Transcript X - Brocklehurst-13SAM

Topic 5: Probability Distributions
Achievement Standard 90646
Solve Probability Distribution
Models to solve straightforward
problems
4 Credits
Externally Assessed
NuLake Pages 278  322
NORMAL DISTRIBUTION
PART 2
Lesson 3: Making a continuity
correction
• Go over 1 of the final 2 qs from HW (combined events).
NuLake p303.
• How to calculate normal distribution probabilities using
your Graphics Calculator.
To practice using GC: Do Sigma (NEW – photocopy): p358
– Ex. 17.01 (Q3 only). Write qs on board as a quiz.
• Continuity corrections – how and when to make them.
Work: Fill in handout on cont. corr., then NuLake p309:
Q42-46. Finish for HW.
*Don’t do Q47.
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
=5
E.g. 1:
If =178, =5, P(175 X  184) = ?
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 =178 184
Using your Graphics Calc. for Standard
Normal problems
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
=5
E.g. 1:
If =178, =5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 =178 184
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
=5
E.g.1:
If =178, =5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 =178
=3.5
E.g.2:
If =30, =3.5, P(X  31) = ?
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
=30 31
• MENU, STAT, DIST,NORM; then there are three options:
* Npd – you will not have to use this option
* Ncd – for calculating probabilities
* InvN – for inverse problems
• NB: On your graphics calculator shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
=5
E.g.1:
If =178, =5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 =178
=3.5
E.g.2:
If =30, =3.5, P(X  31) = 0.61245
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
=30 31
• MENU, STAT, DIST,NORM; then there are three options:
10 minutes
* Npd – you will not have to use this option
Ncd – for
calculating probabilities
Do Sigma *(NEW)
p358.
Ex. 17.01* InvN – for inverse problems
NB:on
Onthe
yourboard
graphics
• •Q3
ascalculator
a quiz. shaded areas are from -∞ to
the point.
–
To enter -∞ you type – EXP 99.
–
To enter +∞ you type
EXP 99.
=5
E.g.1:
If =178, =5, P(175 X  184) = 0.61067
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
175 =178
=3.5
E.g.2:
If =30, =3.5, P(X  31) = 0.61245
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
184
=30 31
Making a Continuity Correction
(USE THE HANDOUT)
17.02A
Heights of Year 13 males in NZ are normally distributed with mean
174 cm and standard deviation 6 cm. If they are measured to the
nearest cm, calculate the probability that the height of a student
is more than 165 cm.
Actual curve for heights of students
(continuous)
Distribution when rounding heights
(discrete histogram)
For example the probability that a
student was 165 cm tall would
have to be represented by a
column with base 164.5 to 165.5
165
174
165
174
164.5 165 165.5
Heights of Year 13 males in NZ are normally distributed with mean
174 cm and standard deviation 6 cm. If they are measured to the
nearest cm, calculate the probability that the height of a student
is more than 165 cm.
Actual curve for heights of students
(continuous)
Distribution when rounding heights
(discrete histogram)
For example the probability that a
student was 165 cm tall would
have to be represented by a
column with base 164.5 to 165.5
165
174
165
174
164.5 165 165.5
164.5 165 165.5
Heights of Year 13 males in NZ are normally distributed with mean
174 cm and standard deviation 6 cm. If they are measured to the
nearest cm, calculate the probability that the height of a student
is more than 165 cm.
Actual curve for heights of students
(continuous)
Distribution when rounding heights
(discrete histogram)
For example the probability that a
student was 165 cm tall would
have to be represented by a
column with base 164.5 to 165.5
because any student with a
height in this interval would be
recorded as having a height of
165 cm.
165
174
165
174
164.5 165 165.5
164.5 165 165.5
Heights of Year 13 males in NZ are normally distributed with mean
