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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.! ΔΖ«+*+)%/Ζ«0!$!.Ζ«¨*8*Ζ«1. +'
The final push for CAO points...
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6th Year
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0(* /ΔΖ«0%((+.#*ΔΖ«+ΔΖ«1(%*Δ
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 donβt 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
ο·
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
ο·
Be clear about what you want to find out.
ο·
Keep each question as simple as possible.
ο·
Never ask a leading question designed to get a particular response.
ο·
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:
ππππ =
ππ’π ππ π‘βπ ππ’πππππ
ππ’ππππ ππ ππ’πππππ
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
β΄ mode = 1
There are 30 football matches
β΄ 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
15π‘β+16π‘β
Median =
=
3+3
=3
iii)
2
2
Finally the mean is found in a similar way to with a group of numbers.
ππ’π ππ (πππππ’ππππ¦ × ππ’ππππ)
ππππ =
ππ’π ππ π‘βπ ππππ’ππππππ
=
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)
ππππ =
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) + π₯(1) + 1(2) + 5(3)
1+π₯+1+5
0 + π₯ + 2 + 15
2=
π₯+7
π₯ + 17
2=
π₯+7
2(π₯ + 7) = π₯ + 17
2π₯ + 14 = π₯ + 17
2π₯ β π₯ = 17 β 14
=> π₯ = 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 β 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 classesβ 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 πππππ
The mean for class 1B:
=
9 + 10 + 11 + 11 + 12 + 12 + 12 + 13 + 15 + 15
10
120
=
10
= 12 πππππ
The range for class 1A:
π
ππππ = πππ₯πππ’π β ππππππ’π
= 18 β 7
= 11
The range for class 1B:
π
ππππ = πππ₯πππ’π β ππππππ’π
= 15 β 9
=6
Comment:
The mean values are the same but class 1B has a smaller range.
Therefore class 1Bβs 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 itβs not affected by outliers.
Lower quartile rule:
ο· List the values in order.
1
ο· Multiply by the number of values.
ο·
ο·
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
ο· This is the same as the lower quartile except we multiply by rather than .
4
4
These are best explained using examples β¦
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
9π‘β value = 11
iii)
The interquartile range = upper quartile β lower quartile
= 11 β 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
3ππ + 4π‘β π£πππ’π 9 + 10
=>
=
2
2
19
=
2
= 9.5
ii)
The upper quartile Q3
iii)
The interquartile range = upper quartile β lower quartile
= Q3 β Q1
= 14.5 β 9.5
=5
©The Dublin School of Grinds
3
× 12 = 9
4
9π‘β + 10π‘β π£πππ’π 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 π 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 letβs 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 β 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 π₯ππ (standard deviation). Finally press =
0
=>The standard deviation π = 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 π = 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 β 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 π = 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 π = 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 βsymmetrical bell shapedβ curve is called a βnormal distribution curveβ.
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 π‘π 21.6]?
Solution
ππππ = 18
π π‘ππππππ πππ£πππ‘πππ = 3.6
14.4 = ππππ β π π‘ππππππ πππ£πππ‘πππ
21.6 = ππππ + π π‘ππππππ πππ£πππ‘πππ
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].
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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.
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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 studentsβ 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.
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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?
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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.
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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:
ππ’ππππ ππ πππππππ =
ππ’ππππ ππ πππβ π πππ‘πππ
× 360°
πππ‘ππ π πππππ π ππ§π
Step 3
Use a compass and a protractor to draw the pie charts
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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
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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.
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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 β 150 β 120 β 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
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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
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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:
ο There are no gaps between the bars in a histogram
ο Histograms can represent discrete or continuous data, while bar charts only represent discrete data.
ο 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 letβs 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
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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.
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2-4
8
4-6
4
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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 β¬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)
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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?
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12. Stem and leaf diagrams
ο·
Stem and leaf diagrams are sometimes called stem plots.
Example of a stem and Leaf Diagram
ο·
ο·
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 β 29, 30 β 39 40 β 49, 50 β 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
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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?
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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
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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.
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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 β 1 and 1
If r = 1, there is a perfect positive correlation between the two variables.
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(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
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Moderately Strong Negative Correlation
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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.
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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.
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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:
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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 donβt
cause shark attacks, or vice-verse. The thing that is causing both these to increase the temperature, this is called a
βlurking variableβ because it is hidden (lurking) in the background.
Question 13.3
Explain with the aid of an example, what is meant by this statement:
βCorrelation dos not imply causalityβ
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14. Hypothesis testing and the line of best fit
We may decide to test a theory for example βeverybody should learn how to drive a carβ.
ο
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 βshould everybody learn how to drive a car?β , 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.
