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Chapter 6: Random Variables
If we tossed a coin 3 times...
• How many tails could we possibly get?
• The possible # of tails (possible values of X)
are 0, 1, 2, 3.
• Every time we toss a coin three times, X will
either be 0, 1, 2, or 3 (and possibly different
each time)
What if…
… we tossed a fair coin 3 times. With a partner,
list all the possible outcomes
HHH HHT HTH THH HTT THT TTH TTT
This is the “sample space” for this chance
process
Probability distribution…
8 equally likely outcomes; probability is 1/8 for each
possible outcome.
Define the variable X = the number of heads
obtained. The value of X will vary from one set of tosses
to another but will always be one of the numbers
0, 1, 2, or 3.
How likely is X to take each of those values? It will be
easier to answer this question if we group the possible
outcomes by the number of heads obtained:
Random Variables ...
• X is called a random variable because its
values vary when the coin tossing is repeated.
• are usually denoted by capital letters near the
end of the alphabet, like X or Y
Discrete Random Variables
• Discrete random variables have a “countable”
number of possible positive outcomes and must
satisfy two requirements. Note: ‘countable’ is
not the same as finite.
• (1) Every probability is a # between 0 and 1;
• (2) The sum of the probabilities is 1
World-Wide 2015 AP Statistics Score Distribution
• Is this a discrete random variable probability
distribution? Why or why not?
1
2
3
4
5
.238
.189
.252
.189
.132
Discrete Random Variables...
• In your groups, discuss examples of discrete
random variables
Other Examples of Discrete Random
Variables...
• Number of times people have seen Fall Out Boy
in concert
• Number of gifts we get on our birthday
• Number of burgers sold at In-N-Out Burger per
day
• Number of stars in the sky
All are whole, countable numbers; all vary; usually
represented by a table or probability histogram
Non-examples of Discrete Random
Variables...
• Your height
• Weight of a candy bar
• Time it takes to run a mile
Mean of Discrete Random Variables
• To find the mean (measure of center) of
discrete random variable X, multiply each
possible value by its probability, then add all
the products
• Mean, Expected Value, μx
• Symbolic: Σxipi
Expected Value for Discrete Random
Variables...
Grade
0
1
2
3
4
Probability
0.01
0.05
0.30
0.43
0.21
μx =
0 (0.01) + 1(0.05) + 2(0.30) + 3(0.43) + 4(0.21)
= 2.78
Expected Value for Discrete Random
Variables...
Find μx for the following discrete random
variable:
Mean =
Expected Value for Discrete Random
Variables...
Find μx for the following discrete random
variable:
Mean = 1.75
Standard Deviation for Discrete
Random Variables...
• Standard deviation measures the spread or
variability of the distribution (remember mean,
SD, variance; median, IQR)
• Symbol: σ
•
(σ)2
= variance;
2
σ =σ
• Don’t recommend using formula to calculate
variance and SD; use calculator & lists
Standard Deviation for Discrete
Random Variables...
L1 random variables X
L2 corresponding probabilities
stat – calc – 1-var stats L1, L2
(L1, L2 is only way calculator knows you have a
frequency distribution)
𝑥 is really μx (or expected value); σx is SD
Standard Deviation & Expected Value
of Discrete Random Variables...
Cars Sold
0
1
2
3
Probabilities
0.3
0.4
0.2
0.1
μx = expected value =
σx = standard deviation =
σ2 = variance =
Standard Deviation & Expected Value
of Discrete Random Variables...
Cars Sold
0
1
2
3
Probabilities
0.3
0.4
0.2
0.1
μx = expected value = 1.1
σx = standard deviation = 0.943
σ2 = variance = 0.890
Go back to grade distribution
• Use calculator to confirm the expected value;
and calculate the standard deviation and
variance using your calculator
Grades
0
1
2
3
4
Probability
0.01
0.05
0.30
0.43
0.21
μx =
σ=
σ2 =
Go back to grade distribution
• Use calculator to confirm the expected value;
and calculate the standard deviation and
variance using your calculator
Grades
0
1
2
3
4
Probability
0.01
0.05
0.30
0.43
0.21
μx = 2.78
σ = .867
σ2 = .752
Continuous Random Variables ...
