6 th 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).
MATHS PAPER 2 BLOCK COURSES One last drive for Maths Paper 2... The Dublin School of Grinds is running four-hour block courses on Sunday 11th June. These courses are designed for those students who feel that they could use an extra push after Maths Paper 1. This exam-focused and structured revision environment could be all you need to help increase your Maths Paper 2 result. Note: At these courses our teachers will predict what questions are most likely to appear on your Maths Paper 2 exam. All these questions will be covered in detail and our teachers will provide you with the techniques to answer each question. FEES: 160 PER COURSE To book your place, call us on: 01 442 4442 or visit: www.dublinschoolofgrinds.ie Maths Paper 2 Block Courses Timetable 6th Year SUBJECT LEVEL DATE TIME Maths H Sunday 11th June 9am - 1pm Maths O Sunday 11th June 9am - 1pm DATE TIME Sunday 11th June 9am - 1pm 3rd Year SUBJECT Maths LEVEL H Note: All courses will take place in Stillorgan Plaza, Lower Kilmacud Road, Stillorgan, Co. Dublin. DSOG January 12pg A4 Brochure.indd 4 20/11/2014 14:32
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 is worth 12% to 17% of The Leaving Cert. It appears on Paper 2. Statistics 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. Univariate 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. Definition: 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 mode The mode is the set of values that occur the most often Definition: The median 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: When to use the mean The mean is only used for numerical data. Mean = 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. Sum of the numbers Number of numbers 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 17 = 340 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 4 th and 5 th numbers. 4 th = 17 5 th =18 Median = 17+18 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, 12, 4, 6, 8 Then find: i) The mode ii) The median iii) The mean The Dublin School of Grinds Page 6
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 5 6 Number of matches 1 7 6 5 2 6 3 Write down the: i) Mode ii) Median iii) Mean Solution: i) 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 ii) iii) There are 30 football matches Looking for the 15 th and 16 th match. Add up the numbers until you reach the 15 th /16 th 1 + 7 + 6 + 5 Median = 15th+16th 2 = 3+3 2 = 3 One of these 5 matches is the 15 th /16 th 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) 1+7+6+5+2+6+3 = 0+7+12+15+8+30+18 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) The mode ii) The median iii) 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) Just like the frequency table it is the number above the largest number on the bottom line. => The modal group = 20-30 ii) There are 40 students so the median is between the 20 th and 21 st students. 4 + 15 + 11 One of these 11 teenagers is the 20 th /21 st iii) =>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 35 45 No. of students 4 15 11 10 We can then find the mean the same we did for the frequency distribution table above using the midinterval values. mean = 4(15)+15(25)+11(35)+10(45) 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) 60-70 70-80 80-90 90-100 100-110 110-120 120-130 Frequency 8 15 12 10 8 3 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 0 1 2 3 Number of families 1 X 1 5 Find the value of x. Solution 1(0) + x(1) + 1(2) + 5(3) 2 = 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 The Dublin School of Grinds Page 10
Question 4.4 The table shows the number of goals a hockey team scored in each match over the season. Goals 0 1 2 3 4 5 Number of matches 1 3 X 9 7 4 The mean number of goals they scored was 3. Find the value of x. The Dublin School of Grinds Page 11
5. The range, quartiles and interquartile range The Range It shows the spread of data. It is useful for comparing sets of data. Range = Maximum Value Minimum Value 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: The mean for class 1B: The range for class 1A: The range for class 1B: Comment: = = 7 + 9 + 9 + 12 + 12 + 12 + 13 + 14 + 14 + 18 10 = 120 10 = 12 Marks 9 + 10 + 11 + 11 + 12 + 12 + 12 + 13 + 15 + 15 10 = 120 10 = 12 Marks Range = Maximum Minimum = 18 7 = 11 Range = Maximum Minimum = 15 9 = 6 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. Multiply 1 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: This is the same as the lower quartile except we multiply by 3 4 rather than 1 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, 2, 3, 9 Find i) The lower quartile ii) The upper quartile iii) The interquartile range 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
