 5.1 – Probability Distributions

Requirements for a Probability Distribution
o
o
0  P(X = x)  1
The sum of the probabilities of a discrete random variable is 1.
 P( X  x)  1

To evaluate the mean and standard deviation of a probability distribution using the
calculator
Enter x into L1
Enter the probabilities into L2
Press STAT
Arrow right to CALC
Select 1: 1-Var Stats L1,L2
Press ENTER

Identifying Unusual Results with the Range Rule of Thumb
Maximum usual value =
Minimum usual value =

  2
  2
Identifying Unusual Results with Probabilities
Unusually high: x successes among n trials is unusually high if P(x or more) is very
small (such as less than 0.05)
Unusually low: x successes among n trials is unusually low if P( or fewer) is very
small (such as less than 0.05)
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M116 – TI 83/84 CALCULATOR – CH 5
Using the TI-83/84 calculator to find the Mean and Standard Deviation of Probability
Distributions
To evaluate the mean and standard deviation using the calculator
Enter x into L1
Enter the probabilities into L2
Press STAT
Arrow right to CALC
Select 1: 1-Var Stats L1,L2
Press ENTER
Example 1) When randomly selecting jail inmates convicted of DWI (driving while
intoxicated), the probability distribution for the number x of prior DWI sentences is as
described in the accompanying table (based on data from the U.S. Department of Justice).
x
0
1
2
3
P(x)
0.512
0.301
0.138
0.049
a) What is the population and the success attribute?
b) Describe in words the random variable. (What are we counting?)
c) What are the possible values of the random variable?
d) Verify that the given table is a probability distribution
e) Use the calculator to find the mean and standard deviation of this distribution.
f) Which values are usual and which are unusual, according to
(i) The probability rule?
(ii) The range rule of thumb?
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M116 – NOTES – CH 5
Section 5.2 & 5.3 – Binomial Experiments

Features of a binomial experiment (5.2)
1)
2)
3)
4)
The experiment has a fixed number of trials (n)
The trials must be independent
Each trial has 2 possible outcomes: success (S) and failure (F)
Probabilities remain constant for each trial.
p is the probability of success, and q is the probability of failure
When sampling without replacement, the events can be treated as if they were
independent if the sample size is no more than 5% of the population size. (That
is, n  0.05 N )

Find binomial probabilities with a shortcut feature of the calculator
To find individual probabilities: Use binompdf(n,p,x)
Press 2nd VARS
Select 0:binompdf(
Type n,p,x)
Press ENTER
To calculate cumulative probabilities from 0 to x, use binomcdf(n,p,x)

