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6th Year Maths Ordinary Level Strand 1 of 5 [ Topics: Statistics Probability ] No part of this publication may be copied, reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without prior written permission from The Dublin School of Grinds. (Notes reference: 6-mat-o-Strand 1). 6-HOUR CRASH COURSES MAY & JUNE 2016 %01.! čƫ+*+)%/ƫ0!$!.ƫ¨*8*ƫ1. +' The final push for CAO points... $!ƫ 1(%*ƫ $++(ƫ +"ƫ .%* /ƫ %/ƫ .1**%*#ƫ ćġ$+1.ƫ ./$ƫ +1./!/ƫ 0ƫ 0$!ƫ !* ƫ +"ƫ 5ƫ * ƫ 0$!ƫ !#%**%*#ƫ +"ƫ 1*!ċƫ $!/!ƫ +1./!/ƫ #%2!ƫ /01 !*0/ƫ 0$!ƫ !/0ƫ ,+//%(!ƫ 2*0#!ƫ /ƫ 0$!5ƫ ,.!,.!ƫ "+.ƫ 0$!ƫ ((ġ%),+.0*0ƫ 00!ƫ 4)%*0%+*/ċƫ *!ƫ (/0ƫ +1*!ƫ +"ƫ !""+.0ƫ +1( ƫ )'!ƫ ((ƫ 0$!ƫ %""!.!*!ċ !.!ƫ%/ƫ$+3ƫ0$!/!ƫ+1./!/ƫ3%((ƫ !*!"%0ƫ5+1č • They will offer students one last opportunity to avail of expert teaching before the State Examinations • They will provide students with a final boost of confidence before exam day • They will give students an exam strategy plan to help them maximise their grade on the day +0!: At these courses our teachers will predict what questions are most likely to appear on your exam paper. These questions will be covered in detail and our teachers will provide you with model A1 answers. čƫĺāćĀƫƫ To book, call us on 01ġ442 4442 or book online at www.dublinschoolofgrinds.ie ./$ƫ+1./!s Timetable 6th Year Subject Date Time Accounting Level H Sunday 29th May 9am - 3pm Biology H Saturday 28th May 9am - 3pm Business H Sunday 29th May 2pm - 8pm Chemistry H Saturday 4th June 9am - 3pm Economics H Saturday 28th May 9am - 3pm English H Sunday 29th May 9am - 3pm English H Saturday 4th June 9am - 3pm French H Saturday 4th June 9am - 3pm Geography H Saturday 28th May 9am - 3pm Irish H Saturday 4th June 9am - 3pm Maths Paper 1 H Saturday 4th June 9am - 3pm Maths Paper 2 H Sunday 5th June 9am - 3pm Maths O Saturday 28th May 9am - 3pm Maths O Saturday 4th June 9am - 3pm Physics H Saturday 28th May 9am - 3pm Spanish H Sunday 5th June 9am - 3pm Date Time 3rd Year Subject Level Business Studies H Sunday 5th June 9am - 3pm English H Sunday 5th June 9am - 3pm French H Sunday 29th May 9am - 3pm Irish H Sunday 29th May 9am - 3pm Maths H Sunday 29th May 9am - 3pm Science H Saturday 4th June 9am - 3pm H = Higher O = Ordinary (!/!ƫ*+0!ƫ0$0ƫ((ƫ+1./!/ƫ3%((ƫ0'!ƫ,(!ƫ0ƫ +1.ƫ!.*%*#ƫ!*0.!ƫ0ƫ$!ƫ.%).5ƫ$++(ƫ%*ƫ 0(* /Čƫ0%((+.#*Čƫ+ċƫ1(%*ċ Strand 1 is worth 20% to 30% of The Leaving Cert. Contents Statistics 1. Types of data ............................................................................................................................................................................ 2 2. Populations and sampling .................................................................................................................................................. 3 3. Collecting data ......................................................................................................................................................................... 3 4. Averages..................................................................................................................................................................................... 4 5. The range, quartiles and interquartile range ......................................................................................................... 12 6. The standard deviation using a calculator............................................................................................................... 16 7. The shape of distributions .............................................................................................................................................. 20 8. Bar charts ............................................................................................................................................................................... 22 9. Line plots ................................................................................................................................................................................ 24 10. Pie charts ................................................................................................................................................................................ 25 11. Histograms ............................................................................................................................................................................. 30 12. Stem and leaf diagrams .................................................................................................................................................... 34 13. Scatter graphs and correlation ..................................................................................................................................... 38 14. Hypothesis testing and the line of best fit ................................................................................................................ 44 15. Past and probable exam questions ............................................................................................................................. 50 16. Solutions to Statistics ........................................................................................................................................................ 81 Probability 1. The fundamental principle of counting ................................................................................................................... 104 2. Arrangements ..................................................................................................................................................................... 105 3. Arrangements with restrictions ................................................................................................................................. 106 4. What is probability? ......................................................................................................................................................... 108 5. The probability of an event not happening ........................................................................................................... 111 6. Two events: sample spaces .......................................................................................................................................... 112 7. Estimating probabilities from an experiment ...................................................................................................... 113 8. Expected frequency.......................................................................................................................................................... 114 9. Addition rule ....................................................................................................................................................................... 115 10. Using Venn diagrams ....................................................................................................................................................... 117 11. The AND rule ....................................................................................................................................................................... 119 12. Bernoulli trials ................................................................................................................................................................... 120 13. Tree diagrams ..................................................................................................................................................................... 122 14. Expected value ................................................................................................................................................................... 124 15. Past and probable exam questions ........................................................................................................................... 125 16. Solutions to Probability .................................................................................................................................................. 143 ©The Dublin School of Grinds Page 1 Statistics Statistics is worth 12% to 17% of The Leaving Cert. It appears on Paper 2. 1. Types of data The Examiner can ask you about 8 types of data. Learn these off by heart. Free marks! Note: you will see below that the syllabus requires you to know examples, advantages and disadvantages for some of the data types. 1. Primary data: This is information that you collect yourself. E.g. from doing an experiment or a survey. Advantage: We know where it comes from. Disadvantage: It can be time consuming to collect. 2. Secondary Data: This is information that you get from existing records. E.g. from the census or internet based sources. Advantage: it can be easy to obtain. Disadvantage: We don’t know how it has been collected. 3. Numerical Data: This is data the can be counted or measured. E.g. heights, masses, lengths etc. 4. Discrete Data: Can only take particular values. E.g. such as goals scored or numbers of cars sold per month. 5. Continuous Data: This can take any value in a particular range. E.g. weight, temperature and length. 6. Categorical Data: This is described using words E.g. favourite sport, country of birth or favourite food. 7. Univarite data consists of one item of information E.g. colour of eyes. 8. Bivariate data contains two items of information E.g. colour of your eyes and your age. ©The Dublin School of Grinds Page 2 2. Populations and sampling The Examiner can ask you about populations and sampling so you must know the following: A population is the entire group being studied E.g All the secondary school students in Ireland A sample is a group selected from the population E.g. Selecting several secondary schools from around Ireland In a simple random sample every member of the population has an equal chance of being chosen. 3. Collecting data The Syllabus requires you to know different ways that information can be collected. Surveys A survey collects primary data. One way of collecting primary data is to design and complete a data collection method or questionnaire. The Examiner can ask you to give examples so you must know the following four examples. 1. Face to face interviews 2. Telephone interviews 3. Questionnaires sent out by post or online 4. Observational studies When designing a questionnaire Be clear about what you want to find out. Keep each question as simple as possible. Never ask a leading question designed to get a particular response. Provide response boxes where possible. Experiments Experiments are another method of collecting data. E.g. Tossing a coin a number of times and recording the outcome. ©The Dublin School of Grinds Page 3 4. Averages The Syllabus requires you to know three different averages the mode, the median and mean. The Examiner can also ask you when each one can be used and the advantages or disadvantages of each one. The mode Definition: The mode is the set of values that occur the most often When to use the mode If the data is categorical E.g. Hair color, favourite subject. Advantage: Easy to find. Disadvantage: May not exist. The median Definition: The median is the middle value when the values are arranged in order. If there is an even number of numbers, we take the average of the two middle numbers when the numbers are arranged in order. When to use the median If there are extreme values, use the median. Advantage: Easy to calculate if the data is ordered. Disadvantage: Not very useful for further analysis. The mean The mean of a set numbers (values) is the sum of all the values divided by the number of values. Definition: 𝑀𝑒𝑎𝑛 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠 When to use the mean The mean is only used for numerical data. If there are not extreme values in the data set, use the mean. Advantage: Uses all the data. Disadvantage: It is not always a given data value. ©The Dublin School of Grinds Page 4 When working with averages the Examiner can ask you to find the mode, median or mean in 3 different ways. The first one is with a list of numbers given to you. Example 1a The ages of a group of college students in a Dublin college are shown below. 