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Topic 5: Probability Distributions
Achievement Standard 90646
Solve Probability Distribution Models to solve straightforward
problems
4 CreditsExternally 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.
175 =178 184
=5
E.g. 1:
If =178, =5, P(175 X 184) = ?
MENU, STAT, DIST, NORM, Ncd
lower: 175, upper: 184, σ: 5, μ: 178
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.
175 =178 184
=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
• 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.
175 =178 184
=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: 178E.g.2:
If =30, =3.5, P(X 31) = ?
MENU, STAT, DIST, NORM, Ncd
lower: -EXP99, upper: 31, σ: 3.5, μ: 30
31 =30
=3.5
• 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.
175 =178 184
=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: 178E.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
31 =30
=3.5
• 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.
175 =178 184
=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: 178E.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
31 =30
=3.5
10 minutesDo Sigma (NEW) p358. Ex. 17.01• Q3 on the board as a quiz.
Making a Continuity Correction
(USE THE HANDOUT)
17.02AHeights 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.
174165
174165
Distribution when rounding heights (discrete histogram)
Actual curve for heights of students (continuous)
165 165.5164.5
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
174165
174165
165 165.5164.5 165 165.5164.5
Distribution when rounding heights (discrete histogram)
Actual curve for heights of students (continuous)
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
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.
174165
174165
because any student with a height in this interval would be recorded as having a height of 165 cm.
165 165.5164.5 165 165.5164.5
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
Distribution when rounding heights (discrete histogram)
Actual curve for heights of students (continuous)
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.
174165
174165
165 165.5164.5
165 165.5164.5
P(X > 165) = P(X > ?) with continuity correction.
To find the cut-off point for continuity corrections, move up or down to the midpoint between two whole-numbers.
In this example the wording is ‘more than 165’, so move up to 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.
174165
174165
165 165.5164.5165 165.5164.5
P(X > 165) ≈ P(X > 165.5) with continuity correction.
To find the cut-off point for continuity corrections, move up or down to the midpoint between two whole-numbers.
In this example the wording is ‘more than 165’, so move up to 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.
≈ 0.9217 (4 sf)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that
is DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) P(3.5 < X < 4.5)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) P(3.5 < X < 4.5)
P(X > 6)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) P(3.5 < X < 4.5)
P(X > 6) P(X > 6.5)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) P(3.5 < X < 4.5)
P(X > 6) P(X > 6.5)
P(X > 6)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) P(3.5 < X < 4.5)
P(X > 6) P(X > 6.5)
P(X > 6) P(X > 5.5)
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) 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)
Do NuLake qs on “Continuity Corrections for a Normal Distribution”:Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) 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)
Do NuLake qs on “Continuity Corrections for a Normal Distribution”:Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) 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)
Do NuLake qs on “Continuity Corrections for a Normal Distribution”:Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
Continuity CorrectionIf we use a Normal Distribution to approximate a variable that is
DISCRETE, we must make a Continuity Correction.
DISCRETE CONTINUOUS
P(X = 4) 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)
Do NuLake qs on “Continuity Corrections for a Normal Distribution”:Pg. 311-313: Q4246 (NOTE: Don’t do Q47).
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 Examples3. 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 zz = ?
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 zz = 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 zz = 1.16
(b)Find the value of z if the area to the right of z is only 0.05.
0.05
0.45
0 zz = ?
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 zz = 1.16
(b)Find the value of z if the area to the right of z is only 0.05.
0.05
0.45
0 zz = 1.645
Inverse problems where you’re given the probability, and , and asked to
find the value of X.
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
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 ?
0.05
0.45
0 z= 24 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.
First find the z cut-off like in the last example
= 4.7
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
0.05
0.45
0 z= 24
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.
zcut-off = 1.645
0.05
0.45
0 z= 24
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.
zcut-off = 1.645
0.05
0.45
0 z
STAT, DIST, NORM, InvNArea: _____ , σ: 4.7, μ: 24
Area: Enter total area to the LEFT of xcut-off .
