1. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward problems 4 Credits Externally Assessed NuLake Pages 278  322

2. NORMAL DISTRIBUTION PART 2

3. 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.

4. 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. 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  =178,  =5, P(175  X  184) = ? MENU, STAT, DIST, NORM, Ncd lower: 175, upper: 184, σ: 5, μ: 178

6. 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  =178,  =5, P(175  X  184) = 0.61067 MENU, STAT, DIST, NORM, Ncd lower: 175, upper: 184, σ: 5, μ: 178

7. 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  =178,  =5, P(175  X  184) = 0.61067 MENU, STAT, DIST, NORM, Ncd lower: 175, upper: 184, σ: 5, μ: 178 E.g.2: If  =30,  =3.5, P(X  31) = ? MENU, STAT, DIST, NORM, Ncd lower: -EXP99, upper: 31, σ: 3.5, μ: 30

8. 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  =178,  =5, P(175  X  184) = 0.61067 MENU, STAT, DIST, NORM, Ncd lower: 175, upper: 184, σ: 5, μ: 178 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

9. 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  =178,  =5, P(175  X  184) = 0.61067 MENU, STAT, DIST, NORM, Ncd lower: 175, upper: 184, σ: 5, μ: 178 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 10 minutes Do Sigma (NEW) p358. Ex. 17.01 Q3 on the board as a quiz.

10. Making a Continuity Correction (USE THE HANDOUT)

11. 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. 174165 174165 Distribution when rounding heights (discrete histogram) Actual curve for heights of students (continuous) 165165.5164.5 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

12. 174165 174165 165.5164.5 165165.5164.5 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.

13. 174165 174165 because any student with a height in this interval would be recorded as having a height of 165 cm. 165165.5164.5 165165.5164.5 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.

14. 174165 174165 165.5164.5 165165.5164.5 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.

15. 174165 174165 165.5164.5 165165.5164.5 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)

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

17. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)

18. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)P(3.5 < X < 4.5)

19. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)P(3.5 < X < 4.5) P(X > 6)

20. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)P(3.5 < X < 4.5) P(X > 6)P(X > 6.5)

21. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)P(3.5 < X < 4.5) P(X > 6)P(X > 6.5) P(X > 6)

22. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS P(X = 4)P(3.5 < X < 4.5) P(X > 6)P(X > 6.5) P(X > 6)P(X > 5.5)

23. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS 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(8 < X < 12) Do NuLake qs on “Continuity Corrections for a Normal Distribution”: Pg. 311-313: Q42  46 (NOTE: Don’t do Q47).

24. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS 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: Q42  46 (NOTE: Don’t do Q47).

25. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS 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: Q42  46 (NOTE: Don’t do Q47).

26. Continuity Correction If we use a Normal Distribution to approximate a variable that is DISCRETE, we must make a Continuity Correction. DISCRETECONTINUOUS 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: Q42  46 (NOTE: Don’t do Q47).

27. 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.

28. 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 = ?

29. P(0 < Z < z) = 0.377 What is z ? Answer (from tables): z = ?

30. P(0 < Z < z) = 0.377 What is z ? Answer (from tables): z = 1.16

31. 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

32. 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 (b)Find the value of z if the area to the right of z is only 0.05. z = ?

33. P(0 < Z < z) = 0.45 What is z ? Answer (from tables): z = ?

34. P(0 < Z < z) = 0.45 What is z ? Answer (from tables): z = 1.645

35. 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 (b)Find the value of z if the area to the right of z is only 0.05. z = 1.645

36. 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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? z = x –   can be re-arranged to solve for x x =  +  z

37. 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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ?

38. 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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ?

39.  = 24 x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-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

40. P(0 < Z < z) = 0.45 What is z ? Answer (from tables): z = ?

41. P(0 < Z < z) = 0.45 What is z ? Answer (from tables): z = 1.645

42.  = 24 x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645

43.  = 24 x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645

44. STAT, DIST, NORM, InvN Area: _____, σ: 4.7, μ: 24 Area: Enter total area to the LEFT of x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645 x cut-off =

45. STAT, DIST, NORM, InvN Area: 1-0.05, σ: 4.7, μ: 24 Area: Enter total area to the LEFT of x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645 x cut-off =

46. STAT, DIST, NORM, InvN Area: 1-0.05, σ: 4.7, μ: 24 = ____ Area: Enter total area to the LEFT of x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645 x cut-off =

47. STAT, DIST, NORM, InvN Area: 1-0.05, σ: 4.7, μ: 24 = 0.95 Area: Enter total area to the LEFT of x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-off ? First do using working (standardise it), then check with G.Calc. z cut-off = 1.645 x cut-off =

48. STAT, DIST, NORM, InvN Area: 1-0.05, σ: 4.7, μ: 24 = 0.95 Area: Enter total area to the LEFT of x cut-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 > x cut-off ) = 0.05. We’re told that  =24 and  =4.7. What is the value, x cut-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 x cut-off = 31.73 z cut-off = 1.645

49. 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.

50. STARTER: Question from what we did last lesson: Inverse Normal: Calculating the x cut-off score.

51. 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  x cut-off ) = 0.02. gives a z value of -2.054 (see tables ) So her guarantee should run for 1.48 years. Answer Note: In practice, what would be a sensible guarantee? x cut-off = ? x cut-off = 1.48 yrs = -2.054 STARTER QUESTION (from what we did last lesson)

52. 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  x cut-off ) = 0.02. gives a z value of -2.054 (see tables ) So her guarantee should run for 1.48 years. Answer Note: In practice, what would be a sensible guarantee? Perhaps 17 months? = -2.054 x cut-off = 1.48 yrs

53. 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.