18, 19, 23, 19, 24, 18, 19, 18, 23, 17, 19, 18, 22, 21, 20, 18, 24, Find the i) Mode ii) Median iii) Mean Solution First write the numbers in order: 17, 18, 18, 18, 18, 18, 19, 19, 19, 19, 20, 21, 22, 23, 23, 24, 24 i) The most commonly occurring number is 18 because it occurs 5 times. The mode is = 18 ii) The median is the middle number. There are seventeen numbers, so the ninth number is the middle number. The median = 19 iii) Mean = 17+18+18+18+18+18+19+19+19+19+20+21+22+23+23+24+24 340 17 = 17 = 20 To find the median when there is an even number of values/numbers we must take the mean of the two numbers in the middle: Example 1b Find the median of the following group of numbers 18, 16, 15, 17, 19, 18, 13, 20 Solution First write the numbers in order: 13, 15, 16, 17, 18, 18, 19, 20 There are 8 numbers, so the median is the average of the 4th and 5th numbers. 4th = 17 5th =18 17+18 Median = 2 = 17.5 ©The Dublin School of Grinds Page 5 Question 4.1 Write the following group of numbers in size order: 4, 10, 18, 6, 4, Then find: i) ii) iii) The mode The median The mean ©The Dublin School of Grinds Page 6 12, 4, 6, 8 The second way you can be asked about averages is when you are given a frequency distribution table. Example 2 The following frequency table shows the number of goals scored in 30 matches. Goals Scored 0 1 2 3 4 Number of matches 1 7 6 5 2 5 6 6 3 Write down the: i) Mode ii) Median iii) Mean Solution: i) ii) The mode in a frequency distribution table like this is the number in the same column as the largest frequency (usually the number on top of the biggest number on the bottom.) The biggest number on the bottom is 7 and above it is 1 ∴ mode = 1 There are 30 football matches ∴ Looking for the 15th and 16th match. Add up the numbers until you reach the 15th/16th 1 + 7 + 6 + 5 One of these 5 matches is the 15th/16th 15𝑡ℎ+16𝑡ℎ Median = = 3+3 =3 iii) 2 2 Finally the mean is found in a similar way to with a group of numbers. 𝑆𝑢𝑚 𝑜𝑓 (𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 × 𝑛𝑢𝑚𝑏𝑒𝑟) 𝑀𝑒𝑎𝑛 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑟𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 = 1(0)+7(1)+6(2)+5(3)+2(4)+6(5)+3(6) = 0+7+12+15+8+30+18 1+7+6+5+2+6+3 30 =3 ©The Dublin School of Grinds Page 7 Question 4.2 A test consisted of six questions with each question worth 1 mark. The following table shows how a class of students scored in the test. Marks 1 2 3 4 5 6 Number of students 7 7 8 9 5 4 Use the table to find i) ii) iii) The mode The median The mean ©The Dublin School of Grinds Page 8 Finally, the Examiner can ask you to find averages using a grouped frequency distribution Example 3: The frequency table below shows the number of text messages a group of teenagers sends per day. Texts sent 10-20 20-30 30-40 40-50 No. of Students 4 15 11 10 Note: 10-20 means 10 is included but 20 is not, etc. i) Which is the modal group? ii) In which interval does the median lie? iii) Find the mean of the frequency distribution. Solution i) ii) iii) Just like the frequency table it is the number above the largest number on the bottom line. => The modal group = 20-30 There are 40 students so the median is between the 20 th and 21st students. 4 + 15 + 11 One of these 11 teenagers is the 20th/21st =>The median lies in the interval 30-40 marks To calculate the mean we need to find values called the mid-interval values. We find these by taking the mean of the two values given in each interval e.g. The interval 10-20 would have a mid-interval value of 10 + 20 2 = 15 = We can use this to find the mid-interval values for all of the intervals Texts sent 15 25 No. of students 4 15 35 45 11 10 We can then find the mean the same we did for the frequency distribution table above using the midinterval values. 4(15)+15(25)+11(35)+10(45) 𝑚𝑒𝑎𝑛 = 40 = 1270 40 = 31.75 ©The Dublin School of Grinds Page 9 Question 4.3 The speed of vehicles passing under a bridge are recorded in the table below: Speed (km/h) Frequency 60-70 8 70-80 15 80-90 12 90-100 10 100-110 8 110-120 3 120-130 4 Calculate the mean, modal group and median group of the frequency distribution using the mid-interval values (correct to one decimal place). The Examiner can also give you the mean and ask you to find an unknown. Example 4 The table below shows the number of cars each family on a street owns. The mean number of cars owned is 2. Number of cars Number of families Find the value of x. Solution 0 1 1 X 1(0) + 𝑥(1) + 1(2) + 5(3) 1+𝑥+1+5 0 + 𝑥 + 2 + 15 2= 𝑥+7 𝑥 + 17 2= 𝑥+7 2(𝑥 + 7) = 𝑥 + 17 2𝑥 + 14 = 𝑥 + 17 2𝑥 − 𝑥 = 17 − 14 => 𝑥 = 3 2= ©The Dublin School of Grinds 2 1 Page 10 3 5 Question 4.4 The table shows the number of goals a hockey team scored in each match over the season. Goals 0 1 2 Number of matches 1 3 X The mean number of goals they scored was 3. Find the value of x. ©The Dublin School of Grinds Page 11 3 9 4 7 5 4 5. The range, quartiles and interquartile range The Range Range = Maximum Value − Minimum Value It shows the spread of data. It is useful for comparing sets of data. Example 1 There are 10 students in a class 1A. Their results in a maths test out of 20 are as follows: 14, 7, 12, 9, 12, 9, 14, 13, 18, 12 There are also 10, students in class 1B. Their marks in the same test are as follows: 11, 15, 12, 15, 11, 12, 9, 13, 10, 12 Find the i) Mean ii) Range Of both classes’ tests and compare the results. Solutions The mean for class 1A: = 7 + 9 + 9 + 12 + 12 + 12 + 13 + 14 + 14 + 18 10 120 = 10 = 12 𝑀𝑎𝑟𝑘𝑠 The mean for class 1B: = 9 + 10 + 11 + 11 + 12 + 12 + 12 + 13 + 15 + 15 10 120 = 10 = 12 𝑀𝑎𝑟𝑘𝑠 The range for class 1A: 𝑅𝑎𝑛𝑔𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 − 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 18 − 7 = 11 The range for class 1B: 𝑅𝑎𝑛𝑔𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 − 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 15 − 9 =6 Comment: The mean values are the same but class 1B has a smaller range. Therefore class 1B’s results are more consistent. Note: When a set of data has a small range it is said to be consistent. ©The Dublin School of Grinds Page 12 Question 5.1 The pulse rates of 7 boys and 7 girls were recorded as follows: Boys: 87 88 84 91 81 85 86 Girls: 86 87 95 72 82 99 88 i) Calculate the mean and the range for each (set data set). ii) Comment briefly on how the two groups compare. Quartiles and Interquartile Range Better measure of spread than the range as it’s not affected by outliers. Lower quartile rule: List the values in order. 1 Multiply by the number of values. 4 If you get a decimal, then round up and use this value. If you get a whole number, then add this value and the next value, then divide by two. Upper quartile rule: 3 1 This is the same as the lower quartile except we multiply by rather than . 4 4 These are best explained using examples … Example 2 (odd number of numbers) Here are the times in minutes it takes 11 students to walk to school: 4, 12, 7, 6, 10, 5, 11, 14, Find i) The lower quartile ii) The upper quartile iii) The interquartile range 2, Solution First list the numbers from smallest to biggest = 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14 (11 numbers) i) The lower quartile Q1 1 × 11 = 2.75 4 ie: a decimal => round up to 3 => Lower Quartile = 4 ii) The upper quartile Q3 3 × 11 = 8.25 4 ie: a decimal => round up to 9 9𝑡ℎ value = 11 iii) The interquartile range = upper quartile – lower quartile = 11 – 4 = 7 ©The Dublin School of Grinds Page 13 3, 9 Question 5.2 Find i) The lower quartile ii) The upper quartile iii) The interquartile range For the following set of data: 15, 7, 9, 12, 9, 12, 19, 6, 11, 16, 8 Example 3 (even number of numbers) Here are the number of minutes it took a group of students to finish their maths homework. 16, 8, 10, 13, 10, 20, 7, 12, 10, 16, Find the i) Lower quartile. ii) Upper quartile. iii) And the interquartile range. Solution First write the numbers down in size order from smallest to biggest 7, 8, 9, 10, 10, 10, 11, 12, 13, 16, 16, 20 (12 numbers) i) The lower quartile Q1 1 × 12 = 3 4 ie: a whole number => add the 3rd and 4th values then divide by 2 3𝑟𝑑 + 4𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 9 + 10 => = 2 2 19 = 2 = 9.5 ii) The upper quartile Q3 iii) The interquartile range = upper quartile – lower quartile = Q3 – Q1 = 14.5 – 9.5 =5 ©The Dublin School of Grinds 3 × 12 = 9 4 9𝑡ℎ + 10𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 13 + 16 => = 2 2 29 = 2 = 14.5 Page 14 9, 11 Question 5.3 The following lists gives the weights of 12 tennis players at a tennis club: 49, 80, 63, 48, 78, 46, 52, 58, Find the i) ii) iii) Lower quartile. The upper quartile. And the interquartile range. Solution ©The Dublin School of Grinds Page 15 71, 64, 73, 55 6. The standard deviation using a calculator The standard deviation measures the average deviation or spread from the mean of all values in a set. A low standard deviation tells us the data is very close to the mean. A high standard deviation tells us that the data points are spread out over a large range of values. The Greek letter 𝜎 is used to denote standard deviation The Syllabus can ask you to use your calculator to find the standard deviation. You can be asked this in two ways: 1. A group of numbers. 2. A frequency distribution table. First let’s look at an example with a group of numbers. Example 1 Using a calculator find the standard deviation of the following numbers correct to two decimal places: 6, 7, 9, 11, 12 Solution How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. 2. 3. Key in Mode , then: 20(STAT) Then, 10 for 1 – Var Then each number followed by = 60 = 0 70 = 0 90 = 0 110 =0 4. 5. 120 = 0 Next, press AC , SHIFT and 1 0 for the menu 0 0 Then, press 4 0 Now, press 30 for 𝑥𝜎𝑛 (standard deviation). Finally press = 0 =>The standard deviation 𝜎 = 2.28 How to do it in a Sharp EL-W531 1. 2. 3. Key in Mode , then: 10 (STAT) Then, 00for Standard Deviation Then each number followed by DATA 60 DATA 0 70 DATA 0 9 0 DATA 0 110 DATA 0 4. 5. 120 DATA 0 Next, press ALPHA then 60 for standard deviation 0 Finally, Press = 0 =>The standard deviation 𝜎 = 2.28 Note: If using a Sharp EL-W531 please reset your calculator after every Standard Deviation calculation. ©The Dublin School of Grinds Page 16 Question 6.1 Calculate the standard deviation of the following set of numbers using a calculator: 18, 26, 22, 34, 25 Question 6.2 Find the standard deviation of the following set of numbers 2, 4, 6, 8, 10 to one decimal place. ©The Dublin School of Grinds Page 17 Example 2 Find the standard deviation of the following frequency distribution correct tot two decimal places. Variable 1 2 3 4 Frequency 1 4 9 6 Solution How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. Key in Mode , then: 20(STAT) 2. Then, 1 0 for 1 – Var 3. Then each number followed by = then scroll across and input the frequency 1 =0 1 = 0 0 0 20 30 4. =0 =0 4 = 0 0 9 = 0 0 40 =0 6 = 0 0 Find the standard deviation AC SHIFT 1 4 3 = =>The standard deviation 𝜎 = 0.84 How to do it on a Sharp EL-W531 1. Key in Mode , then: 10 (STAT) 2. Then, 00 forStandard Deviation 3. Next, input each number followed by (x,y) and then the frequency followed by DATA 1 (x,y) 0 0 20 (x,y) 0 1 DATA 0 4 DATA 0 30 (x,y) 0 4 (x,y) 0 0 9 DATA 0 6 DATA 0 0 0 0 0 4. Next, press ALPHA then 0 60 for standard deviation 5. Finally, Press = 0 =>The standard deviation 𝜎 = 0.84 ©The Dublin School of Grinds Page 18 Question 6.3 A survey asked a group of people how many days a week they exercised. The following frequency distribution table shows the data: Number of days 1 2 3 Number of People 6 9 4 Calculate the standard deviation correct to two decimal places. 