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Example 1
A student in a class had a hypothesis (theory) that βthe larger the size of a television screen was the more expensive it
wasβ. He did some research and this was the data that he found:
Size (inches)
32
37
40
46
50
55
59
Price (β¬)
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 studentβs 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 studentβs 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.
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Question 14.1
A Leaving Cert student wanted to test his hypothesis that βthe taller you are the heavier you areβ. 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 studentβs hypothesis?
iii)
Draw the line of best fit by eye.
i)
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Hypothesis tests
We use a hypothesis tests when a claim is being made about something.
There are 6 steps to follow:
Step 1: π»0 = null hypothesis (write as a decimal) this is the percentage claimed to be true.
π»1 = alternative hypothesis.
Step 2: Write the number of students who support the claim as a fraction of the total sample:
ππππππ πππππππ‘πππ =
ππ’ππππ π€βπ π π’πππππ‘ π‘βπ π πππππ
ππππππ π ππ§π
Step 3: Find the margin of error using this formula:
ππππππ ππ πππππ =
Step 4:
1
βππππππ π ππ§π
Find the confidence interval:
ππππππ πππππππ‘πππ ± ππππππ ππ πΈππππ
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:
ο·
ο·
ο·
β95% confidence intervalβ
β95% confidence levelβ
β5% level of significanceβ
We do not use these in the questions, this is how the Examiner tells us we will use the formula:
ππππππ ππ πππππ =
Always round off to three decimal places.
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1
βππππππ π ππ§π
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 partyβs claim, at the 5% level of
significance.
Solution
Step 1: π»0 = The political party has 23% support of the electorate. (= 0.23)
π»1 = The party does not have 23% support of the electorate.
Step 2:
ππ’ππππ π€βπ π π’πππππ‘ π‘βπ π πππππ
ππππππ π ππ§π
234
=
1111
= 0.210621062
= 0.211
ππππππ πππππππ‘πππ =
Step 3:
ππππππ ππ πππππ =
=
1
βππππππ π ππ§π
1
β1111
= 0.0300015
= 0.030
Step 4: Confidence interval
ππππππ πππππππ‘πππ ± ππππππ ππ πΈππππ
0.211 β 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 π»0
The political party has 23% support of the electorate.
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Question 14.2
Past records from a survey of McDonalds show that 20% of people who buy a βBig Macβ also buy βMcChicken Nuggetsβ.
On a particular day a random sample of 30 people was taken from those that had bought a βBig Macβ and 2 of them were
found to have bought βMcChicken Nuggetsβ. Test at the 5% significance level, whether or not the proportion of people
who bought βMcChicken Nuggetsβ that day had changed. State you your hypothesis clearly.
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15. Past and probable exam questions
Question 1
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Question 2
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Question 3
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Question 4
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Question 5
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Question 6
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Question 7
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Question 8
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Question 9
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Question 10
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Question 11
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Question 12
Note: This questions involves a mixture of topics.
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Question 13
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Question 14
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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
β΄ mode = 4
ii)
There are 40 students in the class.
β΄ 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 =
20π‘β+21π π‘
3+3
2
=
2
=3
iii)
Finally the mean is found in a similar way to with a group of numbers.
ππ’π ππ (πππππ’ππππ¦ × ππ’ππππ)
ππππ =
ππ’π ππ π‘βπ ππππ’ππππππ
=
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)
ππππ =
40
=
5300
60
= 88.3ππ/β
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.
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Question 4.4
1(0) + 3(1) + π₯(2) + 9(3) + 7(4) + 4(5)
1+3+π₯+9+7+4
78 + 2π₯
3=
24 + π₯
3(24 + π₯) = 78 + 2π₯
72 + 3π₯ = 78 + 2π₯
3π₯ β 2π₯ = 78 β 72
=> π₯ = 6
3=
Question 5.1
i)
Boys:
87 + 88 + 84 + 91 + 81 + 85 + 86
7
602
=
7
= 86
π
ππππ = πππ₯πππ’π β ππππππ’π
= 91 β 81
= 10
ππππ =
Girls:
86 + 87 + 95 + 72 + 82 + 99 + 88
7
609
=
7
= 87
π
ππππ = πππ₯πππ’π β ππππππ’π
= 99 β 72
= 27
ππππ =
Comment:
The mean values are similar for boys and girls but the range for boys is smaller.
