Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download X - Brocklehurst-13SAM
Document related concepts
Transcript
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: Q4246 (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: Q4246 (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: Q4246 (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: Q4246 (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= nVar (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= nVar (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)