4 4 5 4 6 3 Question 6.4 The following frequency distribution table gives the goals scored by a team over the entire season. No. of Goals (x) 1 2 3 4 5 6 Number of matches (f) 7 8 4 4 3 4 Find the standard deviation correct to two decimal places. When asked to find the standard deviation of a grouped frequency distribution use the mid-interval values just like when we used them to find the mean. Question 6.5 The number of minutes taken by 20 students to run 1 kilometre in PE class was recorded. The data is shown in the following distribution table: Minutes 2-4 4-6 6-8 8-10 Number of Pupils 6 6 4 1 Find the standard deviation ©The Dublin School of Grinds Page 19 7. The shape of distributions The Examiner can ask you about the shape of a distribution so you need to be able to recognise them and know their characteristics. This ‘symmetrical bell shaped’ curve is called a ‘normal distribution curve’. The Empirical Rule The Syllabus requires you to know the empirical rule and be able to use it. We will look at an example to help explain the rule. i) 68% of the data lies within : mean ± 1 S.D. ii) 95% of the data lies within: mean ± 2 S.D. iii) 99.7% of the data lies within: mean ± 3 S.D. Example 1 The mean number of students absent from a secondary school per week is 18. The standard deviation is 3.6. Using this information, what percentage of the population lies in the range [14.4 𝑡𝑜 21.6]? Solution 𝑚𝑒𝑎𝑛 = 18 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 3.6 14.4 = 𝑚𝑒𝑎𝑛 − 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 21.6 = 𝑚𝑒𝑎𝑛 + 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 According to the empirical rule 68% of the population lies in the range [14.4,21.6] Question 7.1 The mean time for a group of students to find the solution to a problem was 6 minutes. The standard deviation from the mean is 2. What percentage of the population lies in the range [2 minutes to 10 minutes]. ©The Dublin School of Grinds Page 20 Positive skew (Also known as right skew). This is when the data is lower on the right. Positively skewed histogram/ curve Real life example: a) Age of people at 1 Direction concert Negative skew (Also known as left skew). This is when the data is lower on the left. Negatively skewed histogram/curve Real life examples: a) Ages of people at a Rod Stewart concert. ©The Dublin School of Grinds Page 21 8. Bar charts The Syllabus requires you to know when a bar chart can be used. A bar chart can be used to represent categorical data. There are two types of bar charts. You can use whichever one floats your boat. Example 1 Students were given a test containing ten questions, each correct answer was worth 1 mark. The table below shows the students’ marks. Marks 4 5 6 7 8 9 10 No. of Students 1 3 4 5 8 3 Draw a bar chart to represent this data. Solution You must always label the x and y axes. You will lose marks in The Leaving Cert if you do not. ©The Dublin School of Grinds Page 22 1 Question 8.1 The frequency table below shows the scores from a four day golf tournament Score 280-284 285-289 290-294 295-299 300-304 No. of Players 1 3 5 4 2 i) ii) Draw a bar chart to represent this data What percentage of players scored between 290-294? ©The Dublin School of Grinds Page 23 9. Line plots A line plot is used to display small sets of data. It is similar to a bar chart with dots or crosses used instead of bars. Each dot or cross represents one unit of the variable. Example 1 A survey was conducted of how many pets each family in an estate had. The results are shown below: Number of pets 1 2 3 4 5 Number of families 4 3 6 2 5 Represent this data using a line plot. Solution Write the smallest number on the left and with the largest on the right. Then, each unit is represented by one dot e.g. for 1 pet there are 4 families therefore 4 dots. Question 9.1 Paul surveyed some of the students in the class about how many books the students read over the last two weeks. The results are given in the table below: Number of books read 0 1 2 3 4 5 6 Number of students 3 2 0 6 4 2 3 Draw a line plot to represent this data below. ©The Dublin School of Grinds Page 24 10. Pie charts Pie Charts are suitable for displaying categorical data How to draw a pie chart Step 1 Find out the total number of the sample Step 2 Turn the information into degrees. For each section: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑑𝑒𝑔𝑟𝑒𝑒𝑠 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑖𝑛 𝑒𝑎𝑐ℎ 𝑠𝑒𝑐𝑡𝑖𝑜𝑛 × 360° 𝑇𝑜𝑡𝑎𝑙 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 Step 3 Use a compass and a protractor to draw the pie charts ©The Dublin School of Grinds Page 25 Example 1 A shop sold, 70 bicycles in September, 140 in October,175 in November and 315 in December. Illustrate this data using a pie chart. Solution Step 1 Total number of bicycles 70 + 140 + 175 + 315 = 700 Step 2 Turn the information into degrees: Month No. of bicycles September 70 October 140 November 175 December 315 Total 700 Step 3 Draw the pie chart using a compass and a protractor ©The Dublin School of Grinds Page 26 70 × 360 = 36° 700 140 × 360 = 72° 700 175 × 360 = 90° 700 315 × 360 = 162° 700 360° Question 10.1 At a show in the theatre, 2160 people went on a Thursday night, 1890 went on Saturday night and only 810 went on Sunday night. Represent this data using a pie chart. ©The Dublin School of Grinds Page 27 The Examiner may actually give you the pie chart and ask you questions about it! Example 2 Each student in a class plays one of four sports: football, rugby, hockey or basketball. The pie chart represents the number of students that play each sport. i) What is the measure of the angle for basketball? ii) 10 students play football. How many students play hockey? iii) How many students are in the class? Solution i) ii) Degrees in a circle =3600 Basketball = 360 – 150 – 120 – 60 = 300 10 students = 1500 1 student = 150 So every 150 represents 1 student Hockey =1200 120 Number of students who play hockey = 15 =8 iii) All students = 3600 360 Number of students = 15 = 24 ©The Dublin School of Grinds Page 28 Question 10.2 The given pie chart illustrates the grades of 120 Junior Certificate students in Maths. (Note: a fail is an E or an F) i) ii) How many students got an A How many students passed the exam ©The Dublin School of Grinds Page 29 11. Histograms One of the most common ways of representing a frequency distribution is by means of a histogram. Histograms are similar to bar charts, but there are some important differences: There are no gaps between the bars in a histogram Histograms can represent discrete or continuous data, while bar charts only represent discrete data. The data is always grouped. The groups are called classes. The Examiner can ask about histograms in two different ways. 1. 2. You may be asked to draw a histogram when given a frequency distribution. You may be given the histogram and ask questions about it. First let’s look at drawing a histogram. Example 1 The frequency table below shows the time in minutes, spent by a group of teenagers on the internet a day. Represent this data using a histogram. Time 0-20 20-40 40-60 60-80 80-100 No. of teenagers 2 3 8 12 5 Solution Remember: the interval (time) is on the bottom of a histogram ©The Dublin School of Grinds Page 30 Question 11.1 The frequency table given below shows the distance a group of employees at a company have to travel to work every day. Distance (km) 0-2 Number of employees 6 Draw a histogram to represent this data. ©The Dublin School of Grinds 2-4 8 4-6 4 Page 31 6-8 13 8-10 5 10-12 4 Another way the Examiner can ask you about a histogram is when you are given the histogram. Example 2 The histogram below show the amount of money spent by customers buying their groceries in a shop. i) ii) iii) iv) Solution i) How many customers spent more than €80 in the shop? How many customers were included in the survey? What is the modal class? In which interval does the median lie? 30+20 =55 customers ii) Total = 10+25+40+50+35+20 = 180 iii) Modal class (mode) = [60-80] iv) There are 180 customer so the median is between the 90 th and 91st customer. 10 + 25 + 40 + 50 One of these 50 customers is the 90th/91st Median = [60-80] class (90th & 91st customers) ©The Dublin School of Grinds Page 32 Question 11.2 The histogram below gives the numbers of people in the indicated age groups at a cinema. i) ii) iii) iv) How many people over 40 attended the cinema? How people attended the cinema when the information was collected? What is the modal class? In which interval does the median lie? ©The Dublin School of Grinds Page 33 12. Stem and leaf diagrams Stem and leaf diagrams are sometimes called stem plots. Example of a stem and Leaf Diagram There must be a key included to show how the stem and leaf combine. We will now look at some examples to explain how to draw them. Examples 1 The number of customers in a restaurant over a two week period was recorded as follows. 36 43 39 53 29 43 33 47 51 27 42 31 34 22 i) Draw a stem and leaf diagram ii) Find the range Solution i) 1. Find the smallest and largest values and decide on the intervals to use Smallest = 22 Largest = 53 Intervals (groups): 20 – 29, 30 – 39 40 – 49, 50 – 59 2. Draw the stem and leaf and fill it in unordered: (When you fill in the leaves on the diagram, cross the numbers out in the data) Key: 2|9 means 29 customers 3. Now write the leaves in size order Key: 2|9 means 29 customers ii) Range = Maximum Value - Minimum Value Range = 53 - 22 = 31 ©The Dublin School of Grinds Page 34 Question 12.1 The following array of numbers gives the ages of the members of a tennis club. 15 17 12 16 24 29 36 25 38 42 53 44 49 53 29 21 11 38 14 29 i) ii) iii) iv) Draw a stem and leaf diagram to show these ages. What is the lower quartile? Find the upper quartile. What is the interquartile range? ©The Dublin School of Grinds Page 35 17 The Examiner can also ask you to draw back to back stem and leaf diagrams. Example 2 The results of a maths and a science exam are given in the table below. Science 55 27 30 71 45 91 52 53 83 25 59 65 Maths 64 76 45 75 48 51 55 72 85 64 36 65 i) Draw a back to back stem and leaf diagram. ii) Find the median mark in a) Science b) Maths Solution i) First, draw the diagram unordered. Remember: cross the numbers out as you write them in the diagram. 67 67 69 58 38 74 73 47 54 40 Then rewrite the diagram with the numbers in order. Start with the smallest numbers closest to the stem. ii) There are 19 results for each subject, therefore the median (middle) mark must be the 10 th number. Median in Science = 55 Median in Maths = 64 ©The Dublin School of Grinds Page 36 86 83 45 62 Question 12.2 Ten boys and ten girls in a class were asked how long they had studied for a test. The times in minutes are in the table below. Boys Girls i) ii) 67 63 53 93 54 40 66 62 71 95 41 51 42 88 Draw a back to back stem and leaf diagram for these results. Find the interquartile range for the boys and the girls. ©The Dublin School of Grinds Page 37 52 87 43 41 64 75 13. Scatter graphs and correlation Scatter graphs are used to investigate relationships between two sets of data. If the points on a scatter graph are close to a straight line, then we say there is a strong correlation between the two sets of data. The closer the points are to a straight line, the stronger the relationship will be. Examples of sets of data that could be compared are i. ii. iii. Obesity and heart attacks Height and age Drink driving and accidents The strength of the relationship between two sets of data is known as correlation. (i.e.: uphill) (i.e.: downhill) Correlation Coefficient The correlation coefficient, r, is always a number between – 1 and 1 If r = 1, there is a perfect positive correlation between the two variables. ©The Dublin School of Grinds Page 38 (i.e.: all over the gaf) If r = 0, there is no correlation between the two variables. If r = - 1, there is perfect negative correlation between the two variables. . The Examiner can ask you to estimate the correlation (see diagrams below). He can also ask you to describe the correlation (see diagrams below). Moderately Strong Positive Correlation Strong Positive Correlation Strong Negative Correlation ©The Dublin School of Grinds Moderately Strong Negative Correlation Page 39 No Correlation Question 13.1 There are 4 scatter graphs A, B, C and D shown below. Here are six correlation coefficients: 0, 0.3, 0.95, -0.8, 0.7, -0.5 Chose the most likely correlation coefficient from the above to match the scatter graphs A, B, C and D. ©The Dublin School of Grinds Page 40 Example 1 The manager of a theme park thought the number of visitors to the park was dependent on the temperature. Temperature 16 22 31 19 23 26 21 17 24 29 21 25 23 29 (0C) Number of 205 248 298 223 252 280 233 211 258 295 229 252 248 284 visitors He kept a record of the temperature and the numbers of the visitors over a two-week period. i) Plot these points on a scatter graph. ii) Comment on the type of correlation between the two points. Solution i) ii) There is a strong positive correlation between the temperature and the number of visitors in the park. ©The Dublin School of Grinds Page 41 Question 13.2 A Leaving Certificate students wanted to check if there was correlation between the predicted heights of daisies and their actual heights. Predicted height (cm) Actual height (cm) 5.3 6.2 4.9 5.0 4.8 6.6 7.3 7.5 6.8 5.5 4.7 6.8 5.9 7.1 4.7 7.0 5.3 4.5 5.6 5.9 7.2 6.5 7.2 5.8 5.3 5.9 6.8 7.6 i) ii) i) Draw a scatter diagram to illustrate the data. Comment on the correlation between the predicted height and the actual height. ii) Comment: ©The Dublin School of Grinds Page 42 Causality and correlation Just because two variables are correlated does not mean they cause each other. Example It can be seen from the scatter plot that there is a strong positive correlation. But the amount of ice-creams sold don’t cause shark attacks, or vice-verse. The thing that is causing both these to increase the temperature, this is called a ‘lurking variable’ because it is hidden (lurking) in the background. Question 13.3 Explain with the aid of an example, what is meant by this statement: “Correlation dos not imply causality” ©The Dublin School of Grinds Page 43 14. Hypothesis testing and the line of best fit We may decide to test a theory for example ‘everybody should learn how to drive a car’. This theory is called a hypothesis. In testing a hypothesis, data may be i) Collected by using a questionnaire. or ii) Given from a source, such as, the Central Statistics Office or a report. To test our theory/hypothesis we might decide to question 100 people. We call these 100 people a random sample. If 100 random people are ask the question ‘should everybody learn how to drive a car?’ , how many of them are required to say yes for us to claim they agree with the stated hypothesis? Is it 20 or 40 or 60? There is no correct answer! The line of best fit Sometimes it is useful to show the line of best fit on a scatter graph/diagram. We attempt to make sense of the pattern. To help us draw the line of best fit we try to have the same number of points on each side of the line while showing how the pattern is changing. E.g. Note Outliers are values that are not typical of the other values. ©The Dublin School of Grinds Page 44 Example 1 A student in a class had a hypothesis (theory) that ‘the larger the size of a television screen was the more expensive it was’. He did some research and this was the data that he found: Size (inches) 32 37 40 46 50 55 59 Price (€) 450 550 700 1000 1200 1800 2000 i) Plot these points on a scatter diagram and describe the correlation? ii) Does the scatter diagram verify the student’s hypothesis? iii) Draw the line of best fit by eye. i) There is a strong positive correlation. ii) The scatter graph shows strong positive correlation between the size of the television and the price => The student’s hypothesis is proved. iii) The line of best fit is shown on the graph with an equal amount of points above the line and below the line. ©The Dublin School of Grinds Page 45 Question 14.1 A Leaving Cert student wanted to test his hypothesis that ‘the taller you are the heavier you are’. To do this he took the height and weight of 10 of his classmates. The results were as follows: Weight (kg) 85 63 88 60 70 58 74 72 75 77 Height (cm) 176 168 180 171 175 166 173 178 174 177 i) Plot these points on a scatter diagram and describe the correlation? ii) Does the scatter diagram verify the student’s hypothesis? iii) Draw the line of best fit by eye. i) ©The Dublin School of Grinds Page 46 Hypothesis tests We use a hypothesis tests when a claim is being made about something. There are 6 steps to follow: Step 1: 𝐻0 = null hypothesis (write as a decimal) this is the percentage claimed to be true. 𝐻1 = alternative hypothesis. Step 2: Write the number of students who support the claim as a fraction of the total sample: 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑤ℎ𝑜 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 Step 3: Find the margin of error using this formula: 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = Step 4: 1 √𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 Find the confidence interval: 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 ± 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝐸𝑟𝑟𝑜𝑟 Step 5: If the null hypothesis lies within the confidence interval we accept the null hypothesis, if not we reject the null hypothesis. Step 6: Comment on the outcome of step 5. In these questions you may see the following terms: “95% confidence interval” “95% confidence level” “5% level of significance” We do not use these in the questions, this is how the Examiner tells us we will use the formula: 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = Always round off to three decimal places. ©The Dublin School of Grinds Page 47 1 √𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 Example 2 A political party had claimed that it is has the support of 23% of the electorate. Of 1111 the voters sampled, 234 stated that they support the party. Is this sufficient evidence to reject the party’s claim, at the 5% level of significance. Solution Step 1: 𝐻0 = The political party has 23% support of the electorate. (= 0.23) 𝐻1 = The party does not have 23% support of the electorate. Step 2: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑤ℎ𝑜 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 234 = 1111 = 0.210621062 = 0.211 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = Step 3: 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = = 1 √𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 1 √1111 = 0.0300015 = 0.030 Step 4: Confidence interval 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 ± 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝐸𝑟𝑟𝑜𝑟 0.211 − 0.30 = 0.181 0.211 + 0.30 = 0.241 Confidence interval: [0.181, 0.241] Step 5: 0.23 is within : [0.181, 0.241] => accept 𝐻0 The political party has 23% support of the electorate. ©The Dublin School of Grinds Page 48 Question 14.2 Past records from a survey of McDonalds show that 20% of people who buy a ‘Big Mac’ also buy ‘McChicken Nuggets’. On a particular day a random sample of 30 people was taken from those that had bought a ‘Big Mac’ and 2 of them were found to have bought ‘McChicken Nuggets’. Test at the 5% significance level, whether or not the proportion of people who bought ‘McChicken Nuggets’ that day had changed. State you your hypothesis clearly. ©The Dublin School of Grinds Page 49 15. Past and probable exam questions Question 1 ©The Dublin School of Grinds Page 50 Question 2 ©The Dublin School of Grinds Page 51 Question 3 ©The Dublin School of Grinds Page 52 Question 4 ©The Dublin School of Grinds Page 53 Question 5 ©The Dublin School of Grinds Page 54 ©The Dublin School of Grinds Page 55 ©The Dublin School of Grinds Page 56 Question 6 ©The Dublin School of Grinds Page 57 Question 7 ©The Dublin School of Grinds Page 58 ©The Dublin School of Grinds Page 59 Question 8 ©The Dublin School of Grinds Page 60 Question 9 ©The Dublin School of Grinds Page 61 ©The Dublin School of Grinds Page 62 ©The Dublin School of Grinds Page 63 ©The Dublin School of Grinds Page 64 Question 10 ©The Dublin School of Grinds Page 65 ©The Dublin School of Grinds Page 66 ©The Dublin School of Grinds Page 67 ©The Dublin School of Grinds Page 68 Question 11 ©The Dublin School of Grinds Page 69 ©The Dublin School of Grinds Page 70 ©The Dublin School of Grinds Page 71 ©The Dublin School of Grinds Page 72 Question 12 Note: This questions involves a mixture of topics. ©The Dublin School of Grinds Page 73 ©The Dublin School of Grinds Page 74 ©The Dublin School of Grinds Page 75 Question 13 ©The Dublin School of Grinds Page 76 ©The Dublin School of Grinds Page 77 Question 14 ©The Dublin School of Grinds Page 78 ©The Dublin School of Grinds Page 79 ©The Dublin School of Grinds Page 80 16. Solutions to Statistics Question 4.1 First write the numbers in order: 4, 4, 4, 6, 6, 8, 10, 12, 18 i) The most commonly occurring number is 18 because it occurs 5 times. The mode is = 4 ii) The median is the middle number. There are seventeen numbers, so the ninth number is the middle number. The median = 6 iii) Mean = 4+4+4+6+6+8+10+12+18 9 72 = 9 =8 Question 4.2 i) The biggest number on the bottom is nine and above it is 4 ∴ mode = 4 ii) There are 40 students in the class. ∴ Looking for the 20th and 21st match. Add up the numbers until you reach the 20th/21st 7 + 7 + 8 One of these 8 students is the 20th/21st Median = 20𝑡ℎ+21𝑠𝑡 3+3 2 = 2 =3 iii) Finally the mean is found in a similar way to with a group of numbers. 𝑆𝑢𝑚 𝑜𝑓 (𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 × 𝑛𝑢𝑚𝑏𝑒𝑟) 𝑀𝑒𝑎𝑛 = 𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑟𝑞𝑢𝑒𝑛𝑐𝑖𝑒𝑠 = 7(1)+7(2)+8(3)+9(4)+5(5)+4(6) = 7+14+24+36+25+24 7+7+8+9+5+4 40 = 3.25 Question 4.3 We can use this to find the mid-interval values for all of the intervals Speed (km/h) 65 75 85 95 105 Frequency 8 15 12 10 8 115 3 125 4 We can then find the mean the same we did for the frequency distribution table above using the mid-interval values. 8(65)+15(75)+12(85)+10(95)+8(105)+3(115)+4(125) 𝑚𝑒𝑎𝑛 = 40 = 5300 60 = 88.3𝑘𝑚/ℎ Just like the frequency table it is the number above the largest number on the bottom line. => The modal group = 80-90 There are 60 vehicles so the median is between the 30th and 31st vehicles. 8 + 15 + 12 One of these 12 vehicles is the 30th/31st. ©The Dublin School of Grinds Page 81 Question 4.4 1(0) + 3(1) + 𝑥(2) + 9(3) + 7(4) + 4(5) 1+3+𝑥+9+7+4 78 + 2𝑥 3= 24 + 𝑥 3(24 + 𝑥) = 78 + 2𝑥 72 + 3𝑥 = 78 + 2𝑥 3𝑥 − 2𝑥 = 78 − 72 => 𝑥 = 6 3= Question 5.1 i) Boys: 87 + 88 + 84 + 91 + 81 + 85 + 86 7 602 = 7 = 86 𝑅𝑎𝑛𝑔𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 − 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 91 − 81 = 10 𝑀𝑒𝑎𝑛 = Girls: 86 + 87 + 95 + 72 + 82 + 99 + 88 7 609 = 7 = 87 𝑅𝑎𝑛𝑔𝑒 = 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 − 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 99 − 72 = 27 𝑀𝑒𝑎𝑛 = Comment: The mean values are similar for boys and girls but the range for boys is smaller. Therefore the boys’ pulse rates are more consistent. Question 5.2 List the values in order from smallest to biggest = 6, 7, 8, 9, 9, 11, 12, 12, 15, 16, 19 (11 numbers) i) The lower quartile Q1 1 × 11 = 2.75 4 round up = 3rd value 3rd value = 8 ii) The upper quartile Q3 iii) The interquartile range = upper quartile – lower quartile ©The Dublin School of Grinds 3 × 11 = 8.25 4 round up = 9th value 9th value = 15 = 15 – 8 = 7 Page 82 Question 5.3 First write the numbers down in order from smallest to biggest. 46, 48, 49, 52, 55, 58, 63, 64, 71, 73, 78, 80 (12 numbers) i) The lower quartile Q1 ii) The upper quartile Q3 iii) The interquartile range = upper quartile – lower quartile = Q3 – Q1 = 72 – 50.5 = 21.5 1 × 12 = 3 4 3𝑟𝑑 + 4𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 49 + 52 => = 2 2 = 50.5 3 × 12 = 9 4 9𝑡ℎ + 10𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 71 + 73 => = 2 2 = 72 Question 6.1 How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. 2. 3. Key in Mode , then: 2 0(STAT) Then, 10 for 1 – Var The each number followed by = 18 0 =0 26 0 =0 22 0 =0 34 0 =0 250 =0 4. Next, press AC and 10 for the menu 0 0 , SHIFT Then, press 4 0 5. Now, press 3 0 for 𝑥𝜎𝑛 (standard deviation). Finally press = 0 =>The standard deviation 𝜎 = 5.29 How to do it in a Sharp EL-W531 1. 2. 3. Key in Mode , then: 10 (STAT) Then, 00for Standard Deviation The each number followed by DATA 180 DATA 0 260 DATA 0 220 DATA 0 340 DATA 25 0 4. 5. DATA 0 Next, press ALPHA then 60 for standard deviation Finally, Press = 0 =>The standard deviation 𝜎 = 5.29 ©The Dublin School of Grinds Page 83 Question 6.2 How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. Key in Mode , then: 20 (STAT) 2. Then, 10 for 1 – Var 3. The each number followed by = 20 = 0 40 = 0 60 = 0 80 = 0 100 =0 4. Next, press AC and 10 for the menu 0 0 , SHIFT Then, press 40 5. Now, press 30 for 𝑥𝜎𝑛 (standard deviation). Finally press = 0 =>The standard deviation 𝜎 = 2.8 How to do it in a Sharp EL-W531 1. Key in Mode , then: 10 (STAT) 2. Then, 00 for standard deviation 3. The each number followed by DATA 20 DATA 40 DATA 6 0 DATA 8 0 DATA 4. 5. 100 DATA Next, press ALPHA then 60 for standard deviation Finally, Press =0 =>The standard deviation 𝜎 = 2.8 ©The Dublin School of Grinds Page 84 Question 6.3 How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. Key in Mode , then: 20 (STAT) 2. Then, 10 for 1 – Var 3. Then each number followed by = the scroll across and input the frequency 1 =0 6 = 0 0 0 4. 