174 cm and standard deviation 6 cm. If they are measured to the
nearest cm, calculate the probability that the height of a student
is more than 165 cm.
165
To find the cut-off point for
continuity corrections, move up or
down to the midpoint between two
whole-numbers.
174
In this example the wording is ‘more
than 165’, so move up to 165.5.
165
174
164.5 165 165.5
164.5 165 165.5
P(X > 165) = P(X > ?)
with continuity correction.
Heights of Year 13 males in NZ are normally distributed with mean
174 cm and standard deviation 6 cm. If they are measured to the
nearest cm, calculate the probability that the height of a student
is more than 165 cm.
165
To find the cut-off point for
continuity corrections, move up or
down to the midpoint between two
whole-numbers.
174
In this example the wording is ‘more
than 165’, so move up to 165.5.
165
174
164.5 165 165.5
164.5 165 165.5
P(X > 165) ≈ P(X > 165.5) with
≈ 0.9217 (4 sf)
continuity correction.
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
CONTINUOUS
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
CONTINUOUS
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
P(X > 6)
CONTINUOUS
P(3.5 < X < 4.5)
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
Continuity Correction
If we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
P(X > 5.5)
Continuity
Correction
Do NuLake qs on “Continuity Corrections for a
If we use aDistribution”:
Normal Distribution to approximate a variable that is
Normal
DISCRETE, we must make a Continuity Correction.
Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
P(X > 5.5)
P(X < 10)
P(X < 10)
P(8 < X < 12)
Continuity
Correction
Do NuLake qs on “Continuity Corrections for a
Normal
If we use aDistribution”:
Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
P(X > 5.5)
P(X < 10)
P(X < 9.5)
P(X < 10)
P(8 < X < 12)
Continuity
Correction
Do NuLake qs on “Continuity Corrections for a
Normal
If we use aDistribution”:
Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
P(X > 5.5)
P(X < 10)
P(X < 9.5)
P(X < 10)
P(X < 10.5)
P(8 < X < 12)
Continuity
Correction
Do NuLake qs on “Continuity Corrections for a
Normal
If we use aDistribution”:
Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
DISCRETE
P(X = 4)
CONTINUOUS
P(3.5 < X < 4.5)
P(X > 6)
P(X > 6.5)
P(X > 6)
P(X > 5.5)
P(X < 10)
P(X < 9.5)
P(X < 10)
P(X < 10.5)
P(8 < X < 12)
P(7.5 < X < 12.5)
Lesson 4: Inverse normal problems where
you are given the probability and asked to
calculate the x-value.
Learning outcome:
Calculate the x cut-off score based on
given probabilities, and a given mean and
SD.
Work:
1. Inverse calculations using standard normal.
2. Inverse calculations – standardising Examples
3. Do Sigma (new - photocopy): p366 – Ex. 17.03.
Inverse questions - the other way around
Where you’re told the probability and have to find
the z-values.
Examples:
(a) Find the value of z giving the area of 0.3770
between 0 and z.
0.377
0
z
z = ?
P(0 < Z < z) =
0.377
What is z ?
Answer (from tables):
z = ?
P(0 < Z < z) =
0.377
What is z ?
Answer (from tables):
z = 1.16
Inverse questions - the other way around
Where you’re told the probability and have to find
the z-values.
Examples:
(a) Find the value of z giving the area of 0.3770
between 0 and z.
0.377
0
z
z = 1.16
Inverse questions - the other way around
Where you’re told the probability and have to find
the z-values.
Examples:
(a) Find the value of z giving the area of 0.3770
between 0 and z.
0.377
0
(b)Find the value of z if the area
to the right of z is only 0.05.
z
z = 1.16
0.45
0.05
0
z
z = ?
P(0 < Z < z) =
0.45
What is z ?
Answer (from tables):
z = ?
P(0 < Z < z) =
0.45
What is z ?
Answer (from tables):
z = 1.645
Inverse questions - the other way around
Where you’re told the probability and have to find
the z-values.
Examples:
(a) Find the value of z giving the area of 0.3770
between 0 and z.
0.377
0
(b)Find the value of z if the area
to the right of z is only 0.05.
z
z = 1.16
0.45
0.05
0
z
z = 1.645
Inverse problems where you’re given the
probability,  and , and asked to find the
value of X.
z=x–