• take on all values in an interval of numbers
• probability distribution is described by a
density curve
• probability of event is area under the density
curve and above the values of X that make up
the event
Continuous Random Variables...
• are usually measurements
• heights, weights, time
• amount of sugar in a granny smith apple, time
to finish the New York marathon, height of Mt.
Whitney
How can we distinguish between
continuous and discrete?
• Discuss in your groups for a few minutes.
• Ask yourself ‘How many? How much? Are
you sure?’
• For example, # of children, pounds of Captain
Crunch produced each year, # of skittles,
ounces in a bag of skittles
Continuous Random Variables...
• Probability distribution is area under the
density curve, within an interval, above x-axis
Continuous Random Variables...
For continuous RV’s, there is no difference
between > &  and no difference between
< & 

Continuous Random Variables ...
• This is true for all continuous random
variables, but NOT discrete random variables.
Why? Discuss for a few minutes.
• 𝑃 1 < 𝑋 < 4 is the same as 𝑃 1 ≤ 𝑋 ≤ 4
where x = # of Reeses Peanut Butter Cups I
will eat in a given day (yum yum!)
Random Variables...
Consider a six-sided die... What is the probability...
P ( roll less than or equal to a 2) =
P ( roll less than a 2) =
Different probabilities; discrete random variable
Note: possible outcomes are 1, 2, 3, 4, 5, 6; but
probabilities of those outcomes are (often)
fractions/decimals
Continuous random variables ...
• All continuous random variables assign
probabilities to intervals
• All continuous random
variables assign a probability of zero to every
individual outcome. Why?
Continuous Random Variables...
• There is no area under a vertical line (sketch)
• Consider...
0.7900 to 0.8100
0.7990 to 0.8010
0.7999 to 0.8001
P = 0.02
P = 0.002
P = 0.0002
P (an exact value –vs. an interval--) = 0
Another way to think about this is …
For continuous RV’s, there is no area (so probability = 0) at
a given point…
Remember for continuous RV’s each outcome is just one
of an infinite number of possible outcomes, so the
1
probability of any particular outcome is ► 0
∞
1
𝑛
As the value of n increases, the closer gets to zero
Density Curves & Continuous RV’s..
• Can use ANY density curve to assign
probabilities/model continuous RV’s; many
models
• Most familiar density curves are the Normal
density curves (Chapter 2)
Continuous Random Variables...
Remember Normal density curve...
• N ( , )
x
• N (0, 1)
z


• 2nd – stat- normal cdf (low,high, , )


Mean & Standard Deviation of
Continuous Random Variables...
• μx for continuous random variables lies at the
center of a symmetrical (or fairly symmetrical)
density curve (Normal or approximately Normal,
Chapter 2)
• N (μ, σ)
• Calculating σ and/or σ2 for continuous random
variables…. beyond the scope of this course… will
be given this information if needed
Continuous Random Variables...
Practice….
… with discrete and continuous random variables
and general probability
Cheating in school...
• A sample survey asked an SRS of 400 undergraduates:
“You witness two students cheating on a quiz. Do you go
to the professor?” Twelve percent answered ‘yes.’
• This statistic is a random variable, because repeating the
SRS would give a different sample of 400 undergraduates
and a different proportion (other than 12%). This
distribution of a sample proportion is N(0.12, 0.016).
• What is the probability that a given survey result differs
from the truth about the population by more than 2
percentage points?
Cheating in school...
About 21% of sample results will be off by more
than two percentage points.