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, 9, 11 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 4 12 = 3 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 3 4 12 = 9 9th + 10th value => = 2 = 29 2 = 14.5 13 + 16 2 iii) The interquartile range = upper quartile lower quartile = Q3 Q1 = 14.5 9.5 = 5 The Dublin School of Grinds Page 14
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, 71, 64, 73, 55 Find the i) Lower quartile. ii) The upper quartile. iii) And the interquartile range. Solution The Dublin School of Grinds Page 15
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. Key in Mode, then: 20 (STAT) 2. Then, 1 0 for 1 Var 3. Then each number followed by = 6 0 = 0 7 0 = 0 90 = 0 110 = 0 120 = 0 4. Next, press AC, SHIFT and 1 for the menu 0 0 0 Then, press 4 0 5. 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. Key in Mode, then: 1 0 (STAT) 2. Then, 0 0 for Standard Deviation 3. Then each number followed by DATA 60 DATA 0 70 DATA 0 9 0 DATA 0 110 DATA 0 120 DATA 0 4. Next, press ALPHA 0 then 6 0 for standard deviation 5. 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
0 0 0 0 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: 2 0 (STAT) 2. Then, 10 for 1 Var 3. Then each number followed by = then scroll across and input the frequency 1 = 0 1 = 4. Find the standard deviation AC SHIFT 1 4 3 = 20 = 0 4 = 30 = 0 9 = 4 0 = 0 6 = =>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 1 DATA 2 0 (x,y) 0 4 DATA 3 0 (x,y) 0 9 DATA 4 (x,y) 0 0 6 DATA 4. Next, press ALPHA then 6 0 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 4 5 6 Number of People 6 9 4 4 4 3 Calculate the standard deviation correct to two decimal places. 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. Real life example: a) Age of people at 1 Direction concert Positively skewed histogram/ curve Negative skew (Also known as left skew). This is when the data is lower on the left. Real life examples: a) Ages of people at a Rod Stewart concert. Negatively skewed histogram/curve 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 1 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
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) Draw a bar chart to represent this data ii) 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 70 360 = 36 700 140 360 = 72 700 175 360 = 90 700 315 360 = 162 700 Total 700 360 Step 3 Draw the pie chart using a compass and a protractor The Dublin School of Grinds Page 26
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) Degrees in a circle =360 0 Basketball = 360 150 120 60 = 30 0 ii) 10 students = 150 0 1 student = 15 0 So every 15 0 represents 1 student Hockey =120 0 Number of students who play hockey = 120 15 = 8 iii) All students = 360 0 Number of students = 360 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) How many students got an A ii) 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. You may be asked to draw a histogram when given a frequency distribution. 2. 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 2-4 4-6 6-8 8-10 10-12 Number of employees 6 8 4 13 5 4 Draw a histogram to represent this data. The Dublin School of Grinds Page 31
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) How many customers spent more than 80 in the shop? ii) How many customers were included in the survey? iii) What is the modal class? iv) In which interval does the median lie? Solution i) 35+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 91 st customer. 10 + 25 + 40 + 50 One of these 50 customers is the 90 th /91 st Median = [60-80] class (90 th & 91 st 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) How many people over 40 attended the cinema? ii) How many people attended the cinema when the information was collected? iii) What is the modal class? iv) 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) 3. Now write the leaves in size order Key: 2 9 means 29 customers 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 17 53 44 49 53 29 21 11 38 14 29 i) Draw a stem and leaf diagram to show these ages. ii) What is the lower quartile? iii) Find the upper quartile. iv) What is the interquartile range? The Dublin School of Grinds Page 35
