L2 Normal Distribution & Probability M. Guttormson Normal distribution and probability Normal Distribution –IntroductionIntroduction –Normal CurveNormal

L2 Normal Distribution & ProbabilityM. Guttormson

Normal distribution and probability

• Normal Distribution– Introduction– Normal Curve– Mean– Introducing Standard

Deviation– Standard Deviation and the

Normal curve– The Normal Distribution– The Standard Normal Distrib

ution– Expected Value– Inverse Normal Problems– Probability Exam Questions– Probability exam Answers

• Probability• Theoretical Probability

– Equally Likely Outcomes– Probability Trees– Sampling without

replacement– Venn diagrams

• Experimental probability– Long-Run relative

frequency– Simulation– describe a simulation– Simulation Question– Simulation Exam

Question


IntroductionThe chances of something happening “the next time” is not necessarily related to what has already happened, especially in many gambling games such as coin flipping. You flip a fair coin 20 times and it comes up heads every time. What is the probability it will come up tails next time? (Answer: 0.5, although the probability of a coin coming up the same 21 times in a row is only 0.000000477.) A couple already has two daughters. What is the probability that the next child is a son? (Answer: 0.5, assuming the gender of a child is completely random)


Introduction

This link simulates Galton's Board, in which balls are dropped through a triangular array of nails. This device is also called a quincunx. Every time a ball hits a nail it has a probability of 50 percent to fall to the left of the nail and a probability of 50 percent to fall to the right of the nail. The piles of balls, which accumulate in the slots beneath the triangle, will resemble a binomial distribution. To reach the bin at the far left the ball must fall to the left every time it hits a nail.


The Normal Curve

Many sets of data collected in nature, industry, business and other situations fit what is called a normal distribution.

This means:

•If the frequency of each piece of data were graphed, the curve would fit a symmetrical bell shape.•Most measurements are near the middle•The further away from the middle value (mean), the less frequent the occurrence.•The general shape looks like this


The Normal Curve

For example: The speeds of cars at a particular point on a motorway are normally distributed, with a mean of 95km/h.

95km/h The area under the curve represents 100% of the data collected. 50% of the data is on one side of the mean and 50% on the other side (symmetrical).


Mean

The mean is just the average of the numbers.

It is easy to calculate: Just add up all the numbers, then divide by how many numbers there are.


Introducing Standard Deviation

The Standard Deviation (σ) is a measure of how spread out numbers are.

You and your friends have just measured the heights of your dogs (in millimeters):

The heights (at the shoulders) are: 600mm, 470mm, 170mm, 430mm and 300mm.

Let’s find out the Mean and the Standard Deviation.



Answer:

so the average height is 394 mm. Let's plot this on the chart:

€

Mean =600 + 470 +170 + 430 + 300

5

Mean =1970

5Mean = 394



Calculate the Standard Deviation using your calculators.

Now we can show which heights are within one Standard Deviation (147mm) of the Mean:

So, using the Standard Deviation we have a "standard" way of knowing what is normal, and what is extra large or extra small.

Rottweillers are tall dogs. And Dachsunds are a bit short .


Standard Deviation and the Normal Curve

One standard deviation away from the mean in either direction on the horizontal axis (the red area on the graph) accounts for somewhere around 68 percent of the people in this group. “Likely/probable”

Two standard deviations away from the mean (the red and green areas) account for roughly 95 percent of the people.

“Very likely/very probable”

Three standard deviations (the red, green and blue areas) account for about 99 percent of the people. “Almost certain”

The total area under the curve adds to 1 (or 100%)

€

μ


The Normal DistributionNormally distributed data is described by giving the mean and standard deviation. Our car example: The speeds of cars at a particular point on a motorway are normally distributed, with a mean of 95km/h and a standard deviation of 10km/h.

65km/h 75km/h 85km/h 95km/h 105km/h 115km/h 125km/h 3 s.d 2 s.d 1 s.d 0 1 s.d 2 s.d 3 s.d


The Normal Distribution

What percentage of cars are travelling between 75km/h and 115km/h?