Car Ownership
Choose an American household at random and
let the random variable X be the number of
cars (including SUVs and light trucks) they
own. Here is the probability model if we
ignore the few households that own more
than 5 cars:
(a) Verify that this is a legitimate discrete
distribution.
(b) Say in words what the event {X ≥ 1} is; then find
P(X ≥ 1).
(c) A housing company builds houses with two-car
garages. What percent of households have more
cars than the garage can hold?
How Student Fees Are Used...
Weary of the low turnout in student elections, a
college administration decides to choose an SRS of 3
students to form an advisory board that represents
student opinion. Suppose that 40% of all students
oppose the use of student fees to fund student
interest groups and that the opinions of the 3
students on the board are independent. Then the
probability is 0.4 that each opposes the funding of
interest groups.
(a) Call the three students A, B, and C. What is the
probability that A and B support funding and C
opposes it?
How Student Fees Are Used ...
• (a) Call the three students A, B, and C. What is
the probability that A and B support funding
and C opposes it?
• P (Student A Supports & Student B Supports &
Student C opposes)
• P (SSO) = (0.60) (0.60) (0.40) = 0.144
How Student Fees Are Used ...
(b) List all possible combinations of opinions
that can be held by Students A, B, and C. (Hint:
There are eight possibilities.) Then give the
probability of each of these outcomes.
Remember... Support = 0.60... Oppose = 0.40
How Student Fees Are Used ...
• SSS, SSO, SOS, OSS, SOO, OSO, OOS & OOO
SSS
SSO
SOS
OSS
SOO
OSO
OOS
OOO
(.6)3
(.6)2(.4)
(.6)2(.4)
(.6)2(.4)
(.4)2(.6)
(.4)2(.6)
(.4)2(.6)
(.4)3
How Student Fees Are Used...
(c) Let the random variable X be the number of
student representatives who oppose the
funding of interest groups. Give the
probability distribution of X.
How Student Fees Are Used...
(c) Let the random variable X be the number of
student representatives who oppose the
funding of interest groups. Give the
probability distribution of X.
P(X=0)
P(X=1)
P(X=2)
P(X=3)
0.216
0.432
0.288
0.064
How Student Fees Are Used...
(d) Express the event “a majority of the advisory
board opposes funding” in terms of X and find
its probability.
How Student Fees Are Used...
(d) Express the event “a majority of the advisory
board opposes funding” in terms of X and find
its probability.
This probability can be written two different
ways:
P (X > 1)
P (X ≥ 2)
Either equal to 0.352
Homework...
Page 359, #1, 3, 5, 7, 9, 11, 13, 15, 17, 21, 23, 25
MC: all
FRQs: as needed
Transforming & Combining RV’s
(Changing Units of Measure)
• Remember Chapter 2…the last time you went out
to dinner. How much was the bill?
• What is 𝑥 ? What is s? How about other key
values, like Q1, Q3, etc.?
• Tipped chef $10; what did that do to our mean,
standard deviation, spread, shape of distribution?
• Paid 20% more; what did this do to our mean,
standard deviation, spread, shape of distribution?
Linear Transformation
• A linear transformation changes the original
variable x into the new variable xnew given by an
equation of the form xnew = a + bx
• Adding/subtracting the constant a shifts all values
of x upward (increases; to the right) or downward
(decreases; to left) by the same amount.
• Multiplying/dividing by the positive constant b
changes the size of the unit of measurement.
Linear Transformation Effects
• Adding the same number a (either positive, zero,
or negative) to each observation adds a to
measures of center and to quartiles but does not
change measures of spread or shape.
• Multiplying/dividing each observation by a
positive number b multiplies both measures of
center (mean and median) and measures of
spread (interquartile range and standard
deviation) by b but does not change shape.
If you are a formula person…
(Linear Transformations)
• If X and Y are random variables, and a & b are
fixed (constant) numbers, then
• μ a + bx = a + bμx
• σ2a+bX = b2σ2X
RV Means, SDs, & Variances: Example...
