Download X - Brocklehurst-13SAM

Statistics and Modelling Course
2011
Topic: Sample statistics &
expectation
Part of Achievement Standard 90643
Solve straightforward problems
involving probability
4 Credits
Externally Assessed
NuLake Pages 147163
Sigma: Old version – Ch 2.
New version – Ch 7.
LESSON 1 – Probability
distribution
Points of today:
 Learn the meaning of discrete and continuous random variables.
 Use a probability distribution table to display outcomes.
 What is Expected Value and how do you calculate it?
1.
2.
3.
4.
Notes on discrete & continuous random variables.
Do old Sigma (2nd edition) – p23 – Ex. 2.1
Notes on probability distribution tables (handout to fill in).
Introduction to Expected Value.
Discrete and Continous data
Discrete
Continuous
Discrete and Continous data
Discrete
Where does its
value come from?
Continuous
Discrete and Continous data
Discrete
Where does its
value come from?
Counting
Continuous
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
Discrete and Continous data
Where does its
value come from?
What values can it
take?
Discrete
Continuous
Counting
Measurement
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
All real numbers
(anywhere along the
number line – infinite
precision)
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
What question is
being asked?
All real numbers
(anywhere along the
number line – infinite
precision)
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
What question is
being asked?
‘How many ?’
All real numbers
(anywhere along the
number line – infinite
precision)
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
All real numbers
(anywhere along the
number line – infinite
precision)
What question is
being asked?
‘How long ?’, ‘ How
heavy ?’
‘How many ?’
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
All real numbers
(anywhere along the
number line – infinite
precision)
What question is
being asked?
‘How long ?’, ‘ How
heavy ?’
Examples:
‘How many ?’
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
All real numbers
(anywhere along the
number line – infinite
precision)
What question is
being asked?
‘How long ?’, ‘ How
heavy ?’
Examples:
‘How many ?’
• Number of students
who gained Excellence in
last test.
• Money – if to the
nearest dollar/cent.
Discrete and Continous data
Where does its
value come from?
Discrete
Continuous
Counting
Measurement
What values can it Whole numbers or
take?
rounded values
All real numbers
(anywhere along the
number line – infinite
precision)
What question is
being asked?
‘How many ?’
‘How long ?’, ‘ How
heavy ?’
• Number of students
who gained Excellence in
last test.
• Money – if to the
nearest dollar/cent.
• Height
• Distance
• Weight
• Volume
• Time
Examples:
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values.
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured.
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
_______ random variable; Possible values: __________________
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: __________________
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
_______ random variable; Possible values: __________________
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
Discrete random variable; Possible values: __________________
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
Discrete random variable; Possible values: 0, 1, 2,……, 999, 1000
E.g. The volume of tomato sauce in a bottle – varies slightly.
__________random variable; Values will be ________________.
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
Discrete random variable; Possible values: 0, 1, 2,……, 999, 1000
E.g. The volume of tomato sauce in a bottle – varies slightly.
Continuous random variable; Values will be ________________.
Random variables
A random variable is a variable (e.g. weight of apples) that
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
different values taken by a random variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
Discrete random variable; Possible values: 0, 1, 2,……, 999, 1000
E.g. The volume of tomato sauce in a bottle – varies slightly.
Continuous random variable; Values will be positive real numbers.
Random variables
Do Sigma p23 – Ex. 2.1 – Just Q19, and
A random variable is a variable (e.g. weight of apples) that
13 if you’re fast.
can take a range of different values. Its value is the
outcome being measured. We can assign probabilities to
Indifferent
new Sigma:
p121
– Ex.
– all
qs.
values
taken
by a7.01
random
variable.
E.g. Number of days a randomly-selected student was late for school
last week.
Discrete random variable; Possible values: 0, 1, 2, 3, 4, 5.
E.g. A poll of 1000 voters to see who favours John Key as P.M.
Discrete random variable; Possible values: 0, 1, 2,……, 999, 1000
E.g. The volume of tomato sauce in a bottle – varies slightly.
Continuous random variable; Values will be positive real numbers.
Probability Distributions - fill in notes on handout
Probability Distributions
Survey: How many ‘children’ in your family?
Let x represent ‘number of children in family’.
Probability Distributions
Survey: How many ‘children’ in your family?
Let x represent ‘number of children in family’.
x
Frequency
Relative
frequency

f
n
1
2
3
4
5
6
7
Survey: How many ‘children’ in your family?
Let x represent ‘number of children in family’.
x
1
2
3
4
5
6
7
Frequency
Relative
frequency

f
n
Calculate the mean number of kids per family in this class…
And: probability = long-run relative frequency.