20 30 =0 =0 9 = 0 0 4 = 0 0 40 50 =0 4 = 0 0 =0 4 = 60 =0 3 = 00 0 0 Find the standard deviation AC SHIFT 1 4 3 = =>The standard deviation 𝜎 = 1.63 How to do it in a Sharp EL-W531 1. Key in Mode , then: 1 0 (STAT) 2. Then, 00 for standard deviation 3. Next, input each number followed by (x,y) and then the frequency followed by DATA 1 (x,y) 0 0 (x,y) 0 2 0 30 (x,y) 0 4 (x,y) 0 0 5 (x,y) 0 0 6 (x,y) 0 0 6 DATA 0 9 DATA 0 4 DATA 0 4 DATA 0 4 DATA 3 DATA 0 0 4. Next, press ALPHA then 60 for standard deviation 5. Finally, Press =0 =>The standard deviation 𝜎 =1.63 ©The Dublin School of Grinds Page 85 Question 6.4 How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 1. Key in Mode , then: 20(STAT) 2. Then, 1 for 1 – Var 0 3. Then each number followed by = the scroll across and input the frequency 1 =0 7 = 0 0 0 4. 20 30 =0 =0 8 = 0 0 4 = 0 0 4 0 5 0 6 0 =0 4 = 0 0 =0 3 = = 4 = 0 00 0 0 Find the standard deviation AC SHIFT 1 4 3 = =>The standard deviation 𝜎 = 1.71 How to do it in a Sharp EL-W531 1. Key in Mode , then: 10 (STAT) 2. Then, 00 for standard deviation 3. Next, input each number followed by (x,y) and then the frequency followed by DATA 1 (x,y) 0 0 2 (x,y) 0 0 30 (x,y) 0 4 (x,y) 0 0 5 (x,y) 0 0 60 (x,y) 0 7 DATA 0 8 DATA 0 0 4 DATA 0 0 0 0 0 4 DATA 0 3 DATA 4 DATA 0 0 4. Next, press ALPHA then 60 for standard deviation 5. Finally, Press = 0 =>The standard deviation 𝜎 =1.71 ©The Dublin School of Grinds Page 86 Question 6.5 Using the mid interval values: Minutes 3 5 7 9 Number of Pupils 6 6 4 1 How to do it on a CASIO fx-83GT Plus and fx-85GT Plus. 5. Key in Mode , then: 20(STAT) 6. Then, 10 for 1 – Var 7. Then each number followed by = the scroll across and input the frequency 3 =0 6 = 0 0 0 50 70 8. =0 =0 6 = 0 0 4 = 0 0 90 =0 1 = 0 0 Find the standard deviation AC SHIFT 1 4 3 = =>The standard deviation 𝜎 = 1.81 How to do it in a Sharp EL-W531 3. Key in Mode , then: 10 (STAT) 4. Then, 0 0 for standard deviation 4. Next, input each number followed by (x,y) and then the frequency followed by DATA 3 (x,y) 0 0 50 (x,y) 0 70 (x,y) 0 9 (x,y) 0 0 6 DATA 0 6 DATA 0 4 DATA 0 1 DATA 0 5. Next, press ALPHA then 6 for standard deviation 0 6. Finally, Press =0 =>The standard deviation 𝜎 =1.81 ©The Dublin School of Grinds Page 87 Question 7.1 𝑚𝑒𝑎𝑛 = 6 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 2 2 = 𝑚𝑒𝑎𝑛 − 2 × 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 10 = 𝑚𝑒𝑎𝑛 + 2 × 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 According to the empirical rule 95% of the population lies in the range [2, 10] Question 8.1 i) ii) 5 × 100% 15 1 = 33 % 3 % 𝑤ℎ𝑜 𝑠𝑐𝑜𝑟𝑒𝑑 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 290 − 294 = Question 9.1 ©The Dublin School of Grinds Page 88 Question 10.1 Question 10.2 i) 120 students = 3600 1 student = 30 A grade students = 1200 120 Number of students who got an A = 3 = 40 ii) Degrees of students who got an A, B, C or a D = 1200 + 750 + 540 + 450 = 2940 294 Number of students who passed = 3 = 98 Question 11.1 ©The Dublin School of Grinds Page 89 Question 11.2 i) 7 + 5 = 12 people ii) Total = 2 + 5 + 12 + 9 + 7 + 5 = 40 iii) Modal class (mode) = [20 - 30] iv) There are 40 people so the median is between the 20th and 21st person. 2 + 5 + 12 + 9 One of these 9 people is the 20th/21st Median = [30 - 40] class (20th & 21st people) Question 12.1 i) ii) The lower quartile Q1 1 = 5.25 4 6𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 17 21 × iii) The upper quartile Q3 3 = 15.75 4 16𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 38 21 × iv) The interquartile range = upper quartile – lower quartile = 𝑄3 – 𝑄1 = 38 − 17 = 21 ©The Dublin School of Grinds Page 90 Question 12.2 i) ii) Boys: The lower quartile Q1 The upper quartile Q3 1 × 10 = 2.5 4 => 3𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 = 43 3 × 10 = 7.5 4 => 8𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 66 The interquartile range = 𝑄3 − 𝑄1 = 66 − 43 = 23 Girls: The lower quartile Q1 The upper quartile Q3 1 × 10 = 2.5 4 => 3𝑟𝑑 𝑣𝑎𝑙𝑢𝑒 = 51 3 × 10 = 7.5 4 => 8𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 88 The interquartile range = 𝑄3 − 𝑄1 = 88 − 51 = 37 Question 13.1 Correlation coefficient of graph A = 0.95 Correlation coefficient of graph B = 0 Correlation coefficient of graph C = -0.8 Correlation coefficient of graph D = 0.7 ©The Dublin School of Grinds Page 91 Question 13.2 i) ii) There is a moderately strong correlation between predicted height and the actual height. Question 13.3 A correlation may exist between the number of umbrellas sold and the number of traffic accidents that occur but the thing that causes both of these to increase is more rain. Question 14.1 i) The is a moderately strong correlation between height and weight. ii) The student’s hypothesis holds according to the scatter diagram. The line of best fit is shown on the graph with an equal number of points above and below the line. ©The Dublin School of Grinds Page 92 Question 14.2 Step 1: 𝐻0 = 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets (=0.2). 𝐻1 = Either more or less than 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets. Step 2: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑤ℎ𝑜 𝑠𝑢𝑝𝑝𝑜𝑟𝑡 𝑡ℎ𝑒 𝑠𝑎𝑚𝑝𝑙𝑒 𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 2 = 30 = 0.067 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = Step 3: 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟 = = 1 √𝑆𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 1 √30 = 0.183 Step 4: Confidence interval 𝑆𝑎𝑚𝑝𝑙𝑒 𝑝𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 ± 𝑀𝑎𝑟𝑔𝑖𝑛 𝑜𝑓 𝐸𝑟𝑟𝑜𝑟 0.067 − 0.183 = −0.116 0.067 + 0.183 = 0.25 Confidence interval: [-0.116, 0.25] Step 5: 0.2 is within: [-0.116, 0.25] => accept 𝐻0 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets. ©The Dublin School of Grinds Page 93 Solutions to past and probable exam questions. Question 1 i) 21 ii) The shape is almost normal but it is slightly skewed right. iii) 57m iv) The median is the middle value when they are all arranged in increasing order. If there are two ‘middle’ numbers, you use the average of these as the median. Question 2 a) b) Difference: The measurements of oxygen levels are generally lower after the clean-up took place. Similarity: The two sets of measurements have the same range. Before: 25 – 2 = 23 After: 26 – 3 = 23 Question 3 a) Mean = 35 Standard Deviation = = 9.848857802 b) i) Answer: 20 ii) Range = 157cm – 133cm = 24cm iii) 10 20 = 50% ©The Dublin School of Grinds Page 94 Question 4 a) b) Offaly Kerry Difference 1: Kerry temperatures are generally lower. Difference 2: Offaly has a greater range of temperatures. Question 5 a) i) ii) iii) iv) v) Cycle 2 minutes 10 and 12 minutes 25 minutes Cycle i) ii) B Frank is the slowest person in the run so his dot is the highest on the graph. The people slower than Frank in the cycle are the points to the right of Frank on the graph. b) iii) Answer: 6 Brian would have had a very fast run but a very slow cycle. This would have been very different to the other data points. He would have been an outlier. c) i) Range of the females = 29.7 – 13.4 = 16.3 minutes Range of the males = 23.0 – 14.9 = 8.1 minutes The female distribution is more spread out. ii) Yes, the shapes are more or less the same. The only difference is in the range. Maybe with a larger sample her theory may be right. ©The Dublin School of Grinds Page 95 Question 6 a) Answer: 28 sweets b) Range = 32 – 25 = 7 sweets c) The lower quartile Q1 The upper quartile Q3 1 × 19 = 4.75 4 => 5𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 28 3 × 19 = 14.25 4 => 15𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 30 The interquartile range = upper quartile – lower quartile = 𝑄3 – 𝑄1 = 30 − 28 = 2 sweets d) “This is a set of univariate data. The data are discrete.” Question 7 a) i) Answer: 27,098 ii) 44,139 × 7 = 308,973 iii) 𝑀𝑒𝑎𝑛 = (326,134 × 1) + (413,786 × 2) + (264,438 × 3) + ⋯ + (1,719 × 9) + (1,128 × 10) 1,462,296 4,105,973 = 1,462,296 ≈ 2.8 𝑝𝑒𝑜𝑝𝑙𝑒 b) Conor was trying to show… Number of households has more than doubled in the given time period. Fiona was trying to show … The gradual reduction in the number of people per household. Ray was trying to show…. The number of households has more than doubled in the time period, and the move has been towards smaller household sizes. ©The Dublin School of Grinds Page 96 Question 8 a) The set that contains more numbers than any other is A and the set that contains fewer numbers than any other is D. b) On average, the data in set C are the biggest numbers and the data in set A are the smallest numbers. c) The data in set B are more spread out than the data in the other sets. d) The set that must contain some negative numbers is set A. e) If the four sets are combined, the median is most likely to be a value in set A. Question 9 a) i) ii) Shape of distribution: The shape of the distribution is approximately normal. Location of data (central tendency/ average): The central tendency (average) should be about 170cm. Spread of data (dispersion): There is a small standard deviation. Most of the data is between 160 cm and 180 cm. iii) You would need the population mean of all the Leaving Cert. 2012 students. b) i) Key: 14 | 9 means 149 ii) Difference: There are different means. The males have a higher mean. Similarity: The two sets of data have a similar range. c) i) Mean: μ =178 8. Spread: σ = 7 9. 95% are between μ - 2σ and μ + 2σ . μ - 2σ = 178.8 – 2(7.9) =163cm μ + 2σ = 178.8 + 2(7.9) =194.6cm “95% of nineteen-year-old Irish men are between 163 cm and 194.6 cm in height.” ii) 68% are between μ - σ and μ + σ . μ - σ = 178.8 – 7.9 =170.9cm μ + 2σ = 178.8 + 7.9 =186.7cm “68% of nineteen-year-old Irish men are between 170.9 cm and 186.7 cm in height.” ©The Dublin School of Grinds Page 97 d) Answer: No Reason: It is not a random sample of the Irish male population. The population are 19 year old males. The Leaving Cert sample could be 17/18/19 year olds. e) Average height of the males in the class = 175.5 cm 95% of the population taken here have heights between 163 cm and 194.6 cm. So although the sample mean is smaller than the population mean it falls well within these extremes and so in general is not less than the average. f) i) ii) There is a moderate correlation between the years spent in full time education and annual income of adults. iii) The sample is not random as it excludes those with no landline or who are ex-directory. This may bias the results in favour of people who own phones. Question 10 a) i) ii) 𝑀𝑒𝑎𝑛 = iii) ©The Dublin School of Grinds 173,273 + 180,754 + 146,470 + 54,432 + 84,907 + 86,932 6 726768 = 6 = 121,128 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 = 86,932 − 54,432 = 32,500 32,500 % 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 = 54,432 = 59.71% Page 98 iv) Aoife’s argument does not recognise that this increase is from a very low base, and that sales had been much better before 2009. Paul’s argument does not recognise that, although sales are much lower now than in 2007, they have recovered a lot since their lowest point in 2009. v) Sales fell dramatically from 2007 to 2009; they recovered a lot since then, but are still much lower than they were at the start. i) Highest quarterly sales are in the first quarter and decrease significantly in each subsequent quarter. ii) People like to buy new cars early in the year so that they have a new year number plate. iii) In 2011 sales in the 1st quarter where 39,484 which is approximately 45% of the sales for the entire year. Assume that the first quarter in 2012 is also 45%. 