Between which two speeds is a car almost certain to be moving?

What percentage of cars are travelling more than 105km/h?

Below what speed is a car likely to be moving?


The Standard Normal Distribution

In the car example, how easy would it be to calculate the percentage of cars t ravelling more than 97km/h? We could easily estimate to be less than 50% but more than 16%. How can we be exact? Every set of data will have a different mean and standard deviation so we convert to what is called the standard normal distribution. We use the equation:

€

z =x − μ

σ

where

€

μ is the mean of the data set, and

€

σ is the standard deviation. z represents the number of standard deviations x is away from the mean.



The Standard Normal Curve has: A mean of zero (

€

μ=0) A standard deviation of one (

€

σ=1) It is symmetrical about the mean zero The area under the curve is equal to one The curve approaches the x-axis asymptotically. (i.e. it never touches) In theory the curve extends to infinity in either direction Its highest point is approximately (0, 0.4)



So now we can calculate the percentage of cars travelling more than 97km/h as follows,

€

z =97 −95

10

€

z = 0.2 Therefore, 97km/h is 0.2 standard deviations away from the mean of 95km/h



97km/h 0.2 65km/h 75km/h 85km/h 95km/h 105km/h 115km/h 125km/h 3 s.d 2 s.d 1 s.d 0 1 s.d 2 s.d 3 s.d Now we revert to the Normal Distribution Table to see what percentage this standard deviation equates to.





So the probability from the mean to a s.d of 0.2 is 0.0793 or 7.93% If we read the question again, we only want the area from the z value onwards (right). “…percent of cars travelling faster than 97km/h…”

0.0793

0.5-0.0793 = 0.4207



Notation

For the same question, calculate the percentage of cars travelling more than 97km/h?

What is the probability that X is greater than 97km/h

It is the probability that Z is greater than 0.2

This is equal to 1 – probability that Z is greater than 0.2

This is equal to 1 – 0.5793

Which is 42.07%

€

P(X > 97) = P(Z >97− 95

10)

€

=P(Z > 0.2)

€

P(Z > 0.2) = 1− P(Z < 0.2)

€

=1− 0.5793

€

=0.4207


Expected Value

Here, we are simply adding extra information to the question in the form of sample/population size. Example: In the car example, what is the expected value of cars exceeding 97km/h if 1500 cars were recorded?

€

1500×0.4207 = 631.05 Therefore 631 cars are expected to be travelling over 97km/h


Inverse Normal ProblemsWe can also look at these problems where we know the probability or percentage in advance, and have to find the cut off point that gives the probability. Consider this, We know that the mean speed of the cars is 95km/h and the standard deviation is 10km/h. Under what speed do 30% of cars travel?

Lets do this using the tables and equation. The bottom 30% will relate to finding the z value for 20% of the data. In the body of the z tables we find 0.20 to be equivalent to 0.524 s.d

65km/h 75km/h 85km/h 95km/h 105km/h 115km/h 125km/h 3 s.d 2 s.d 1 s.d 0 1 s.d 2 s.d 3 s.d


Inverse Normal Problems


Inverse Normal ProblemsTherefore entering this information into our rearranged equation,

€

z= x−μσ

€

x = zσ + μ

€

x=−(0.524×10)+95 (Because it is on the left hand side of the curve).

€

x =89.76 Therefore the speed that 30% of the cars travel less than is 89.8km/h


Equally Likely Outcomes

Much of probability is based on the assumption that in many situations events are equally likely to occur if random choices are made. Examples: A coin is equally likely to fall either “heads” or “tails” up. A six-sided die is equally likely to give any number from the set {1, 2, 3, 4, 5, 6} When outcomes are equally likely the probability of an event is:

Number of outcomes favourable for the event Total number of outcomes


Probability Trees

A probability tree is a branching diagram used to show alternative outcomes. Let’s examine the following example. A plant nursery specialising in selling native trees has a money back guarantee on two species, Kauri and Rimu. 40% of the trees sold under guarantee are kauri, and 60% are rimu. The nursery finds that 10% of the kauri trees are returned for a refund, but only 5% of the rimu trees are returned.