So we can draw a Probability Distribution table from this data:
x
P(X = x)
1
2
3
4
5
6
7
X : variable – the
outcome from a given
trial.
x : values that X can
take.
Survey: How many ‘children’ in your family?
Let x represent ‘number of children in family’. .
x
1
2
3
4
5
6
7
Frequency
Relative
frequency

f
n
And: probability = long-run relative frequency.
So we can draw a Probability Distribution table from this data:
x
P(X = x)
1
2
3
4
5
6
7
X : variable – the
outcome from a given
trial.
x : values that X can
take.
Properties of a probability distribution :
1.For each value x, probabilities must be between 0&1:
2.The probabilities must add to exactly 1 (100%):
n
0 < P(xi) < 1
 P( x )  1
i 1
i
And: probability = long-run relative frequency.
So we can draw a Probability Distribution table from this data:
x
1
2
3
4
5
6
P(X = x)
7
X : variable – the
outcome from a given
trial.
x : values that X can
take.
Properties of a probability distribution :
1.For each value x, probabilities must be between 0&1: 0 < P(xi) < 1
n
2.The probabilities must sumto exactly 1 (100%):
 P( xi )  1
i 1
Expected Value
-
E(X)
The Expected Value E(X) is what you’d expect something to average out
to in the long run.
x
P(X = x)
1
2
3
4
5
6
7
X : variable – the
outcome from a given
trial.
x : values that X can
take.
Properties of a probability distribution :
1.For each value x, probabilities must be between 0&1: 0 < P(xi) < 1
n
2.The probabilities must add to exactly 1 (100%):
 P( xi )  1
i 1
Expected Value - E(X)
The Expected Value E(X) is what you’d expect something to average out
to in the long run.
We can get the mean by calculating proportions (relative frequencies).
Hence we can get the mean by calculating probabilities.
x
P(X = x)
1
2
3
4
5
6
7
X : variable – the
outcome from a given
trial.
x : values that X can
take.
Properties of a probability distribution :
1.For each value x, probabilities must be between 0&1: 0 < P(xi) < 1
n
2.The probabilities must add to exactly 1 (100%):
 P( xi )  1
i 1
Expected Value - E(X)
The Expected Value E(X) is what you’d expect something to average out
to in the long run.
We can get the mean by calculating proportions (relative frequencies).
Hence we can get the mean by calculating probabilities.
In probability, since we are looking at what would happen rather than
what has happened, we call the mean the Expected Value.
i.e. Mean = E(X)
Expected Value - E(X)
The Expected Value E(X) is what you’d expect something to average out
to in the long run.
We can get the mean by calculating proportions (relative frequencies).
Hence we can get the mean by calculating probabilities.
In probability, since we are looking at what would happen rather than
what has happened, we call the mean the Expected Value.
i.e. Mean = E(X)
To calculate the Expected Value:
1. Multiply each possible value that X can take by its
probability of occurring.
2. The sum of all of these gives the Expected Value of X.
RULE:
E[X] =
n
 x
i 1
i
. P( xi )

We can get the mean by calculating proportions (relative frequencies).
Hence we can get the mean by calculating probabilities.
In probability, since we are looking at what would happen rather than
what has happened, we call the mean the Expected Value.
i.e. Mean = E(X)
To calculate the Expected Value:
1. Multiply each possible value that X can take by its
probability of occurring.
2. The sum of all of these gives the Expected Value of X.
n
RULE:
E[X] =
 x
i 1
i
. P ( xi )

Example: I pick a member of this class at random. Use the
formula above to find the expected number of children in that
person’s family (including the person chosen).
Write your answer in your book. Show working using the formula
above.
i.e. Mean = E(X)
To calculate the Expected Value:
HW: Do NuLake pg. 154  156
1. Multiply each possible value that X can take by its probability of
occurring.
(Q154

156).
Finish
for
HW.
2. The sum of all of these gives the Expected Value of X.
n
RULE:
E[X] =
 x
i 1
i
. P ( xi )

Example: I pick a member of this class at random. Use the
formula above to find the expected number of children in that
person’s family (including the person chosen).
Write your answer in your book. Show working using the formula above.
x
P(X = x)
1
2
3
4
5
6
7
LESSON 1A (if time) – Expected
Value applications
Points of today:
 Calculate expected values.
 Calculate expected gains / losses when there is a price/reward
associated with the outcomes.
1.
2.
Go over a couple of the HW problems – NuLake p154 & 156.
