Topic 5: Probability Distributions

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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.

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Topic 5: Probability Distributions

Achievement Standard 90646

Solve Probability Distribution Models to solve straightforward problems

4 Credits

Externally Assessed

NuLake Pages 278  322

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.

E.g. 1:

If m=178, s=5, P(175X 184) = ?

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.

E.g. 1:

If m=178, s=5, P(175X 184) = 0.61067

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.

E.g.1:

If m=178, s=5, P(175X 184) = 0.61067

lower: 175, upper: 184, σ: 5, μ: 178

E.g.2:

If m=30, s=3.5, P(X 31) = ?

lower: -EXP99, upper: 31, σ: 3.5, μ: 30

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.

E.g.1:

If m=178, s=5, P(175X 184) = 0.61067

lower: 175, upper: 184, σ: 5, μ: 178

E.g.2:

If m=30, s=3.5, P(X 31) = 0.61245

lower: -EXP99, upper: 31, σ: 3.5, μ: 30

10 minutes

• Do Sigma (NEW) p358.
• Ex. 17.01
• Q3 on the board as a quiz.
• 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.

E.g.1:

If m=178, s=5, P(175X 184) = 0.61067

lower: 175, upper: 184, σ: 5, μ: 178

E.g.2:

If m=30, s=3.5, P(X 31) = 0.61245

lower: -EXP99, upper: 31, σ: 3.5, μ: 30

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.

17.02A

165

174

165

174

164.5

165

165.5

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

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

174

165

174

164.5

164.5

165

165

165.5

165.5

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

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

174

165

174

164.5

164.5

165

165

165.5

165.5

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 thisintervalwould be recorded as having a height of165 cm.

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

174

165

174

164.5

164.5

165

165

165.5

165.5

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.

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

174

165

174

164.5

164.5

165

165

165.5

165.5

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.

P(X > 165)≈P(X > 165.5) with continuity correction.

≈0.9217 (4 sf)

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

Do NuLake qs on “Continuity Corrections for a Normal Distribution”:

Pg. 311-313: Q4246 (NOTE: Don’t do Q47).

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

Do NuLake qs on “Continuity Corrections for a Normal Distribution”:

Pg. 311-313: Q4246 (NOTE: Don’t do Q47).

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

Do NuLake qs on “Continuity Corrections for a Normal Distribution”:

Pg. 311-313: Q4246 (NOTE: Don’t do Q47).

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

Continuity Correction

Do NuLake qs on “Continuity Corrections for a Normal Distribution”:

Pg. 311-313: Q4246 (NOTE: Don’t do Q47).

If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction.

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:

• Inverse calculations using standard normal.
• Inverse calculations – standardising Examples
• 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.

z = ?

P(0 < Z <z) = 0.377

What isz?

Answer (from tables): z = ?

P(0 < Z <z) = 0.377

What isz?

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.

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.

z = 1.16

• Find the value of z if the area
• to the right of z is only 0.05.

z = ?

P(0 < Z <z) = 0.45

What isz?

Answer (from tables): z = ?

P(0 < Z <z) = 0.45

What isz?

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.

z = 1.16

• Find the value of z if the area
• to the right of z is only 0.05.

z = 1.645

Inverse problems where you’re given the probability, m and s, and asked to find the value of X.

x=  + z

z= x–m

s

can be re-arranged to solve for 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 m =24 and s =4.7. What is the value, xcut-off?

z= x–m

s

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 m =24 and s =4.7. What is the value, xcut-off?

z= x–m

s

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 m =24 and s =4.7. What is the value, xcut-off ?

z= x–m

s

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 m =24 and s =4.7. What is the value, xcut-off ?

First do using working (standardise it), then check with G.Calc.

s= 4.7

First find the z cut-off like in the last example

m= 24

xcut-off = ?

P(0 < Z <z) = 0.45

What isz?

Answer (from tables): z = ?

P(0 < Z <z) = 0.45

What isz?

Answer (from tables): z = 1.645

z= x–m

s

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 m =24 and s =4.7. What is the value, xcut-off ?

First do using working (standardise it), then check with G.Calc.

zcut-off =1.645

m= 24

xcut-off = ?

z= x–m

s

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 m =24 and s =4.7. What is the value, xcut-off ?

