Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

Statistics and Modelling Course 2011

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

PART 1

Lesson 1 Overview of probability distributions (discrete and continuous). The Standard Normal Distribution. HANDOUT: Just the first page of the Normal Distribution handouts – on the std. normal distribution. HW: –1. In NuLake: Do pages 296 & 297 (intro to Normal Distn). –2. In Sigma – match-up task Q5  7: In OLD edition this is on p98 (Ex. 7.1). In NEW edition it’s on p440 (exercise A.01).

Discrete Probability Distributions:

Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis.

Discrete Probability Distributions: Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis. Probability = Frequency Total Continuous Probability Distributions:

Discrete Probability Distributions: Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis. Probability = Frequency Total Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve.

Discrete Probability Distributions: Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis. Probability = Frequency Total Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve. A bar graph is not appropriate because continuous data contains no gaps!

Discrete Probability Distributions: Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis. Probability = Frequency Total Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve. A bar graph is not appropriate because continuous data contains no gaps! Probabilities are for a domain of values. A person is very unlikely to be exactly 170.0…cm tall, or exactly 17 years old. No limit to precision.

Discrete Probability Distributions: Discrete Data is usually modelled using a Bar Graph with Frequency on the vertical axis. Probability = Frequency Total Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve. A bar graph is not appropriate because continuous data contains no gaps! Probabilities are for a domain of values. A person is very unlikely to be exactly 170.0…cm tall, or exactly 17 years old. No limit to precision. Instead we’d ask for, say: P(169.5cm < X < 170.5cm)

In this topic, we will use 3 types of probability distribution. 2. The Binomial Distribution 3. The Poisson Distribution 1.The Normal Distribution Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve. A bar graph is not appropriate because continuous data contains no gaps! Probabilities are for a domain of values. A person is very unlikely to be exactly 170.0…cm tall, or exactly 17 years old. No limit to precision. Instead we’d ask for, say: P(169.5cm < X < 170.5cm)

In this topic, we will use 3 types of probability distribution. FOR DISCRETE DATA: 2. The Binomial Distribution 3. The Poisson Distribution FOR CONTINUOUS DATA: 1.The Normal Distribution Continuous Probability Distributions: Continuous Data is modelled by a Histogram (using grouped values) or a Probability Curve. A bar graph is not appropriate because continuous data contains no gaps! Probabilities are for a domain of values. A person is very unlikely to be exactly 170.0…cm tall, or exactly 17 years old. No limit to precision. Instead we’d ask for, say: P(169.5cm < X < 170.5cm)

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for ____________ data (hence it is a _________curve).

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a ___________curve).

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures _____________, NOT probability.

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution.

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. ____________________________________________

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. height, weight, dimensions of the human body. 2 Parameters:  : _______  : _______ ________

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. height, weight, dimensions of the human body. 2 Parameters:  : _______  : _______ ________

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. height, weight, dimensions of the human body. 2 Parameters:  : Mean  : _______ ________

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. height, weight, dimensions of the human body. 2 Parameters:  : Mean  : _______ ________

The NORMAL DISTRIBUTION  “Bell-shaped Curve”  Used for continuous data (hence it is a continuous curve).  Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution. E.g. height, weight, dimensions of the human body. 2 Parameters:  : Mean  : Standard Deviation

 Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because ____ _______________________________________________. 2 Parameters:  : Mean  : Standard Deviation

 Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision. 2 Parameters:  : Mean  : Standard Deviation

 Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a _________ ___ _________. 2 Parameters:  : Mean  : Standard Deviation

 Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. ____________________________________________. 2 Parameters:  : Mean  : Standard Deviation

 Height of the curve (vertical axis) measures FREQUENCY, NOT probability.  Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. 2 Parameters:  : Mean  : Standard Deviation

 Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. PROBABILITY=______ ______ ___ _____ 2 Parameters:  : Mean  : Standard Deviation

 Measurements which occur in nature often fit a Normal Distribution – symmetrical, greatest frequency is in the middle. E.g. height, weight, dimensions of the human body.  Cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. PROBABILITY=AREA UNDER THE CURVE 2 Parameters:  : Mean  : Standard Deviation

 We cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. Standard Deviations,  On a Normal Distribution Curve, about…  ___% of the population lies within 1 standard deviation either side of the mean, . PROBABILITY=AREA UNDER THE CURVE X 

 We cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. Standard Deviations,  On a Normal Distribution Curve, about…  68% of the population lies within 1 standard deviation either side of the mean, .  __% within 2 standard deviations of . PROBABILITY=AREA UNDER THE CURVE X 