= 24
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.
zcut-off = 1.645
xcut-off =
0.05
0.45
0 z
STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24
Area: Enter total area to the LEFT of xcut-off .
= 24
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.
zcut-off = 1.645
xcut-off =
0.05
0.45
0 z
STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = ____
Area: Enter total area to the LEFT of xcut-off .
= 24
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.
zcut-off = 1.645
xcut-off =
0.05
0.45
0 z
STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = 0.95
Area: Enter total area to the LEFT of xcut-off .
= 24
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.
zcut-off = 1.645
xcut-off =
0.05
0.45
0 z
STAT, DIST, NORM, InvNArea: 1-0.05 , σ: 4.7, μ: 24 = 0.95
Area: Enter total area to the LEFT of xcut-off .
= 24
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.
Once you’ve copied down the e.g. & working:
Do Sigma (NEW version): p366 – Ex. 17.03 Complete for HW.
Extension (after you’ve finished this): NuLake p307 & 308
xcut-off = 31.73
zcut-off = 1.645
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.
0.02
0.48
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.
gives a z value of -2.054 (see tables )
So her guarantee should run for 1.48 years. AnswerNote: In practice, what would be a sensible guarantee?
xcut-off = ?xcut-off = 1.48 yrs
02.0)( zZP
= -2.054 4.0
3.2 offcutx
z
STARTER QUESTION(from what we did last lesson)
0.02
0.48
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.
gives a z value of -2.054 (see tables )
So her guarantee should run for 1.48 years. AnswerNote: In practice, what would be a sensible guarantee? Perhaps 17 months?
02.0)( zZP
= -2.054 4.0
3.2 offcutx
z
xcut-off = 1.48 yrs
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
903
X
0.657
Inverse Normal Problems where you’re asked to calculate the MEAN or STANDARD DEVIATION
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.
2.0537
Z
0
Standard normal distribution 37
X
45
0.02
0.02
Calculate z using your graphics calc.Use the Standard Normal Distribution,so use InvN and enter:Area :0.02 :1 :0
z = 2.0537
P( Z ) = 0.02
We want such that P( X 37 ) = 0.02
P( Z ) = 0.02
4537 8
So P( Z -2.0537) = 0.02
37
X
45
0.02
Calculate z using your graphics calc.Use the Standard Normal Distribution,so enter:Area :0.02 :1 :0
z = 2.0537
P( Z ) = 0.02
We want such that P( X 37 ) = 0.02
P( Z ) = 0.02
4537 8
P( Z -2.0537) = 0.02
So = -2.0537
8
= 3.895 (4 sf)
Re-arrange to solve for .= 8 2.0537
Do Sigma (new edition) - p369, Ex 17.04
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 & differences2. 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.(b) The standard deviation of T.
(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.
If a boy and girl are both chosen at random, calculate the probability that the girl finishes before the boy.
Boys Girls
Mean 28 35
Standard Deviation 5 4
Do Sigma (new version): pg. 377 – Ex. 18.01
Let the total passenger load be T
E(T) = E(X1) + E(X2) + . . . + E(X25)Write the formula for the mean.
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
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
E(T) = E(X1) + E(X2) + . . . + E(X25)
E(T) = 25×E(X) Since the 25 distributions are identical.
E(T) = 25
E(T) = n
E(T) = n
= 25 65
= 1625 kg
In general, for n items with identical distributions, the Expected Value of the distribution of the total is given by:
3.01
Let the total passenger load be T
E(T) = n
= 25 65
= 1625 kg
Write the formula for the mean.
Substitute and calculate.
To find the std. deviation, first work through the VARIANCE.Let the total passenger load be T
Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)
Var(T) = 25×Var(X)
T = )(25 XVarTo find the Standard Deviation, take the square root of the Variance.