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 67 69 38 73 54 86 45 Maths 64 76 45 75 48 51 55 72 85 64 36 65 67 58 74 47 40 83 62 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. 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
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 67 53 54 66 71 41 42 52 43 64 Girls 63 93 40 62 95 51 88 87 41 75 i) Draw a back to back stem and leaf diagram for these results. ii) Find the interquartile range for the boys and the girls. The Dublin School of Grinds Page 37
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. Obesity and heart attacks ii. Height and age iii. Drink driving and accidents The strength of the relationship between two sets of data is known as correlation. (i.e.: uphill) (i.e.: downhill) (i.e.: all over the gaf) 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
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). Strong Positive Correlation Moderately Strong Positive Correlation Strong Negative Correlation Moderately Strong Negative Correlation No Correlation The Dublin School of Grinds Page 39
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 Choose 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 ( 0 C) 16 22 31 19 23 26 21 17 24 29 21 25 23 29 Number of visitors 205 248 298 223 252 280 233 211 258 295 229 252 248 284 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) Draw a scatter diagram to illustrate the data. ii) Comment on the correlation between the predicted height and the actual height. i) 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 does 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 asked 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) iii) The scatter graph shows strong positive correlation between the size of the television and the price => The student s hypothesis is proved. 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: H 0 = null hypothesis (write as a decimal) this is the percentage claimed to be true. H 1 = alternative hypothesis. Step 2: Write the number of students who support the claim as a fraction of the total sample: Sample proportion = Step 3: Find the margin of error using this formula: Number who support the sample Sample size 1 Margin of error = Sample size Step 4: 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: 1 Margin of error = Sample size Always round off to three decimal places. The Dublin School of Grinds Page 47
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: H 0 = The political party has 23% support of the electorate. (= 0.23) H 1 = The party does not have 23% support of the electorate. Step 2: Step 3: Number who support the sample Sample proportion = Sample size = 234 1111 = 0.210621062 = 0.211 1 Margin of error = Sample size = 1 1111 = 0.0300015 = 0.030 Step 4: Confidence interval Sample proportion ± Margin of Error 0.211 0.030 = 0.181 0.211 + 0.030 = 0.241 Confidence interval: [0.181, 0.241] Step 5: 0.23 is within : [0.181, 0.241] => accept H 0 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 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
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Question 12 Note: This questions involves a mixture of topics. The Dublin School of Grinds Page 73
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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) iii) There are 40 students in the class. Looking for the 20 th and 21 st match. Add up the numbers until you reach the 20 th /21 st 7 + 7 + 8 One of these 8 students is the 20 th /21 st Median = 20th+21st 2 = 3+3 2 = 3 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+7+8+9+5+4 = 7+14+24+36+25+24 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 95 105 115 125 Frequency 8 15 12 10 8 3 4 We can then find the mean the same we did for the frequency distribution table above using the mid-interval values. mean = 8(65)+15(75)+12(85)+10(95)+8(105)+3(115)+4(125) 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 30 th and 31 st vehicles. 8 + 15 + 12 One of these 12 vehicles is the 30 th /31 st. The Dublin School of Grinds Page 81