0.4

0.6

Kauri

Rimu

0.1

0.9

0.05

0.95

Returned

Not Returned

Returned

Not Returned

0.04

0.36

0.03

0.57


Probability Trees

If a tree is chosen at random, calculate the probability that it is a Kauri tree and is returned.

0.4

0.6

Kauri

Rimu

0.1

0.9

0.05

0.95

Returned

Not Returned

Returned

Not Returned

0.04

0.36

0.03

0.57

Multiply the probabilities along the kauri branch and the returned branch to get

€

0.4×0.1=0.04


Probability Trees

0.4

0.6

Kauri

Rimu

0.1

0.9

0.05

0.95

Returned

Not Returned

Returned

Not Returned

0.04

0.36

0.03

0.57

Calculate the probability that a tree chosen at random is returned.

There are two possible ways in which a tree can be returned. It could be Kauri or Rimu. Add the probabilities of these two outcomes to get

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0.04+0.03=0.07


Probability Trees

Be very clear as to when probabilities should be multiplied or added. Multiply along branches. Add probabilities when combining different outcomes, which are contributing to the desired result.

Action Where Key Word

Multiply Sideways along branches

“and”

Add Vertically at ends of branches

“or”


Sampling Without Replacement

This situation is found when the sample is finite, or there are a fixed number of outcomes. This is seen in situations where say a coloured marble is chosen at random from a bag and not replaced. What is the probability of choosing two consecutive blue marbles from a bag containing 4 blue and 3 red?

3/7

Blue

Red

1/2

1/2

2/3

1/3

Blue

Red

Blue

Red

2/7

2/7

2/7

1/7

4/7


Venn DiagramsIn some situations we cannot multiply probabilities (tree diagrams) because the events are not independent. For this reason we use sets. So events that occur in more than one set can be represented by the intersection of two or more sets. Ve nn diagrams show this clearly. A large rectangle is used to represent all possibilities Smaller groups or sets are represented by circles inside the rectangle. Overlapping circles show that the smaller groups have items in common.

€

A∪B

€

A

€

B

€

A

€

B

€

A ∩ B

Union (one or the other or both) Intersection (both only)


Venn Diagrams

€

A

€

A'

€

A

€

B

€

A

€

B

Complimentary EventsP(A) is the probability inside the circle.P(A’) is the probability outside the circle.P(A) + P(A’)=1Example: Probability that it rains tomorrow is 0.4.The probability that it does not rain tomorrow is 0.6

Mutually Exclusive EventsThis is when it is not possible to be in both groups at the same time.

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A∩B=∅ To find the probability of A or B, we add the individual probabilities of A and B.

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P(A∪B)=P(A)+P(B)−P(A∩B)

Intersecting EventsTo find the probability of

€

A∪B, add the probabilities of A and B. This counts the intersection twice, so subtract the probability of the intersection.

€

P(A∪B)=P(A)+P(B)−P(A∩B) Example:

€

P(A)=0.4,P(B)=0.2,P(A∩B)=0.1P(A∪B)=0.4+0.2−0.1


Long-run relative frequencyWhen the underlying probability of an event cannot be found exactly, we can still estimate it by taking many observations. The long-run relative frequency is given by: Number of times the event occurred total number of observations For example, we know from theoretical probability that the probability of getting a head from a fair coin is 0.5 The long-run relative frequency approaches 0.5, but at any given point, even after many simulated flips, we cannot expect it to be 0.5 exactly.


Simulation

What is a flight simulator? Why is it used? What are some of the advantages of using a flight simulator instead of flying an aircraft? Advantages of simulating probability experiments: Often there is no cost Because a simulation can be carried out quickly there is a saving in time and effort. Disadvantages of simulating probability experiments: It may be difficult to check you are actually using the right formulae A simulation imitates a real situation, and is supposed to give similar results, and so acts as a predictor of what should actually happen. It is a model in which repeated experiments are carried out for the purpose of estimating what might occur in real life.


