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CIVE2602 - Engineering Mathematics 2.2 (20 credits) Statistics and Probability Lecture 4 Probability distributions -Poisson (discrete events) -Binomial.

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Presentation on theme: "CIVE2602 - Engineering Mathematics 2.2 (20 credits) Statistics and Probability Lecture 4 Probability distributions -Poisson (discrete events) -Binomial."— Presentation transcript:

1 CIVE2602 - Engineering Mathematics 2.2 (20 credits) Statistics and Probability Lecture 4 Probability distributions -Poisson (discrete events) -Binomial (discrete events) -Normal Distribution (continuous) Z- values Using Normal tables ©Claudio Nunez 2010, sourced from http://commons.wikimedia.org/wiki/File:2010_Chile_earthquake_- _Building_destroyed_in_Concepci%C3%B3n.jpg?uselang=en-gb Available under creative commons license

2 - Individual probabilities can be calculated from the distribution Probability distributions -different distributions exist for discrete or continuous data p(x) 60 mph Vehicle Speed on M62(x) Normal distribution Poisson distribution Binomial distribution (e.g. number of goals scored in football match) (e.g. proportion of defective items in a small sample Relative frequency is directly analogous to probability, so when we plot a histogram showing the relative frequency we are plotting a probability distribution. Total area =1.0 AREA under curve = 1

3 Can you use Normal tables? (A) YES (B) NO (C) maybe but I’ve forgotten

4 Probability Distributions - Normal distribution Clickers- Estimate how many miles you walk/run per week?

5 Probability Distributions - Normal distribution -Clickers – What is your height in cm (have a guess if not sure) I’m just under 6 foot tall and that’s about 183cm 5 foot is 152 cm

6 Measuring things

7

8 Frequency Distribution Function A graph showing the frequency of something – can be easily converted to show relative frequency/probability of an event -when we calculate the probability it becomes a probability distribution function

9 Measuring things Failure strength of a steel reinforcement sample ©Chris Willis 2009, sourced from http://www.flickr.com/photos/arcticpuppy/3783923342/ Available under creative commons license

10 Measuring things Percentage change in daily stock prices ©Tax_Rebate 2010, Sourced from http://www.flickr.com/photos/59937401@N07/5474194477/ Available under creative commons license

11 What do they all have in common?

12 ‘Normal’ distribution (continuous) Lots of datasets show variation which is ‘normal’ or ‘Gaussian’ “Bell shaped curve” DO NOT NEED TO USE THIS AS WE ALWAYS USE TABLES!! is the population variance of X, is the population mean is a mathematical constant (3.1415..).

13 f(X) Mean (μ) Cube strength (x) Approximately 95% of the data lies within the range given by -2.0σ to +2.0σ Probability Distribution Curves for the strength of Concrete Samples (Normally distributed) σ =7.0 σ =0.7 μ=4 1. Mean μ. 2. Variance σ 2 3. It is symmetric about μ. 4.Its range is from -infinity to +infinity 5.It is normalised so that 6. The notation is N(Mean, variance) e.g. N(3,9) i.e. the areas under the curve is one.

14 What is this useful for? If a dataset is normally distributed you can ask questions like – What is the probability that the height of a student will be greater than 170cm –What is the probability that the failure strength of a concrete sample will be above 200KN –What is the probability the wind speed will be between 5 -10 m/s Normal Distribution

15 μ=0 Standard normal distribution (special case with mean =0, standard deviation =1) All the datasets considered so far have different means (μ) and standard deviations But they can all be converted into a standard form by -Subtracting the mean (μ) -Dividing by the standard deviation (σ) The Standard Normal distribution is customarily denoted as Z. If we have a variable X which is distributed “N(μ, σ 2 )” we can convert it to a Z-value by the Z-transform operator: σ =7.0 μ=4 N(4, 7 2 )

16 It is therefore possible to convert from any Normal distribution with a given mean and standard deviation to the standard Normal distribution with mean=0 and standard deviation=1. This means that it is only necessary to have a copy of probability tables for the standard Normal distribution to be able to work out probabilities for any Normal distribution. μ = 4, σ = 2 So, X~N(4, 2 2 ) Example – The amount of pollution in a river (X) is normally distributed with mean 4, sd=2 Q) Find the required z-value to calculate p(X>6) USE NORMAL TABLES (z-tables) μ = 0, σ = 1 Z~N(0, 1)

17 It is therefore possible to convert from any Normal distribution with a given mean and standard deviation to the standard Normal distribution with mean=0 and standard deviation=1. This means that it is only necessary to have a copy of probability tables for the standard Normal distribution to be able to work out probabilities for any Normal distribution. μ = 4, σ = 2 So, X~N(4, 2 2 ) Example – The amount of pollution in a river (X) is normally distributed with mean 4, sd=2 Q) Find the required z-value to calculate p(X>6) USE NORMAL TABLES (z-tables) μ = 0, σ = 1 Z~N(0, 1)

18 -use tables to look up probabilities for the standard Normal distribution (as opposed to using the complex formula seen earlier). Being able to read from Normal tables is essential!! A SET OF NORMAL TABLES ARE AT THE BACK OF THE NOTES NORMAL TABLES The particular tables provided give the area, A(Z), to the right of a given Z-value (The first decimal of Z is given in the vertical margin, and the second decimal on the horizontal.) The body of the table reports the A(Z) values, equivalent to the probability of obtaining a value greater than the given Z value. QUESTION Find the required z-value to calculate a) p(X>5), b) p(X>2) (mean 4, sd=2)