can be re-arranged to solve for x
x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
= 4.7
0.45
0.05
= 240
First
xcut-off
z find= ?the z cut-off
like in the last example
P(0 < Z < z) =
0.45
What is z ?
Answer (from tables):
z = ?
P(0 < Z < z) =
0.45
What is z ?
Answer (from tables):
z = 1.645
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
0.45
0.05
= 240
zcut-off
z = 1.645
xcut-off = ?
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
0.45
0.05
= 240
zcut-off
z = 1.645
xcut-off =
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
STAT, DIST, NORM, InvN
Area: _____ , σ: 4.7, μ: 24
0.45
0.05
= 240
Area: Enter total area
to the LEFT of xcut-off .
zcut-off
z = 1.645
xcut-off =
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
STAT, DIST, NORM, InvN
Area: 1-0.05 , σ: 4.7, μ: 24
0.45
0.05
= 240
Area: Enter total area
to the LEFT of xcut-off .
zcut-off
z = 1.645
xcut-off =
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
STAT, DIST, NORM, InvN
Area: 1-0.05 , σ: 4.7, μ: 24
= ____
0.45
0.05
= 240
Area: Enter total area
to the LEFT of xcut-off .
zcut-off
z = 1.645
xcut-off =
z=x–

can be re-arranged to solve for x x =  + z
E.g. A normally distributed random variable has a mean of 24 & std.
deviation of 4.7. What value has only 5% of the distribution above it?
i.e. P(X > xcut-off ) = 0.05.
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
STAT, DIST, NORM, InvN
Area: 1-0.05 , σ: 4.7, μ: 24
= 0.95
0.45
0.05
= 240
Area: Enter total area
to the LEFT of xcut-off .
zcut-off
z = 1.645
xcut-off =
z=x–

can be re-arranged to solve for x x =  + z
Once you’ve copied down the e.g. & working:
A normally
has a meanComplete
of 24 & std.for
Do E.g.
Sigma
(NEWdistributed
version):random
p366 variable
– Ex. 17.03
deviation of 4.7. What value has only 5% of the distribution above it?
HW.
i.e. P(X > xcut-off ) = 0.05.
Extension (after
you’ve finished this): NuLake p307 & 308
We’re told that  =24 and  =4.7. What is the value, xcut-off ?
First do using working (standardise it), then check with G.Calc.
STAT, DIST, NORM, InvN
Area: 1-0.05 , σ: 4.7, μ: 24
= 0.95
0.45
0.05
= 240
Area: Enter total area
to the LEFT of xcut-off .
zcut-off
z = 1.645
xcut-off = 31.73
Lesson 5: Inverse normal problems where
you must calculate the mean or SD
• Calculate the mean if given the SD and the
probability of X taking a certain domain of
values.
• Calculate the SD if given the mean and the
probability of X taking a certain domain of
values.
Sigma (new - PHOTOCOPY): p369, Ex 17.04.
STARTER:
Question from what we did last lesson:
Inverse Normal: Calculating the x cut-off score.
STARTER QUESTION
(from what we did last lesson)
Inverse normal question: A manufacturer of car tyres knows that her product has
a mean life of 2.3 years with a standard deviation of 0.4 years. Assuming that
the lifetime of a tyre is normally distributed what guarantee should she
offer if she only wants to pay out on 2% of tyres produced.
Solution:
Let X be a random variable representing the life of a tyre.
X is normal with μ = 2.3 and σ = 0.4
We want an x value such that P(X xcut-off ) = 0.02.
P( Z  z )  0.02
xcut off  2.3
z
0.4
gives a z value of -2.054 (see tables )
= -2.054
0.48
0.02
xcut-off
= ? yrs
xcut-off
= 1.48
So her guarantee should run for 1.48 years. Answer
Note: In practice, what would be a sensible guarantee?
Inverse normal question: A manufacturer of car tyres knows that her product has
a mean life of 2.3 years with a standard deviation of 0.4 years. Assuming that
the lifetime of a tyre is normally distributed what guarantee should she
offer if she only wants to pay out on 2% of tyres produced.
Solution:
Let X be a random variable representing the life of a tyre.
X is normal with μ = 2.3 and σ = 0.4
We want an x value such that P(X xcut-off ) = 0.02.
P( Z  z )  0.02
xcut off  2.3
z
0.4
gives a z value of -2.054 (see tables )
= -2.054
0.48
0.02
xcut-off = 1.48 yrs
So her guarantee should run for 1.48 years. Answer
Note: In practice, what would be a sensible guarantee? Perhaps 17 months?
Inverse Normal Problems where you’re asked to
calculate the MEAN or STANDARD DEVIATION
E.g. 1:
P(X < 903) = 0.657. The standard deviation is 17.3. Calculate the mean.
Calculate the value of z from the information P(Z < z ) = 0.657
Note: z must be above the mean as the probability is > 0.5
X
0.657
 903
E.g. 2:
X is a normally distributed random variable with mean of 45. The
probability that X is less than 37 is 0.02.
Estimate the standard deviation of X.
E.g. 2: X is a normally distributed random variable with mean of 45.
The probability that X is less than 37 is 0.02.
Estimate the standard deviation of X.
We want  such that
P( Z 
P( X  37 ) = 0.02
37  45

