For use only in Badminton School November 2011 S1 Note. S1 Notes (Edexcel)

Transcription

1 For use only in Badminton School November 011 s (Edexcel) Copyright - For AS, A notes and IGCSE / GCSE worksheets 1

2 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets

3 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 3

4 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 4

5 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 5

6 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 6

7 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 7

8 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 8

9 For use only in Badminton School November 011 Copyright - For AS, A notes and IGCSE / GCSE worksheets 9

10 For use only in Badminton School November 011 Definitions for S1 Statistical Experiment A test/investigation/process adopted for collecting data to provide evidence for or against a hypothesis. Explain briefly why mathematical models can help to improve our understanding of real world problems Simplifies a real world problem; enables us to gain a quicker / cheaper understanding of a real world problem Advantage and disadvantage of statistical model Advantage : cheaper and quicker Disadvantage : not fully accurate Write down two reasons for using statistical models To simplify a real world problem To improve understanding / describe / analyse a real world problem Quicker and cheaper than using real thing To predict possible future outcomes Refine model / change parameters possible Any Statistical models can be used to describe real world problems. Explain the process involved in the formulation of a statistical model. Observe real-world problem Devise a statistical model and collect data (Experimental) data collected Model used to make predictions Compare and observe against expected outcomes and test model; Statistical concepts are used to test how well the model describes the real-world problem Refine model if necessary. A sample space A list of all possible outcomes of an experiment Event Sub-set of possible outcomes of an experiment. Normal Distribution Bell shaped curve symmetrical about mean; mean = mode = median 95% of data lies within standard deviations of mean 68.3% between one standard deviation of mean Independent Events PA ( B) PA ( ) PB ( ) Mutually Exclusive Events PA ( B) 0 Copyright - For AS, A notes and IGCSE / GCSE worksheets 10

11 For use only in Badminton School November 011 Explanatory and response variables The response variable is the dependent variable. It depends on the explanatory variable (also called the independent variable). So in a graph of length of life versus number of cigarettes smoked per week, the dependent variable would be length of life. It depends (or may do) on the number of cigarettes smoked per week. Give two reasons to justify the use of statistical models Used to simplify or represent a real world problem Cheaper or quicker or easier (than the real situation) or more easily modified (any two lines) To improve understanding of the real world problem B1 Used to predict outcomes from a real world problem (idea of predictions) Describe the main features and uses of a box plot. Two tests for skewness Positive skew if Q Q Q Q and if Mean > Median > Mode Negative skew if Q Q Q Q and if Mean < Median < Mode A good way to remember the condition for skewness involving mean, median and mode. Write down the three averages in alphabetical order, that is Mean Median Mode For positive skew fill the blank with a > sign to give Mean > Median > Mode For negative skew fill the blank with a < sign to give Mean < Median < Mode Which sort of average and which sort of measure of dispersion should you use? If there are outliers or if the data is skewed then use median and IQR, If the are not outliers then use mean and standard deviation. Copyright - For AS, A notes and IGCSE / GCSE worksheets 11

12 For use only in Badminton School November 011 Data Discrete Discrete data can only take certain values in any given range. Number of cars in a household is an example of discrete data. The values do not have to be whole numbers (e.g. shoe size is discrete). Continuous Continuous data can take any value in a given range. So a person s height is continuous since it could be any value within set limits. Categorical Categorical data is data which is not numerical, such as choice of breakfast cereal etc. Data may be displayed as grouped data or ungrouped data. We say that data is grouped when we present it in the following way: Weight (w) Frequency Or Score (s) Frequency NB: We can group discrete data or continuous data. We must know how to interpret these groups, So that Weight (w) w w 75 Or Score (s) s s 14.5 Copyright - For AS, A notes and IGCSE / GCSE worksheets 1

13 For use only in Badminton School November 011 Representation of Data Histograms, stem and leaf diagrams, box plots. Use to compare distributions. Back-to-back stem and leaf diagrams may be required. Stem and Leaf Diagrams The stem and leaf diagram is a very useful way of grouping data whilst retaining the original data. For example suppose we had the following scores from children in a Maths test: 85, 18, 38, 67, 43, 75, 78, 81, 9, 71, 5, 6, 49, 6, 8, 69, 55, 57, 95, 6, 37 We see that the smallest value is 18 and the largest is 95. The classes of stem and leaf diagrams must be of equal width and so it would seem sensible to choose classes 10-19, 0-9, etc. The stem in this case represents the tens and the leaf represents the units so we have the following: Scores in Maths Test Stem (Tens) Leaf (Units) We then arrange these in numerical order to give the following: Scores in Maths Test Stem (Tens) Leaf (Units) NB : the data must be in order in a Stem and Leaf Diagram. We m also include a key with the diagram, so we say 1 8 means 18 This diagram tells us the basic shape of the distribution. We can easily see the smallest and largest values and we can see that the mode is 6. We can also use it to calculate Q 1, Q and Q 3. Copyright - For AS, A notes and IGCSE / GCSE worksheets 13

14 For use only in Badminton School November 011 NB: If we wanted to represent the interval 18- on a stem & leaf we could not make 1 the stem since not all the numbers would begin with 1. What we could do is have a stem of 18 and then make the leaf the number we add on to the stem. In this case our key would be: 18 0 means 18 and 18 4 means Back to back stem diagrams We can use these to compare two samples by using a back to back stem plot. In this we put stems down the middle and then one set of data on the left and the on the on the right. So we might end up with a diagram as follows: Physics Maths Our key here would be In Physics 7 1 means 17 In Maths 1 8 means 18 Copyright - For AS, A notes and IGCSE / GCSE worksheets 14

15 For use only in Badminton School November 011 Histograms Data that has been grouped can be represented using a histogram. A histogram is made up of rectangles of varying widths and heights there are no gaps between the blocks. The key feature of a histogram is that the area of each block is proportional to the frequency In order for the area to be equal (or proportional) to the frequency we plot frequency density on the frequency vertical axis, where frequency density. The class width is the width of the interval class width (i.e. it runs from the lower boundary to the upper boundary). Remember Frequency Density is Frequency Divided by class width. Example Plot a histogram for the following: Length (h) Frequency Class width Frequency Density So the first block runs from 650 to 670 and has height 0.15 etc. FD Length NB: If there are gaps between the stated upper limit of one class interval and the lower limit of the next class interval then we need to fill those gaps as shown below. For example, When question says give a reason to justify the use of a histogram to represent these data. The answer is Data is continuous Length (m) x x x 9.5 So the class width is 5. Not NB: Be careful with age since would mean 15 x 0 since one is 19 until the moment before one s 0 th birthday. The shape of the histogram gives us information about the mean and the dispersion Copyright - For AS, A notes and IGCSE / GCSE worksheets 15