Therefore the boysβ 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 β lower quartile
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3
× 11 = 8.25
4
round up = 9th value
9th value = 15
= 15 β 8
= 7
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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 β lower quartile
= Q3 β Q1
= 72 β 50.5
= 21.5
1
× 12 = 3
4
3ππ + 4π‘β π£πππ’π 49 + 52
=>
=
2
2
= 50.5
3
× 12 = 9
4
9π‘β + 10π‘β π£πππ’π 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 β 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 π₯ππ (standard deviation). Finally press = 0
=>The standard deviation π = 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 π = 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 β 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 π₯ππ (standard deviation). Finally press = 0
=>The standard deviation π = 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 π = 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 β 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 π = 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 π =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 β 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 π = 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 π =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 β 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 π = 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 π =1.81
©The Dublin School of Grinds
Page 87
Question 7.1
ππππ = 6
π π‘ππππππ πππ£πππ‘πππ = 2
2 = ππππ β 2 × π π‘ππππππ πππ£πππ‘πππ
10 = ππππ + 2 × π π‘ππππππ πππ£πππ‘πππ
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
% π€βπ π πππππ πππ‘π€πππ 290 β 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
6π‘β π£πππ’π = 17
21 ×
iii)
The upper quartile Q3
3
= 15.75
4
16π‘β π£πππ’π = 38
21 ×
iv)
The interquartile range = upper quartile β lower quartile
= π3 β π1
= 38 β 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
=> 3ππ π£πππ’π = 43
3
× 10 = 7.5
4
=> 8π‘β π£πππ’π = 66
The interquartile range = π3 β π1
= 66 β 43
= 23
Girls:
The lower quartile Q1
The upper quartile Q3
1
× 10 = 2.5
4
=> 3ππ π£πππ’π = 51
3
× 10 = 7.5
4
=> 8π‘β π£πππ’π = 88
The interquartile range = π3 β π1
= 88 β 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 studentβs 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: π»0 = 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets (=0.2).
π»1 = Either more or less than 20% of people who buy a Big Mac in McDonalds also buy McChicken Nuggets.
Step 2:
ππ’ππππ π€βπ π π’πππππ‘ π‘βπ π πππππ
ππππππ π ππ§π
2
=
30
= 0.067
ππππππ πππππππ‘πππ =
Step 3:
ππππππ ππ πππππ =
=
1
βππππππ π ππ§π
1
β30
= 0.183
Step 4: Confidence interval
ππππππ πππππππ‘πππ ± ππππππ ππ πΈππππ
0.067 β 0.183 = β0.116
0.067 + 0.183 = 0.25
Confidence interval: [-0.116, 0.25]
Step 5: 0.2 is within: [-0.116, 0.25]
=> accept π»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)
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 βmiddleβ
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 β 2 = 23
After: 26 β 3 = 23
Question 3
a)
Mean = 35
Standard Deviation = = 9.848857802
b)
i) Answer: 20
ii) Range = 157cm β 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 β 13.4 = 16.3 minutes
Range of the males = 23.0 β 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 β 25
= 7 sweets
c)
The lower quartile Q1
The upper quartile Q3
1
× 19 = 4.75
4
=> 5π‘β π£πππ’π = 28
3
× 19 = 14.25
4
=> 15π‘β π£πππ’π = 30
The interquartile range = upper quartile β lower quartile
= π3 β π1
= 30 β 28
= 2 sweets
d) βThis is a set of univariate data. The data are discrete.β
Question 7
a)
i)
Answer: 27,098
ii)
44,139 × 7 = 308,973
iii)
ππππ =
(326,134 × 1) + (413,786 × 2) + (264,438 × 3) + β― + (1,719 × 9) + (1,128 × 10)
1,462,296
4,105,973
=
1,462,296
β 2.8 ππππππ
b) Conor was trying to showβ¦
Number of households has more than doubled in the given time period.
Fiona was trying to show β¦
The gradual reduction in the number of people per household.
Ray was trying to showβ¦.
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: ΞΌ =178 8.
Spread: Ο = 7 9.
95% are between ΞΌ - 2Ο and ΞΌ + 2Ο .
ΞΌ - 2Ο = 178.8 β 2(7.9) =163cm
ΞΌ + 2Ο = 178.8 + 2(7.9) =194.6cm
β95% of nineteen-year-old Irish men are between 163 cm and 194.6 cm in height.β
ii)
68% are between ΞΌ - Ο and ΞΌ + Ο .
ΞΌ - Ο = 178.8 β 7.9 =170.9cm
ΞΌ + 2Ο = 178.8 + 7.9 =186.7cm
β68% of nineteen-year-old Irish men are between 170.9 cm and 186.7 cm in height.β
©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)
ππππ =
iii)
©The Dublin School of Grinds
173,273 + 180,754 + 146,470 + 54,432 + 84,907 + 86,932
6
726768
=
6
= 121,128
πΌππππππ π = 86,932 β 54,432
= 32,500
32,500
% πππππππ π =
54,432
= 59.71%
Page 98
iv)
Aoifeβs argument does not recognise that this increase is from a very low base, and that sales had
been much better before 2009.