) = 0.02
Calculate z using your graphics calc.
Use the Standard Normal Distribution,
so use InvN and enter:
Area :0.02
z = 2.0537
 :1
0.02
 :0 Z
Standard normal distribution
0.02
2.0537
So
0
P( Z  -2.0537) = 0.02
P( Z 
8

X
37
45
) = 0.02
We want  such that
P( X  37 ) = 0.02
37  45
P( Z  8 ) = 0.02
Do Sigma
edition)
P( Z (new

) = 0.02 - p369, Ex 17.04


Calculate z using your graphics calc.
Use the Standard Normal Distribution,
so enter:
Area :0.02
0.02
z = 2.0537
 :1
 :0
P( Z  -2.0537) = 0.02
So
8 =

=
X
37
45
-2.0537
8
2.0537
Re-arrange to solve for .
 = 3.895 (4 sf)
Lessons 6 : Sums & differences of 2 or
more normally-distributed variables.
Learning outcome:
Calculate probabilities of outcomes that
involve sums or differences of 2 or more
normally-distributed random variables.
Work:
1. Notes & examples on sums & differences
2. Spend 15 mins on Sigma (NEW version): Ex. 18.01
(p377)
3. Notes & example on totals of n identical
independent random variables.
4. Finish Sigma Ex. 18.01 (complete for HW)
Probabilities when variables are combined
1.
X and Y have independent normal distributions with means 70 and 100
and standard deviations 5 and 12, respectively. If T = X + Y,
calculate:
(a) The mean of T.
Do Sigma (new
(b) The standard deviation of T.
version): pg. 377 –
Ex. 18.01
(c) P(T < 180)
2.
A large high school holds a cross-country race for both boys and girls
on the same course. The times taken in minutes can be modelled by
normal distributions, as given in the table.
Boys
Girls
Mean
28
35
Standard Deviation
5
4
If a boy and girl are both chosen at random, calculate the probability
that the girl finishes before the boy.
Suppose there is a total of 25 people in a lift. Each person, if chosen at
random, has a mean weight of 65 kg with standard deviation 7 kg.
(a) Find the mean and standard deviation of the total passenger load.
Let the total passenger load be T
formula for the mean.
E(T) = E(X1) + E(X2) + . . .Write
+ E(Xthe
25)
E(T) = 25×E(X)
Since the 25 distributions are identical.
E(T) = 25
In general, for n items with identical distributions, the
Expected Value of the distribution of the total is given by:
E(T) = n
Let the total passenger load be T
E(T) = E(X1) + E(X2) + . . . + E(X25)
E(T) = 25×E(X)
Since the 25 distributions are identical.
E(T) = 25
In general, for n items with identical distributions, the
Expected Value of the distribution of the total is given by:
E(T) = n
E(T) = n
= 25  65
= 1625 kg
3.01
Suppose there is a total of 25 people in a lift. Each person, if chosen at
random, has a mean weight of 65 kg with standard deviation 7 kg.
(a) Find the mean and standard deviation of the total passenger load.
Let the total passenger load be T
Write the formula for the mean.
E(T) = n
= 25  65
Substitute and calculate.
= 1625 kg
To find the std. deviation, first work through
Let the total passenger
load
be T
the
VARIANCE.
Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)
Var(T) = 25×Var(X)
T =
To find the Standard Deviation, take the
25 Var( X ) square root of the Variance.
Let the total passenger load be T
Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)
Var(T) = 25×Var(X)
T =
To find the Standard Deviation, take the
25 Var( X ) square root of the Variance.
In general, for n items with identical distributions, the
Standard Deviation of the distribution of the total is given by:
 T=
nVar (X )
3.01
Suppose there is a total of 25 people in a lift. Each person, if chosen at
random, has a mean weight of 65 kg with standard deviation 7 kg.
(a) Find the mean and standard deviation of the total passenger load.
T =
25 Var( X )
In general, for n items with identical distributions, the
Standard Deviation of the distribution of the total is given by:
 T=
nVar (X )
 T  n   2 Write the formula for the standard deviation.
X
 25 7
2
 25 49
= 35 kg
Substitute and calculate.
3.01
Suppose there is a total of 25 people in a lift. Each person, if chosen at
random, has a mean weight of 65 kg with standard deviation 7 kg.
(a) Find the mean and standard deviation of the total passenger load.
Let the total passenger load be T
2
E(T) = n
= 25  65
= 1625 kg
 T  n 
X
 25 7
2
 25 49
= 35 kg
So the total weight of the passengers, T, is approximately normally
distributed with  of 1625 kg and  of 35 kg.
3.03
(b) The lift is overloaded when the total passenger load exceeds 1700
kg. Calculate the probability that the lift is overloaded, assuming that
the lift is carrying 25 passengers.
Strategy: Find the mean and standard deviation of T, the total load.
As the distribution of T is approximately normal, use this
information to calculate the probability of overload.
(b) The lift
is overloaded
the totalComplete
passenger for
load HW
exceeds 1700
Continue
through
Sigmawhen
Ex. 18.01.
kg. Calculate the probability that the lift is overloaded, assuming that
the lift is carrying 25 passengers.
 T  n  2 X
E(T) = n
= 25  65
= 1625 kg
 25 7
2
 25 49
= 35 kg
So the total weight of the passengers, T, is approximately normally
distributed with  of 1625 kg and  of 35 kg.