Go through expected gain problem on the following slides
3. More practice of basics? Do Sigma (old) p25 – Ex. 2.2.
OR
Got it. Ready to do some more applications problems? Do Sigma (old) p26 –
Ex. 2.3 – complete for HW.
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Let x: expected winnings form 1 ticket.
List the four possible outcomes for the buyer of one ticket
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Let x: expected winnings form 1 ticket.
The four possible outcomes for the buyer of one ticket are:
Win $2000
Win $500 Write down the amount the buyer wins each case.
Win $100
Lose
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Let x: expected winnings form 1 ticket.
The four possible outcomes for the buyer of one ticket are:
Win $2000
Win $500
Win $100
Lose
Winnings= 2000
500
100
0
Express this as a probability distribution in a table.
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Let x: expected winnings form 1 ticket.
The four possible outcomes for the buyer of one ticket are:
Winnings = 2000
500
100
0
Win $2000
Win $500
Win $100
Lose
Buyer’s
winnings, x
P(X = x)
2000
500
100
1
1300
1
1300
1
1300
0
1297
1300
As there 3 winning tickets, the number of losing tickets is 1297
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Buyer’s
winnings, x
P(X = x)
2000
500
100
1
1300
1
1300
1
1300
0
1297
1300
Write down an expression for the expected value.
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
tickets, and assuming all tickets are sold, what should the organiser
charge for each ticket if the lottery is to be ‘fair’?
Buyer’s
winnings, x
P(X = x)
2000
500
100
1
1300
1
1300
1
1300
0
1297
1300
E( X) 
If the lottery is fair the ticket price must be equal
to the expected winnings per ticketet
i.e. on average, nobody gains and nobody loses
So the tickets should sell for $___
7.03B
A lottery has three prizes: $2000, $500 and $100. If there are 1300
Need
practice
of thearebasics?
tickets,more
and assuming
all tickets
sold, what should the organiser
Do
Sigma
(old)
– pg.
– Ex.
charge
for each
ticket
if the25
lottery
is to 2.2
be ‘fair’?
2000 to do more
500 applied
100
0
Buyer’s
Got it. Now ready
problems like
winnings,
this one:x Do Sigma pg. 26 – Ex. 2.3.
1
1
1
1297
P(X = x)
1300
1300
1300
1300
E( X) 
If the lottery is fair the ticket price must be equal
to the expected winnings per ticketet
i.e. on average, nobody gains and nobody loses
So the tickets should sell for $2
LESSON 2 – Calculate the Variance from
a probability distribution table 1
Points of today:
 Calculate the expected value of a function of a variable.
 Learn how to calculate the variance from a probability distribution
table.
1.
2.
3.
4.
Go over a couple of the HW problems – NuLake pg. 154156 (focus on
application qs as this is the emphasis in NCEA).
Briefly go through how to calculate the expected value of a function of a
variable – examples on next slide.
Notes on how to calculate the variance of a random variable (most of lesson).
Do NuLake pg. 158161 (finish for HW).
A random variable X has E(X) = 9.
a
Calculate:
E(4X) = 4  E(X)
=49
= 36
b
E(X – 2) = E(X) – 2
=9–2
=7
c
E(2X + 3) = 2  E(X) + 3
=29+3
= 21
Do Sigma pg. 29 – Ex. 2.4 – 10 mins.
a E(4X)
b E(X – 2)
c E(2X + 3)
Variance
“The average of the squared distances from the
mean.”
Var(X)
 ( x  x ) 
2
=
n
Var(X)
E[(X – μ)2]
=
The same as saying the
“Expected” squared distance
from the population mean.
This can be re-arranged to get
Var(X)
=
E(X2)
–
[E(X)]
2
Variance
“The average of the squared distances from the
mean.”
Var(X)
 ( x  x ) 
2
=
n
Var(X)
E[(X – μ)2]
=
The same as saying the
“Expected” squared distance
from the population mean.
This can be re-arranged to get
Var(X)
=
E(X2)
–
μ
2
Variance
“The average of the squared distances from the
mean.”
Var(X)
 ( x  x ) 
2
=
n
Var(X)
E[(X – μ)2]
=
The same as saying the
“Expected” squared distance
from the population mean.
This can be re-arranged to get
Var(X)
=
E(X2)
–
[E(X)]
2
Calculate the mean, variance and standard deviation for this
Do NuLake
pg. 158
the
161
probability
distribution.
Use
formula VAR(X) = E(X2) – µ2
for the variance.
7.05D
(Q164176). Finish for HW.
x
P(X = x)
10
0.18
20
0.32
30
0.15
40
0.35
Calculate
μ = E(X) = 10  0.18 + 20  0.32 + 30  0.15 +
40  0.35E[X].