Question 4.4 1(0) + 3(1) + x(2) + 9(3) + 7(4) + 4(5) 3 = 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 Question 5.1 i) Boys: Comment: Girls: 87 + 88 + 84 + 91 + 81 + 85 + 86 Mean = 7 = 602 7 = 86 Range = Maximum Minimum = 91 81 = 10 86 + 87 + 95 + 72 + 82 + 99 + 88 Mean = 7 = 609 7 = 87 Range = Maximum Minimum = 99 72 = 27 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 3 11 = 8.25 4 round up = 9th value 9th value = 15 iii) The interquartile range = upper quartile lower quartile = 15 8 = 7 The Dublin School of Grinds 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 1 4 12 = 3 3rd + 4th value => = 2 = 50.5 3 4 12 = 9 9th + 10th value => = 2 = 72 49 + 52 2 71 + 73 2 iii) The interquartile range = upper quartile lower quartile = Q3 Q1 = 72 50.5 = 21.5 Question 6.1 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 = 180 = 0 26 0 = 0 220 = 0 340 = 0 250 = 0 4. Next, press AC 0, SHIFT 0 and 1 0 for the menu Then, press 4 0 5. Now, press 30 for xσn (standard deviation). Finally press = 0 =>The standard deviation σ = 5.29 How to do it in a Sharp EL-W531 1. Key in Mode, then: 1 0 (STAT) 2. Then, 0 0 for Standard Deviation 3. The each number followed by DATA 180 DATA 0 260 DATA 0 220 DATA 0 340 DATA 25 0 DATA 0 4. Next, press ALPHA then 60 for standard deviation 5. 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 0, SHIFT 0 and 10 for the menu 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: 1 0 (STAT) 2. Then, 00 for standard deviation 3. The each number followed by DATA 20 DATA 40 DATA 60 DATA 80 DATA 100 DATA 4. Next, press ALPHA then 60 for standard deviation 5. 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 = 4. Find the standard deviation AC SHIFT 1 4 3 = 20 = 0 9 = 30 = 0 4 = 4 0 = 0 4 = 5 0 = 0 4 = 6 0 = 0 3 = =>The standard deviation σ = 1.63 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 6 DATA 2 0 (x,y) 0 9 DATA 3 0 (x,y) 0 4 DATA 4 (x,y) 0 0 4 DATA 5 (x,y) 0 0 4 DATA 6 (x,y) 0 0 3 DATA 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
0 0 0 0 Question 6.4 How to do it on a CASIO fx-83gt Plus and fx-85gt Plus. 1. Key in Mode, then: 2 0 (STAT) 2. Then, 1 for 1 Var 0 3. Then each number followed by = the scroll across and input the frequency 1 = 0 7 = 20 = 0 8 = 4. Find the standard deviation AC SHIFT 1 4 3 = 30 = 0 4 = 4 = 0 0 4 = 5 = 0 0 3 = 6 = 0 0 4 = =>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 7 DATA 2 (x,y) 0 0 8 DATA 3 0 (x,y) 0 4 DATA 4 (x,y) 0 0 4 DATA 5 (x,y) 0 0 3 DATA 6 0 (x,y) 0 4 DATA 4. Next, press ALPHA then 6 0 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: 2 0 (STAT) 6. Then, 10 for 1 Var 7. Then each number followed by = the scroll across and input the frequency 3 = 0 6 = 50 = 0 6 = 7 0 = 0 4 = 9 0 = 0 1 = 8. 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, 00 for standard deviation 4. Next, input each number followed by (x,y) and then the frequency followed by DATA 3 (x,y) 0 6 DATA 5 0 (x,y) 0 6 DATA 7 0 (x,y) 0 4 DATA 9 (x,y) 0 0 1 DATA 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) % who scored between 290 294 = 5 15 100% = 33 1 3 % Question 9.1 The Dublin School of Grinds Page 88
Question 10.1 Question 10.2 i) 120 students = 360 0 1 student = 3 0 A grade students = 120 0 Number of students who got an A = 120 3 = 40 ii) Degrees of students who got an A, B, C or a D = 120 0 + 75 0 + 54 0 + 45 0 = 294 0 Number of students who passed = 294 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 20 th and 21 st person. 2 + 5 + 12 + 9 One of these 9 people is the 20 th /21 st Median = [30-40] class (20 th & 21 st people) Question 12.1 i) ii) iii) The lower quartile Q1 The upper quartile Q3 21 1 4 = 5.25 6th value = 17 21 3 4 = 15.75 16th value = 38 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 = Q 3 Q 1 = 66 43 = 23 Girls: The lower quartile Q1 The upper quartile Q3 The interquartile range = Q 3 Q 1 = 88 51 = 37 1 10 = 2.5 4 => 3rd value = 51 3 10 = 7.5 4 => 8th value = 88 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: H 0 = 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets (=0.2). H 1 = Either more or less than 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets. Step 2: Step 3: Number who support the sample Sample proportion = Sample size = 2 30 = 0.067 1 Margin of error = 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 H 0 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) iii) iv) The shape is almost normal but it is slightly skewed right. 