16 For use only in Badminton School November 011 Box Plot Diagram (or Box and whisker diagram) This is a diagram used to illustrate the dispersion of data. There is a box which runs from the lower quartile, Q 1 to the upper quartile Q 3 with the median, Q marked on it. The whisker then goes from this box to the lowest value in one direction and to the highest value in the other. We end up with a diagram as follows: Lowest Value Q1 Q Q 3 Highest Value NB : It must have a horizontal axis with a scale on it. Skewness. Concepts outliers. Any rule to identify outliers will be specified in the question. If the question refers to outliers then use a refined box plot where the maximum length for the whisker is to, for example, 1.5Q3 Q1 at the end where an outlier lies. In this case outliers (which lie outside of the whiskers) are marked with crosses. Note that the whisker does not need to take the its maximum length. It goes only to the lowest value on this side. Lowest Value Q1 Q Q 3 Outlier Q 1.5 Q Q NB We can tell something about the skewness / symmetry of the distribution from the box plot. For example, Lowest Value Q1 Q Q 3 Highest Value Copyright - For AS, A notes and IGCSE / GCSE worksheets 16

17 For use only in Badminton School November 011 Q Q Q Q with a long tail of We can see from the above that this is positively skewed high values. Similarly, Lowest Value Q1 Q Q 3 Highest Value The above is negatively skewed since Q Q Q Q Copyright - For AS, A notes and IGCSE / GCSE worksheets 17

18 For use only in Badminton School November 011 Measures of Location Measures of location - mean, median, mode. Data may be discrete, continuous, grouped or ungrouped. Understanding and use of coding. Measures of dispersion variance, standard deviation, range and interpercentile ranges. Simple interpolation may be required. Interpretation of measures of location and dispersion. Ungrouped Data Mean sum of all the values The mean of a set of data is, which can be written as x number of values x fx n f. When the data is grouped so that it appears in class intervals we must use the midpoints of those intervals to calculate the mean (in which case it is only an estimate of the mean) Example Find the mean of the following weights (in kg): 75.6, 81.4, 67.3, 81.5, 84.0, 6.8, 79.1, 85.6, 79.3, 85., 69.1, 75. Also given the letter. x kg 1 Example Find the mean of the following scores: Score Frequency x Median, Quartiles, and Percentiles The lower quartile of a set of data is the value one quarter of the way into the data set, when written in ascending order. It is represented by the symbol Q 1. The media n of a set of data is the value one half of the way into the data set, when written in ascending order. It is represented by the symbol Q. The upper quartile of a set of data is the value three-quarters of the way into the data set, when written in ascending order. It is represented by the symbol Q 3. There are many methods for calculating Q 1, Q and Q 3 but the following method is recommended by Edexcel and will get full credit in exams. The x th percentile, P x, to be the value which is x% of the way through the distribution. If there are n observations then the x th percentile is calculated as follows: xn If 100 is an integer r then the x th percentile is the midpoint of the rth and r 1 th values. xn If 100 lies between the integers r and r 1 then the x th percentile is the r 1 th value Copyright - For AS, A notes and IGCSE / GCSE worksheets 18

19 For use only in Badminton School November 011 We use the same method to find Q 1, Q and Q 3, treating them as the 5th, 50th and 75th percentiles respectively. If there are n observations then. the lower quartile is calculated as follows: n If 4 is an integer r then the lower quartile is the midpoint of the rth and r 1 th values. n If 4 lies between the integers r and r 1 then the lower quartile is the r 1 th value. the median is calculated as follows: n If is an integer r then the median is the midpoint of the rth and r 1 th values. n If lies between the integers r and r 1 then the median is the r 1 th value. the upper quartile is calculated as follows: If 3 n 4 is an integer r then the upper quartile is the midpoint of the rth and r 1 th values. If 3 n 4 lies between the integers r and r 1 then the upper quartile is the r 1 th value. Example Find the median and quartiles of the following weights (in kg): 75.6, 81.4, 67.3, 81.5, 84.0, 6.8, 79.1, 85.6, 79.3, 85., 69.1, 75. First of all list them in order: 6.8, 67.3, 69.1, 75., 75.6, 79.1, 79.3, 81.4, 81.5, 84, 85., 85.6 There are 1 pieces of data so the lower quartile is the midpoint of the 3rd and 4th values, that is kg 1 6 so the median is the midpoint of the 6th and 7th values, that is 79.kg so the median is the midpoint of the 9th and 10th values, that is 8.75kg The interquartile range is Q3 Q1. Example Find the median of the following scores: Score Frequency There are 1 pieces of data: so the lower quartile is 6th value, that is so the median is the 11th value, that is so the upper quartile is the 16th value, that is 15. Copyright - For AS, A notes and IGCSE / GCSE worksheets 19

20 For use only in Badminton School November 011 Grouped Data Mean Use the midpoints of each category and use the same method for ungrouped data. Example Find an estimate of the mean of the following data: Height, h (cm) Frequency x 153 cm (to 3sf) Median When dealing with grouped data we have to interpolate to find estimates for Q 1, Q and Q 3. Example Find the median of the following data: Height, h (cm) Frequency There are 1115 pieces of data so the median is the height associated with the th value. This lies somewhere between 150cm and155cm. The first 385 (that is ) values have a height less than 150cm The first 739 (that is ) have a height less than 155cm cm Median 155cm So the median lies 17.5 of the way into the interval so our estimate for the median is cm (to 1dp). 354 With grouped data, work with n th value for the median (eg ), do not round up (e.g. to 558) as with ungrouped data. Use exactly the same method for finding lower quartile and upper quartile So the lower quartile is the height associated with the th value. This lies somewhere between 145cm and150cm The lower quartile lies of the way into the interval so our estimate for the lower quartile is cm (to 1dp). 45 Copyright - For AS, A notes and IGCSE / GCSE worksheets 0

21 For use only in Badminton School November 011 Mode The mode is the most frequently occurring value. Example Find the mode of the following weights (in kg): 75.6, 81.4, 67.3, 81.5, 84.0 There is no mode because all values occur with equal frequency. Example Find the mode of the following scores: Score Frequency The mode is 14. Example Find the modal class of the following data: Height, h (cm) Frequency Frequency Density The modal class is the class interval with the highest frequency density and so is Variance Suppose that we have the following frequency table: Value x 1 x x 3 x 4 Frequency f 1 f f 3 f 4 x r f r The variance is a measure of spread, or dispersion. It is the average of the square of the difference of each piece of data from the mean. That is, it is the average of all the values ( x i x). ( x x) f So we have, Variance. f This can be rearranged so that it becomes easier to use and we have x f Variance x. f We tend not to put in the f and to write f as n, giving Variance x x n mean of the squares minus the square of the mean Copyright - For AS, A notes and IGCSE / GCSE worksheets 1