First do using working (standardise it), then check with G.Calc.

zcut-off =1.645

m= 24

xcut-off =

z= x–m

s

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 m =24 and s =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

zcut-off =1.645

m= 24

xcut-off =

Area: Enter total area to the LEFT of xcut-off.

z= x–m

s

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 m =24 and s =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

zcut-off =1.645

m= 24

xcut-off =

Area: Enter total area to the LEFT of xcut-off.

z= x–m

s

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 m =24 and s =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

= ____

zcut-off =1.645

m= 24

xcut-off =

Area: Enter total area to the LEFT of xcut-off.

z= x–m

s

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 m =24 and s =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

zcut-off =1.645

m= 24

xcut-off =

Area: Enter total area to the LEFT of xcut-off.

z= x–m

s

can be re-arranged to solve for x

x=  + z

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

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 m =24 and s =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

zcut-off =1.645

m= 24

xcut-off = 31.73

Area: Enter total area to the LEFT of xcut-off.

• 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.

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.

STARTER QUESTION

(from what we did last lesson)

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 )

= -2.054

xcut-off = ?

xcut-off = 1.48 yrs

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.

gives a z value of -2.054 (see tables )

= -2.054

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?

X

903

0.657

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

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.

X

0.02

0.02

37

45

We want  such that P( X  37 ) =0.02

P( Z ) =0.02

P( Z ) =0.02

Calculate z using your graphics calc.

Use the Standard Normal Distribution,

so use InvN and enter:

Area :0.02

s :1

m :0

z = 2.0537

Z

Standard normal distribution

2.0537

0

So P( Z  -2.0537) =0.02

X

0.02

37

45

We want  such that P( X  37 ) =0.02

P( Z ) =0.02

P( Z ) =0.02

Do Sigma (new edition) - p369, Ex 17.04

Calculate z using your graphics calc.

Use the Standard Normal Distribution,

so enter:

Area :0.02

s :1

m :0

z = 2.0537

P( Z  -2.0537) =0.02

So =-2.0537

s= 8

2.0537

Re-arrange to solve for s.

 = 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:

• Notes & examples on sums & differences
• Spend 15 mins on Sigma (NEW version): Ex. 18.01 (p377)
• Notes & example on totals of n identical independent random variables.
• Finish Sigma Ex. 18.01 (complete for HW)
Probabilities when variables are combined
• 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 thatthe girl finishes before the boy.

Do Sigma (new version): pg. 377 – Ex. 18.01

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) = E(X1) + E(X2) + . . . + E(X25)

E(T) = 25×E(X)

Since the 25 distributions are identical.

E(T) = 25m

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) = 25m

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

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.

3.01

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 the VARIANCE.

Let the total passenger load be T

Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)

Var(T) = 25×Var(X)

To find the Standard Deviation, take the square root of the Variance.

sT =

Let the total passenger load be T

Var(T) = Var(X1) + Var(X2) + . . . + Var(X25)

Var(T) = 25×Var(X)

To find the Standard Deviation, take the square root of the Variance.

sT =

In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:

sT=

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.

3.01

sT =

In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:

sT=

Write the formula for the standard deviation.

Substitute and calculate.

= 35 kg

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.

3.01

Let the total passenger load be T

E(T) = n

= 25  65

= 1625 kg

= 35 kg

So the total weight of the passengers, T, is approximately normally distributed with m of 1625 kg and s 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 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

= 1625 kg

= 35 kg

So the total weight of the passengers, T, is approximately normally distributed with m of 1625 kg and s of 35 kg.

Calculate the probability that T > 1700.

P(T > 1700) =

= 0.01606 (4sf)

Learning outcome:

Calculate probabilities of outcomes that involve a linear function of a random variable or a linear combination of 2 random variables.

Work:

• Notes on linear combinations (re-cap of expectation)
• Sigma (NEW) – Ex. 18.02 (pg. 380) – do 1st 2 qs.
• Handout – distinguishing between totals & a linear functions.
• 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 =

Linear Combination of 2independent 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 =

Linear Combination of 2 independentrandom 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 =

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 =

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.
• 1. X has a normal distribution with mean 40 & standard dev. of 3.
• Calculate the mean & standard deviation of W where W = 6X + 15.
• 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.
• Calculate the mean and SD of W.
• Calculate P(W<160)

### 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?

Continue through Sigma Ex. 18.02. Complete for HW

Probability that one jobcost > \$70 is 0.1056 (4SF)

Probability that total length required for 5 jobs exceeds 70m is 0.002394 (4SF)