36,081 ∴ × 100 = 80,180 45 => 45% = 36,081 1% = 801.8 100% = 80,180 b) c) i) ii) d) Answer: D i) Diesel Petrol ii) Yes. The diesel engines grouped at the top of the plot have a smaller median [/mean] value. iii) No. The emissions for the petrol engines are more spread out than for the diesel ones. The range for the petrol engines is greater than that for the diesel engines. ©The Dublin School of Grinds Page 99 Question 11 a) b) The marriage rates range from 43 to 52 and are grouped at the top of the plot. The death rates range from 61 to 90 and are grouped at the bottom of the plot. c) Median: 50 The lower quartile Q1 1 × 21 = 5.25 4 => 6𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 46 The upper quartile Q3 3 × 19 = 15.75 4 => 16𝑡ℎ 𝑣𝑎𝑙𝑢𝑒 = 51 The interquartile range = upper quartile – lower quartile = 𝑄3 – 𝑄1 = 51 − 46 = 5 d) i) 𝑀𝑒𝑎𝑛 43 + 43 + 45 + 45 + 46 + 46 + 47 + 47 + 48 + 49 + 50 + 50 + 50 + 51 + 51 + 51 + 52 + 52 + 52 + 52 + 52 = 21 1645 = 21 = 78.3 ii) Range of 1 standard deviation about the mean: [78⋅3−10⋅3, 78⋅3+10⋅3] =[68, 88⋅6] 68, 71, 73, 76, 79, 83, 87, 85, 86, 87, 86, 87 e) f) 75,174 × 10,000 = 4,556,000 165 4,556,000 × 61 = 27,791 10,000 g) The birth rates given are per 10000 of the population. If the population in 2000 was greater than in 1990, more children could have been born in 2000 than in 1990 even though the birth rate in 2000 was lower. h) 1990 Ratio: 151:90 2010 Ratio: 165:61 Prediction: The population of the country is expected to increase. Reason: The increase in the ratio from 1990 to 2010 suggests that more children are being born for each person that dies. i) Strong negative correlation With the increasing birth rate, the population is getting younger and the death rate is declining. ©The Dublin School of Grinds Page 100 Question 12 a) (i) Registered properties = 1.9𝑚𝑖𝑙𝑙𝑖𝑜𝑛 × 90% = 1.9𝑚𝑖𝑙𝑙𝑖𝑜𝑛 × 0.9 = 1.71 𝑚𝑖𝑙𝑙𝑖𝑜𝑛 (ii) Some of the people may have been boycotting the property tax. b) €0 - €100,000 24.9 × 360 = 89.6 100 €100,001 €150,000 28.6 × 360 = 102.96 100 €150,001 €200,000 21.9 × 360 = 78.84 100 €200,001 €250,000 10.4 × 360 = 37.44 100 €250,001 €300,000 4.9 × 360 = 17.64 100 Over €300,000 9.3 × 360 = 33.48 100 c) (i) to find the number of properties in each valuation band multiply 1,710,000 by the percentage in each band given in part b) (ii) 𝑇𝑎𝑥 𝑑𝑢𝑒 𝑢𝑝 𝑡𝑜 €300 000 = 19 160 550 + 54 774 720 + 58 794 930 + 35 923 680 + 20 696 130 = €189 350 010 (iii) 𝑇𝑎𝑥 𝑑𝑢𝑒 𝑜𝑣𝑒𝑟 €300 000 = 241 000 000 − 189 350 010 = €51 649 990 51 649 990 (iv) 𝑀𝑒𝑎𝑛 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦 𝑡𝑎𝑥 𝑜𝑣𝑒𝑟 €300 000 = 159 030 = €328.78 (v) Find 20% of the properties valued €100 001 – €150 000 489 060 × 20% = 489 060 × 0.20 = 97 812 𝑇𝑎𝑥 𝑝𝑎𝑖𝑑 𝑜𝑛 𝑡ℎ𝑒𝑠𝑒 97,812 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 = 97 812 × 112 = €10 954 944 𝑇𝑎𝑥 𝑡ℎ𝑎𝑡 𝑠ℎ𝑜𝑢𝑙𝑑 𝑏𝑒 𝑝𝑎𝑖𝑑 𝑜𝑛 𝑡ℎ𝑒𝑠𝑒 97 812 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑖𝑒𝑠 = 97 812 × 157 = €15 356 484 𝐸𝑥𝑡𝑖𝑚𝑎𝑡𝑒 𝑜𝑓 𝑒𝑥𝑡𝑟𝑎 𝑡𝑎𝑧 𝑡ℎ𝑎𝑡 𝑤𝑜𝑢𝑙𝑑 𝑏𝑒 𝑟𝑎𝑖𝑠𝑒𝑑 = 15 356 484 − 10 954 944 = €4 401 540 ©The Dublin School of Grinds Page 101 Question 13 (a) (b) (i) 𝑀. 𝑂. 𝐸. = = 1 √𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 1 √1000 = 0.032 (ii) 𝑁𝑜. 𝑤ℎ𝑜 𝑠𝑖𝑝𝑝𝑜𝑟𝑡 𝑐𝑙𝑎𝑖𝑚 𝑠𝑎𝑚𝑝𝑙𝑒 𝑠𝑖𝑧𝑒 540 = 1000 = 0.54 𝑆𝑎𝑚𝑝𝑙𝑒 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 = 𝐶𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = 𝑆𝑎𝑚𝑝𝑙𝑒 𝑃𝑟𝑜𝑝𝑜𝑟𝑡𝑖𝑜𝑛 ± 𝑀. 𝑂. 𝐸. = 0.54 + 0.032 = 0.572 = 0.54 + 0.032 𝑜𝑟 = 0.54 − 0.032 = 0.572 = 0.508 𝐶𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝐼𝑛𝑡𝑒𝑟𝑣𝑎𝑙 = [0.508, 0.572] ©The Dublin School of Grinds Page 102 Question 14 (a) Numerical Continuous Explanation: The data collected is a number i.e. heights, most likely in meters. Heights can have any value inside some range so therefore it’s numerical continuous. (b) (i) 𝑥 𝑓 𝑓𝑥 147.5 15 2212.5 152.5 48 7320 157.5 80 12600 ∑ 𝑓𝑥 82215 162.5 112 18200 = 167.5 125 20937.5 ∑𝑥 500 172.5 81 13972.5 = 164.43 m 177.5 29 5147.5 182.5 10 1825 500 82215 (ii) 185 − 145 = 40 (iii) The median is the middle value when the values are arranged in order. If you were to arrange all the girls in order depending on their height, the middle girls height ((250th + 251st) divided by 2) would be 164.5cm. (c) 3 (ii) 9.6 16 16.2 5.8 2 81 × 100 = 16.2% 500 29 × 100 = 5.8% 500 10 × 100 = 2% 500 15 × 100 = 3% 100 48 × 100 = 9.6% 500 80 × 100 = 16% 500 (iii) Answer: Yes Reason: Girls between 175 -180 = 29% Boys between 175 – 180 = 11.5% 11.5% of 500 = 58 boyd 2(29) = 58 (iv) Answer: No Reason: If you take the last two bars which are the tallest you will see: Girls Boys: 175 - 180: 175 - 180: 5.8% = 29 11.5% = 58 180 – 185 180 – 185 2% = 10 9.5% = 48 => There are more tall boys. (d) (i) 166.7 + 2(𝑆. 𝐷. ) = 166.7 + (8.9) 166.7 − 2(𝑆. 𝐷. ) = 166.7 − 2(8.9) = 184.5 = 148.9 148.9 − 184.5 (ii) The girls heights are closer to the mean than the boys heights. The boys height are more spread out. ©The Dublin School of Grinds Page 103 Probability Probability is worth 8% to 13% of the Leaving Cert. It appears on Paper 2. 1. The fundamental principle of counting The examiner can ask you to state the fundamental principal of counting so you need to know the following: Definition: If one option can be selected in x ways and a second option can be selected in y ways then the amount of different ways the two options can be selected is x times y. We can also use this idea for three or more options. Example 1 A tractor company sells three different tractor models shown below: Each models comes in four different colours: red, green, blue and yellow. If a farmer wants to choose a tractor, how many variations does he/she have to choose from? Solution Here we have two options: the model (option 1) and the colour (option 2) When working on The model and The Colour counting problems we write the number of 3 X 4 = 12 options in boxes => There are 12 options for the farmer. Question 1.1 There are 5 roads from Dublin to Mullingar and 3 roads from Mullingar to Ballinasloe. Then there are 2 roads to Galway City from Ballinasloe. In how many ways can a person travel from Dublin to Galway city (via Mullingar then Ballinasloe)? ©The Dublin School of Grinds Page 104 2. Arrangements An arrangement (also known as a permutation) is the way a number of things (letters, numbers, people etc.) can be arranged. For example the letters X, Y and Z can be arranged in six different ways. XYZ XZY YXZ YZX ZXY ZYX We had three choices for the first option. Then, two for the second option and one for the third option, i.e.: 3 × 2 × 1 = 6 On your calculator you can do this by typing 3! Note: 3! Is pronounced ‘3 factorial’. Example 1 In how many ways can the letters in the word JACKET be arranged? Solution There are 6 letters in the word table so, 6 × 5 × 4 × 3 × 2 × 1 ( which can be typed in as 6!) = 720 Question 2.1 There are 5 students sitting on a bench. In how many different ways can the students be seated on the bench? ©The Dublin School of Grinds Page 105 3. Arrangements with restrictions When you are asked to find out how many ways objects can be arranged there will often be restrictions on the way the objects can be arranged. Example 1 How many different arrangements can be made using all of the letters in the word IRELAND with the following restrictions? i) How many of these begin with L and end with D ii) How many of these arrangements contain the A and the N together? Solution i) If L is in the first box and D is in the last box we cannot chose these letters. So there are only five choices left for the other boxes. L 5 4 3 2 1 D So that gives : = 5! = 120 ii) If A and N must come together we treat them as one unit at the start. AN So there are six boxes to be arranged, or 6! ways. But for each of these arrangements A and N can be arranged two ways (AN or NA), which can be written as 2! => the number of arrangements is 6! × 2! = 1440 Question 3.1 The code for a safe is made up of 4 numbers from 1-9. How many ways can a code be created if the numbers cannot be repeated (only using each number once)? Question 3.2 The pin code for a door is made up of 3 letters from the alphabet followed by 2 numbers from 1-9. How many different codes can be created if the letters or numbers cannot be repeated? (Note: There are 26 letters in the alphabet). ©The Dublin School of Grinds Page 106 Question 3.3 8 cars are in a race. All the cars finish the race and no cars finish the race at the same time. i) ii) How many different ways can the cars finish the race? Two of the cars are faster than the rest and always win or come second. In how many ways can the cars finish the race in this situation? ©The Dublin School of Grinds Page 107 4. What is probability? Probability is how likely it is that something is going to happen. In maths when we measure probability it ranges in value: A probability of 1 means the event is certain. A probability of 0 means the event is impossible. The probability of an event P(E) is given by 𝑃(𝐸) = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑜𝑢𝑡𝑐𝑜𝑚𝑒𝑠 We can write probability as a fraction, decimal or a percentage. Probability can be represented on a line called the probability scale. The Examiner requires you to know definitions in this section: A Trial: This is the act of doing an experiment e.g. tossing a coin or drawing a card at random from a deck. Outcome: This is the possible result from a trial e.g. the outcome from tossing a coin is heads or tails. Event: This is the outcome you want to happen e.g. rolling a six on a dice. Equally likely: you have the same chance of getting any of the outcomes. Example 1 Ten discs are placed in a bag and are numbered 1 to 10. If one of the discs is randomly selected, what is the probability of getting i) The number 7 ii) An odd number iii) A number divisible by 4 Solution i) There is one disc labelled 7 and ten discs: 1 => 𝑃(7) = 10 ii) The odd numbers are 1, 3 5, 7 and 9, so there are 5 odd numbers: 5 => 𝑃(𝑜𝑑𝑑 𝑛𝑢𝑚𝑏𝑒𝑟) = 10 1 = 2 iii) The numbers divisible by four are 4 and 8: 2 => 𝑃(𝑛𝑢𝑚𝑏𝑒𝑟 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4) = 10 1 = 5 ©The Dublin School of Grinds Page 108 Note: You are required to know what is in a deck of cards: Standard Deck of 52 Playing Cards: Diamonds (Red): 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10 ♦ J ♦ Q♦ K♦ A♦ Hearts (Red): 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥ A♥ Clubs (Black): 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣ A♣ Spades(Black): 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♠ Picture cards What is the probability of choosing, at random, one of the following cards from a normal pack of 52 playing cards? 1. A red card 26 2. A black card or ‘not a red card’ 26 3. A spade 13 Not a spade 39 4. 5. 52 52 52 52 4 An ace 52 48 6. Not an ace 7. The ace of spades 8. A picture card 12 9. A number card or ‘not a picture card’ 40 A card that is either a heart or a club 26 A card that is neither a heart nor a club 26 12. A 4 or 5 8 13. A 4 or 5 but not a spade 14. An even numbered card 10. 11. 52 = 1 2 2 . . = 1 = 3 = 1 = 4 4 13 12 13 1 52 52 52 52 52 6 52 20 52 If a deck of cards contains 52 cards, find the probability it is: an ace a diamond a black card ©The Dublin School of Grinds 1 52 Question 4.1 i) ii) iii) = Page 109 = 3 13 = 10 = 1 = 1 = = = 13 2 2 2 13 3 26 5 13 Question 4.2 In a school 30 boys and 30 girls were asked their favourite sports and the results were as follows: Boys Girls Hockey 5 14 Football 15 4 If one person was randomly selected what is the probability it was i) ii) iii) A girl A boy who said football was his favourite sport. Basketball as his or her favourite sport. ©The Dublin School of Grinds Page 110 Basketball 10 12 5. The probability of an event not happening If A is an event then: P(A not happening) = 1 - P(A happening) Example 1 A fair spinner has four letters A, B, C and D. What is the probability that the spinner will not land on C? Solution 𝑃(𝑛𝑜𝑡 𝐶) = 1 − 𝑃(𝐶) 1 𝑃(𝑛𝑜𝑡 𝐶) = 1 − 4 3 𝑃(𝑛𝑜𝑡 𝐶) = 4 Question 5.1 Sarah has a bag of sweets. There are 7 sweets in a bag and 2 of those sweets are lemon. i) ii) What is the probability she takes out a lemon sweet at random? What is the probability the she does not select a lemon sweet? ©The Dublin School of Grinds Page 111 6. Two events: sample spaces Sample space for 2 dice Sometimes the Examiner tries to help us with probability by asking us to write out all the outcomes, in a thing called a sample space. For example if two six sided die are thrown we can write out all the possible results for when you add the numbers shown on each die (see sample space on the right). From this we can easily work out probabilities. e.g. 𝑃(7) = = 6 36 1 6 Example 1 A coin is tossed and a six sided die is thrown. Write down all the possible outcomes. i) Find the probability of getting a tail and a 3. ii) Find the probability of getting an odd number and heads. Solution The Examiner has asked us to “Write down all the possible outcomes”. For this we can just use a list (If you want you could draw a sample space such as the one above, but a list is quicker in this case). {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6} There are 12 different outcomes. 