= 26.7 The mean, based on the above probability
distribution, is μ = 26.7.
Calculate
E(X2) = 102  0.18 + 202  0.32 + 302  0.15
+ 402 E[X
0.352].
= 18 + 128 + 135 + 560
= 841
Var(X) = E(X2) – µ2
Use VAR(X) = E(X2) – µ2
= 841 – (26.7)2
to calculate the variance.
= 128.11
The standard deviation, based on the
above probability
probability distribution, is
The variance, based on the above
Hence calculate SD(X).
128.11
 = Var(X ) =
distribution, is 128.11
= 11.32 (to 4SF) σ = 11.32 (to 4 S.F.).
LESSON 3 – Calculate the Variance from
a prob. distn.table 2
Point of today – practice!:
 Get confident at calculating the variance from a probability
distribution table.
1. Any Qs from the HW – NuLake – pg. 158161?
2. Another e.g. on board (following slides) for those needing it
(others carry on).
3. Do Sigma: pg. 33 (Ex. 2.5) in old, or pg. 135 (Ex. 7.05) in new:
MUST do Q 1 & 2, (* 35 are optional as extension).
4. Then on to next exercise (2.6 in old / 7.06 in new): Do all.
Complete for HW.
Variance
“The average of the squared distances from the
mean.”
Var(X)
 ( x  x ) 
2
=
n
Var(X)
E[(X – m)2]
=
The same as saying the
“Expected” squared distance
from the population mean.
This can be re-arranged to get
Var(X)
=
E(X2)
–
μ
2
Variance
“The average of the squared distances from the
mean.”
Var(X)
 ( x  x ) 
2
=
n
Var(X)
E[(X – m)2]
=
The same as saying the
“Expected” squared distance
from the population mean.
This can be re-arranged to get
Var(X)
=
E(X2)
–
[E(X)]
2
For the probability distribution table below:
CalculateNuLake
the mean E(X)
= m.158161 (ask for
1. 1.)Finish
pages
2.) Calculate m2
with any you’re stuck on.)
3.) Calculate E(X2).
4.) Explain in your own words why E(X2) ≠ m2
5.) Use the formula VAR(X) = E(X2) – µ 2
2. Doto Sigma
(old) pg. 33 – Ex. 2.5 – JUST
calculate the variance & standard deviation of X.
x
2. (* Q35 are optional as extension).
5
7
8
help
Q1 &
14
= x) on 0.2
0.4(pg. 34).0.3Finish for
0.1 HW.
3. P(X
Then
to Ex. 2.6
the above
+ 7mean,
 0.4based
+ 8 on
 0.3
+ 14 probability
0.1 Calculate E[X].
μ = E(X) = 5  0.2 The
distribution, is μ = 7.6. So μ2 = 7.62 = ____
= 7.6
2  above
The
based
E(X
) Calculate
= 52  0.2
+on2].
7the
0.4 probability
+ 82  0.3 + 142  0.1
Now2variance,
E[X
2 = 5.64
distribution,
is
σ
= 63.4
2) – µ2
Var(X) = E(X2) – µ2
Use
VAR(X)
=
E(X
= 63.4 – 7.62
to calculate
the
variance.
The
standard
deviation,
based
on the
= 5.64
above probability distribution, is
Now calculate σ
5.64
(4 S.F.).
Standard Deviation)
= 2.375 (4 S.F.) σ = 2.375(the
  Var(X ) 
Variance formula proof using Expectation:
Link between the 2 variance formulas: Prove that
Left hand side =
E( X  m )2 = E ( X 2 )  m 2
E( X  m )2
= E ( X 2  2 X .m  m 2 )
= E ( X 2 )  E (2 Xm )  E ( m 2 )
=
E( X 2 )  2E( X )m  m 2
=
E ( X 2 )  2m 2  m 2
=
E( X 2 )  m 2
=
Right hand side
Since E(m) = m. i.e. the expected
value of the mean is the mean.
Since E(X) = m

LESSON 4 – Variance of a
function of a variable
Point of today:
 How to calculate the variance (and standard deviation) of a
function of a variable.
1.
2.
3.
Notes & examples.
Do NuLake pg. 164 & 166.
Sigma (old – 2nd edition) – pg. 36: Ex. 2.7.
Linear Functions of Random Variables
Sometimes we deal with random variables whose behaviour
is modelled by a linear function:
Y = aX +c
(like y = mx + c)
E.g. A taxi service charges $3 per kilometer travelled
plus a flat fee of $5.