 We cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. Standard Deviations,  On a Normal Distribution Curve, about…  68% of the population lies within 1 standard deviation either side of the mean, .  95% within 2 standard deviations of .  __% within 3 standard deviations of . PROBABILITY=AREA UNDER THE CURVE X 

 We cannot calculate probabilities for specific values because the distribution is continuous – infinite degree of precision.  We calculate probabilities for a domain of values. E.g. Year 13s in NZ whose height is between 170 and 180cm. Standard Deviations,  On a Normal Distribution Curve, about…  68% of the population lies within 1 standard deviation either side of the mean, .  95% within 2 standard deviations of .  99% within 3 standard deviations of . PROBABILITY=AREA UNDER THE CURVE X 

The Standard Normal Distribution Properties * Horizontal axis measures “z”, the number of standard deviations away from the mean. * Mean,  = 0 * Standard deviation,  = 1 * P(a < Z < b) = shaded area Use of Standard Normal Tables - The tables give the value P(0 < Z < a). - Diagrams are essential. z = Number of Standard Deviations from the mean, .

Use of Standard Normal Tables - The tables give the value P(0 < Z < a). -Diagrams are essential. z = Number of Standard Deviations from the mean, . Examples: (a) P(0  Z  1) = ? Properties * Horizontal axis measures “z”, the number of standard deviations away from the mean. * Mean,  = 0 * Standard deviation,  = 1 * P(a < Z < b) = shaded area

(c) P(0  Z  1) = ?

(c) P(0  Z  1) = 0.3413

Properties * Mean,  = 0 * Standard deviation,  = 1 * The curve is symmetrical P(a < Z < b) = shaded area Use of Standard Normal Tables - The tables give the value P(0 < Z < a). -Diagrams are essential. z : Number of Standard Deviations from the mean, . Examples: (a) P(0  Z  1) = 0.3413

(c) P(0  Z  1) = 0.3413

Properties * Mean,  = 0 * Standard deviation,  = 1 * The curve is symmetrical P(a < Z < b) = shaded area Use of Standard Normal Tables - The tables give the value P(0 < Z < a). -Diagrams are essential. z : Number of Standard Deviations from the mean, . Examples: (a) P(0  Z  1) = 0.3413

Use of Standard Normal Tables - The tables give the value P(0 < Z < a). -Diagrams are essential. z : Number of Standard Deviations from the mean, . Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = ?

Use of Standard Normal Tables - The tables give the value P(0 < Z < a). -Diagrams are essential. z : Number of Standard Deviations from the mean, . Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = ?

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = P(0 <Z  3.2) – P(0<Z<0.3)

(c) P(0  Z  3.2) = ?

(c) P(0  Z  3.2) = 0.4993

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - ?

(c) P(0  Z  0.3) = ?

(c) P(0  Z  0.3) = 0.1179

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - ?

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179

Examples: (a) P(0  Z  1) = 0.3413 (b) P(-1  Z  1) = 2  0.3413 = 0.6826 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(b) P(-1  Z  1) = 2  0.3413 = 0.6826 (d) P(-0.326  Z  2.215) = ? (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(b) P(-1  Z  1) = 2  0.3413 = 0.6826 (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(c) P(-0.326  Z  0) = ?

(c) P(-0.326  Z  0) = 0.1255 + 0.0022

(c) P(-0.326  Z  0) = 0.1277

(b) P(-1  Z  1) = 2  0.3413 = 0.6826 (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) = 0.1277 + ? (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(c) P(0  Z  2.215) = ?

(c) P(0  Z  2.215) = 0.4864 + 0.0002

(c) P(0  Z  2.215) = 0.4866

(b) P(-1  Z  1) = 2  0.3413 = 0.6826 (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) = 0.1277 + 0.4866 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(b) P(-1  Z  1) = 2  0.3413 = 0.6826 (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) = 0.1277 + 0.4866 = 0.6143 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(e) P(Z  0.342 ) = ? (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) = 0.1277 + 0.4866 = 0.6143 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(e) P(Z > 0.342) = 0.5 – P(0 < Z < 0.342) = 0.5 - ? (d) P(-0.326  Z  2.215) = P(-0.326<Z  0) + P(0<Z<2.215) = 0.1277 + 0.4866 = 0.6143 (c) P(0.3  Z  3.2) = P(0<Z  3.2) – P(0<Z<0.3) = 0.4993 - 0.1179 = 0.3814

(e) P(Z > 0.342) = 0.5 – P(0 < Z < 0.342) = 0.5 - =

(c) P(0.3  Z  3.2) = 0.4993 – 0.1179 = 0.3814 (d) P(-0.326  Z  2.215) = 0.1278 + 0.4866 = 0.6144 (e) P(Z > 0.342) = 0.5 – P(0 < Z < 0.342) = 0.5 - 0.1339 = 0.3661 HW: 1.In NuLake: Do pages 296 & 297 (intro to Normal Distn). 2. In Sigma – match-up task: OLD edition - p98 (Ex. 7.1) – Q5  7 only. Or NEW edition – p 440 (exercise A.01) – Q5  7 only.