1700  1625
P(T > 1700) = P Z 
35
= 0.01606 (4sf)

Calculate the probability
that T > 1700.
Lessons 7 : Linear combinations of
normally-distributed variables.
Learning outcome:
Calculate probabilities of outcomes that involve
a linear function of a random variable or a
linear combination of 2 random variables.
Work:
1. Notes on linear combinations (re-cap of expectation)
2. Sigma (NEW) – Ex. 18.02 (pg. 380) – do 1st 2 qs.
3. Handout – distinguishing between totals & a linear
functions.
4. Finish Sigma Ex. 18.02 (complete for HW). To Q4
compulsory. Q5 on extension.
Linear Function of a Random Variable, X
aX + c,
(e.g. taxi fares: hourly rate per km + fixed cost)
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 Combination of 2 independent random
variables, X & Y
Distribution of aX + bY where a & b are constants
Its mean E(aX + bY)
=
a×E(X) + b×E(Y)
Its variance Var(aX+c)
=
a2 × Var(X)
Its std. deviation σaX+c
=
a 2Var ( X )
Linear Combination of 2 independent random
variables, X & Y
Distribution of aX + bY where a & b are constants
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 )
Linear Combination of 2 independent random
variables, X & Y
Distribution of aX + bY where a & b are constants
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 )
Linear Combination of 2 independent random
variables, X & Y
Distribution of aX + bY where a & b are constants
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 )
Do Sigma (NEW): pg. 380 – Ex. 18.02 - Q1 and 2.
1. X has a normal distribution with mean 40 & standard dev. of 3.
(a) Calculate the mean & standard deviation of W where W = 6X + 15.
(a) Calculate P(W>250)
Linear Combination of 2 independent random
variables, X & Y
Distribution of aX + bY where a & b are constants
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
=
Do Sigma (NEW): pg. 380 – Ex. 18.02 - Q1 and 2.
2. X has a normal distribution with mean 18 & SD of 2.4, and Y has a
norm. distn. with mean 22 & SD of 1.5. W=3X + 5Y.
(a) Calculate the mean and SD of W.
(a) Calculate P(W<160)
HANDOUT ON
DISTINGUISHING BETWEEN
LINEAR FUNCTIONS AND
TOTALS
Distinguishing between Linear
Functions and Totals of
Identically Distributed
Variables.
A telephone contractor installs cable from the street to
the nearest jackpoint inside a house. The length of
cable installed for each job in a particular new
subdivision can be modelled by a normal distribution X
with a mean of 12m and a standard deviation of 1.6m.
What is the difference between the following 2
questions?
Situation 1 -
Linear Function
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
Situation 2 - TOTAL of identicallydistributed variables.
What is the probability that a group of
5 jobs require a total of more than
70m?
Situation 1
- Linear Function
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
Multiplying our variable (nbr metres) by a
co-efficient ($5 per metre).
Situation 2 - Total
What is the probability that a group of
5 jobs require a total of more than
70m?
Total of 5 different random variables. It
just so happens they have the same
So it’s a Linear Function of ONE variable. distribution.
Situation 1
- Linear Function
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
Situation 2 - Total
What is the probability that a group of
5 jobs require a total of more than
70m?
Multiplying our variable (nbr metres) by a
co-efficient ($5 per metre).