• In a process for manufacturing glassware,
glass stems are sealed by heating them in a
flame. The temperature of the flame varies a
bit. Below is the distribution of the
temperature X measured in degrees Celsius:
RV Means & SD/Variances
The mean temperature μx = 550 degrees Celsius &
standard deviation, σx = 5.7 degrees Celsius
A manager asks for results in degrees Fahrenheit. The
conversion of X into degrees Fahrenheit is y = 9/5 X + 32.
Find the mean μx & standard deviation σx in Fahrenheit
μ a + bx = a + bμx
σ2a+bX = b2σ2X
RV Means & SD/Variance
The conversion of X into degrees Fahrenheit is
y = 9/5 X + 32. Find the mean μx in Fahrenheit.
Mean = 1022 degrees Fahrenheit
SD = 10.26 degrees F.
RV Means, etc.
• Can choose to use the formula or...
• L3= (L1) (( 9/5) X + 32), then do 1-var stats with
L3 and L2
• Let’s try it
You try it...
• Hart Partners is planning a major investment. The
amount of profit X is uncertain, but an estimate gives
the following distribution (in millions of dollars):
• Find the mean profit μx and the standard deviation σx
• Hart Partners owes its source of capital a fee of
$200,000 plus 10% of the profits X. So, the firm
actually retains Y = 0.9X - 0.2 from the investment.
Find the mean and the standard deviation of what it
expects to actually retain
You try it...
• Hart Partners is planning a major investment.
The amount of profit X is uncertain, but an
estimate gives the following distribution (in
millions of dollars):
• Find the mean profit μx = $3 million; standard
deviation $2.52 million
• The firm actually retains Y = 0.9X -0.2 from the
investment. Find the mean of what it expects to
actually retain = $2.5 million; SD = $2.27 million
RVs: Combining Means & Standard
Deviations/Variances
If X and Y are random variables, then
μX+Y = μX + μY … remember, means play nice
The mean of the sum of the random variables is the
sum of their means
You earn an average of $20/week allowance μX =
$20
Your sister earns on average $16/week allowance μY
= $16
So, μX + μY = $36
RVs: Combining Means & Standard
Deviations/Variances
• If X and Y are independent random variables,
then
σ2X+Y = σ2X + σ2Y
σ2X-Y = σ2X + σ2Y
… variances? They don’t play so nice
Notes: (1) Go through variances always; (2) always
add variances; (3) constant “a” doesn’t change/
influence standard deviation/variance
Combining Normal Random Variables
• Any linear combination of independent
Normal random variables is also going to be
Normally distributed
• If X and Y are Normal independent random
variables, and if a and b are fixed numbers,
then aX plus ± minus bY is also Normally
distributed
Gold Medalist Carly Patterson
• Carly Patterson won the gold medal in gymnastics at the
2004 Olympics in Athens, Greece. A competitor's total
score is determined by adding the scores for four events:
vault, parallel bars, balance beam, and floor exercise.
• Suppose we know that the eventual silver medalist,
Russia's Svetlana Khorkina, earned a total score of 38.211
(her actual total in Athens). Suppose also that Carly's
scores in 100 previous meets leading up to the Olympics
have been approximately Normally distributed.
We will further assume that Carly's perfomance
at the Olympics will follow the same pattern
as her pre-Olympic meets. This is a reasonable
assumption for world-class athletes like Carly.
We want to know Carly's chances of beating the
total score of the current leader, Khorkina.