If the distance, X, required for the next job is a random
variable, then the price, Y, is given by Y = 3X+5
Its mean is given by
E(3X + 5) =
3 × μ + 5
Linear Functions of Random Variables
Sometimes we deal with random variables whose behaviour
is modelled by a linear function:
Y = aX +c
E.g. A taxi service charges $3 per kilometer travelled
plus a flat fee of $5.
If the distance, X, required for the next job is a random
variable, then the price, Y, is given by Y = 3X+5
Its mean is given by
Its variance,
Its std. deviation,
E(3X + 5) =
Var(3X+5) =
σ3X+5
=
3 × E(X) + 5
32 × Var(X)
32 Var ( X )
Linear Functions of Random Variables
E.g. A taxi service charges $3 per kilometer travelled plus a flat fee of $5.
If the distance, X, required for the next job is a random variable, then the price,
Y, is given by Y = 3X+5
Its mean is given by
E(3X + 5) = 3 × E(X) + 5
32 × Var(X)
Its variance,
Var(3X+5) =
Its std. deviation,
σ3X+5
=
32 Var ( X )
RULES: Linear Function, Y, of a Random
Variable, X
Y = aX + c
Its mean E(aX+c)
=
a × E(X) + c
Its variance Var(aX+c)
=
a2 × Var(X)
Its std. deviation σaX+c
=
a 2Var ( X )
Linear Functions of Random Variables
RULES: Linear Function, Y, of a Random
Variable, X
Y = aX + c
Its mean E(aX+c)
=
a × E(X) + c
Its variance Var(aX+c)
=
a2 × Var(X)
Its std. deviation σaX+c
=
2
a Var ( X )
E.g. The taxi service needs to estimate its earnings per job.
Calculate the mean, variance and standard deviation of the amount
charged for one job, if the mean distance for a job is 8km, with a
variance of 6.25.
Linear Functions of Random Variables
RULES: Linear Function, Y, of a Random
Variable, X
Y = aX + c
Its mean E(aX+c)
=
a × E(X) + c
Its variance Var(aX+c)
=
a2 × Var(X)
Its std. deviation σaX+c
=
2
a Var ( X )
E.g. The taxi service needs to estimate its earnings per job.
Calculate the mean, variance and standard deviation of the amount
charged for one job, if the mean distance for a job is 8km, with a
variance of 6.25.
Remember the taxi service charges $3 per kilometer plus a flat fee of $5.
E.g. The taxi service needs to estimate its earnings per job.
Calculate the mean, variance and standard deviation of the amount
charged for one job, if the mean distance for a job is 8km, with a
variance of 6.25.
Remember the taxi service charges $3 per kilometer plus a flat fee of $5.
E.g. The taxi service needs to estimate its earnings per job.
Calculate the mean, variance and standard deviation of the amount
charged for one job, if the mean distance for a job is 8km, with a
variance of 6.25.
Remember the taxi service charges $3 per kilometer plus a flat fee of $5.
Let
X: Distance for a job.
Y: Price of a job.
Then Y = 3X + 5
So the mean price charged
for a job is $29.
μY = E(Y) =E(3X + 5) = 3× 8 + 5
= $29
σ2Y = Var(Y) = Var(3X+5) =
•Do NuLake pg. 164-166
•Then Sigma (old) – pg. 36
Further extension:
Sigma (old): Ex. 2.7
σY
32  6.25
= 56.25 So
=
56.25
= $7.50
the random variable, Y
(price charged) has a
variance of 56.25
and a std. deviation of
$7.50.
LESSONS 5 & 6 – Sums & differences of
2 random variables
Point of today:
 How to calculate the mean, variance (and standard deviation) of
the sum of 2 independent random variables.
 How to calculate the mean, variance (and standard deviation) of
the difference between 2 independent random variables.
1.
2.
3.
4.
5.
6.
Warm-up quiz (re-cap from last lesson – function of a random variable).
Examples & notes.
Do NuLake pg. 168 (just Q187 & 188)
Then do Sigma (old – 2nd edition) – pg. 41: Ex. 2.9 (skip Q2).
Do Sigma (old – 2nd edition) – pg. 42 & 43: Ex. 2.10.
Extension: NuLake p169, 170: Questions 189-194 only.
RULES: Linear Function, Y, of a Random
Variable, X
Y = aX + c
Its mean E(aX+c)
=
a × E(X) + c
Its variance Var(aX+c)
=
a2 × Var(X)
Its std. deviation σaX+c
=
a 2Var ( X )
WARM-UP QUIZ: X is a random variable with mean 20 and variance 2 = 16.