22 For use only in Badminton School November 011 Standard Deviation The standard deviation is simply the square root of the variance.. ( x x) x Hence the standard deviation,, is given by x n n Example Find the standard deviation of the following weights (in kg): 75.6, 81.4, 67.3, 81.5, 84. First of all we calculate the mean as x We use this to get that the standard deviation is kg (to 3sf) 5 Example Find the standard deviation of the following scores: Score Frequency First of all we calculate the mean as x We use this to get that the standard deviation is Example Find an estimate of the standard deviation of the following data: Height, h (cm) Frequency The only difference between this and the previous example is that we have to use midpoints. First of all we calculate the mean as x cm (to 3sf) We use this to get that the standard deviation is cm (to sf) 1115 Copyright - For AS, A notes and IGCSE / GCSE worksheets

23 For use only in Badminton School November 011 Understanding and use of coding. Coding Suppose we had to find the mean and standard deviation of , , 00.04, Suppose now that we now consider t 1000 x 00. Our values of t are 37, 39, 4, 54 which are easier to deal with than the initial data. The summary statistics for t are : n 5, t 17 and t It follows from this that t 43 and t (to 3sf). 4 4 Shifting numbers along by 00 has no effect on the dispersion (i.e. the standard deviation) but it does shift the mean by 00. Dividing by 1000 has the effect of reducing both the mean and the standard deviation by a factor of t Since x 00 it follows that 1000 t t x 00 and that x Hence we see that 43 x and that x (to 3sf) In general if we have x at b then x at b and x a t. Copyright - For AS, A notes and IGCSE / GCSE worksheets 3

24 For use only in Badminton School November 011 On Texas Using calculator to find mean and standard deviation Height (cm) Frequency Press to get: Now press to get Type in the following: Press and choose CALC, 1-Var Stats to see this: Now type in to get Copyright - For AS, A notes and IGCSE / GCSE worksheets 4

25 For use only in Badminton School November 011 Press to see the following Write down: 800 x (to 3sf) sd 40 IGNORE Sx for S1. Copyright - For AS, A notes and IGCSE / GCSE worksheets 5

26 For use only in Badminton School November 011 On Casio Scroll down using to get Press 3 to get Press 1 Press then (STAT) and then 1 (1-VAR) to get Type in all the data to get Press For x and x Press to give Choose 4 (sum) to get Choosing 1 gives Choosing again. Choose 4 (sum) and then gives 800. Copyright - For AS, A notes and IGCSE / GCSE worksheets 6

27 For use only in Badminton School November 011 For the value of n, standard deviation and mean Press to give Choose 5 (Var) to give Choosing 1 gives n to be 40. For standard deviation Choose 5 (Var) and then 3 to get So standard deviation us 6.3 (to 3sf) Copyright - For AS, A notes and IGCSE / GCSE worksheets 7

28 For use only in Badminton School November 011 Skewness and Outliers Skewness. Concepts outliers. Any rule to identify outliers will be specified in the question. Skewness A distribution that is not symmetric is said to be skewed. Positive Skew If the distribution has a long tail of high values then it is said to be positively skewed. In this case the median is less than the mean. Mode Median Mean Negative Skew If the distribution has a long tail of low values then it is said to be negatively skewed. In this case the median is greater than the mean. Mean Median Mode Copyright - For AS, A notes and IGCSE / GCSE worksheets 8

29 For use only in Badminton School November 011 Two tests for skewness Positive skew if Q Q Q Q and if Mean > Median > Mode Negative skew if Q Q Q Q and if Mean < Median < Mode A good way to remember the condition for skewness involving mean, median and mode. Write down the three averages in alphabetical order, that is Mean Median Mode For positive skew fill the blank with a > sign to give Mean > Median > Mode For negative skew fill the blank with a < sign to give Mean < Median < Mode Outliers Values which are at one of the extremes of the distribution are said to be outliers. How extreme a value has to be in order to be classified as an outlier will be stated in the question. One measure of an outlier would be if the value was greater than Q3 1.5Q3 Q1 it was less than Q 1.5Q Q at the other end at one end or if Copyright - For AS, A notes and IGCSE / GCSE worksheets 9

30 For use only in Badminton School November 011 Probability Elementary probability. Sample space. Exclusive and complementary events. Sum and product laws. The probability of Tottenham Hotspur winning their next game is The probability of scoring a six when a die is thrown is 1 6. The probability of a part made in a factory being faulty is 0.0. Probabilities can be expressed either as a fraction or a decimal. All probabilities lie between 0 and 1. If an event has a probability of 0 then it will certainly not occur. It is more likely that an event occurs as its probability moves along this line. If an event has a probability of 1 then it will certainly occur. 0 p event not happening1 pevent happening Sets and Venn Diagrams A set is a collection of items such as numbers, people, letters etc. For example Set A is the set of pupils in year 11 who wear glasses Set B is the set of pupils in year 11 who have blonde hair A Venn diagram is a diagram which represents various sets. It is made up of a large rectangle with various oblongs inside to represent the various sets. These may or may not overlap. A B By the side of the large rectangle is the symbol. This represents the universal set and so could be, for example, all year 11 pupils. In this example only year 11 pupils in this school would feature in the Venn diagram. If an element is in set A and also in set B then it lies in the intersection of the two oblongs. If an element is neither in A or B then it lies inside the rectangle but outside both oblongs. Copyright - For AS, A notes and IGCSE / GCSE worksheets 30

31 For use only in Badminton School November 011 Example The diagram shown below is a Venn diagram indicating who enjoy swimming and running in a class Running Swimming (a) How many students enjoy both running and swimming? (b) How many enjoy running? (c) How many enjoy neither running nor swimming? (a) (b) (c) There are 5 students who enjoy both running and swimming as there are 5 students lying in the intersection of the two oblongs. There are 13 who enjoy running. That is the 8 who enjoy running but not swimming and the 5 who enjoy both running and swimming. There are 10 students who enjoy neither running nor swimming. These are the 10 who lie outside both oblongs. Example In a class of 33 pupils, 0 like chess, 1 like draughts and 5 like neither. Represent this on a Venn diagram. Those who like chess will be represented by one oblong, those who like draughts will be represented by another oblong. Those who like both are in the overlap of the oblongs whilst those who like neither will lie outside both oblongs. So the 4 who like neither go outside the two oblongs. Chess Draughts 5 These 5 pupils do not lie in either oblong so that leaves 8 to be put in the three positions of the oblongs. How is the rest of the Venn diagram to be filled in? Think of the two oblongs are being two rugs which can overlap. One rug has area 0, one has area 1. The combined area has to be 8. If the oblongs did not overlap their combined area would be 3 so the overlap must be 4. So There are 0 who like chess in all so 16 must go in here. Chess Draughts 5 There are 1 who like draughts in all so 8 must go in here Copyright - For AS, A notes and IGCSE / GCSE worksheets 31

