# “Teach A Level Maths” Vol. 3: S1

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“Teach A Level Maths” Vol. 3: S1

Volume 3 of “Teach A level Maths” covers the work on Probability and Statistics for the A/AS level Option Module S1. All topics for the 4 specifications offered by the English examining bodies are covered. Where a topic relates to some specifications only, this is indicated in a contents file and also at the start of the presentation.

Explanation of Clip-art images
An important result, example or summary that students might want to note. It would be a good idea for students to check they can use their calculators correctly to get the result shown. An exercise for students to do without help.

The slides that follow are samples from 9 of the 40 presentations.
23: Binomial Problems 4: Box and Whisker Diagrams 6: Histograms 26: Hypothesis Testing 10: Introduction to Probability 28: Standardizing to Z 14: Discrete Random Variables 36: Calculating Residuals 16: Linear Functions of a Discrete Random Variable

4: Box and Whisker Diagrams
Demo version note: The S1 specifications require students to be familiar with topics covered in Data Handling at GCSE. The first few presentations revise and extend the GCSE work. By the time the students reach this 4th presentation they have been reminded about cumulative frequency diagrams and have met the age data referred to on the slide.

Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative frequency diagram. The projected population of the U.K. for 2005 ( by age ) I’ll use the age data that we met earlier. The box can be any depth. minimum age One whisker The box The other whisker median maximum age lower quartile upper quartile

Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative frequency diagram. The projected population of the U.K. for 2005 ( by age ) I’ll use the age data that we met earlier. median minimum age maximum age lower quartile upper quartile

Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative frequency diagram. The projected population of the U.K. for 2005 ( by age ) I’ll use the age data that we met earlier. We need a scale. 100 50 Age (years)

6: Histograms Histograms
Demo version note: As well as explaining theory, the presentations show worked examples and set introductory exercises. The 6th presentation reminds students about the rules for drawing Histograms. The exercise shown here reinforces these rules without the students needing to spend time drawing a diagram.

Histograms Exercise 0-19 20-29 30-39 40-44 45-49 50-59 60-89 5 8 16 22
95 components are tested until they fail. The table gives the times taken ( hours ) until failure. Time to failure (hours) 0-19 20-29 30-39 40-44 45-49 50-59 60-89 Number of components 5 8 16 22 18 10 Find 3 things wrong with the histogram which represents the data in the table.

Histograms Answer: 0-19 20-29 30-39 40-44 45-49 50-59 60-89 5 8 16 22
Time to failure (hours) 0-19 20-29 30-39 40-44 45-49 50-59 60-89 Number of components 5 8 16 22 18 10 Frequency has been plotted instead of frequency density. There is no title. There are no units on the x-axis.

Histograms Incorrect diagram Time taken for 95 components to fail

Introduction to Probability
Demo version note: This presentation covers the introductory ideas of probability and leads to a later one on conditional probability. Summaries are given from time to time which teachers may want students to note down. This slide shows an example of a summary.

Introduction to Probability
SUMMARY Outcomes are the results of trials or experiments. An event is a particular result or set of results. A possibility space is the set of all possible outcomes. For equally likely outcomes, the probability of an event, E, is given by P (E) = number of ways E can occur number of possible outcomes

Discrete Random Variables
Demo version note: The presentations all contain worked examples of the straightforward type of questions found in exams. This is the first of the examples in the presentation on Discrete Random Variables.

Discrete Random Variables
e.g. 1. A random variable X has the probability distribution P x (X = ) p Find (a) the value of p and (b) the mean of X. Solution: (a) Since X is a discrete r.v., (b) mean, Tip: Always check that your value of the mean lies within the range of the given values of x. Here, or 5·25, does lie between 1 and 10.

Linear Functions of a Discrete Random Variable
Demo version note: Some topics are not required by all the specifications. The contents file shows which topics are needed by each of the specifications and contains hyperlinks to the files. The topic Linear Functions of a Discrete Random Variable is only required in S1 by Edexcel.

Linear Functions of a Discrete Random Variable
The results we have found can be generalised to give E(aX + b) = aE(X) + b e.g. The probability distribution for the r.v. X is given by 10 8 6 4 2 x Find (a) E(X), (b) Hence find E(2X - 3) Solution: (a) “Hence” in part (b) of the question means that we must use the answer to part (a) rather than using the values and probabilities of 2X - 3. (b)

23: Binomial Problems Binomial Problems
Demo version note: The Binomial Distribution is covered by AQA, MEI/OCR and OCR. Having learnt to carry out Binomial Calculations, students practise recognising the conditions for using the model and also learn the importance of defining a random variable and writing down its distribution.