Calculate the mean, variance and standard deviation of Y, if:
a
Y = 6X
b
Y = 3X -4
c
Y = -X
WARM-UP QUIZ: X is a random variable with mean 20 and variance 2 = 16.
Calculate the mean, variance and standard deviation of Y, if:
a
Y = 6X
b
Y = 3X -4
c
Y = -X
Simply multiply E(X) by 6.
μ = E(Y) = E(6X) = 6  E(X)
Y
= 6  20
= 120
= 62  Var(X) Square the coefficient of X.
= 36  16
= 576
σ2Y = Var(Y) =Var(6X)
σY =
=
576
24
Square root of the variance.
WARM-UP QUIZ: X is a random variable with mean 20 and variance 2 = 16. Calculate
the mean, variance and standard deviation of Y, if:
a
b
c
Y = 6X
Y = 3X -4
Y = -X
μY = E(Y) = E(3X - 4) = 3 E(X) - 4
= 320 - 4
Multiply E(X) by 3 and
subtract 4.
= 56
σ Y =Var(Y) =Var(3X - 4)
2
=  Var(X)
= 9  16
32
Square the coefficient
of X. IGNORE THE
-4
= 144
σY =
=
144
12
Square root of the variance.
7.07
WARM-UP QUIZ: X is a random variable with mean 20 and variance 2 = 16. Calculate
the mean, variance and standard deviation of Y, if:
a
b
c
Y = 6X
Y = 3X -4
Y = -X
μY = E(Y) = E(-X) = -1  E(X)
Simply multiply E(X) by -1.
= -20
σ2Y = Var(Y) =Var(-X)
= Var(X)
= 16
σY =
=
16
4
This has no effect on the
SPREAD of the data.
So of course the standard
deviation is also unchanged.
SUMS OF 2 INDEPENDENT RANDOM
VARIABLES
Distribution of X + Y
Its mean E(X + Y)
=
E(X) + E(Y)
Its variance Var(X + Y)
=
Var(X) + Var(Y)
Its std. deviation σX+Y
=
Var ( X )  Var (Y )
E.g. 1. The mean weight of Year 13 males at a school is 72Kg with a standard
deviation of 5Kg. The mean weight of Year 13 females at the same school is 56Kg
with a standard deviation of 4Kg.
One boy and one girl are chosen at random and their individual weights are added
together.
What would be the mean, variance and standard deviation of their combined
weight?
E.g. 1. The mean weight of Year 13 males at a school is 72Kg with a
standard deviation of 5Kg. The mean weight of Year 13 females at the same
school is 56Kg with a standard deviation of 4Kg.
One boy and one girl are chosen at random and their individual weights are
added together.
What would be the mean, variance and standard deviation of their combined
weight?
Let T: Combined Weight;
X: Weight of a randomly chosen boy.
Y: Weight of a randomly chosen girl.
Where T = X + Y
μT = E(T) = E(X+Y) = E(X) + E(Y) Var(T)
 72  56
 128kg
= Var(X)+Var(Y)
= 52 + 42
Always work through the variances
rather than the standard deviations.
We square the boys’ and girls’ standard
deviations to get their variances.
Assumption: That the marks for the 2 tests are independent.
Then we can add the variances.
E.g. 1. The mean weight of Year 13 males at a school is 72Kg with a
standard deviation of 5Kg. The mean weight of Year 13 females at the same
school is 56Kg with a standard deviation of 4Kg.
One boy and one girl are chosen at random and their individual weights are
added together.
What would be the mean, variance and standard deviation of their combined
weight?
Let T: Combined Weight;
X: Weight of a randomly chosen boy.
Y: Weight of a randomly chosen girl.
Where T = X + Y
μT = E(T) = E(X+Y) = E(X) + E(Y) Var(T)
 72  56
 128kg
= Var(X)+Var(Y)
= 52 + 42
= 41
 T  Var(T )  41 = 6.403kg (4sf)
So, assuming that the weights of boys and girls are independent, we
would expect the combined weight of a randomly chosen boy & girl
to be 128kg, with a standard deviation of 6.403kg (4sf).
DIFFERENCE BETWEEN 2 INDEPENDENT
RANDOM VARIABLES
Distribution of X – Y
Its mean E(X - Y)
=
E(X) - E(Y)
Its variance Var(X - Y)
=
Var(X)
Its std. deviation σX-Y
=
Var ( X )  Var (Y )
+ Var(Y)
We SUBTRACT the MEANS, but still ADD the VARIANCES.