32 For use only in Badminton School November 011 Notation A B represents all the elements A B represents all the elements which belong to A or B or both. which belong to A and B. A B A B A B represents the fact that A is a subset of B A represents all the elements outside of A. A B A B NB When A is a subset of B, A B B. A B represents the fact that A is not a subset of B From this we see that P AB P A P B P A B A B Mutually Exclusive In this example AB since the two oblongs do not intersect. A B This is an example of two events being mutually exclusive. A and B are said to be mutually exclusive if they cannot both occur at the same time, for example getting a six and a five from one throw of the die. If A and B are mutually exclusive then P( A B) P( A) P( B). Copyright - For AS, A notes and IGCSE / GCSE worksheets 3

33 For use only in Badminton School November 011 Example If there are 7 red balls, blue balls and 6 yellow balls in a bag. A ball is chosen out of the bag then what is the probability that the ball is: (a) blue (b) red (c) either red or blue (a) out of 15 of the balls are blue so the probability of picking a blue ball is 15 (b) 7 out of 15 of the balls are blue so the probability of picking a red ball is 7 15 (c) The events of being red and blue are mutually exclusive so: Probability of being red or blue = Probability of being red + Probability of being red BE CAREFUL Example A spinner is equally likely to land on the numbers 1,, 3, 4, 5 or 6. Find the probability that the spinner lands on: (a) an even number (b) a number greater than 4 (c) an even number or a number greater than 4 (a) There are three even numbers (that is, 4, 6) so the probability is (b) There are two numbers greater than 4 (that is 5 and 6) so the probability is (c) The event of landing on an even number and the event of landing on a number greater than 4 are not mutually exclusive so the probabilities cannot be added. The numbers which are even or greater than 4 are, 4, 5 or 6. Hence the probability is Example The probability of a girl in a school wearing glasses is 0.3 and the probability of a girl having blond hair is 0.4. Joshua thinks that the probability of wearing glasses or having blond hair is 0.7. Is he right? The answer is that he is not right. Some people have blond hair and wear glasses. In other words, wearing glasses and having blond hair are not mutually exclusive, so their probabilities cannot be added together. Copyright - For AS, A notes and IGCSE / GCSE worksheets 33

34 For use only in Badminton School November 011 Set notation Set A might be all the blonde haired pupils in year 11 in a school. It could also be the set of even numbers less than 10. In which case curly brackets are used and A is written as A, 4,6,8. 4 A means that 4 is an element of A. 9 A means that 9 is not an element of A. ( ) A, 4,6,8 then na ( ) 4. na represents the number of elements in A. If Example is all whole numbers less than 10. A x: x is an even number (a) Express this on a Venn diagram. (b) List all the elements of A. (c) What is na B? (d) Find A B. B x: x is a prime number (a) Since is all whole numbers less than 10, the only numbers appearing on the Venn diagram are 1,, 3, 4, 5, 6, 7, 8 and 9 1, 9 (b) A represents all the elements outside of A. A 1, 3, 5, 7, 9 So (c) A B represents all the elements which belong to A or B or both. A B,3, 4,5,6,7,8 4, 6, 8 3, 5, 7 A B So. There are 8 elements in this set so n A B 8. (d) A B represents all the elements which belong to A and B. So AB since is the only number which is both even and prime. Shading Unions Example A B includes any region that is shaded in A or B (or both). = A B A B Area represented by A Area represented by B Area represented by A B Copyright - For AS, A notes and IGCSE / GCSE worksheets 34

35 For use only in Badminton School November 011 Example A B includes any region that is shaded in A or B (or both). A B A C A C = C B B Area represented by A Area represented by B Area represented by A B Shading Intersections Example A B includes any region that is shaded in both A and B. = A B A Area represented by A Area represented by B Area represented by A B B A B Some important regions to know A B A B A B A B A B A B A B A B A B A B A B B Copyright - For AS, A notes and IGCSE / GCSE worksheets 35

36 For use only in Badminton School November 011 Combined Events P A B means the probability of A happening, given that B has happened. A B P A B means the probability of A happening, given that B has happened. P A B Referring to the Venn diagram, represents the probability of being in the shaded area PA ( B) So PAB ( ) P B The probability of A happening given that B happens is PA ( B) represented by PA B where PAB ( ). P B Independent Events Two events A and B are said to be independent if they have no effect on each other (e.g. being late to school in Kenya and raining in Bolivia are independent). If A and B are independent then B happening has no effect on A happening so PA B PA If A and B are independent then B then, since PA ( ) PA ( B) and so PA B PA PB PB. PA ( B) PAB ( ), it follows that P B If A and B are independent then PAB PAPB. Copyright - For AS, A notes and IGCSE / GCSE worksheets 36

37 For use only in Badminton School November 011 Exam Question (S1, January 006) For the events A and B, P(A B) = 0.3, P(A B) = 0.11 and P(A B) = (a) Draw a Venn diagram to illustrate the complete sample space for the events A and B. (b) Write down the value of P(A) and the value of P(B). (c) Find P(AB). (d) Determine whether or not A and B are independent. (a) (3) (3) () (3) A B (b) PA PB (c) If PB 0.33 then PB P A B PA B 0.478to 3dp P B (d) PAPB PAB 0. So PAB P A PB and so A and B are not independent. P A and PA B 0.478to 3dp so P A B P A OR not independent. and so A and B are Copyright - For AS, A notes and IGCSE / GCSE worksheets 37

38 For use only in Badminton School November 011 Example If a coin is tossed and a die is rolled then find the probability of getting a head on the coin and a five on the die. Suppose the event A is getting a head on the coin and that the event B is getting a five on the die. 1 1 pa and pb These are independent events so p A and B pa pb Example Alexander fires three arrows at a target. With each arrow, the probability that he hits the target is 0.3. Whether he hits with any given arrow is independent of what happened to the previous arrows he fired. Find the probability that he hits with the first two but misses with the third. The probability that he hits the target with his first arrow is 0.3. The probability that he hits the target with his second arrow is 0.3. The probability that he misses the target with his third arrow is Since the events are assumed to be independent, the probability that he hits with the first and hits with the second and misses with the third is obtained by multiplying the three numbers given above. So the probability that he hits with the first two but misses with the third is Copyright - For AS, A notes and IGCSE / GCSE worksheets 38

39 For use only in Badminton School November 011 Tree Diagrams Tree diagrams are a very clear way of representing the possible outcomes of combined events. On tree diagrams: as we move across we use a cross () and so multiply probabilities as we move Down we add probabilities Questions often involve choosing two or more balls from a bag. In these questions it is vital to establish whether or not the first ball has been replaced before the next one has been chosen. Read the question carefully. The case when the ball is replaced. Example A bag contains 10 balls, seven of which are red and the rest are green. A boy takes out a ball from the bag, notes its colour and puts it back. He then repeats this process. (a) Draw a tree diagram to represent this information. (b) Find the probability that he chose two red balls. (c) Find the probability that he chose two balls of different colours. (a) Notice that the probabilities on the branches that start from the same point always add up to 1. 7/10 R 7/10 3/10 R 49/100 G 1/100 as we move across we use a cross (). 7/10 R 1/100 3/10 G 3/10 G 9/100 (b) Probability of two reds is (as we move across we use a cross ()) (c) p( RG) and p( GR) As we move Down we add so probability is Copyright - For AS, A notes and IGCSE / GCSE worksheets 39