Mean, Variance, and Standard Deviation for the Binomial Distribution (5.3)
If we have the probability distribution in the editor of the calculator we can use the
calculator by doing STAT – CALC, 1-VarStat L1, L2
Otherwise we can use these formulas for binomial distributions.
  npq
  np
Remember that the variance is the square of the standard deviation:
Variance =
 2  ( npq )2  npq
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M116 – TI 83/84 CALCULATOR – CH 5
OPTIONAL (ITP) - Using the TI-83/84 calculator to obtain and graph Binomial
Probability Distributions
To graph a probability distribution follow the steps outlined below:
a) For a binomial experiment with n = 4 and p = ¼
Get into the editor of the calculator and clear two lists
Place the possible values of the random variable into one of the lists, let’s say L1
(In this case the possible values of x are from 0 to 4)
Go to L2, and arrow up until you “sit” on the name of the list
Press 2nd VARS[DISTR]
Arrow down to select 0:binompdf(
Indicate choices of n,p (It should read: binompdf(4,1/4))
Press ENTER
The probabilities should show in L2
Sketch a HISTOGRAM for L1, L2 (GO to STAT PLOT)
Select appropriate WINDOW values
x-min = 0
x-max = 5 (n + 1)
x-scale = 1
y-min = - 0.2
y-max = 0.6
Press GRAPH
The graph of the distribution should show.
Press TRACE and arrow right to see the values of the random variable along
with the probabilities.
Comment on the shape of this distribution.
b) For a binomial experiment with n = 6 and p = ½
Sketch the graph of the distribution and comment on its shape. Work on L3, L4.
Remember to have only one plot ON.
c) For a binomial experiment with n = 10 and p = 4/5
Sketch the graph of the distribution and comment on its shape. Work on L5, L6.
d) Relationship between the value of p and the shape of the binomial distribution
If p < 0.5
the shape is
If p = 0.5
the shape is
If p > 0.5
the shape is
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M116 – TI 83/84 CALCULATOR – CH 5
Binomial Distributions and Simulations (Chapter 5)
Example 2) – Booking tickets:
Air America has a policy of booking as many as 15 persons on an airplane that can seat only
14. Past studies have revealed that only 85% of the booked passengers actually arrive for the
flight. Find the probability that if Air America books 15 persons, not enough seats will be
available.
a) Describe the random variable and success attribute. Give the possible values of the
random variable. Give the number of trials and the probability of success.
b) Use the calculator to find the probability that if Air America books 15 persons, not
enough seats will be available.
c) Is it unusual to find that there are not enough sits available? Should overbooking be a
concern for passengers?
d) SIMULATION (ITP)
Now we are going to simulate this situation by repeating the experiment 20 times.
Use MATH PRB 7:randBin(n,p) and press ENTER 20 times.
Record results in a table, and then use your table to answer the question to the problem.
e) Use class results and answer the question again.
f) OPTIONAL (OYO)
Here we have another simulation technique. Use the calculator to generate 50 numbers
that come from a binomial distribution with n = 15 and p = 0.85
(We’ll clear List 1, generate the numbers and store them into List 1, we’ll sort
the list and then explore the editor)
STAT 4:ClrList L1 :
MATH PRB 7:randBin(n,p,50) STO L1 :
STAT 3:SortA(L1)
Go to the editor, explore the list and count how many times we had 15 passengers
showing up. Then determine the probability, and compare with the theoretical results
from part (a). Comment on the law of large numbers.
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M116 – NOTES – CH 6
Normal Distributions (Chapter 6)

Finding probabilities (areas)
Step 1: Draw the graph shading the area desired. Label the mean and the specific xvalues being considered.
Step 2: Find the z-Score for each x-value involved.
Z-score = (score – mean) / standard deviation
Z-SCORE:
z
x

Step 3: Use table 5 to find the cumulative left area bounded by z.
Step 4: Answer the problem.

Using the TI-83/84 to obtain probabilities, percentages, areas
Press
2nd, VARS
Select
2:normalcdf(
Type
left endpoint, right endpoint, , )

Finding Values from Know Areas (Probabilities)
Working BACKWARDS
Step 1: Draw the graph, shade and label the given area, and identify the location
of the x-value being sought.
Step 2: Find the cumulative left area bounded by x.
Step 3: Use table 5 to find the z-score.
(Go backwards! From the main body of the table to the z-score)
Step 4 Find the score, x, by using the formula
x    z