How to describe a simulationTool: Definition of the probability tool

Statement of how the tool models the situation (Assign)

“Generate random numbers using the random number key on a calculator and truncate. 12ran#+1=”

“Designate numbers to colours of tee-shirts in the appropriate proportions.

1, 2 = blue3, 4, 5 = red6 = green7, 8 = purple9, 10 = yellow11, 12 = black

Trial: Definition of a trialDefinition of a successful outcome of the trial

“One trial would involve generating random numbers until there are seven outcomes representing each day of the week. Days need to be labelled Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, and Sunday.” A successful trial would be if 3 or more of the same colour shirt appears in one trial (one week).


How to describe a simulation

Results: Statement of how the results will be tabulated giving an example of a successful outcome and an unsuccessful outcome.

Statement of how many trials should be carried out.

“ A tick in the results column will indicate when 3 of the 7 random numbers generated represent the same colour tee-shirt. Repeat a minimum of 20 times.”

Calculation: Statement of how the calculation needed for the conclusion will be done:

Long run relative frequency = Number of successful results Number of trials

Mean = Sum of trial results Number of trials

The probabilities will be calculated using: Number of successful results 20


Simulation Question

The jump course in a horse paddock has 5 hurdles of equal difficulty. From past experience it is known that the mare Black Blaze has a chance of 1 in 3 to clear a hurdle. What is the probability that Black Blaze clears at least 3 of the 5 hurdles in the paddock? Describe your simulation


Simulation Exam Question “T-shirts”

Andrew has 12 t-shirts in six different colours. Each day he randomly chooses one of his t-shirts to wear. They are all available each day. He has 2 blue, 3 red, 1 green, 2 purple, 2 yellow, and 2 black shirts from which to make his choice each day. Describe your simulation (see notes example) What is the probability that Andrew chooses the same colour shirt on three or more days of a seven day week? How many weeks in a year (52) would he randomly choose to wear the same colour shirt on both Saturday and Sunday? How many weeks in a year (52) would he randomly choose to wear a purple shirt on at least one of the two days of the weekend?


Simulation Exam Question “T-shirts”Trial No.

MondayTuesdayWednesdayThursdayFridaySaturdaySunday

Chooses same colour

T-shirt at least 3 times

Chooses a purple T-shirt at least

once on the weekend

Chooses the same colour T-shirt on Saturday & Sunday

1 3 11 9 9 9 3 5 Yes No 12 7 2 10 7 7 1 3 Yes No 03 6 4 7 2 5 11 7 No Yes 04 5 4 3 2 3 5 2 Yes No 05 11 3 8 5 12 12 6 Yes No 06 10 8 7 5 6 6 3 No No 07 8 4 1 4 5 10 10 Yes No 18 9 4 8 5 1 4 5 Yes No 19 12 7 12 3 9 2 1 No No 1

10 6 7 4 6 11 9 4 No No 011 10 11 11 7 3 8 7 Yes Yes 112 6 7 1 8 10 7 6 Yes Yes 013 6 7 8 1 3 6 9 No No 014 12 1 4 4 5 5 5 Yes No 115 4 8 2 7 7 8 5 Yes Yes 016 6 3 1 2 2 6 5 Yes No 017 9 11 4 3 7 4 6 Yes No 018 9 7 3 2 9 3 10 Yes No 019 5 12 7 6 5 5 11 Yes No 020 1 4 1 9 9 9 7 Yes Yes 0

15 5 6

0.75 0.25 0.3 13 15.6

Result in next 52 weekends

Results

TotalPercentage

Outcomes


Simulation Exam Question “Fair Cop”In New Zealand, cars driven on the road must have a current warrant of fitness and be registered. A recent newspaper article contained the following statement: “80% of cars on the road in New Zealand have a current Warrant of Fitness and 90% and currently registered.” In response to the article the local police decided to investigate the situation in your town. They consider having a registration and having a warrant of fitness are independent. A police officer parked on the main street and gave out fines to motorists found breaching either of these requirements. Design a way of simulating the traffic flow which passes the police officer to find out how many of the next 50 drivers will get a fine. You need to:

1. Describe the method you use in suffi cient detail so that another person could repeat it again without your help.

2. Carry out at least 50 trials of your simulation. 3. Record the result of each trial of the simulation, e.g. in a table. 4. Use the results of your simulation to find the number of drivers who will be fined for

a breach of these requirements. 5. Use the results of your simulation to find the number of drivers out of the next 200

that pass the police offi cer that would be fined for both off ences (not being registered and not having a warrant of fi tness).