57m 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) Kerry Offaly b) Difference 1: Kerry temperatures are generally lower. Difference 2: Offaly has a greater range of temperatures. Question 5 a) b) i) Cycle ii) 2 minutes iii) 10 and 12 minutes iv) 25 minutes v) Cycle i) B ii) 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. 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 1 19 = 4.75 4 => 5th value = 28 The upper quartile Q3 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) (326,134 1) + (413,786 2) + (264,438 3) + + (1,719 9) + (1,128 10) Mean = 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. b) iii) i) 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. You would need the population mean of all the Leaving Cert. 2012 students. ii) Difference: There are different means. The males have a higher mean. Key: 14 9 means 149 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) iii) Mean = 173,273 + 180,754 + 146,470 + 54,432 + 84,907 + 86,932 6 = 726768 6 = 121,128 Increase = 86,932 54,432 = 32,500 % increase = 32,500 54,432 = 59.71% The Dublin School of Grinds 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. b) 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) iii) People like to buy new cars early in the year so that they have a new year number plate. In 2011 sales in the 1 st 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 c) i) d) ii) i) Answer: D Diesel Petrol ii) iii) Yes. The diesel engines grouped at the top of the plot have a smaller median [/mean] value. 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) 75,174 10,000 = 4,556,000 165 f) 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) 0-100,000 100,001-150,000 150,001-200,000 200,001-250,000 250,001-300,000 Over 300,000 24.9 360 = 89.6 100 28.6 360 = 102.96 100 21.9 360 = 78.84 100 10.4 360 = 37.44 100 4.9 360 = 17.64 100 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) (ii) 1 M. O. E. = sample size = 1 1000 = 3.2% No. who sipport claim Sample Proportion = sample size = 540 1000 = 0.54 Confidence Interval = Sample Proportion ± M. O. E. = 0.54 + 0.032 = 0.572 = 0.54 + 0.032 = 57.2% or = 0.54 0.032 = 50.8% Confidence Interval = [50.8%, 57.2%] 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 162.5 112 18200 167.5 125 20937.5 172.5 81 13972.5 177.5 29 5147.5 182.5 10 1825 500 82215 (ii) fx x = 82215 500 = 164.43 m 185 145 = 40 (c) (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. 3 9.6 16 16.2 5.8 2 (ii) 15 100 100 = 3% 81 100 = 16.2% 500 48 500 100 = 9.6% 29 100 = 5.8% 500 80 500 100 = 16% 10 100 = 2% 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: (d) (i) Girls 175-180: 5.8% = 29 180 185 2% = 10 166.7 + 2(S. D. ) = 166.7 + (8.9) = 184.5 Boys: => There are more tall boys. 148.9 184.5 (ii) The girls heights are closer to the mean than the boys heights. The boys height are more spread out. 175-180: 11.5% = 58 180 185 9.5% = 48 166.7 2(S. D. ) = 166.7 2(8.9) = 148.9 The Dublin School of Grinds Page 103
Probability is worth 8% to 13% of the Leaving Cert. It appears on Paper 2. Probability 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) The model and The Colour 3 X 4 = 12 => There are 12 options for the farmer. When working on counting problems we write the number of options in boxes 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) How many different ways can the cars finish the race? ii) 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) = We can write probability as a fraction, decimal or a percentage. Number of successful outcomes Number of possible outcomes 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: => P(7) = 1 10 ii) The odd numbers are 1, 3 5, 7 and 9, so there are 5 odd numbers: => P(odd number) = 5 10 = 1 2 iii) The numbers divisible by four are 4 and 8: => P(number divisible by 4) = 2 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): Clubs (Black): 2 3 4 5 6 7 8 9 10 J Q K A 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 52 = 1 2. 2. A black card or not a red card 26 52 = 1 2. 3. A spade 13 52 = 1 4 4. Not a spade 39 52 = 3 4 5. An ace 4 52 = 1 13 6. Not an ace 48 52 = 12 13 7. The ace of spades 1 52 8. A picture card 12 52 = 3 13 9. A number card or not a picture card 40 52 = 10 13 10. A card that is either a heart or a club 26 52 = 1 2 11. A card that is neither a heart nor a club 26 52 = 1 2 12. A 4 or 5 8 52 = 2 13 13. A 4 or 5 but not a spade 6 52 = 3 26 14. An even numbered card 20 52 = 5 13 Question 4.1 If a deck of cards contains 52 cards, find the probability it is: i) an ace ii) a diamond iii) a black card The Dublin School of Grinds Page 109