40 For use only in Badminton School November 011 The case when the ball is not replaced. Example 3 red balls and green balls in a bag. One ball is chosen at random, its colour is noted and it is not replaced. A second ball is then chosen and its colour is also noted. Find the probability that: (a) Both balls are red. (b) The second ball is green (c) At least one of the balls is green. (d) The first ball was red, given that the second ball was green. When the 5 balls are in the bag, the probability of choosing a red is 3 5, since 3 of the 5 balls are red. 4 R If a red ball is chosen first, then there are now 4 balls in the bag and only of them are red. The probability of choosing a red is R 4 G R G 1 4 G 0 On tree diagrams: as we move across () we multiply. as we move Down we add Use the notation that 1 R represents the first ball being red etc. (a) PR ( R) (b) P(G ) P(R G ) P(G G ) (c) Either At least one of the balls is green or both of them are red. So the probabilities of these two events add up to 1. P(at least one green) 1 P( R R ) So 1 The tree diagram shows us that PR ( R) And so we have that P(at least one green) 1 P( R R ) Questions involving At least one This is an example of an important short cut. You DO NOT need to think of all the cases when there is at least one green ball. (d) We need to find PR1 R. PR G 6 PR G P 8 G 0 Copyright - For AS, A notes and IGCSE / GCSE worksheets 40

41 For use only in Badminton School November 011 Example 60% of the population of a certain town are vaccinated against flu. The probability of someone getting the flu given that they have had the vaccination is 0. but the probability of someone getting flu given that they have not had the vaccination is (a) Find the probability that a person chosen at random gets flu. (b) Write down the probability that a person chosen at random doesn t get flu given that he didn t have the vaccination. (c) Find, as a fraction, the probability that that a person chosen at random didn t have the vaccination given that he did get flu V NV F 0.1 NF 0.48 F 0.6 NF 0.14 Notice that the probabilities on the branches that come from the same point always add up to 1. So (a) There are two groups of people who get flu. Either a person has a vaccination and gets flu or a person does not have a vaccination and gets flu. The probability of a person has a vaccination and gets flu is 0.1 The probability of a person does not have a vaccination and gets flu is 0.6 As we move Down we add so probability of getting flu is P ( F) So P ( F) (b) P ( NF NV ) (straight from tree diagram) (c) PNV ( F) PNVF ( ) PF ( ) Copyright - For AS, A notes and IGCSE / GCSE worksheets 41

42 For use only in Badminton School November 011 Correlation Scatter Diagram Many statistical investigations have to do with the relationship between several variables. We will now examine ways of handling data with two variables. Suppose we want to compare two sets of data x and y. For example x might be the boy s mark in a Maths exam and y might be the boy s mark in an English exam. Boy A B C D E F G H I J Maths Mark, x English Mark, y We plot these values on a graph in the form of a scatter diagram. Scattter Diagram of M aths and English M arks English Mark y th 1st 3rd nd x Maths Mark In the diagram we have marked on the dotted lines through x and y. This splits the diagram into four quadrants. If all the points of our scatter diagram lie near a straight line then we say that there is linear correlation between x and y. (a) If y tends to increase as x increases then we say that there is a positive linear correlation. In this case most points will be in either the first or the third quadrants. (b) If y tends to decrease as x increases then we say that there is a negative linear correlation. In this case most points will be in either the second or the fourth quadrants. (c) If there is no relationship between x and y then we say that there is no correlation. In this case the points will be randomly distributed in all the quadrants. Copyright - For AS, A notes and IGCSE / GCSE worksheets 4

43 For use only in Badminton School November 011 How do we measure the strength of linear correlation? Suppose, for example that we look at the case above. If x, y represented the general point then because most points lie in either the first or the third quadrant we would find that xi x yi y was positive most of the time and so we would find that x x y y i i i i would be positive. However the problem with x xy y i i is that it varies with the size of x and y. To avoid this we divide by x i x and by y i y moment correlation coefficient. and we end up with the product Product Moment Correlation Coefficient (PMCC) The product moment correlation coefficient, its use, interpretation and limitations. Derivations and tests of significance will not be required. As stated above this measures the degree of linear correlation between x and y. The coefficient r, known as the PMCC, is defined as r y y x x x x y y. This can also be written as: r x y xy n x y x y n n FORMULA BOOK HAS Copyright - For AS, A notes and IGCSE / GCSE worksheets 43

44 For use only in Badminton School November 011 We now define three further quantities, which are, S xx, S xy and S yy. These are defined as follows: x S xx xx x n xy x y S x x y y xy n y yy S y y y n Using these we see that Sxy r. S S xx yy Copyright - For AS, A notes and IGCSE / GCSE worksheets 44

45 For use only in Badminton School November 011 Now consider the properties of this coefficient. We know that xx y y 0 for all real. y y y yx x x x Hence 0 for all real This quadratic in is always postive and so " 4 ". Hence y yxx x x y y 4 4 x x y y y yx x x x y y y yxx 1 1 1r 1 xx y y b ac NB1 r 1 if all the points lie on a straight line with positive gradient, in which case we say there is perfect positive correlation. NB r 1 if all the points lie on a straight line with negative gradient, in which case we say there is perfect negative correlation. NB 3 NB4 The lower the value of r the less linear correlation there is. r is only a measure of a linear relationship between x and y. They may be related in another way but r would still be very low. NB5 Coding the values of x and y does not change the product moment correlation coefficient, r. So if we make X ax b and Y cx d then the product moment correlation coefficient of x and y is the same as the product moment correlation coefficient of X and Y. NB6 Suppose we told to make X ax b and Y cx d and are then asked to find the regression line of Y on X having been given the summary statistics for x and y. First off find the regression line of y on x then replace y and x in that equation with X b Y b x and y. a a Copyright - For AS, A notes and IGCSE / GCSE worksheets 45

46 For use only in Badminton School November 011 Regression interpret Scatter diagrams. Linear regression. Explanatory (independent) and response (dependent) variables. Applications and ations. Use to make predictions within the range of values of the explanatory variable and the dangers of extrapolation. Derivations will not be required. Variables other than x and y may be used. Linear change of variable may be required. When we looked at the product moment correlation coefficient we were looking to measure the amount of linear association between two quantities x and y. We now want to go further and find the relationship between x and y when we know that there is this linear association. Once have plotted our data on a scatter diagram we look for a relationship y a bx (since the relationship is linear). We work back or regress from the points we have plotted to find the equation of the line and hence the line is called the regression line of y on x. If we had a set of data such as x y then x has been controlled and so it is called the explanatory (or independent) variable. y is called the response (or dependent) variable. The explanatory (or independent) variable always goes on the horizontal axis If there is a linear relationship between y and x then for each value of y i there will be an experimental error associated with it. We denote this error with i which can be either positive or negative. Therefore we can write y i x i i. The mean value of the errors, i, will be about zero. So if we are given a set of points how do we determine the equation of that line? How do we find a and b? We could simply try to draw a line of best fit by eye. There is, however, a standard line that we draw, called least-squares regression line which minimises the sum of squares of the distances shown below. y x, y 1 1 y abx x, a bx 1 1 x Copyright - For AS, A notes and IGCSE / GCSE worksheets 46