Are the conditions met for using the Binomial model?
Binomial Problems e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect? Are the conditions met for using the Binomial model? A trial has 2 possible outcomes, success and failure. Yes: Each chip is either defective or not. The trial is repeated n times. Yes: 20 chips are selected so n = 20. The probability of success in one trial is p and p is constant for all the trials. Yes: We are given 15% (so p = 0·15 ) and we can assume it is constant. The trials are independent. Yes: The probability of selecting a defective chip does not depend on whether one has already been selected.

Binomial Problems e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect? Solution: Let X be the r.v. “number of defective chips” We must never miss out this stage since it reminds us that (i) X represents a number ( that can be 0, 1, 2, n ), and (ii) we have to make the decision as to whether to count the number of defective chips or perfect ones. So, Writing the distribution of X in this way makes us check that we have the p that fits our definition of the r.v., defective rather than perfect.

Binomial Problems e.g. 1. A factory produces a particular type of computer chip. Over a long period the number that are defective has been found to be 15%. What is the probability that in a sample of 20 taken at random, 19 are perfect? Solution: Let X be the r.v. “number of defective chips” So, The solution is now straightforward. We want We need to be very careful here and not use by mistake. I had set up the Binomial for the number of defective chips, because I had the proportion for defective. However, the question asked for the probability of 19 perfect ones. If I had written Let X be the r.v. “ number of perfect chips” Then, and I would want

26: Hypothesis Testing Hypothesis Testing
Demo version note: In the presentations extensive use is made of snapshots from the software package “Autograph”. Here Autograph is used to illustrate an example on Hypothesis Testing in the presentation for the MEI/OCR specification.

Hypothesis Testing e.g. 2. In a trial, 16 seeds are sown and only 11 germinate. Use a 10% significance level to test the supplier’s claim that 85% germinate. Find the critical region for the test. Solution: Let X be the random variable ”the number of seeds that germinate” Test at 10% level of significance. To test the supplier’s claim, the alternative hypothesis is that fewer than 85% germinate. This is again a 1-tailed test but this time we need to test the bottom end of the distribution. There is a probability of 0·0791 ( less than 10% ) that 11 or fewer seeds will germinate. We reject the null hypothesis at the 10% level of significance and conclude that the germination rate is below 85% .

( 21·01% ), so the critical region for the test is
Hypothesis Testing The Autograph illustration is as follows: The probability of 12 or fewer germinating is 0·2101 ( 21·01% ), so the critical region for the test is 0, 1, 2, , 11.

Standardizing to Z 28: Standardizing to Z Demo version note: Students are encouraged to use their Formulae and Statistical Tables even when worked examples are being developed. This presentation is part of a series to be used by AQA and Edexcel students on the Normal Distribution.

x = 400, so Standardizing to Z
e.g.1 If X is a random variable with distribution find (a) (b) Solution: (a) x = 400, so Tables only give 2 d.p. for z so this is all we need. So,

Standardizing to Z e.g.1 If X is a random variable with distribution find (a) (b) Solution: (b) There are 2 values to convert so we use subscripts for z. N.B. This is left of the mean so the z value will be negative. So,

Standardizing to Z e.g.1 If X is a random variable with distribution find (a) (b) Solution: (b)

Calculating Residuals 36: Calculating Residuals
Demo version note: Throughout the module, students are encouraged to use their calculators efficiently and this is particularly important in the topic for AQA, Edexcel and OCR on Least Squares Regression. In the following slides, however, the emphasis is on the effect of outliers on the equation of a regression line rather than on calculating the line itself.

Calculating Residuals
e.g. This is a scatter diagram of the data shown in the table. x y 1 5 2 18 3 12 4 14 5 12 If we were to draw the line “by eye”, the 1st point . . . 6 11 would lie well away from the line we would want to draw. 7 7 8 3 However, the calculation of the regression line includes the 1st point and distorts the position of the line.

Calculating Residuals
e.g. This is a scatter diagram of the data shown in the table. 3 8 7 11 6 12 5 14 4 18 2 1 y x The diagram shows the y on x regression line for all the data. The residuals are shown by the red lines. The left-hand end of the line is further down than it would be without the 1st point.

Calculating Residuals
e.g. This is a scatter diagram of the data shown in the table. 3 8 7 11 6 12 5 14 4 18 2 1 y x Removing the 1st point . . .

Calculating Residuals
e.g. This is a scatter diagram of the data shown in the table. 3 8 7 11 6 12 5 14 4 18 2 1 y x Removing the 1st point gives

Calculating Residuals
e.g. This is a scatter diagram of the data shown in the table. The sum of the squares of the residuals, Removing the 1st point gives The sum of the squares of the residuals, Without the 1st point, we have a regression line that is a much better fit.

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