1 i) 𝑃(𝑡𝑎𝑖𝑙 𝑎𝑛𝑑 3) = ii) 𝑃(𝑜𝑑𝑑 𝑎𝑛𝑑 ℎ𝑒𝑎𝑑𝑠) = = 12 3 12 1 4 Question 6.1 The arrows on two spinners are spun. List all the possible outcomes, for example, (1, 4), (1, 5), and write down the total number of possible outcomes. What is the probability that: i) ii) iii) both arrows point to an even number? both arrows point to an odd number? the numbers on both the numbers add up to 7? ©The Dublin School of Grinds Page 112 7. Estimating probabilities from an experiment So far we have calculated probabilities assuming all outcomes are equally likely. But in reality this is not always the case. In these circumstances we can estimate the probability from the result of an experiment or trial. We call this estimated probability the relative frequency: 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑡𝑟𝑖𝑎𝑙𝑠 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 Example 1 In an experiment David suspects that a coin is biased, so he tosses the coin 500 times. He records the number of tails every 100 hundred tosses. Number of Number of tails Relative The results are shown in the table. tosses Frequency (Heads ÷tosses) Comment on the results. Solution As the number of tosses (trials) increases, the relative frequency gets closer to 0.5 e.g. even chance. From the results of the experiment David can conclude that the coin is not biased. Note: 100 59 0.59 200 107 0.535 300 155 0.517 400 205 0.513 500 251 0.502 ‘Relative Frequency’ can also be called ‘Experimental Probability’ by the Examiner. Question 7.1 Sophie wanted to know if a dice she had was biased or not so she performed an experiment where she threw the dice 300 times. Her results are shown below. Number on die 1 2 3 4 5 6 Frequency 30 55 40 65 50 60 i) For this die what is the experimental probability of getting a) a 1 b) a 5 ii) From these results does this seem like a fair die? Give reasons for your answer. ©The Dublin School of Grinds Page 113 8. Expected frequency To find the expected frequency of an event you must: 1. Find the probability of the event. 2. Multiply the number of times the trial (experiment) is done by the probability. 𝑬𝒙𝒑𝒆𝒄𝒕𝒆𝒅 𝒇𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚 = 𝑷𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 × 𝑵𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒕𝒓𝒊𝒂𝒍𝒔 Example 1 The probabilities of a biased spinner are given in the table below Number 1 2 3 4 5 Frequency 0.2 x 0.1 0.25 0.15 i) Find the value of x. ii) If the spinner is spun 400 hundred times how many times would you expect it to land on 5. Solution i) The sum of all probabilities = 1 => 𝟎. 𝟓 + 𝒙 + 𝟎. 𝟏 + 𝟎. 𝟐𝟓 + 𝟎. 𝟏𝟓 = 𝟏 => 𝟎. 𝟕 + 𝒙 = 𝟏 => 𝒙 = 𝟏 − 𝟎. 𝟕 => 𝒙 = 𝟎. 𝟑 Expected frequency = expected number of 5’s = P(5) X number of trials = 0.15 X 400 = 60 ii) Question 8.1 Michael and Steven play 45 games of chess. Michael wins 20 of these games. i) ii) What is the probability Michael wins the next game based on the results of the previous 45 games? If they play 18 more games how many games would you expect Michael to win? ©The Dublin School of Grinds Page 114 9. Addition rule The addition rule is sometimes known as the OR rule. The Syllabus requires you to know what mutually exclusive events are so learn the following definition: Mutually exclusive events are events that cannot happen at the same time. For example: you cannot roll a four and an odd number on a die, therefore it is a mutually exclusive event. So, if two events cannot happen together then, 𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) This is what we call the addition rule (sometimes known as the OR rule) Example 1 If a die is rolled what is the probability of rolling a 3 or an even number from one roll of the die. Solution Obviously you cannot roll a 3 and an even number at the same time because 3 is odd so the events are mutually exclusive. There are three possible even numbers 2, 4 and 6. P(3 or Even) = P(3) + P(even) 1 3 = + 6 4 = 6 6 2 = 3 Question 9.1 A bag contains 12 discs numbered 1 to 12 inclusive. When a disc is chosen at random what is the probability that it is: i) odd ii) ©The Dublin School of Grinds divisible by 4 iii) odd or divisible by 4 Page 115 When events are not mutually exclusive i.e. they can occur together, we use: 𝑷(𝑨 𝒐𝒓 𝑩) = 𝑷(𝑨) + 𝑷(𝑩) − 𝑷(𝑨 𝒂𝒏𝒅 𝑩) Example 1 What is the probability of rolling a number less than 3 or an even number on a die? Solution The probability of a number less than three or an even number are not mutually exclusive since two is less than three and is also even. Therefore, the events are not mutually exclusive. 𝑷(< 𝟑𝒐𝒓 𝒆𝒗𝒆𝒏) = 𝑷(< 𝟑) + 𝑷(𝒆𝒗𝒆𝒏) − 𝑷(𝟐) 𝟐 𝟑 𝟏 𝟔 𝟔 𝟔 = + − = 𝟒 = 𝟐 𝟔 𝟑 Question 9.2 A card is drawn at random from a pack of 52. What is the probability that the card is: i) A spade ii) A jack iii) A spade or a jack iv) A red card v) A ten vi) A red card or a ten ©The Dublin School of Grinds Page 116 10. Using Venn diagrams We can use Venn diagrams to help us solve problems involving probability. Venn diagrams can be used to represent sample spaces. The Examiner can ask a question where you must use a given Venn diagram or they can ask to fill in the information on a Venn diagram. Example 1 In the Venn diagram: U = the students in 6th year in a school I = the number of students who study Irish F = The number of students who study French i) How many students are in 6th year? Using the result from i) find the probability that if a student is selected at random that they i) Study Irish ii) Study neither Irish nor French iii) Study French only iv) Study both French and Irish and illustrate this on a Venn diagram. Solution i) Total number of students = 70 ii) 𝑃(𝐼𝑟𝑖𝑠ℎ) = = iii) 58 70 29 35 = iv) 𝑃(𝐹𝑟𝑒𝑛𝑐ℎ 𝑜𝑛𝑙𝑦) = = v) 5 𝑃(𝑛𝑒𝑖𝑡ℎ𝑒𝑟 𝐼𝑟𝑖𝑠ℎ 𝑜𝑟 𝐹𝑟𝑒𝑛𝑐ℎ) = 14 7 70 1 10 𝑃(𝑏𝑜𝑡ℎ 𝐹𝑟𝑒𝑛𝑐ℎ 𝑎𝑛𝑑 𝐼𝑟𝑖𝑠ℎ) = = ©The Dublin School of Grinds 70 1 40 70 4 7 Page 117 Question 10.1 The Venn diagram shows how a group of 100 children like to spend their free time. S = play sports R = read V = play video games If a child is selected at random, what is the probability they: i) ii) iii) iv) v) vi) play video games play sports and video games do none of these activities read only read or play sports do at least two of these activities ©The Dublin School of Grinds Page 118 11. The AND rule The AND rule is also known as the multiplication rule. The AND rule is when we multiply the probability of two events to find the probability of both occurring in that order: 𝑷(𝑨 𝒂𝒏𝒅 𝑩) = 𝑷(𝑨) × 𝑷(𝑩) Example 1 Two unbiased dice are thrown. What is the probability of rolling two 5s? Solution 1 6 1 𝑃(5 𝑜𝑛 𝑡ℎ𝑒 2𝑛𝑑 𝑑𝑖𝑐𝑒) = 6 𝑃(𝑡𝑤𝑜 5𝑠) = 𝑃(5 𝑜𝑛 𝑡ℎ𝑒 1𝑠𝑡 𝑑𝑖𝑐𝑒) × 𝑃(5 𝑜𝑛 𝑡ℎ𝑒 2𝑛𝑑 𝑑𝑖𝑐𝑒) 1 1 = × 6 6 1 = 36 𝑃(5 𝑜𝑛 𝑡ℎ𝑒 1𝑠𝑡 𝑑𝑖𝑐𝑒) = Question 11.1 The letters of the word SCHOOL are written on cards and the cards are placed in a bag. A card is selected at random then put back. A second card is then selected. Find the probability of obtaining: i) The letter O twice ii) The letters S and L in that order iii) The letter C twice ©The Dublin School of Grinds Page 119 12. Bernoulli trials When an experiment consists of repeated trials and follows the condition below it is known as a Bernoulli trial. The Syllabus requires you to know the definition below. A Bernoulli trial is an experiment that has two outcomes (often called success or failure) Example 1 A bag contains 5 green balls and 4 red balls. Each time a ball is selected it is then replaced. What is the probability the first red ball is selected at the third attempt. Solution Success means a red ball is selected so: 𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = Failure means a green ball is selected so: 4 9 5 9 If the first red ball comes out at the third attempt the first two attempt must be green (two failures) then a red (success) P (F, F, S) = P ( F and F and S) 5 5 4 = × × 𝑃(𝑓𝑎𝑖𝑙𝑢𝑟𝑒) = = 9 9 100 9 729 Example 2 There are four cards in a bag numbered 1-4. What is the probability that a 2 is chosen only once in three attempts? Solution Success means a 2 is selected so 𝑃(𝑠𝑢𝑐𝑐𝑒𝑠𝑠) = Failure means a card numbered 1, 3 or 4 is selected so 𝑃(𝑓𝑎𝑖𝑙𝑢𝑟𝑒) = 1 4 3 4 A 2 can be drawn only once in three attempts so it can be selected first, second or third. P (at least one 2) = P(S and F and F) OR P(F and S and F) OR P( F and F and S) 1 3 3 4 9 4 9 4 =( × × ) = = 64 27 + 64 + 9 + 3 1 3 4 4 4 ( × × ) + 3 3 1 4 4 4 ( × × ) 64 64 ©The Dublin School of Grinds Page 120 Question 12.1 A fair die is rolled. i) ii) What is the probability the first six is rolled on the third attempt? What is the probability that a six is rolled once in three attempts? ©The Dublin School of Grinds Page 121 13. Tree diagrams The possible outcomes of two or more events can be shown in a particular type of diagram called a tree diagram. This is sometimes referred to as a probability tree. Example 1 Using a tree diagram find out the probabilities of all the possible outcomes of tossing two coins. Solutions The tree diagram below shows the outcomes and probabilities when a coin is tossed twice. We write the probability of each event along the branch. Note 1: Notice how the sum of the probabilities at the end of the four outcomes adds up to 1. Note 2: Sometimes the trial will change for the second event. E.g. There is a biscuit tin with 2 chocolate cookies and 2 digestives (4 to choose from). A child chooses one at random and eats it. If he/she chooses a second one, he/she no longer has 4 to choose from. ©The Dublin School of Grinds Page 122 Question 13.1 There are 24 marbles in a bag. Six of the marbles are black. a) A marble is drawn at random from the bag and replaced. A second marble is the taken from the bag. Find the probability that: i) ii) iii) neither of the balls are black only the second ball is black both balls are black b) If the first ball is not replaced. Find the probability that: i) ii) both are black one is black and one is not, in any order. ©The Dublin School of Grinds Page 123 14. Expected value The expected value is also known as average value. The expected value is the same as the average of the results. We find it by adding up the outcomes multiplied by their probabilities Example 1 Find the expected value for the roll of a die Solution X P(x) 1 1/6 2 1/6 3 1/6 4 1/6 5 1/6 6 1/6 1 1 1 1 1 1 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑉𝑎𝑙𝑢𝑒 = (1). ( ) + (2). ( ) + (3). ( ) + (4). ( ) + (5). ( ) + (6). ( ) 6 6 6 6 6 6 1 2 3 4 5 6 = + + + + + 6 6 6 6 6 6 21 = 6 = 3.5 Note 1: The expected value does not need to be one of the outcomes. We can use the expected value to tell us whether a game is fair or not and whether a bet is good or not. Note 2: if you have to pay for a game you must take this into account when finding the expected value. Question 14.1 The spinner shown in the diagram is used to play a game. The game costs €5 to play. A player wins whatever amount the spinner lands. i) ii) Calculate the expected value. Would you advise a person to play this game? Justify your answer. ©The Dublin School of Grinds Page 124 15. Past and probable exam questions Question 1 ©The Dublin School of Grinds Page 125 Question 2 ©The Dublin School of Grinds Page 126 Question 3 ©The Dublin School of Grinds Page 127 Question 4 ©The Dublin School of Grinds Page 128 Question 5 ©The Dublin School of Grinds Page 129 Question 6 ©The Dublin School of Grinds Page 130 Question 7 ©The Dublin School of Grinds Page 131 ©The Dublin School of Grinds Page 132 Question 8 ©The Dublin School of Grinds Page 133 Question 9 ©The Dublin School of Grinds Page 134 Question 10 ©The Dublin School of Grinds Page 135 Question 11 ©The Dublin School of Grinds Page 136 ©The Dublin School of Grinds Page 137 Question 12 ©The Dublin School of Grinds Page 138 Question 13 ©The Dublin School of Grinds Page 139 Question 14 ©The Dublin School of Grinds Page 140 Question 15 ©The Dublin School of Grinds Page 141 Question 16 ©The Dublin School of Grinds Page 142 16. Solutions to Probability Question 1.1 Dublin Mullingar Ballinasloe → Mullingar: → Ballinasloe: → Galway City: 5 options 3 options 2 options 5 × 3 × = 30 2 Question 2.1 5 students 5 × 4 × 3 × = 120 2 × 1 Question 3.1 9 numbers from 1-9: 9 × 8 × 7 × = 3024 6 Question 3.2 26 letters in the alphabet. 