47 For use only in Badminton School November 011 These distances could be denoted by d i where di yi a bxi. We are, therefore, trying to minimise y i abxi. Sxy We can derive the values of a and b using calculus and when we do so we find that b where, Sxx as we have already seen, x y S xy x xy y x xy and S xx x x x n n Once we have the value of b we can find a using the fact that the line passes through the point ( x, y ). It follows from this that y a bx and so that a y bx. So we say that the equation of the regression line of y on x is y a bx where a y bx. Sxy b and S NB : When drawing the line of best fit always mark the point x, y on your diagram and make sure that the line of best fit passes through this point. xx x p y m When using coding we might make X and Y. We could then use the values of X q n and Y to find the line of regression of Y on X. This would be Y a bx. What we have left to do, is, therefore to replace X with x p and Y with y q y m x p ab. We can then rearrange this to get y in terms of x. n q m n and so we get: We can use this find to find an estimate for the mean value of y (the response variable) when we have been given a value of x (the explanatory variable). This process is called interpolation. NB: When finding a value of y from a value of x we must be careful only to use values of x which lie in outside the range of x values which have been given. If we predict values of y using values of x which lie outside the range of given values then we are using extrapolation. We should take great care when extrapolating to see if this can be justified. A word on interpretation If the x values were the lengths of salmon and the y values were the weights of salmon and if r was close to 1 then the interpretation is not positive correlation. It is as the length of the salmon increases, so the weight of the salmon increases. The January 011 markscheme suggests that if r is greater than 0.4 then this is evidence of positive correlation though this is not explained until S. Copyright - For AS, A notes and IGCSE / GCSE worksheets 47

48 For use only in Badminton School November 011 Using calculators to find product moment correlation coefficient and line of regression For example find the summary statistics for the data shown below On Texas. 1. Press. Press and type in all the data so that it look as follows 3. Press Copyright - For AS, A notes and IGCSE / GCSE worksheets 48

49 For use only in Badminton School November Highlight CALC 5. Highlight -Var Stats and press twice and you get this: Scrolling down gives us all we need to use in the formulae x 70 y 101 xy 1388 x y n Copyright - For AS, A notes and IGCSE / GCSE worksheets 49

50 For use only in Badminton School November 011 On Casio. 1. Press. Press to get 3. Press (A+BX) to get 4. Type in all the data so that it look as follows 5. Press to get NB : To go back to Data on Casio choose to get 6. Press to get Press to get back to 7. Press to get Copyright - For AS, A notes and IGCSE / GCSE worksheets 50

51 For use only in Badminton School November Using the different options gives us x 70 y 101 xy 1388 x y n Using the values from the calculator. Formula book tells us, for example, that 70 So we see that Sxx x. Sxx x x x n x. Formula book tells us, for example, that Sxx x x x n 70 So we see that Sxx x y Similarly the formula book tells us that Sxy xy so we see that n S xy In the same way the formula book tells us that that 101 S yy S yy y y so we see that n Copyright - For AS, A notes and IGCSE / GCSE worksheets 51

52 For use only in Badminton School November 011 PMCC Sxy 378 We could calculate directly using r to give r (to 3sf). S S On Texas xx yy Type, choosing 5: Statistics, choosing EQ and then 7 to get On Casio Press and then to get Press to get Copyright - For AS, A notes and IGCSE / GCSE worksheets 5

53 For use only in Badminton School November 011 Regression Sxy We are also given that the regression line of y on x is y a bx where b. Sxx 378 So we can calculate b We then use the fact that y a bx x, y to see that a y bx and so a (to 3sf). 7 7 On Texas passes through the point Press, choose CALC and press 4 to get the following NB Formula book refers to y a bx, whilst Texas uses y ax b On Casio Press to get Press to get NB Casio uses the same notation as the formula book, that is y a bx. Press to get Press to get Copyright - For AS, A notes and IGCSE / GCSE worksheets 53

54 For use only in Badminton School November 011 Discrete Random Variables The concept of a discrete random variable. A random variable is a variable, taking numerical values, which represents the value obtained when we take a measurement from a real-life experiment. There are two kinds of random variable - discrete and continuous. A continuous random variable can take any value in a given range, whereas a discrete random variable can take only certain values. The probability function and the cumulative distribution function for a discrete random variable. Simple uses of the probability function p(x) where p( x) P( X x). The set of all possible values of a random variable along with the associated probabilities is called the probability distribution. Example Suppose a coins is tossed three times and the number of heads is the random variable X. The probability distribution for X is as follows: Number of Heads (X) Probability Sometimes the probabilities are expressed as a function as opposed to a table. This function is called the probability function. Example Find the probability function for X, the score on a die We know that PX ( 1), PX ( ), PX ( 3) etc. and so we get the formula for x 1,,3, 4,5,6 PX ( x) 0 otherwise NB p( x ) (or simply p) is a short way of writing P(X x). If X is a continuous random variable we cannot talk about the probability of X taking particular values but only of the probability of it taking a range of values, for example PX ( 3). We cannot, therefore, use a simple probability function to define X. Example Find p in the following table NB: px ( ) 1 x PX x p 0.55 Since px ( ) 1, it follows that p and so p Copyright - For AS, A notes and IGCSE / GCSE worksheets 54

55 For use only in Badminton School November 011 Use of the cumulative distribution function F x0 P X x0 p( x) Just as we have met cumulative frequency graphs in GCSE so we know meet a cumulative distribution function for a random variable, X. The cumulative distribution function of X is defined as F x P X x so that the probability that X is less than or equal to x 0. If X is discrete then we can also write F x 0 as F x 0 p( x). xx0 xx0 0 0 If r is the largest value which X can take then Fr 1. If s is the smallest value which X can take then F s P X s. If X takes consecutive integer values then Fr 1 Fr Pr 1 F x means 0 Example Find F 1 and 3 F in the following Number of Heads (X) Probability F(1) P X 0 P X is the largest value that X can take so F(3) 1. k x Example The random variable X is such that F x for x 0,1,,3. Find the positive 16 integer k is positive and draw the probability distribution table for X. 3 is the largest value that X can take so (3) 1 k 3 4 so k 7 or k 1. k is positive, so k 1. k 3 F. F 3 1. So 16 k Using F x 1 x we have: 16 x F x and 0 1 so F3 F P X 3 F P X P X P X P X F P X P X P X PX So Copyright - For AS, A notes and IGCSE / GCSE worksheets 55