Its mean E(X - Y)
=
E(X) - E(Y)
Its variance Var(X - Y)
=
Var(X)
Its std. deviation σX-Y
=
Var ( X )  Var (Y )
+ Var(Y)
We SUBTRACT the MEANS, but still ADD the VARIANCES.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60 g
and standard deviation 4 g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60 g
and standard deviation 4 g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60g
and standard deviation 4g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
Let the random variables involved be
X = volume of tin contents Write an expression for
Y = volume on spoon
the volume of food left in
the tin once the spoonful is
removed.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60g
and standard deviation 4g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
Let the random variables involved be
X = volume of tin contents
Y = volume on spoon
X–Y = volume of cat food remaining in tin.
Calculate the expected volume of cat food remaining – i.e. E[X–Y].
E(X–Y) = E(X) – E(Y)
= 500 – 60
= 440 g
Mean = expected value
So the mean volume
remaining is 440g.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60g
and standard deviation 4g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
Let the random variables involved be
X = volume of tin contents
Y = volume on spoon
X–Y = volume of cat food remaining in tin.
Calculate Var(X–Y).
E(X–Y) = E(X) – E(Y)
= 500 – 60
= 440 g
Mean = expected value
So the mean volume
remaining is 440g.
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60g
and standard deviation 4g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
Let the random variables involved be
X = volume of tin contents
Y = volume on spoon
X–Y = volume of cat food remaining in tin.
Var(X–Y) = Var(X) + Var(Y)
E(X–Y) = E(X) – E(Y)
= 500 – 60
= 440 g
Mean = expected value
So the mean volume
remaining is 440g.
= 62 + 42
= 52
Calculate σX–Y
7.09
The contents of a tin of cat food have a distribution with mean 500 g
and standard deviation 6 g. A spoon is used to remove the food from
the tin. The amount of food removed has a distribution with mean 60g
and standard deviation 4g. Calculate the mean and standard deviation
of the amount of food left in the tin after one spoonful is removed.
Let the random variables involved be
•Do NuLake pg. 169 (Q187, 188 only)
X = volume of tin contents •Then Sigma (old) – pg. 41
Y = volume on spoon
Ex. 2.9 (skip Q2) – FINISH FOR HW
X–Y = volume of cat food remaining in tin.
Var(X–Y) = Var(X) + Var(Y)
E(X–Y) = E(X) – E(Y)
= 500 – 60
= 440 g
Mean = expected value
So the mean volume
remaining is 440g.
= 62 + 42
= 52
σX–Y = 52
= 7.211 g (4 sf)
So the standard deviation of the
remaining volume is 7.211g (4SF).
Extension work– Linear combinations of 2
random variables
Point of today:
 How to calculate the mean, variance (and standard deviation) of a
linear combination of 2 independent random variables.
1.
2.
3.
4.
Go over HW qs.
Notes & examples.
Do Sigma (old – 2nd edition) – pg. 42 & 43: Ex. 2.10.
Extension: NuLake p169 & 170: Questions 189-194.
DO NOT DO Q195 on (not covered yet).
LINEAR COMBINATIONS OF
INDEPENDENT RANDOM VARIABLES
Distribution of aX + bY
Its mean E(aX + bY) =
a.E(X) + b.E(Y)
Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)
Its std. deviation σaX+bY
=
a 2Var ( X )  b 2Var (Y )
X and Y are independent random variables. E(X) = 20, E(Y) =30.
Var(X) = 3 and Var(Y) = 4.
Calculate: a E(3X + Y)
b Var(3X + Y)
c E(4X – Y)
d Var(4X – Y)
LINEAR COMBINATIONS OF
INDEPENDENT RANDOM VARIABLES
Distribution of aX + bY
Its mean E(aX + bY) =
a.E(X) + b.E(Y)
Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)
Its std. deviation σaX+bY
7.10
=
a 2Var ( X )  b 2Var (Y )
E.g. A battery retailer sells surplus trade-in batteries to a scrap-metal
manufacturer. The retailer receives $50 per kg for the lead content
and $20 per litre for the sulphuric acid content of each battery. For a
battery, the quantities of recoverable lead and sulphuric acid are
independent random variables, with standard deviations 30 g (0.03 kg)
and 10 mL (0.01 L) respectively. What is the standard deviation of
the amount received for each battery?
7.10
E.g. A battery retailer sells surplus trade-in batteries to a scrap-metal
manufacturer. The retailer receives $50 per kg for the lead content
and $20 per litre for the sulphuric acid content of each battery. For a
battery, the quantities of recoverable lead and sulphuric acid are
independent random variables, with standard deviations 30 g (0.03 kg)
and 10 mL (0.01 L) respectively. What is the standard deviation of
the amount received for each battery?