6. Use the results of your simulation to find the expected number of drivers who should be fined if 200 cars pass the police officer.


Simulation Exam Question “Fair Cop”Trial No. Trial No.

Random NumberWOFRandon NumberRegistrationFinedFined for both Offences Random NumberWOFRandon NumberRegistrationFinedFined for both Offences1 6 Yes 6 Yes No No 26 2 Yes 7 Yes No No2 8 Yes 9 Yes No No 27 8 Yes 10 No Yes No3 7 Yes 7 Yes No No 28 9 No 4 Yes Yes No4 4 Yes 4 Yes No No 29 2 Yes 4 Yes No No5 8 Yes 2 Yes No No 30 7 Yes 2 Yes No No6 4 Yes 9 Yes No No 31 1 Yes 2 Yes No No7 7 Yes 10 No Yes No 32 6 Yes 5 Yes No No8 7 Yes 9 Yes No No 33 5 Yes 3 Yes No No9 2 Yes 9 Yes No No 34 8 Yes 4 Yes No No

10 3 Yes 2 Yes No No 35 9 No 7 Yes Yes No11 6 Yes 10 No Yes No 36 6 Yes 4 Yes No No12 6 Yes 9 Yes No No 37 6 Yes 3 Yes No No13 1 Yes 10 No Yes No 38 3 Yes 6 Yes No No14 4 Yes 4 Yes No No 39 10 No 4 Yes Yes No15 2 Yes 6 Yes No No 40 7 Yes 9 Yes No No16 7 Yes 8 Yes No No 41 9 No 7 Yes Yes No17 8 Yes 5 Yes No No 42 2 Yes 6 Yes No No18 2 Yes 6 Yes No No 43 3 Yes 2 Yes No No19 5 Yes 2 Yes No No 44 3 Yes 4 Yes No No20 9 No 10 No Yes Yes 45 9 No 6 Yes Yes No21 4 Yes 8 Yes No No 46 8 Yes 7 Yes No No22 4 Yes 6 Yes No No 47 10 No 2 Yes Yes No23 8 Yes 2 Yes No No 48 5 Yes 5 Yes No No24 6 Yes 6 Yes No No 49 8 Yes 5 Yes No No25 7 Yes 6 Yes No No 50 6 Yes 5 Yes No No

4 1 7 011 1

444

Expected Value for both offences (200 drivers)Expected Value for a fine (200 drivers)

Subtotal fines

OutcomesOutcomes

Subtotal finesTotal Fines


Probability exam Questions

QUESTION ONE

Andrew collected a large sample of chest measurements from students of his year level.

He found the chest measurements to be approximately normally distributed with a mean chest measurement of 86cm and a standard deviation of 3.5cm.

(a) Calculate the probability that a randomly selected student from Andrew’s year level

(i) would have a chest measurement between 86cm and 90cm(ii) would have a chest measurement less than 87.5cm

(b) Andrew’s class has 34 students in it. How many of these students would you expect to have chest measurements of more than 89cm?


Probability exam Questions

QUESTION TWO Andrew finds that 12.7% of the students surveyed have chest measurements over 90cm. He also finds that if the students have chest measurements over 90cm, 75% of them prefer to wear black or purple tee-shirts at the weekend. If the students have chest measurements of 90cm or less, 30% of them prefer to wear black or purple tee-shirts at the weekend. Calculate the probability that a randomly selected student from Andrew’s year level would prefer to wear black or purple tee-shirts at the weekend.


Probability exam answers

Documents

L2 Normal Distribution & Probability M. Guttormson Normal distribution and probability Normal Distribution –IntroductionIntroduction –Normal CurveNormal