56 For use only in Badminton School November Similarly FF1 PX so PX And F 1 F0 PX 1so 1 And F0 PX 0 so 0.So we have the following table P X P X 16 x PX x Mean and variance of a discrete random variable. Use of variance of X. Knowledge and use of EaX b ae X b and Var ax b a Var X E X, E X for calculating the. The mean of a probability distribution is called the expected value of the random variable X. It is written as E(X) for short and its symbol is. We have already met the formula for the mean of a sample, i.e. x relative frequency f f with p we have the formula: E(X) xf f xp for the random variable X and by replacing the We have also met the formula variance x f f x. f If we replace the relative frequency f we have the formula, Var( X ) x pe( X). Now for any function g(x) we have E( g X) E( X ) x p. with p as we did before and replace x with E X then g x p and so we see from this that Hence we have the following formula for the variance of X: Var( X ) E X E X Copyright - For AS, A notes and IGCSE / GCSE worksheets 56

57 For use only in Badminton School November 011 In an exam if all the scores were doubled then the mean score would clearly double. Since every value of x would increase by a factor of 4 it follows that both E X and E X increase by a factor of 4. Hence Var( X ) EX E X So E X E X and Var X 4Var X increases by a factor of 4.. Similarly if in exam if all the scores were increased by 10 then the mean would increase by 10. However, since the spread of the scores has not changed, the variance remains the same. So E X 10 EX 10 and Var X 10 Var X. Generalising the results shown above gives the following: E E ax ae X ax b aex b ax a Var X ax b a Var X Var Var Example The random variable X has the following probability distribution : Find (a) E X (b) Var X (c) E3X 1 (d) Var 10X 7 X 1 5 P X x (a) & (b) E X E X Var( X) E( X ) E( X) Copyright - For AS, A notes and IGCSE / GCSE worksheets 57

58 For use only in Badminton School November 011 On Casio Scroll down using to get Press 3 to get Press 1 Press then (STAT) and then 1 (1-VAR) to get Type in all the data to get Press For mean and variance Press to give Choose 5 (Var) to get Choosing gives the mean to be 4. Copyright - For AS, A notes and IGCSE / GCSE worksheets 58

59 For use only in Badminton School November 011 Choosing again. Choose 4 (Sum) and then 1 to get with probabilities not frequencies. This is 18.4 x, that is E X when dealing Choosing variance to be.4. again. Choose 5 (Var) and then 3 to get standard deviation. Square to give the If using the calculator, write down X X X Var E E (c) Use X X E 3 1 3E DO NOT USE Y 3X 1 y PY y (d) Use X X Var Var Copyright - For AS, A notes and IGCSE / GCSE worksheets 59

60 For use only in Badminton School November 011 The discrete uniform distribution. This is the distribution in which the random variable can take a set of, say, n values and X is equally likely to take each value (e.g. X being the score on the die). In general, then, if X can take the values 1 x1, x, x3,... x n we have PX ( x r ) for r 1,, 3,... n. n The mean and variance of this distribution. In the above example we see that: n n 1 1 E X xr xr n n and Var r1 r1 n n n n r r r r r 1 n r1 n n r1 n r1 X x x x x If the random variable X took the values 1,, 3, 4, n then n n xr r r1 r1 n n 1n1 (these are standard results). 6 n n n n1 x r and r r1 r1 In which case we see that n 1 1 n n1 n1 E X xr n n r1 And, in a similar way we see that Var n n 1 1 X xr x r n r1 n r n n n 1 n n n n n n n n1 n1 4n3n1 n1 1 1 n Copyright - For AS, A notes and IGCSE / GCSE worksheets 60

61 For use only in Badminton School November 011 So for the random variable X which has discrete uniform distribution over the integers 1,, 3,, n n 1 n 1 we have E X and Var X 1 From this we can, using coding, find E X and Var( X ) if X has discrete uniform distribution over, for example, the integers 3, 5, 7,, 19. We say that X Y 1 and so Y has discrete uniform distribution over the integers 1,, 3,, We know, therefore, that EY 5 and that Var( Y ) Using the coding we see that : 80 E X=EY 1 EY1 11 and that Var( X) Var(Y 1) 4Var( Y) 3 Copyright - For AS, A notes and IGCSE / GCSE worksheets 61

62 For use only in Badminton School November 011 Normal Distribution The Normal distribution including the mean, variance and use of tables of the cumulative distribution function. Knowledge of the shape and the symmetry of the distribution is required. Knowledge of the probability density function is not required. Derivation of the mean, variance and cumulative distribution function is not required. Interpolation is not necessary. Questions may involve the solution of simultaneous equations. The normal distribution provides us with a suitable model for a very large number of distributions and so is the most important distribution in statistical theory. Heights or weights of men or women, time taken to run the 00m etc. all follow the normal distribution. It has some key features, notably that the vast majority of data is centred around the mean and that very small or large values are rare. It is symmetrical about the mean (an hence the mean is also equal to the mode and the median). We describe the normal distribution using two parameters, its mean and its variance. So if the random variable X has normal distribution with mean and variance we write X~ N, for short. Z is the letter given to the random variable which has mean 0 and variance 1. In fact the random variable Z is called the standard normal distribution and it and so is written as N 0,1. A sketch of the distribution of Z~ N0,1 is shown below: The equation that generates 1 1 x this curve is y e. NB: The total area under this curve is t So e dt 1 We have looked at the cumulative distribution functions of discrete random variables. We are now dealing with a continuous random variable and so we cannot talk about the probability of X taking any particular value. We can only talk about the probability of X lying between two values a and b. Copyright - For AS, A notes and IGCSE / GCSE worksheets 6

63 For use only in Badminton School November 011 The cumulative distribution function for the random variable Z is written as. z z 1 1 t In other words z P( Z z) e dt. So z is the area shown in the diagram. From the tables we have 0 0.5, , 0.977, etc. It is always true that PZ z z 1 (1) z The symmetry of the graph gives us the following: z z z These two areas combine to give 1 and so we have the following: -z z z 1 () P Z z P Z z -z In the same way Combining (3) and (1) these gives PZ z PZ z 1 (3) PZ z z (4) z Copyright - For AS, A notes and IGCSE / GCSE worksheets 63

64 For use only in Badminton School November 011 We need to be able to use the tables in the formula book. Three stages: Finding a probability when given a value of z a. Turn it into a less than problem, using PZ z 1 z b. Ensure that the z value is positive, using z 1 z c. Read off from tables We use the large table to find probabilities when we have been given values of z: Case 1: Find 1.7. a. Already is less than problem. b. z value is already positive. c. Read off from tables, to give Case : Find 0.8. a. Already is less than problem z 1 z b. Use to get c. Read off from tables Case 3: Find PZ 0.3 a. Use PZ z 1 z to get PZ b. z value is already positive c. Read off from tables. and so P Z Case 4: Find PZ 1.7. a. Use PZ z 1 z to get PZ b. Use z 1 z to get c. Read off from tables NB : We could have used P Z z z to give PZ Copyright - For AS, A notes and IGCSE / GCSE worksheets 64