9 numbers from 1-9. numbers letters 26 × 25 × 24 × 9 × 4 × 8 = 1,123,200 Question 3.3 i) 8 × 7 × 6 × 5 × 3 × 2 × = 8! = 40,320 ii) Let the two fastest cars be called A& B. A & B are always 1st or 2nd followed by the other 6. 𝐴𝐵 × 6 × 5 × 4 2! × × 6! = 1,440 ©The Dublin School of Grinds Page 143 3 × 2 × 1 1 Question 4.1 i) 4 Aces in a deck: 4 52 1 = 13 𝑃(𝑎𝑛 𝐴𝑐𝑒) = ii) 13 diamonds in a deck: 13 52 1 = 4 𝑃(𝑎 𝑑𝑖𝑎𝑚𝑜𝑛𝑑) = iii) 26 black cards in a deck: 26 52 1 = 2 𝑃(𝑏𝑙𝑎𝑐𝑘) = Question 4.2 There are 60 students in total: i) 30 girls: 30 60 1 = 2 𝑃(𝑔𝑖𝑟𝑙) = ii) 15 boys like football: 15 60 1 = 4 𝑃(𝑏𝑜𝑦 𝑤ℎ𝑜 𝑙𝑖𝑘𝑒𝑠 𝑓𝑜𝑜𝑡𝑏𝑎𝑙𝑙) = iii) 22 students favourite sport is basketball: 22 60 11 = 30 𝑃(𝑓𝑎𝑣𝑜𝑟𝑖𝑡𝑒 𝑠𝑝𝑜𝑟𝑡 𝑖𝑠 𝑏𝑎𝑠𝑘𝑒𝑏𝑎𝑙𝑙) = Question 5.1 There are 7 sweets in total: i) There 2 lemon sweets 𝑃(𝑙𝑒𝑚𝑜𝑛) = ii) Not lemon ©The Dublin School of Grinds 2 7 𝑃(𝑛𝑜𝑡 𝑙𝑒𝑚𝑜𝑛) = 1 − 𝑃(𝑙𝑒𝑚𝑜𝑛) 2 = 1− 7 5 = 7 Page 144 Question 6.1 List all the outcomes: (1,4), (1,5), (1, 6), (1, 7), (2, 4), (2, 5), (2, 6), (2, 7), (3, 4), (3, 5), (3, 6), (3, 7) There are 12 possible out comes i) Both even: (2, 4) & (2, 6) 2 𝑃(𝑏𝑜𝑡ℎ 𝑒𝑣𝑒𝑛) = 12 1 = 6 ii) Both odd: (1, 5), (1, 7), (3, 5) & (3, 7) 4 12 1 = 3 𝑃(𝑏𝑜𝑡ℎ 𝑜𝑑𝑑) = iii) Both numbers add up to seven (1, 6), (2, 5) & (3, 4) 3 12 1 = 4 𝑃(𝑎𝑑𝑑 𝑢𝑝 𝑡𝑜 𝑠𝑒𝑣𝑒𝑛) = Question 7.1 Experimental probability is another name for relative frequency: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑓𝑢𝑙 𝑡𝑟𝑖𝑎𝑙𝑠 𝑅𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 i) a) 𝑟𝑒𝑎𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 1 = 30 300 = b) 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝑜𝑓 5 = 1 10 50 300 = ii) 1 6 No, it does not seen like a fair dice because it is not equally likely to roll each number. Question 8.1 i) Find the probability using the previous results: 20 45 4 = 9 𝑃(𝑀𝑖𝑐ℎ𝑎𝑒𝑙 𝑤𝑖𝑛𝑠) = ii) Expected wins =probability ×number of trials: 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑤𝑖𝑛𝑠 = 18 × =8 ©The Dublin School of Grinds Page 145 4 9 Question 9.1 i) Odd numbers from 1-12: 1, 3, 5, 7, 9 & 11 6 12 1 = 2 𝑃(𝑜𝑑𝑑) = ii) Numbers divisible by 4 from 1-12: 4, 8 & 12 3 12 1 = 4 𝑃(𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4) = iii) P(A or B) = P(A) + P(B) The events are not mutually exclusive because a number cannot be odd and divisible by 4: 𝑃(𝑜𝑑𝑑 𝒐𝒓 𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4) = 𝑃(𝑜𝑑𝑑) + 𝑃(𝑑𝑖𝑣𝑖𝑠𝑖𝑏𝑙𝑒 𝑏𝑦 4) 6 3 = + 12 12 9 = 12 3 = 4 Question 9.2 i) 13 52 1 = 4 𝑃(𝑠𝑝𝑎𝑑𝑒) = ii) 4 52 1 = 13 The probability of a spade or a jack is not a mutually exclusive event since there is the jack of spades. Remember when events are not mutually exclusive: P(A or B) = P(A) + P(B)-P(A and B) 𝑃(𝐽𝑎𝑐𝑘) = iii) iv) 𝑃(𝐽𝑎𝑐𝑘 𝑜𝑟 𝑎 𝑠𝑝𝑎𝑑𝑒) = 𝑃(𝐽𝑎𝑐𝑘) + 𝑃(𝑠𝑝𝑎𝑑𝑒) − 𝑃(𝐽𝑎𝑐𝑘 𝑜𝑓 𝑠𝑝𝑎𝑑𝑒𝑠) 4 13 1 = + − 52 52 52 16 = 52 4 = 13 26 52 1 = 2 𝑃(𝑟𝑒𝑑) = v) 4 52 1 = 13 𝑃(𝑡𝑒𝑛) = vi) ©The Dublin School of Grinds 𝑃(𝑟𝑒𝑑 𝑜𝑟 𝑎 𝑡𝑒𝑛) = 𝑃(𝑟𝑒𝑑) + 𝑃(𝑡𝑒𝑛) − 𝑃(𝑟𝑒𝑑 𝑎𝑛𝑑 𝑡𝑒𝑛) 26 4 2 = + − 52 52 52 2 red tens: ten of hearts & ten of 28 diamonds = 52 7 = 13 Page 146 Question 10.1 There are 100 students in total: i) This is everyone in the set V 20 + 5 + 27 + 3 100 55 = 100 11 = 20 𝑃(𝑝𝑙𝑎𝑦 𝑣𝑖𝑑𝑒𝑜 𝑔𝑎𝑚𝑒𝑠) = ii) This is everyone in the intersection between the sets V and S: 27 + 5 𝑃(𝑠𝑝𝑜𝑟𝑡𝑠 & 𝑣𝑖𝑑𝑒𝑜 𝑔𝑎𝑚𝑒𝑠) = 100 32 = 100 8 = 25 iii) None of the activities is everything outside the sets R, S and V: 𝑃(𝑑𝑜 𝑛𝑜𝑛𝑒 𝑜𝑓 𝑡ℎ𝑒𝑠𝑒 𝑎𝑐𝑡𝑖𝑣𝑖𝑡𝑒𝑠) = 1 100 iv) Read only is what is in the set R and not the intersections: 2 𝑃(𝑟𝑒𝑎𝑑 𝑜𝑛𝑙𝑦) = 100 1 = 50 v) Reads or plays sports is everything in the sets R and S: 2 + 20 + 5 + 12 + 30 + 27 𝑃(𝑟𝑒𝑎𝑑𝑠 𝑜𝑟 𝑝𝑙𝑎𝑦𝑠 𝑠𝑝𝑜𝑟𝑡𝑠) = 100 96 = 100 24 = 25 vi) At least two of these activities is all the intersections all of the set: 12 + 4 + 20 + 27 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑎𝑠𝑡 𝑡𝑤𝑜 𝑠𝑝𝑜𝑟𝑡𝑠) = 100 64 = 100 16 = 25 ©The Dublin School of Grinds Page 147 Question 11.1 Note: O is the letter not zero: 2 i) 𝑃(𝑂) = = 6 1 3 𝑃(𝑂 𝑡𝑤𝑖𝑐𝑒) = 𝑃(𝑂 𝑎𝑛𝑑 𝑂) = 𝑃(𝑂) × 𝑃(𝑂) 1 1 = × 3 3 1 = 9 ii) 𝑃(𝑆 𝑎𝑛𝑑 𝐿) = 𝑃(𝑆) × 𝑃(𝐿) 1 1 = × 6 6 1 = 36 iii) 𝑃(𝐶 𝑡𝑤𝑖𝑐𝑒) = 𝑃(𝐶 𝑎𝑛𝑑 𝐶) = 𝑃(𝐶) × 𝑃(𝑂) 1 1 = × 6 6 1 = 36 Question 12.1 Success (S) = a six rolled 𝑃(𝑆) = 5 1 6 Failure (F) = not a six 𝑃(𝐹) = 6 i) If a six is rolled on the third attempt that means the first two rolls are failures. 𝑃(𝐹, 𝐹, 𝑆) = 𝑃(𝐹 𝑎𝑛𝑑 𝐹 𝑎𝑛𝑑 𝑆) = 𝑃(𝐹) × 𝑃(𝐹) × 𝑃(𝑆) ii) = 5 5 1 × × 6 6 6 = 25 216 If a 6 is rolled once in three attempts it can be rolled first, second or third. 𝑃(𝑜𝑛𝑒 𝑠𝑖𝑥 𝑖𝑛 𝑡ℎ𝑟𝑒𝑒 𝑎𝑡𝑡𝑒𝑚𝑝𝑡𝑠) = 𝑃(𝑆, 𝐹, 𝐹)𝑜𝑟 𝑃(𝐹, 𝑆, 𝐹)𝑜𝑟 𝑃(𝐹, 𝐹, 𝑆) 1 5 5 5 1 5 5 5 1 = ( × × )+( × × )+( × × ) 6 6 6 6 6 6 6 6 6 = 25 25 25 + + 216 216 216 75 216 25 = 72 = ©The Dublin School of Grinds Page 148 Question 13.1 a) 𝑃(𝑏𝑙𝑎𝑐𝑘) = = 6 24 1 4 𝑃(𝑛𝑜𝑡 𝑏𝑙𝑎𝑐𝑘) = = 18 24 3 4 i) 𝑃(𝑛𝑜𝑡, 𝑛𝑜𝑡) = ii) 9 16 𝑃(𝑛𝑜𝑡, 𝑏𝑙𝑎𝑐𝑘) = iii) 3 16 1 16 The total number of balls for the second event changes by 1 because the ball is not replaced. 𝑃(𝑏𝑙𝑎𝑐𝑘, 𝑏𝑙𝑎𝑐𝑘) = ©The Dublin School of Grinds Page 149 i) 𝑃(𝑏𝑙𝑎𝑐𝑘, 𝑏𝑙𝑎𝑐𝑘) = = ii) 𝑃(𝑏𝑙𝑎𝑐𝑘, 𝑛𝑜𝑡) 𝑜𝑟𝑃(𝑛𝑜𝑡, 𝑏𝑙𝑎𝑐𝑘) = 30 552 5 92 108 108 + 552 552 216 552 9 = 23 = Question 14.1 i) 1 1 1 1 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = (1) ( ) + (4) ( ) + (5) ( ) + (7) ( ) 4 4 4 4 1 4 5 7 = + + + 4 4 4 4 17 = 4 = €4.25 The game coast €5 to play so, we must take this away from expected value we calculated. 𝐴𝑐𝑡𝑢𝑎𝑙 𝑒𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = €4.25 − €5.00 = −€0.75 ii) No, the game is not fair you would be expected to lose money in the long run. ©The Dublin School of Grinds Page 150 Past and probable exam questions solutions: Question 1 a) If one option can be selected in X ways and a second option can be selected in Y ways then, the amount of different ways the two options can be selected is X times Y. b) 5 × 4 × 3 × = 5! = 120 2 × 1 c) The novels can be arranged in 5! ways The poetry books can be arranged 3! ways The novels can go on the left or right so that is 2! ways. 𝑇𝑜𝑡𝑎𝑙 = 5! × 3! × 2! = 1440 Question 2 a) A list of all the possible outcomes (1, 1), (1, 2),(1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3,2), (3, 3), (3,4) (4, 1), (4, 2), (4,3), (4,4) b) 𝑃(4,4) = 1 16 c) Spin the spinner a large number of times and see if the relative frequency of the numbers is approximately equal to ¼ ©The Dublin School of Grinds Page 151 Question 3 3 8 1 4 1 6 1 4 7 12 1/24 3 8 (A, B) 3 8 1 4 (A, N) (B, A) 3 8 1/16 1/16 (B, B) 3/32 3 8 (B, N) 3/32 1 4 (N, A) 7/48 (N, B) 7/32 (N, N,) 7/32 3 8 3 8 𝑃(𝐴𝑙𝑒𝑥 𝑜𝑟 𝐵𝑜𝑏𝑏𝑦 𝑤𝑖𝑛 𝑜𝑛𝑒 𝑟𝑎𝑐𝑒) = = ©The Dublin School of Grinds 1/16 1 1 + 16 16 1 8 Page 152 Question 4 a) Option 1: 10,000×1 = €10,000 Option 2: 50,000×0.5 = €25,000 Option 3: 75,000×0.3 = €22,500 Option 4: 100,000×0.2 = €20,000 b) Take the €10,000 euro because it is certain. OR Option 2 because it has the highest expected value. Question 5 a) Expected value for biased die 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = (1)(0.25) + (2)(0.25) + (3)(0.15) + (4)(0.15) + (5)(0.1) + (6)(0.1) = 2.9 b) Expected value of a fair die: 1 1 1 1 1 1 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = (1) ( ) + (2) ( ) + (3) ( ) + (4) ( ) + (5) ( ) + (6) ( ) 6 6 6 6 6 6 = 3.5 Expected value of the game with the fair die costing €3 €3.50 − €3.00 = €0.50 Expected value of the game with the biased die costing €3 €2.90 − €3.00 = −€0.10 50 cent 10 cent Question 6 a) i) 7 × 6 × 5 × 5 = 210 ii) 1 × 6 = 30 b) i) ii) 𝑃(ℎ𝑒𝑎𝑑𝑠 𝑎𝑛𝑑 ℎ𝑒𝑎𝑑𝑠) = 𝑃(ℎ𝑒𝑎𝑑𝑠) × 𝑃(ℎ𝑒𝑎𝑑𝑠) 1 1 = × 2 2 1 = 4 1000 × 1 = 250 4 c) 50-50 Chance It is a fair coin so it is not influenced by previous tosses. ©The Dublin School of Grinds Page 153 Question 7 a) 𝑃(𝑚𝑎𝑙𝑒 𝑤ℎ𝑜 𝑐𝑙𝑎𝑖𝑚𝑒𝑑) = 977 9634 = 0.101 b) 𝑃(𝑓𝑒𝑚𝑎𝑙𝑒 𝑤ℎ𝑜 𝑐𝑙𝑎𝑖𝑚𝑒𝑑) = 581 6743 = 0.086 c) (Using the probability from a) 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = €6108 × 0.101 = €616.91 d) (Using the probability from b) 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑣𝑎𝑙𝑢𝑒 = €6051 × 0.0.86 = €520.39 e) Male: (Using the answer from c) €1688 − €616.91 = €1071.09 Female: (Using the answer from c) €1024 − €520.39 = €503.61 Comment: Insurance companies are making more money from male drivers. f) 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑐𝑙𝑎𝑖𝑚 𝑣𝑎𝑙𝑢𝑒 = €3900 × 0.07 = €273 𝐴𝑚𝑜𝑢𝑛𝑡 𝑡𝑜 𝑐ℎ𝑎𝑟𝑔𝑒 = €273 + €175 = €448 ©The Dublin School of Grinds Page 154 Question 8 a) The different shaped surfaces could affect the outcome so the outcomes do not have to be equally likely. b) i) Group B Generally the greater number of trials, the better the estimate you get will be. ii) 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑡𝑟𝑖𝑎𝑙𝑠 × 𝑝𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 = 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑠𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 500 × 0.812 = 406 𝑡𝑖𝑚𝑒𝑠 iii) First, find the number of ‘successes’ for Group A: 100 × 0.7 = 76 𝑡𝑖𝑚𝑒𝑠 Then use the total number of success and trial: 𝑡𝑜𝑡𝑎𝑙 ′𝑠 𝑢𝑐𝑐𝑒𝑠𝑠𝑒𝑠 ′ = 76 + 406 = 482 𝑡𝑜𝑡𝑎𝑙 𝑡𝑟𝑖𝑎𝑙𝑠 = 100 + 500 = 600 𝐵𝑒𝑠𝑡 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒 = 482 600 = 0.8 Question 9 a) 𝑃(𝑚𝑎𝑙𝑒) = b) i) MMMM MMMF MMFM MFMM FMMM MMFF MFFM FFMM 1 2 MFFF FFFM FFFF MFMF FMFM FMFF FFMF FMMF ii) four males 1 16 three males; one female 4 16 two males; two females 6 16 Answer: No Justification: From the results table: one male; three females 4 16 four females 6 16 3 = 8 𝑃(2 𝑚𝑎𝑙𝑒𝑠, 2 𝑓𝑒𝑚𝑎𝑙𝑒𝑠) = 𝑃(𝑁𝑂𝑇 2 𝑚𝑎𝑙𝑒𝑠, 2 𝑓𝑒𝑚𝑎𝑙𝑒𝑠) = 1 − 𝑃(2 𝑚𝑎𝑙𝑒𝑠, 2 𝑓𝑒𝑚𝑎𝑙𝑒𝑠) 3 =1− 8 5 = 8 => Both events are not equally likely ©The Dublin School of Grinds Page 155 1 16 Question 10 a) Each player throws more than 50% heads b) Total number of heads thrown Total number of throws = 109 + 238 + 291 = 638 = 200 + 400 + 500 =1100 638 1100 = 0.58 𝑃(𝐻𝑒𝑎𝑑) = c) P(H) = 0.58 P(T) = 0.42 (P(T)=1-058=0.42) 𝑃(𝐻 𝑡ℎ𝑟𝑒𝑒 𝑡𝑖𝑚𝑒𝑠) = 𝑃(𝐻 𝑎𝑛𝑑 𝐻 𝑎𝑛𝑑 𝐻) = 0.58 × 0.58 × 0.58 = 0.195112 𝑃(𝑇 𝑡𝑤𝑖𝑐𝑒) = 𝑃(𝑇 𝑎𝑛𝑑 𝑇) = 0.42 × 0.42 = 0.1764 Joe’s claim is NOT true. Question 11 a) i) 𝑃(𝑔𝑟𝑒𝑒𝑛) = 2 5 ii) RRR RGG RRG GRG RGR GGR GRR GGG b) Player winds €0 €1 €2 €3 Required outcomes RRR RRG RGR GRR RGG GRG GGR GGG c) A Bernoulli trial is an experiment whose outcome is random and can be either of two possibilities: “success” or “failure”. ©The Dublin School of Grinds Page 156 Question 12 a) BBB GBB GGB BBG BGG GGG BGB GBG b) The number of boys and the number of girls in the school. c) P(GGB) = d) Peter: 1 8 1 3 + 8 8 4 = 8 1 = 2 𝑃(𝑡ℎ𝑟𝑒𝑒 𝑏𝑜𝑦𝑠 𝑜𝑟 𝑡𝑤𝑜 𝑏𝑜𝑦𝑠 𝑎𝑛𝑑 𝑎 𝑔𝑖𝑟𝑙) = Niamh: 7 8 Niamh is more likely to be correct because of the greater probability. 𝑃(𝑎𝑡 𝑙𝑒𝑎𝑠𝑡 𝑜𝑛𝑒 𝑔𝑖𝑟𝑙) = (𝑓𝑟𝑜𝑚 𝑎𝑏𝑜𝑣𝑒) Question 13 a) 13 clubs 12 picture cards 3 clubs that are picture cards 52-(10 + 3 + 9) = 30 10 3 9 30 b) i) 𝑃(𝑘𝑖𝑛𝑔 𝑜𝑓 𝑐𝑙𝑢𝑏𝑠) = ii) 𝑃(𝑐𝑙𝑢𝑏 𝑜𝑟 𝑝𝑖𝑐𝑡𝑢𝑟𝑒 𝑐𝑎𝑟𝑑) = 1 52 10 + 3 + 9 52 = 22 52 = 11 26 iii) 𝑃(𝑛𝑜𝑡 𝑎 𝑐𝑙𝑢𝑏 𝑜𝑟 𝑎 𝑝𝑖𝑐𝑡𝑢𝑟𝑒 𝑐𝑎𝑟𝑑) = = 30 29 × 52 51 870 2652 = 0.33 ©The Dublin School of Grinds Page 157 Question 14 a) (i) 𝑃(𝑏𝑙𝑎𝑐𝑘) = 5 24 (ii) 𝑃(𝑏𝑙𝑎𝑐𝑘 𝑜𝑟 𝑟𝑒𝑑) = = = 5 24 14 + 9 24 24 7 12 b) The total of number of cars will be 24 for the 1st choice and 23 for the second choice. 10 5 (i) 𝑃(1𝑠𝑡 𝑐𝑎𝑟 𝑖𝑠 𝑠𝑖𝑙𝑣𝑒𝑟 𝑎𝑛𝑑 2𝑛𝑑 𝑐𝑎𝑟 𝑖𝑠 𝑏𝑙𝑎𝑐𝑘) = × = (ii) 24 50 23 552 25 = 276 𝑃(1𝑠𝑡 𝑐𝑎𝑟 𝑖𝑠 𝑟𝑒𝑑 𝑎𝑛𝑑 2𝑛𝑑 𝑐𝑎𝑟 𝑖𝑠 𝑏𝑙𝑎𝑐𝑘) or 𝑃(1𝑠𝑡 𝑐𝑎𝑟 𝑖𝑠 𝑏𝑙𝑎𝑐𝑘 𝑎𝑛𝑑 2𝑛𝑑 𝑐𝑎𝑟 𝑖𝑠 𝑟𝑒𝑑) 9 5 5 9 = × + × = = = c) 𝑃(𝑟𝑒𝑑 𝑜𝑟 𝑑𝑖𝑒𝑠𝑒𝑙) = = = = = 9 24 9 24 18 24 16 + + − 2 3+2+4 9 24 24 − − 2 2 24 24 24 24 2 24 45 23 552 90 + 45 24 23 552 552 15 92 Note: Two of the red cars are diesel and we must subtract these. 3 Question 15 a) 𝑃(𝑑𝑜𝑒𝑠 𝑁𝑂𝑇 𝑠𝑐𝑜𝑟𝑒) = 1 − 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠) 3 =1− 4 1 = 4 b) 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠 𝑏𝑜𝑡ℎ) = 𝑃(𝑠𝑐𝑟𝑜𝑒𝑠) 𝑎𝑛𝑑 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠) 3 3 = × 4 4 9 = 16 c) 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠 𝑒𝑥𝑎𝑐𝑡𝑙𝑦 𝑡𝑤𝑖𝑐𝑒) = 𝑃(𝑠𝑐𝑜𝑟𝑒 𝑎𝑛𝑑 𝑠𝑐𝑜𝑟𝑒 𝑎𝑛𝑑 𝑚𝑖𝑠𝑠) 𝑜𝑟 𝑃(𝑠𝑐𝑜𝑟𝑒 𝑎𝑛𝑑 𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑠𝑐𝑜𝑟𝑒)𝑜𝑟 𝑃(𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑠𝑐𝑜𝑟𝑒 𝑎𝑛𝑑 𝑠𝑐𝑜𝑟𝑒) 3 3 1 3 1 3 1 3 3 =( × × )+( × × )+( × × ) 4 4 4 4 4 4 4 4 4 9 9 9 = + + 64 64 64 27 = 64 d) 𝑃(𝑠𝑐𝑜𝑟𝑒𝑠 5𝑡ℎ 𝑝𝑒𝑛𝑎𝑙𝑡𝑦) = 𝑃(𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑚𝑖𝑠𝑠 𝑎𝑛𝑑 𝑠𝑐𝑜𝑟𝑒) 1 1 1 1 3 = × × × × 4 4 4 4 4 3 = 1024 ©The Dublin School of Grinds Page 158 Question 16 (a) 10 × 10 × 10 × 10 × 10 × 10 = 1,000,000 (b) (i) 9 × 9 × 9 × 9 × 9 × 9 = 531,441 (ii) 1,000,000 − 531,441 = 468,559 468,559 1,000,000 (c) Not on course ©The Dublin School of Grinds 6×5×3 3×2×1 = 20 6𝐶3 = Page 159 Carl Brien 6th Year Maths Ordinary Level Having worked for The State Examination Commission, Carl brings his reputation as an authority on the Maths Syllabus to The Dublin School of Grinds. 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