Lesson 2: Solve normal distribution problems Calculate probabilities using the normal distribution – standardise and use tables. Notes, NuLake pg. 300 Q30-37. HW: Finish NuLake Qs (30-37) *Extension (once NuLake Q30-37 done): Sigma (new edition): p358 – Ex. 17.01: Q4, 5, 10, 11, 14 & 15.

P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) Calculating probabilities for ANY normally- distributed random variable X, with mean,  Any normal random variable, X, can be transformed into a standard normal random variable by the formula: Example 1 If X is normally distributed with μ = 8 and σ = 2, find the probability that X takes a value greater than 11, STANDARDISING IT:  z =0  =1 z= 

P(0  Z  1.5) = ?

P(0  Z  1.5) = 0.4332

Calculating probabilities for ANY normally- distributed random variable X, with mean,  Any normal random variable, X, can be transformed into a standard normal random variable by the formula: Example 1 If X is normally distributed with μ = 8 and σ = 2, find the probability that X takes a value greater than 11, STANDARDISING IT:  z =0  =1 z=  P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer

Example 1 If X is normally distributed with μ = 8 and σ = 2, find the probability that X takes a value greater than 11, STANDARDISING IT:  z =0  =1 z=  Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer

P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer STANDARDISING IT:  z =0  =1 z=  Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) STANDARDISING IT:  =1    z =0

P(-0.6  Z  0) = ?

P(-0.6  Z  0) = 0.2258

P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer  z =0  =1 z=  Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) STANDARDISING IT:  =1   STANDARDISING IT:  z =0

P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) = 0.2258 + ?  z =0  =1 z=  Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. STANDARDISING IT:  =1   STANDARDISING IT:  z =0

P(0  Z  1.2) = ?

P(0  Z  1.2) = 0.3849

P( X > 11) = P( Z > ) = P(Z > 1.5) = 0.5 – P(0  Z  1.5) = 0.5 – 0.4332 = 0.0668 answer P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) = 0.2258 + 0.3849  z =0  =1 z=  Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. STANDARDISING IT:  =1   STANDARDISING IT:  z =0

P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) = 0.2258 + 0.3849 Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. STANDARDISING IT:  =1    z =0

P(175  X  184) = P(  Z  ) = P( -0.6  Z  1.2) = P(-0.6  Z  0) + P( 0  Z  1.2) = 0.2258 + 0.3849 = 0.6107 answer Example 2: The heights of a class of Y13 males are normally distributed with a mean of 178cm and standard deviation of 5cm. Find the probability that a randomly picked student from the class is between 175cm & 184cm tall. STANDARDISING IT:  z =0  =1  

17.05A Eg3: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. Find the probability that two consecutive passengers both have to pay for excess baggage.

P(excess) = 0.2266 = P(X > 20) Eg3: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. Find the probability that two consecutive passengers both have to pay for excess baggage.

Draw a tree diagram to show the possibilities for two suitcases. P(X > 20) = 0.2266 Eg3: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. Find the probability that two consecutive passengers both have to pay for excess baggage.

Eg3: The weights of suitcases received by an airline at check-in can be modelled by a normal distribution with mean 17 kg and standard deviation 4 kg. The airline charges for excess baggage if a suitcase is ‘overweight’—defined as weighing more than 20 kg. Find the probability that two consecutive passengers both have to pay for excess baggage. P(two cases overweight) = 0.2266  0.2266 = 0.0513 P(X > 20) = 0.2266 Over weight Under weight Over weight Under weight Over weight Under weight 0.2266 0.7734 0.2266 0.7734 * Do Nulake pg. 300-303 Q30  37 (MUST do). Extension: Sigma (new edition): p358 – Ex. 17.01: Q5, 10, 11, 14 & 15.

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  =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. 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. 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

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

Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

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Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward.

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Presentation on theme: "Statistics and Modelling Course 2011. Topic 5: Probability Distributions Achievement Standard 90646 Solve Probability Distribution Models to solve straightforward."— Presentation transcript:

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