7.10
E.g. A battery retailer sells surplus trade-in batteries to a scrap-metal
•Do Sigma
pg.
42 &$5043:
manufacturer.
The(old)
retailer
receives
per Ex.
kg for2.10
the lead content
and $20 per litre for the sulphuric acid content of each battery. For a
battery,
the quantities
of recoverable
lead and sulphuric
•Extension:
Once
you’ve finished
this: acid are
independent random variables, with standard deviations 30 g (0.03 kg)
Do NuLake p169170 – Q189-194.
and 10 mL (0.01 L) respectively. What is the standard deviation of
the amount received for each battery?
Let X be the content of lead in kg
Y be the content in litres of acid recoverable from each battery
50X+20Y gives the total refund.
Var(50X+20Y) =
σ50X +20Y
502Var(X) + 202Var(Y)
Calculate Var(50X + 20Y).
= 2500 Var(X) + 400 Var(Y)
= 2500  0.032 + 400  0.012
= 2.29
= 2.29
Calculate σ50X + 20Y
= $1.51 (nearest cent)
LESSON 7 – Difference between Sums and
Linear Combinations of independent random
variables
To do today:
 STARTER: HANDOUT TO GO WITH IT - Sigma coin-tossing
investigation (new – Ex. 7.08, old – Ex. 2.8) + Chicken examples.
 Finish Sigma Ex. 2.10 (7.10 in new).
 NuLake pg. 169, 170 – Q188194 only.
 Then past Probability exams (AS90643).
Linear functions of a random variable vs sums of identical variables
EXAMPLE 1:
Consider tossing a fair six-sided die. X is the number on the top face.
1.
Calculate the mean and variance of X.
x
1
P(X = x)
1
6
3.5
m  X) = __________
2
1
6
3
4
5
1
6
1
6
1
6
6
1
6
1
E(X2) = __________
6
15
E( X 2 )  m 2
Var(X) = __________________________________ (write down the formula)
1
2
15

3
.
5
= __________________________________
(sub in the values)
6
2.9166
= _________________
(answer)
Linear functions of a random variable vs sums of identical variables
x
1
2
3
4
5
P(X = x)
1
6
1
6
1
6
1
6
1
6
3.5
m  X) = __________
6
1
6
1
15
E(X2) = __________
6
E( X 2 )  m 2
Var(X) = __________________________________ (write down the formula)
1
15  3.5 2
= __________________________________
(sub in the values)
6
2.9166
= _________________ (answer)
These two situations are different:
• Doubling the number on the top face (case 1),
• Tossing the die twice and adding the two results together (case 2).
Linear functions of a random variable vs sums of identical variables
3.5
m  X) = __________
1
15
E(X2) = __________
6
E( X 2 )  m 2
Var(X) = __________________________________ (write down the formula)
1
15  3.5 2
= __________________________________
(sub in the values)
6
2.9166
= _________________ (answer)
These two situations are different:
• Doubling the number on the top face (case 1),
• Tossing the die twice and adding the two results together (case 2).
2. a)Calculate the variance in case 1. b) Calculate the variance in case 2.
The values of X are all doubled.
So there is still just one variable X.
This is just a linear function of it.
i.e. The
distribution
of 2X.
Here there are two independent variables:
X1: Result of 1st toss; X2: Result of 2nd
i.e. The
Hence Var(2X) = 22 Var(X)
distribution
2
= 2 (2.91667) of X +X :
1
2
= 11.67 (4sf)
Var(X1+X2) = Var(X1)+Var(X2)
= 2.91667+ 2.91667
= 5.833 (4sf)
Linear functions of a random variable vs sums of identical variables
EXAMPLE
2: this
The mean
weightthen
of chicken
thighs ispg.
200g
with
standard
Copy
example
do NuLake
169
& a170:
deviation
of 10g.
Q188194.
NOTE: Stop after Q194.
Chicken thighs are sold for 3c per gram.
What is the difference between the following two questions?
Question A:
Calculate the mean and standard deviation
of the price of one chicken thigh.
Question B:
Calculate the mean and standard deviation
of the total weight of 3 chicken thighs.
LESSON 8 – Revision of Achievement Std
90643 - Probability
Point of today:
 Work through Merit & Excellence level problems from all parts of
the probability topic – continue in preparation for test (next
lesson). Sort out any areas of weakness.
1.
2.
Work through NuLake practice assessment for Probability (pg. 172-174)
Continue for HW as study for test.
Past NCEA papers (AS90643): 2006, 2007.
TEST
FORMATIVE
ASSESSMENT
(Achievement Std 90643)