(score = mean + z-score * standard deviation)
Using the TI-83/84 to obtain normal scores, percentiles
Press
2nd, VARS
Select
3:invNorm(
Type
total area to the left of the desired value, , )
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
Example 3) - The United States Air Force ACES-II ejection seat used in fighter jets have been
originally designed for men whose weight is between 140 and 211 pounds. Nowadays many
women are joining the air force and we wonder if it is necessary to re-design the ejection sits.
We’ll take into consideration that weights of women are normally distributed with a mean of
143 and a standard deviation of 29 pounds, and that weights of men are normally distributed
with a mean of 170 and a standard deviation of 40 pounds.
(i) If a woman is randomly selected, what is the probability that her weight is between
140 and 211 pounds?
a) Show all your work, along with the diagram and the shading of the area you are
calculating.
b) Use a feature of the calculator to find the answer.
c) OPTIONAL (ITP) Now we’ll simulate the problem by generating 50 numbers that
come from a normal distribution with a mean of 143 and a standard deviation of 29
pounds.
(We’ll clear List 1, generate the numbers and store them into List 1, we’ll sort the
list and then explore the editor)
STAT 4:ClrList L1 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L1 :
STAT 3:SortA(L1)
Go to the editor, explore the list and count how many women have weights in the given
interval. Then determine the probability.
How does the experimental probability compare with the theoretical probability from
part (a)? Comment on the law of large numbers.
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
Example 3 Continued (ii) If a man is randomly selected, what is the probability that his
weight is between 140 and 211 pounds?
a) Show all your work, along with the diagram and the shading of the area you are
calculating.
b) Use a feature of the calculator to find the answer.
c) OPTIONAL (ITP) Now we’ll simulate the problem by generating 50 numbers that
come from a normal distribution with a mean of 170 and a standard deviation of 40
pounds.
(We’ll clear List 2, generate the numbers and store them into List 2, we’ll sort the
list and then explore the editor)
STAT 4:ClrList L2 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L2 :
STAT 3:SortA(L2)
Go to the editor, explore the list and count how many men have weights in the given
interval. Then determine the probability.
How does the experimental probability compare with the theoretical probability of part
(a)? Comment on the law of large numbers.
d) Are women at greater risk of being injured? Do you think it is necessary to redesign the
ejection seats?
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
Example 4) Designing Helmets
Engineers must consider the breadths of male heads when designing motorcycle helmets.
Men have head breaths that are normally distributed with a mean of 6.0 in. and a standard
deviation of 1.0 in. Due to financial constraints, the helmets will be designed to fit all men
except those with head breaths that are in the smallest 2.5% or largest 2.5 %. Find the
minimum and maximum head breaths that will fit men.
a) Show all steps to get the answer.
b) Now use a feature of the calculator to answer.
c) OPTIONAL (ITP) We are going to use simulation to find the probability that a man
selected at random has a head breath smaller than the MINIMUM found in part (a).
Generate 50 numbers that come from a normal distribution with a mean of 6 and a
standard deviation of 1.
STAT 4:ClrList L3 :
MATH PRB 6:randNorm(mean,st-dev,50) STO L3 :
STAT 3:SortA(L3)
Explore the list to count how many men in the sample have head breaths smaller than
the minimum found in part (a). What percent of the sample is this? How does it
compare to 2.5%?
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M116 – TI 83/84 CALCULATOR – CH 6
Normal Distributions and Simulation (Section 6.3)
How do we fit in?
ERGONOMICS is the study of people fitting into their environment
Designing cars
The sitting height of drivers must be considered in the design of a new car model.
Sitting heights of men are normally distributed with a mu of 36.0 in. and a sigma of 1.4 in.
A car that accommodates men with sitting heights up to 38.8 in. is being designed.
If a man is randomly selected, find the probability that he has a sitting height less than 38.8 in.
What percentage will be “left out” (will not seat comfortably in the car?)
Hip breadths and airplane seats
In designing seats to be installed in commercial aircraft, engineers want to make the seat wide
enough to fit 98% of all males.
Men hip breadths are normally distributed with a mean of 14.4 inches and a standard deviation
of 1.0 inches.
Find the _________th percentile of the distribution which is the score that separates the ones
who will fit on the seat from the ones who will not fit.
Designing car dashboards
When designing the placement of a CD player in a new model car, engineers must consider the
forward grip reach of the driver. Design engineers decide that the CD should be placed so that
it is within the forward grip reach of 95% of women.
Women have forward grip reaches that are normally distributed with a mean of 27.0 in. and a
standard deviation of 1.3 in.
Find the _____ th percentile of this distribution which is the score that separates the ones who
will reach from the ones who will not reach.
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M116 – NOTES – CH 6
Normal Approximation to Binomial Distributions (Section 6.4)
When can we use the Normal distribution to approximate a Binomial distribution?

Normal Distributions as Approximation to Binomial Distributions
If
np  5 and nq  5 , then the binomial random variable has a probability distribution
that can be approximated by a normal distribution with the mean and standard deviation
given as
  n* p
  n* p*q
Example 5) For each of the following problems indicate whether the normal distribution is
appropriate to approximate the binomial distribution. If so, calculate the probability of exactly
5 successes using the binomial distribution and then estimate the probability using the normal
distribution.
a) Case I: n = 12, p = 0.5
b) Case II: n = 12, p = 0.2
c) Case III: n = 12, p = 0.9
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