μv + μp + μb + μf =
σ2v + σ2p + σ2b + σ2f =
so σv+p+b+f =
μv + μp + μb + μf = 37.871 points
σ2v + σ2p + σ2b + σ2f = 0.1126, so σv+p+b+f = 0.3355
Carly’s total score has a distribution that can be
described as …
μv + μp + μb + μf = 37.871 points
σ2v + σ2p + σ2b + σ2f = 0.1126, so σv+p+b+f = 0.3355
Carly’s total score will be N (37.871, 0.3355)
P (Carly’s total > Khorkina) =
μv + μp + μb + μf = 37.871 points
σ2v + σ2p + σ2b + σ2f = 0.1126, so σv+p+b+f = 0.3355
Carly’s total score will be N (37.871, 0.3355)
P (Carly’s Score > 38.211) = 0.1554
Speed Dating…
To save time and money, many single people have
decided to try speed dating. At a speed dating
event, women sit in a circle and men spend about
10 minutes getting to know a woman before
moving on to the next one.
Suppose that the height M of male speed daters
follows a Normal distribution with a mean of 69.5
inches and a standard deviation of 4 inches, and
that the height F of female speed daters follows a
Normal distribution with a mean of 65 inches and a
standard deviation of 3 inches.
Speed dating…
• What is the probability that a randomly
selected male speed dater is taller than the
randomly selected female speed dater with
whom he is paired?
• There is about an 82% chance that a randomly
selected male speed dater will be taller than
the randomly selected female speed dater
with whom he is paired.
Homework
Page 382, # 35, 37, 39, 41, 43, 45, 47, 49, 51, 53,
55, 61
MC
FRQs
Binomial & Geometric Random
Variables
Many types of distributions...
• What is the main type of distribution that we
have worked with (and is really valuable to us)
so far?
• Normal distribution
• Characteristics of Normal distribution?
• (fairly) Symmetric, uni-modal, 68-95-99.7,
area of one (all density curves), NPP is (fairly)
linear.
Many other important, “special” types
of distributions...
• If certain criteria is met, easier to calculate
probabilities in specific situations
• Next types of distributions we will examine
are situations where there are only two
outcomes
• Win or lose; make a basket or not; boy or girl
...
Discuss situations where there are
only two outcomes...
•
•
•
•
•
•
Yes or no
Open or closed
Patient has a disease or doesn’t
Something is alive or dead
Person has a job or doesn’t
A part is defective or not
that is what this section is all about...
• ... two classes of distributions that are
concerned about events that can only have 2
outcomes
• Binomial Distribution &
• Geometric Distribution
Binary; Independent; fixed Number;
probability of Successes
The Binomial setting is:
1. Each observation is either a success or a failure
(i.e., it’s binary)
2. All n observations are independent
3. Fixed # (n) of observations
4. Probability of success, p, is the same for each
observation
“BINS”
Binomial Distribution: practice…
I roll a die 3 times and observe each roll to see if it
is even or odd. Is:
1.
2.
3.
4.
each observation is either a success or a failure?
all n observations are independent?
fixed # (n) of observations?
probability of success, p, is the same for each
observation ?
BINS
Binomial Distribution
• If BINS is satisfied, then the distribution can be
described as B (n, p)
• B
binomial
• n
the fixed number of observations
• p
probability of success
• Note: This is a discrete probability distribution.
• Remember N (μ, σ)… is this discrete??
Binomial Distribution
• Most important: being able to recognize
situations and then use appropriate tools for
that situation
• Let’s practice...
Are these binomial distributions? Why
or why not?
• Toss a coin 20 times to see how many tails
occur.
• Asking 200 people if they watch ABC news
• Rolling a die until a 6 appears
Are these binomial distributions or
not? Why or why not?
• Asking 20 people how old they are
• Drawing 5 cards from a deck for a poker hand
• Rolling a die until a 5 appears
How could we change the situations to make
these/force these to be binomial distributions?
• Asking 20 people how old they are
• Drawing 5 cards from a deck for a poker hand
• Rolling a die until a 5 appears
Binomial Distribution... Is the situation
‘independent enough’?
An engineer chooses a SRS of 20 switches from a
shipment of 10,000 switches. Suppose (unknown
to the engineer) 12% of switches in the shipment
are bad.
Not quite a binomial setting. Why?