54. 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 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 So P( Z  -2.0537) = 0.02

55. 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 P( Z  -2.0537) = 0.02 So = -2.0537  = 3.895 (4 sf) Re-arrange to solve for .   = 8 2.0537 Do Sigma (new edition) - p369, Ex 17.04

56. 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)

57. 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. BoysGirls Mean2835 Standard Deviation54 Do Sigma (new version): pg. 377 – Ex. 18.01

58. Let the total passenger load be T E(T) = E(X 1 ) + E(X 2 ) +... + E(X 25 ) 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.

59. Let the total passenger load be T E(T) = E(X 1 ) + E(X 2 ) +... + E(X 25 ) E(T) = 25×E(X) Since the 25 distributions are identical. E(T) = 25  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:

60. 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(X 1 ) + Var(X 2 ) +... + Var(X 25 ) Var(T) = 25×Var(X)  T = To 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.

61. Let the total passenger load be T Var(T) = Var(X 1 ) + Var(X 2 ) +... + Var(X 25 ) Var(T) = 25×Var(X) To 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 =

62. 3.01 = 35 kg Write the formula for the standard deviation. Substitute and calculate.  T = 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 = In general, for n items with identical distributions, the Standard Deviation of the distribution of the total is given by:

63. 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 E(T) = n  = 25  65 = 35 kg = 1625 kg So the total weight of the passengers, T, is approximately normally distributed with  of 1625 kg and  of 35 kg.

64. 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.

65. P(T > 1700) = 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 So the total weight of the passengers, T, is approximately normally distributed with  of 1625 kg and  of 35 kg.

66. 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 1 st 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.

67. 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)=a 2 × Var(X) Its std. deviation σ aX+c = 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)

68. Its variance Var(aX+c)=a 2 × Var(X) Its std. deviation σ aX+c = 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)= a 2 Var(X) + b 2 Var(Y) Its std. deviation σ aX+bY =

69. 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)= a 2 Var(X) + b 2 Var(Y) Its std. deviation σ aX+bY =

70. 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)= a 2 Var(X) + b 2 Var(Y) Its std. deviation σ aX+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. (a)Calculate the mean & standard deviation of W where W = 6X + 15. (a)Calculate P(W>250)

71. 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)= a 2 Var(X) + b 2 Var(Y) Its std. deviation σ aX+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)

72. HANDOUT ON DISTINGUISHING BETWEEN LINEAR FUNCTIONS AND TOTALS

73. 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?

74. Situation 1 - Linear FunctionSituation 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?

75. Situation 1 - Linear FunctionSituation 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.

76. Situation 1 - Linear FunctionSituation 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: X 1, X 2, X 3, X 4, X 5. The length of cabling is the sum of all 5. Each variable has a mean 12 & Std. Dev 1.6.

77. Situation 1 - Linear FunctionSituation 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: X 1, X 2, X 3, X 4, X 5. 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) = a 2 × VAR(X), so σ C = √ (a 2 × σ 2 X ) The Model is: T = X 1 +X 2 + X 3 +X 4 + X 5. E(T) = E(X 1 )+E(X 2 )+E(X 3 )+E(X 4 )+E(X 5 ). VAR(T) = VAR(X 1 )+VAR(X 2 )+…+VAR(X 5 ). σ T = √ (σ 2 X1 + σ 2 X2 +…+ σ 2 X5 )

78. Situation 1 - Linear FunctionSituation 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: X 1, X 2, X 3, X 4, X 5. 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) = a 2 × VAR(X), so σ C = √ (a 2 × σ 2 X ) The Model is: T = X 1 +X 2 + X 3 +X 4 + X 5. E(T) = E(X 1 )+E(X 2 )+E(X 3 )+E(X 4 )+E(X 5 ). VAR(T) = VAR(X 1 )+VAR(X 2 )+…+VAR(X 5 ). σ T = √ (σ 2 X1 + σ 2 X2 +…+ σ 2 X5 ) E(C) = 5 × 12 = $60 σ C = √ (5 2 × 1.6 2 ) = 8 E(T) = 12 + 12 + 12 + 12 + 12 = 60m. σ T = √ (1.6 2 + 1.6 2 +1.6 2 + 1.6 2 +1.6 2 ) = 3.5777

79. Situation 1 - Linear FunctionSituation 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: X 1, X 2, X 3, X 4, X 5. 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) = a 2 × VAR(X), so σ C = √ (a 2 × σ 2 X ) The Model is: T = X 1 +X 2 + X 3 +X 4 + X 5. E(T) = E(X 1 )+E(X 2 )+E(X 3 )+E(X 4 )+E(X 5 ). VAR(T) = VAR(X 1 )+VAR(X 2 )+…+VAR(X 5 ). σ T = √ (σ 2 X1 + σ 2 X2 +…+ σ 2 X5 ) Parameters: E(C) = 5 × 12 = $60 σ C = √ (5 2 × 1.6 2 ) = $8 Parameters: E(T) = 5 × 12 = 60m. σ T = √ (5 × 1.6 2 ) = 3.5777m With μ C = E(C) = $60 and σ C = $8, we get P(C > 70) = 0.1056 (4sf) With μ T = E(T) = 60m and σ T = 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