65 For use only in Badminton School November 011 With our knowledge of these four cases we can find the area between any limits: ba P a Z b P Z b P Z a a b Example Find P0.7 Z.1. P0.7 Z.1 PZ.1PZ (to 3dp) Example Find P1. Z 1.8. P1. Z 1.8 PZ 1.8PZ (to 3dp) Example Find P0.9 Z 0.1. P0.9 Z 0.1 PZ 0.1 PZ (to 3dp) NB : It is worth knowing that P a Z b P b Z a Copyright - For AS, A notes and IGCSE / GCSE worksheets 65

66 For use only in Badminton School November 011 Finding a value of z when given a neat probability (0.3, 0.8, 0.05 etc.) We use the small table to find probabilities when we have been given values of z: Three stages: P Z z 1 z a. Turn it into a greater than problem, using b. Ensure that the probability is less than 1, using PZ z PZ z 1 c. Read off from tables Case 1 : Solve PZ z 0. a. Already is a greater than problem. b. The probability is already less than 1 c. Read off from tables to give z Case : Solve PZ z 0.7 a. Already is a greater than problem. b. Use PZ z PZ z 1 P Z z c. Read off from tables to give z and so z to give Case 3 : Solve z 0.9 a. Use PZ z 1 z to give PZ z 0.1 b. The probability is already less than 1. c. Read off from tables to give z Case 4 : Solve z 0.15 a. Use PZ z 1 z to give PZ z b. Use PZ z PZ z 1 to give PZ z c. Read off from tables to give z and so z Copyright - For AS, A notes and IGCSE / GCSE worksheets 66

67 For use only in Badminton School November 011 We have done all the working so far on Z~ N0,1. We have seen how to find probabilities and z values. However, heights and weights and so on are not normally distributed with mean 0 and variance 1. It would be impossible to have a set of tables showing us all possible areas for all possible values of and. So we need some way of turning the random variable X ~N, into the standard normal distribution Z~ N0,1. The transformation we need to make is Z X. From the work we did on E ax b and Var ax b we see that Z Var( X ) Var( Z) 1. E 0 and that E X All we have done is shift and stretch the curve shown above and so Z still has the normal distribution. Thus Z X For example may be used to transform N, into N0,1. X~ 10,5 N Z~ N 0,1 Copyright - For AS, A notes and IGCSE / GCSE worksheets 67

68 For use only in Badminton School Finding probability when given a value of x November 011 Five stages: Write it out using the random variable X Turn it into a less than problem Turn it into a Z problem z z... is positive. Use 1 to ensure that the value inside Read off from tables Example The amount of jam in a jar is normally distributed with mean 140g and standard deviation 5.g. Find the probability that the amount of jam in a jar chosen at random is: (a) Less than 143g (b) More than 135g (c) Less than 138g (d) More than 146g (e) Between 139g and 145g (a) P X This value must rounded to dp and stated. The z values in the table are given to dp. (b) 1P X P X (c) P X (d) 1P X P X 146 (e) P X P X P X Copyright - For AS, A notes and IGCSE / GCSE worksheets 68

69 For use only in Badminton School November 011 Finding a value of x given a percentage Five stages: Write it out using the random variable X Turn it into a greater than problem Turn it into a Z problem P Z z P Z z to ensure that Use 1 the probability on the right hand side is less than ½. Use tables (REVERSE SIDE) Example The heights of a certain animal is normally distributed with mean 60cm and standard deviation cm. (a) Find the height above which 0% of animals lie. (b) Find the height above which 60% of animals lie. (c) Find the height below which 10% of animals lie. (d) Find the height below which 70% of animals lie. (a) x 0. P X x 60 PZ 0. x x cm (to 3sf) (b) x 0.6 P X x 60 PZ x PZ x x cm (to 3sf) (c) P X x P X x x 60 PZ x PZ x x cm (to 3sf) (d) P X x P X x x 60 PZ 0.3 x x cm (to 3sf) SHORTCUT : All these problems come down to finding a value which is either z or z, where the z value is the value linked in the tables above with p (if p 0.5 ) or 1 p (if p 0.5 ). It is easy to work out whether it is z or z. This is explained overleaf. Copyright - For AS, A notes and IGCSE / GCSE worksheets 69

70 For use only in Badminton School November 011 SHORTCUT Example The heights of a certain animal is normally distributed with mean 60cm and standard deviation cm. (a) Find the height above which 0% of animals lie. (b) Find the height above which 60% of animals lie. (c) Find the height below which 10% of animals lie. (d) Find the height below which 70% of animals lie. (a) Using the shortcut above, the z value to use is (corresponding to 0.) Since 0% of animals lie above this value, the value is clearly greater than the mean, so the value is cm (to 3sf). (b) Using the shortcut above, the z value to use is (corresponding to ). Since 60% of animals lie above this value, the value is clearly less than the mean, so the value is cm (to 3sf). (c) Using the shortcut above, the z value to use is (corresponding to 0.1). Since 10% of animals lie below this value, the value is clearly less than the mean, so the value is cm (to 3sf). (d) Using the shortcut above, the z value to use is (corresponding to ). Since 70% of animals lie below this value, the value is clearly greater than the mean, so the value is cm (to 3sf). NB : A question may give us a probability which is not neat, so not one of the values listed on the back table. We then need to use the front table. How to do this is explained below. Example If X~ 87,3 N then find x such that P X x clearly does not appear on the reverse table so we have to use the front. x 87 x which gives.10 and so x 93.3 (to 3sf). 3 3 NB : A question may give us a probability which is not one of the values listed on the back table and which is not listed on the front. We need to use the front table and linearly interpolate between the relevant values to find an x value. Example If X~ 160,35 N then find x such that P X x clearly does not appear on the reverse table so we have to use the front. It lies between and lies th of the way between and So the z value is (to 4dp) x 160 x which gives and so x 1640 (to 3sf) Copyright - For AS, A notes and IGCSE / GCSE worksheets 70

71 For use only in Badminton School November 011 Find mean or standard deviation Five stages: Write it out using the random variable X Turn it into a greater than problem Turn it into a Z problem Ensure that the probability on the right hand side is less than ½. Use tables (REVERSE SIDE) Example The amount of jam in a jar is normally distributed with mean and standard deviation 5g. Find the mean if the probability that a jam chosen at random contains: (a) more than 14g is 0.05 (b) less than 14g is 0.01 (a) Using the shortcut outlined earlier, , since 14 is greater than g (to 1dp). (b) Using the shortcut outlined earlier, , since 14 is less than g (to 1dp). Example X is normally distributed with mean 140g and standard deviation. Find given that the probability of being more than 150g is 0.. Using the shortcut outlined earlier, since 150 is greater than the mean, Hence 11.9g (to 1dp). Alternatively P X PZ g (to 1dp) Copyright - For AS, A notes and IGCSE / GCSE worksheets 71