For practical purposes, this behaves like a binomial
setting; ‘close enough’ to independence; as long
as sample size is small compared to population.
Rule of thumb: sample ≤ 10% of population size
Binomial Formula...you don’t need to
memorize... Just know what ‘n’ (# of
observations) and ‘k’ (value of random
variable)
n k
n k
P(X  k)   p (1  p)
k 
An engineer chooses an SRS of 20 switches from a
shipment of 10,000 switches. Suppose that (unknown to
the engineer) 12% of switches in the shipment are bad.
The engineer wants to find the probability of exactly 1
switch failing. Situation ≈ Binomial
k=1
n= 20
p = 0.12
n k
n k
P(X  k)   p (1  p)
k 
Binomial probabilities: PDF
• 2nd-vars
binompdf (trials, p, X value)
• Use for exact value (i.e., probability that a family
has exactly 2 boys)
• Remember, these are discrete random variables
• Can calculate probabilities of exact values (unlike
continuous random variables)
Binomial probabilities: CDF
• 2nd-vars binomcdf (trials, p, X value)
•
•
•
•
Use for cumulative values
Works like table A; cumulative to left
If want area to right, need to do “1 minus ....”
Pay attention to < vs. ≤; and > vs. ≥
Practice...
Each child born to a particular set of parents has probability
0.25 of having blood type O. If these parents have 5
children, what is the probability that exactly 2 of them
have type O blood?
Binomial setting? Check for BINS.
p = 0.25
n=5
X=2
binompdf (n, p, X) = binompdf (5, .25, 2) =
= 0.2636; context, always!
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
exactly 4 of them have type O blood? Binomial
setting? Check for BINS.
p = 0.25
n=5
X=4
binompdf (n, p, X) = binompdf (5, .25, 4) =
= 0.0146; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
exactly 1 of them have type O blood? Binomial
setting? Check for BINS.
p = 0.25
n=5
X=1
binompdf (n, p, X) = binompdf (5, .25, 1) =
= 0.3955; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
at most 2 of them have type O blood? Binomial
setting? Check for BINS.
p = 0.25
n=5
X=2
binomcdf (n, p, X) = binomcdf (5, .25, 2) =
= 0.8965; context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
at most 4 of them have type O blood? Binomial
setting? Check for BINS.
p = 0.25
n=5
X=4
binomcdf (n, p, X) = binomcdf (5, .25, 4) =
= 0.9990, context, always
Practice...
Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
at most 1 of them have type O blood? Binomial
setting? Check for BINS.
p = 0.25
n=5
X=1
binomcdf (n, p, X) = binomcdf (5, .25, 1) =
= 0.6328, context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
at least 2 (meaning 2, 3, 4, or 5) of them have type
O blood?
p = 0.25
n=5
X = 2, 3, 4, or 5
1 – (binomcdf (5, .25, 1)) =
1 - 0.6328 = 0.3672; context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
at least 3 (meaning 3, 4, or 5) of them have type O
blood?
p = 0.25
n=5
X = 3, 4, or 5
1 – (binomcdf (5, .25, 2)) =
1 - 0.8965 = 0.1035; context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
more than 3 (meaning 4 or or 5) of them have type
O blood?
p = 0.25
n=5
X = 4 or 5
1 – (binomcdf (5, .25, 3)) =
1 - 0.9844 = 0.0156; context, always
Practice...
... probability 0.25 of having blood type O. If these
parents have 5 children, what is the probability that
… of them have type O blood?
p = 0.25
n=5
a) …at most 3 of them…
b) …at least 4 of them…
c) …more than 1 of them…
d) …exceeds 3 of them…
e) … below 2 of them…
f) … 0 of them…
Practice...
Suppose the mean number of children in a given
household who have type O blood is 2.1 with a
standard deviation of 0.8. If this distribution is
approximately Normally distributed, what is the
probability that a household has more than 4
children with type O blood?