Total of 5 different random variables. It
just so happens they have the same
So it’s a Linear Function of ONE variable. distribution.
There is one random variable, X, and the
cost of the job is 5X.
There are 5 random variables:
X1, X2, X3, X4, X5.
The length of cabling is the sum of all 5.
Each variable has a mean 12 & Std. Dev
1.6.
Situation 1
- Linear Function
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
Situation 2 - Total
What is the probability that a group of
5 jobs require a total of more than
70m?
Multiplying our variable (nbr metres) by a
co-efficient ($5 per metre).
Total of 5 different random variables. It
just so happens they have the same
So it’s a Linear Function of ONE variable. distribution.
There is one random variable, X, and the
cost of the job is 5X.
There are 5 random variables:
X1, X2, X3, X4, X5.
The length of cabling is the sum of all 5.
Each variable has a mean 12 & Std. Dev
1.6.
The Model is: C = aX.
E(C) = a × E(X) and
VAR(C) = a2 × VAR(X), so
The Model is: T = X1+X2+ X3+X4+ X5.
E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).
VAR(T) = VAR(X1)+VAR(X2)+…+VAR(X5).
σC = √ (a2 × σ2X)
σT = √ (σ2X1+ σ2X2+…+ σ2X5)
Situation 1
- Linear Function
Situation 2 - Total
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
What is the probability that a group
of 5 jobs require a total of more than
70m?
Multiplying our variable (nbr metres) by a
co-efficient ($5 per metre).
Total of 5 different random variables.
It just so happens they have the same
distribution.
So it’s a Linear Function of ONE variable.
There is one random variable, X, and the
cost of the job is 5X.
There are 5 random variables:
X1, X2, X3, X4, X5.
The length of cabling is the sum of all 5.
Each variable has a mean 12 & Std. Dev
1.6.
The Model is: C = aX.
E(C) = a × E(X) and
VAR(C) = a2 × VAR(X), so
The Model is: T = X1+X2+ X3+X4+ X5.
E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).
VAR(T) =
VAR(X1)+VAR(X2)+…+VAR(X5).
σC = √ (a2 × σ2X)
σT = √ (σ2X1+ σ2X2+…+ σ2X5)
E(C) = 5 × 12 = $60
σC = √ (52 × 1.62)
E(T) = 12 + 12 + 12 + 12 + 12
= 8
= 60m.
σT = √ (1.62+ 1.62+1.62+ 1.62+1.62)
= 3.5777
Situation 1
- Linear Function
There is a charge of $5 per metre.
What is the probability that the job
exceeds $70?
Situation 2 - Total
What is the probability that a group of 5
jobs require a total of more than 70m?
Multiplying our variable (nbr metres) by a
Total of 5 different random variables. It just
co-efficient
($5 per
metre). Sigma Ex. 18.02.
so happens
they have thefor
same
distribution.
Continue
through
Complete
HW
So it’s a Linear Function of ONE variable.
There is one random variable, X, and the
cost of the job is 5X.
There are five random variables:
X1, X2, X3, X4, X5.
The length of cabling is the sum of all 5.
Each variable has a mean 12 & Std. Dev 1.6.
The Model is: C = aX.
E(C) = a × E(X) and
VAR(C) = a2 × VAR(X), so
The Model is: T = X1+X2+ X3+X4+ X5.
E(T) = E(X1)+E(X2)+E(X3)+E(X4)+E(X5).
VAR(T) = VAR(X1)+VAR(X2)+…+VAR(X5).
σC = √ (a2 × σ2X)
Parameters:
E(C) = 5 × 12 = $60
σC = √ (52 × 1.62)
= $8
Probability
that
oneand
jobσcost
> $70
With
μC = E(C)
= $60
C = $8,
is 0.1056
we
get P(C(4SF)
> 70) = 0.1056 (4sf)
σT = √ (σ2X1+ σ2X2+…+ σ2X5)
Parameters:
E(T) = 5 × 12 = 60m.
σT = √ (5 × 1.62) = 3.5777m
Probability
that
totaland
length
With
μT = E(T)
= 60m
σT= required
3.577709,for
5 jobs
is 0.002394
we
get exceeds
P(T > 70) 70m
= 0.002594
(4sf) (4SF)