Question 4.2 In a school 30 boys and 30 girls were asked their favourite sports and the results were as follows: Hockey Football Basketball Boys 5 15 10 Girls 14 4 12 If one person was randomly selected what is the probability it was i) A girl ii) A boy who said football was his favourite sport. iii) Basketball as his or her favourite sport. The Dublin School of Grinds Page 110
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) P(not C) = 1 1 4 P(not C) = 3 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) What is the probability she takes out a lemon sweet at random? ii) What is the probability the she does not select a lemon sweet? The Dublin School of Grinds Page 111
6. Two events: sample spaces 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. Sample space for 2 dice 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. i) P(tail and 3) = 1 12 ii) P(odd and heads) = 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) both arrows point to an even number? ii) both arrows point to an odd number? iii) 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. The results are shown in the table. Comment on the results. Number of tosses Number of tails Relative Frequency (Heads tosses) Solution As the number of tosses (trials) increases, the relative frequency gets closer to 0.5 i.e. even chance. From the results of the experiment David can conclude that the coin is not biased. 100 59 0.59 200 107 0.535 300 155 0.517 400 205 0.513 500 251 0.502 Note: 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 ii) Expected frequency = expected number of 5 s = P(5) X number of trials = 0.15 X 400 = 60 Question 8.1 Michael and Steven play 45 games of chess. Michael wins 20 of these games. i) What is the probability Michael wins the next game based on the results of the previous 45 games? ii) 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 2 3 6 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) divisible by 4 iii) odd or divisible by 4 The Dublin School of Grinds 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 6 + 3 6 1 6 = 4 6 = 2 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 6 th 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 6 th 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) = 58 70 = 29 35 iii) P(neither Irish or French) = 5 70 = 1 14 iv) P(French only) = 7 70 = 1 10 v) P(both French and Irish) = 40 70 = 4 7 The Dublin School of Grinds 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) play video games ii) play sports and video games iii) do none of these activities iv) read only v) read or play sports vi) 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 P(5 on the 1st dice) = 1 6 P(5 on the 2nd dice) = 1 6 P(two 5s) = P(5 on the 1st dice) P(5 on the 2nd dice) = 1 6 1 6 = 1 36 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: Failure means a green ball is selected so: P(success) = 4 9 P(failure) = 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 9 5 9 4 9 = 100 729 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 Failure means a card numbered 1, 3 or 4 is selected so P(success) = 1 4 P(failure) = 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 4 3 4 3 4 ) + (3 4 1 4 3 4 ) + (3 4 3 4 1 4 ) = 9 64 + 9 64 + 9 64 = 27 64 The Dublin School of Grinds Page 120
Question 12.1 A fair die is rolled. i) What is the probability the first six is rolled on the third attempt? ii) 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) neither of the balls are black ii) only the second ball is black iii) both balls are black b) If the first ball is not replaced. Find the probability that: i) both are black ii) 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 1 2 3 4 5 6 P(x) 1/6 1/6 1/6 1/6 1/6 1/6 Expected Value = (1). ( 1 6 ) + (2). (1 6 ) + (3). (1 6 ) + (4). (1 6 ) + (5). (1 6 ) + (6). (1 6 ) = 1 6 + 2 6 + 3 6 + 4 6 + 5 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) Calculate the expected value. ii) 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
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