μ= 2.1
σ= 0.8
normalcdf (4, hi, 2.1, 0.8) = 0.0088; context,
always
Caution... binomcdf (n, p, X). What is P(X)
if...
Suppose X is B (23, 0.7). . .
P (X < 17) = binomcdf (23, 0.7, 16)
P (X ≤ 17) = binomcdf (23, 0.7, 17)
Helpful hint... make a list of values you are
interested in...
Caution... binomcdf (n, p, X). What is
P(X) if...
Supposed X is B (23, 0.7). . .
P (X > 21) =
1 - binomcdf (23, 0.7, 21)
interested in 22, 23
P (X ≥ 21) =
1 - binomcdf (23, 0.7, 20)
interested in 21, 22, 23
Binomial Distribution: Mean and
Standard Deviation
• If a basketball player makes 75% (“p”) of her
free throws, what do you think the mean
number of baskets made will be in 12 tries?
Discuss.
• (0.75) ( 12) = 9; we expect she should make 9
baskets in 12 tries
• We expect her 𝑥= 9 baskets
Binomial Mean & Standard Deviation
• If a count X is a binomial distribution with
number of observations n and probability of
success p, then
• μ= np
and
σ=
np(1  p)
• Only for use with binomial distributions;
remember criteria...
BINS

Practice...
• If a basketball player makes 75% (“p”) of her
free throws, we expect her to make 9 baskets
in 12 tries.
• What is the SD of this distribution?
• SD = np(1  p) =
= 1.5
(12)(0.75)(1 0.75)
Practice...
• Each child born to a particular set of parents has
probability 0.25 of having blood type O. If these parents
have 5 children, what is the mean and standard
deviation for this distribution?
• mean = np = (5)(.25) = 1.25; the family should expect to
have 1.25 children with type O blood
• SD = np(1  p) =
= 0.9682
(5)(0.25)(1 0.25)
Remember & Caution...
• Binomial distribution is a special case of a
probability distribution for a discrete random
variable
• All binomials distributions are discrete random
variable distributions BUT not all discrete random
variable distributions are binomial distributions
• Don’t assume; don’t apply binomial tools to all
discrete RV distributions; must meet binomial
distribution criteria (BINS)
Normal Approximation to Binomial
Distribution...
• But as ‘n’ gets larger, binomial distribution ≈
Normal distribution (weird... binomial is
discrete & Normal is continuous...)
• General rule: np ≥ 10 & n(1-p) ≥ 10, the
binomial distribution is ≈ Normal, i.e., the
expected number of successes and failures are
both at least 10; ‘large count condition’
Normal ≈ to Binomial Distribution...
• Attitudes towards shopping...
• SRS of 2,500 (out of 218 million U.S. adults)
asked if they found shopping frustrating or not
• Suppose in fact 60% of all U.S. adults say
shopping is frustrating.
• Is this a binomial distribution? Check BINS.
Binomial, Simulation, or Normal
Distribution...
BINS: 2 (frustrated or not),
i (independent enough), n = 2,500 p = 0.60;
confirmed binomial
• Say we want to find P(X ≥ 1520), we could:
use binomcdf
do a simulation
normalcdf
P(X ≥ 1520), Using Binomial
Distribution...
• binomcdf: P( X ≥ 1520) =
P( X ≥ 1520) = 1 – binomcdf (2500, 0.60, 1519)
= 1 – 0.7869 = 0.2131; context, always
Why is our x value 1519? Discuss.
P(X ≥ 1520), Using Simulation…
n = 2500; p = 0.60
• How would you suggest using a simulation to
model this situation and calculate a
probability?
• Discuss for 2 minutes, then share out
P(X ≥ 1520), Using Normal ≈ to
Binomial Distribution...
• The binomial distribution ≈ N (μ, σ) due to
np ≥ 10 & n(1-p) ≥ 10 (expected values for
success & failures both ≥ 10)
• Calculate μ and σ
μ =1500
σ = 24.49
P(X ≥ 1520), Using Normal ≈ to
Binomial Distribution...