19 QUESTION Find the required z-value to calculate a) p(X>5), mean 4, sd=2

20 QUESTION Find the required z-value to calculate a) p(X>5), mean 4, sd=2

21 What is the probability that the height of a student will be greater than 170cm? Need to calculate the z-score for 170cm given the information we know about the distribution of heights. Mean ( μ)= 162.43 cm Standard deviation) (S heights )= 6.28 cm Z score (170 cm)=

22 Find P(X>1.41) using Normal tables 1) D For a sample: mean = 0, standard deviation = 1

23 Multiple choice Choose A,B,C or D for each of these: NOW find P(X>4.3) using Normal tables 2) A B C D For a sample: mean = 3, standard deviation = 2 What you are looking for is P(Z>0.65)

24 NOW find P(X>4.6) using Normal tables 3) For a sample: mean = 3, standard deviation = 2 What you are looking for is P(Z>0.80)

25 Multiple choice Choose A,B,C or D for each of these: NOW find P(X>9) using Normal tables 4) A B C D For a sample: mean = 10.1, standard deviation = 2 What you are looking for is 1-P(Z>0.55)

26 What is the probability that a sample will failure strength below 20KN? Mean(µ) = 50.3 kN Std Dev (σ) = 10.2 kN

27 i.e. Pr(X 1 < X< X 2 ) = Pr(Z 1 < Z< Z 2 ) and X ~ N (μ,σ) Z ~ N(0,1) Mean X 1 X 2 μ Concrete Strength (X) TRANSFORM TO STANDARD NORMAL Z (standard normal distribution) Z 1 Z 2 Finding a probability within a range using Normal tables e.g. Pr(2 < X< 4)

28 Example 2 The crushing strength of a concrete beam is normally distributed, mean μ= 50kN, σ= 4.0 47 49 μ=50 What is the probability that a particular beam has strength between 47 and 49? Solution Transform to Z distribution We need Pr(-0.75 < Z < -0.25) WE NEED THIS WHICH IS EQUAL IN AREA TO THIS = -0.75 -0.25 0 0 0.25 0.75 - A(Z))=0.2263 Pr(-0.75 0.25) – P(Z > 0.75) = 0.40129 - 0.22663 = 0.17466 A(Z))=0.40129

29 What are the chances you will pass this year? Mean=54.3 S = 13.6

30 Things to remember Z-scores only work if your dataset if normally distributed. If you are assuming that data is normally distributed you should always state this ASSUMPTION (e.g. in course work). The distribution function looks like a bell shaped curve For other distributions similar techniques exist. ( we talk about some of them later in the course)

31 The compressive strength (X) of a number of concrete samples was found. For the samples: Mean(µ) = 50 kN and Std Dev (s) = 10 kN Assume the results are normally distributed- find 1)P(X>60) 2)P(X<40) 3)P(40<X<60) 4)P(X>30)

32 The compressive strength (X) of a number of concrete samples was found. For the samples: Mean(µ) = 50 kN and Std Dev (s) = 10 kN Assume the results are normally distributed- find 1)P(X>60) 2)P(X<40) 3)P(40<X<60) 4)P(X>30)

33 The following table describes when each of the distributions we have discussed so far is appropriate DistributionTypeMeanVarianceDescribes probability of Binomial Discrete np np(1-p)Selecting X ‘successes’ in a sample of n, with probability of success constant (e.g. infinite population, or with replacement). Poisson Discrete m mObserving X ‘rare’ events in a given time knowing the expected number of events in that time Normal Continuous μ σ 2 X occurring between two stated values, knowing μ and σ 2. Approximation Under certain conditions, one distribution can be used as an approximation to another. This is very useful, since it can be very tedious to evaluate probabilities for some distributions. The above ordering of distribution is by decreasing tedium, such that it is always desirable to approximate by a Normal distribution whenever possible. Approximating One Distribution by Another

34 When a continuous distribution, such as the Normal, is used as an approximation to a discrete distribution, such as the Binomial or Poisson, one must apply a continuity correction. This simply involves remembering that the continuous curve is fitted through the mid-points of the equivalent (discrete) histogram columns. Continuity Corrections (when approximating) E.g if the number of passengers on double-deck buses is approximately Normally distributed with mean 40 and standard deviation 15, what is the probability of a bus carrying more than 60 passengers (i.e. 61 or more passengers)? The variable X (no. of passengers) is discrete, (we cannot have half a passenger on a bus). 60.5 When standardizing we would therefore use X=60.5 as follows: Etc…

35 Question The tensile strength of individual re-enforced bars made of a certain manufacturing process are known to be normally distributed with mean 32KN and variance 16KN. 1) What is the probability that a bar will have a tensile strength of at least 24KN 2) If the customer requires 99% of the bars to be stronger than 22KN, would the process meet the specification?

36 Can you use Normal tables? (A) YES (B) NO

37 CIVE2602 - Engineering Mathematics 2.2 Lecture 4- Summary Normal Distribution Standard Normal Distribution Z-scores Using standard Normal tables to find probabilities EXAMPLES class and Mathlab will develop Normal dist


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