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
Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)
Var(T) = 25×Var(X)
)(25 XVarTo find the Standard Deviation, take the square root of the Variance.T =
In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:
T= )(XVarn
3.01
= 35 kg
Write the formula for the standard deviation.
Substitute and calculate.
T =
XnT2
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= )(XVarn
2725
)(25 XVar
4925
In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:
3.01Suppose 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
E(T) = n= 25 65
= 35 kg
= 1625 kg
XT n 2 2725
4925
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.
P(T > 1700) = 1700 1625P 35Z Calculate the probability that T > 1700.
= 0.01606 (4sf)
(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.
Continue through Sigma Ex. 18.01. Complete for HW
E(T) = n= 25 65
= 35 kg
= 1625 kg
XT n 2 2725
4925
So the total weight of the passengers, T, is approximately normally distributed with of 1625 kg and of 35 kg.
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 = )(2 XVara
Linear Combination of 2 independent random variables, X & Y
Distribution of aX + bY where a & b are
constantsIts mean E(aX + bY) = a×E(X) +
b×E(Y)
Its variance Var(aX+c) = a2 ×
Var(X)Its std. deviation σaX+c = )(2 XVara
Linear Combination of 2 independent random variables, X & Y
Distribution of aX + bY where a & b are
constantsIts mean E(aX + bY) = a×E(X) +
b×E(Y)Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)
Its std. deviation σaX+bY = )()( 22 YVarbXVara
Linear Combination of 2 independent random variables, X & Y
Distribution of aX + bY where a & b are
constantsIts mean E(aX + bY) = a×E(X) +
b×E(Y)Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)
Its std. deviation σaX+bY = )()( 22 YVarbXVara
Linear Combination of 2 independent random variables, X & Y
Distribution of aX + bY where a & b are
constantsIts mean E(aX + bY) = a×E(X) +
b×E(Y)Its variance Var(aX + bY) = a2Var(X) + b2Var(Y)
Its std. deviation σaX+bY = )()( 22 YVarbXVara
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
constantsIts 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 Situation 2 - TOTAL of identically-distributed variables.
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?
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).So it’s a Linear Function of ONE variable.
Total of 5 different random variables. It just so happens they have the same distribution.
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).So it’s a Linear Function of ONE variable.
Total of 5 different random variables. It just so happens they have the same 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 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).So it’s a Linear Function of ONE variable.
Total of 5 different random variables. It just so happens they have the same 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
σC = √ (a2 × σ2X)
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).
σT = √ (σ2X1+ σ2
X2+…+ 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).So it’s a Linear Function of ONE variable.
Total of 5 different random variables. It just so happens they have the same 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
σC = √ (a2 × σ2X)
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).
σT = √ (σ2X1+ σ2
X2+…+ σ2X5)
E(C) = 5 × 12 = $60
σC = √ (52 × 1.62) = 8
E(T) = 12 + 12 + 12 + 12 + 12 = 60m.
σT = √ (1.62+ 1.62+1.62+ 1.62+1.62)
= 3.5777
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).
So it’s a Linear Function of ONE variable.
Total of 5 different random variables. It just so happens they have the same distribution.
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
σC = √ (a2 × σ2X)
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).
σT = √ (σ2X1+ σ2
X2+…+ σ2X5)
Parameters:
E(C) = 5 × 12 = $60
σC = √ (52 × 1.62) = $8
Parameters:
E(T) = 5 × 12 = 60m.
σT = √ (5 × 1.62) = 3.5777m
With μC = E(C) = $60 and σC = $8,
we get P(C > 70) = 0.1056 (4sf)
With μT = E(T) = 60m and σT= 3.577709,
we get P(T > 70) = 0.002594 (4sf)
Probability that one job cost > $70 is 0.1056 (4SF)
Probability that total length required for 5 jobs exceeds 70m is 0.002394 (4SF)
Continue through Sigma Ex. 18.02. Complete for HW
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