• Binomial ≈ N (μ, σ) = N(1500, 24.49)
• P (X ≥ 1520) = P (X > 1520) (in Normal setting)
normalcdf (1520, hi, 1500, 24.49)
= 0.2070; context, always
(can also use table A... but lots of work)
Normal ≈ to Binomial Distribution...
Histogram: B(2500, 0.60)
Density Curve: N(np, √np(1-p))= N(1500, 24.49)
Normal ≈ to Binomial Distribution...
• binomcdf = 0.2131
• normalcdf = 0.2070
• Table A, simulation – different but similar
• All are acceptable answers; clearly explain
your process, justify work, answer in context,
etc.
Normal ≈ to Binomial Distribution...
• B (n, p) 
n large  ≈ Normal (np, √np(1-p))
• Rule of thumb: can use Normal approximation
when np ≥ 10 & n(1-p) ≥ 10
• Accuracy of Normal ≈ improves as n increases
• Most accurate for any fixed n when p ≈ 0.50
• Least accurate when p is near 0 or 1 (skewed
distribution)
The Geometric Distribution...
• Remember binomial distribution
• BINS
• Geometric: BI_S; no ‘n;’ no fixed number of
trials
• In Geometric setting, it is the number of trials
required to obtain first success
Geometric Distribution...
geometpdf (p, X)
(exactly) X =
first success will happen
on ___ trial
geometcdf (p, X)
(cumulatively) X ≤ first success will happen
at most on ___ trial
Geometric Distribution ...
• Rolling a die until a 5 appears
• BI_S?
•
•
•
•
it either is a 5 or not
rolls are independent
no fixed n
p = 1/6
Geometric Distribution...
What is the probability we roll exactly 3 times to get
our first 5?
P(X = 3) = geometpdf (1/6, 3) = 0.1157, context
What is the probability we roll exactly 100 times to
get our first 5?
P(X = 100) = geometpdf (1/6, 100) ≈ 0, context
What is the probability we roll exactly 6 times to get
our first 5?
P(X = 6) = geometpdf (1/6, 6) = 0.0670, context
Geometric Distribution...
What is the probability it takes us up to 3 times to
get our first 5?
P(X ≤ 3) = geometcdf (1/6, 3) = 0.4213, context
... up to 100 times....
P (X ≤ 100) = geometcdf (1/6, 100) = 0.9999, context
... up to 6 times....
P (X ≤ 6) = geometcdf (1/6, 6) = 0.6651, context
Geometric Distribution...
What is the probability it takes us more than 3 rolls to get
our first 5?
P(X > 3) = 1 - geometcdf (1/6, 3)
= 1 - 0.4213 = 0.5787, context
... more than100 times....
P (X > 100) = 1 - geometcdf (1/6, 100)
= 1 - 0.9999 = 0.0001 ≈ 0, context
... more than 6 times....
P (X > 6) = 1 - geometcdf (1/6, 6)
= 1 - 0.6651 = 0.3349, context
Mean & SD for Geometric
Distributions...
Mean = Expected Value = μX = E (X)= 1/p
Example: What is the theoretical probability of
rolling a 2?
1/6 ≈ 0.1666
E(X) = 1/p = 1/(0.16666) = 6; makes sense; we
would expect it to take (on average, in the long
run) six rolls of a die to roll a 2.

Mean & SD for Geometric
Distributions...
Standard deviation =
1 p
p2
What is the standard deviation of this
distribution?

SD =
1 p
p2

= 1  0.1666
2
0.1666
= 0.8975
Homework
Remember, always use technology (your
calculator) to calculate probabilities
Page 410, #69, 71, 73, 75, 77, 79, 81, 83, 85, 87,
89, 91, 95, 97
MC, FRQs, FRAPPY, Chapter Review, AP Practice
Test, etc.