Paulβs 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
β΄
× 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
=> 6π‘β π£πππ’π = 46
The upper quartile Q3
3
× 19 = 15.75
4
=> 16π‘β π£πππ’π = 51
The interquartile range = upper quartile β lower quartile
= π3 β π1
= 51 β 46
= 5
d) i)
ππππ
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:
[78β
3β10β
3, 78β
3+10β
3] =[68, 88β
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 = 1.9πππππππ × 90%
= 1.9πππππππ × 0.9
= 1.71 πππππππ
(ii) Some of the people may have been boycotting the property tax.
b)
β¬0 - β¬100,000
24.9
× 360 = 89.6
100
β¬100,001 β¬150,000
28.6
× 360 = 102.96
100
β¬150,001 β¬200,000
21.9
× 360 = 78.84
100
β¬200,001 β¬250,000
10.4
× 360 = 37.44
100
β¬250,001 β¬300,000
4.9
× 360 = 17.64
100
Over β¬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) πππ₯ ππ’π π’π π‘π β¬300 000 = 19 160 550 + 54 774 720 + 58 794 930 + 35 923 680 + 20 696 130
= β¬189 350 010
(iii) πππ₯ ππ’π ππ£ππ β¬300 000 = 241 000 000 β 189 350 010
= β¬51 649 990
51 649 990
(iv) ππππ πππππππ‘π¦ π‘ππ₯ ππ£ππ β¬300 000 =
159 030
= β¬328.78
(v) Find 20% of the properties valued β¬100 001 β β¬150 000
489 060 × 20% = 489 060 × 0.20
= 97 812
πππ₯ ππππ ππ π‘βππ π 97,812 πππππππ‘πππ = 97 812 × 112
= β¬10 954 944
πππ₯ π‘βππ‘ π βππ’ππ ππ ππππ ππ π‘βππ π 97 812 πππππππ‘πππ = 97 812 × 157
= β¬15 356 484
πΈπ₯π‘ππππ‘π ππ ππ₯π‘ππ π‘ππ§ π‘βππ‘ π€ππ’ππ ππ ππππ ππ = 15 356 484 β 10 954 944
= β¬4 401 540
©The Dublin School of Grinds
Page 101
Question 13
(a)
(b) (i)
π. π. πΈ. =
=
1
βπ πππππ π ππ§π
1
β1000
= 0.032
(ii)
ππ. π€βπ π ππππππ‘ πππππ
π πππππ π ππ§π
540
=
1000
= 0.54
ππππππ πππππππ‘πππ =
πΆπππππππππ πΌππ‘πππ£ππ = ππππππ πππππππ‘πππ ± π. π. πΈ. = 0.54 + 0.032 = 0.572
= 0.54 + 0.032
ππ
= 0.54 β 0.032
= 0.572
= 0.508
πΆπππππππππ πΌππ‘πππ£ππ = [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 itβs numerical continuous.
(b) (i)
π₯
π
ππ₯
147.5
15
2212.5
152.5
48
7320
157.5
80
12600
β ππ₯ 82215
162.5
112
18200
=
167.5
125
20937.5
βπ₯
500
172.5
81
13972.5
= 164.43 m
177.5
29
5147.5
182.5
10
1825
500
82215
(ii)
185 β 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 β 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 β 185
180 β 185
2% = 10
9.5% = 48
=> There are more tall boys.
(d) (i)
166.7 + 2(π. π·. ) = 166.7 + (8.9)
166.7 β 2(π. π·. ) = 166.7 β 2(8.9)
= 184.5
= 148.9
148.9 β 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 β3 factorialβ.
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
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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
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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
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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:
ο· A probability of 1 means the event is certain.
ο· A probability of 0 means the event is impossible.
The probability of an event P(E) is given by
π(πΈ) =
ππ’ππππ ππ π π’ππππ π ππ’π ππ’π‘πππππ
ππ’ππππ ππ πππ π ππππ ππ’π‘πππππ
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
=> π(7) =
10
ii)
The odd numbers are 1, 3 5, 7 and 9, so there are 5 odd numbers:
5
=> π(πππ ππ’ππππ) =
10
1
=
2
iii)
The numbers divisible by four are 4 and 8:
2
=> π(ππ’ππππ πππ£ππ ππππ ππ¦ 4) =
10
1
=
5
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Note: You are required to know what is in a deck of cards:
Standard Deck of 52 Playing Cards:
Diamonds (Red): 2β¦ 3β¦ 4β¦ 5β¦ 6β¦ 7β¦ 8β¦ 9β¦ 10 β¦ J β¦ Qβ¦ Kβ¦ Aβ¦
Hearts (Red):
2β₯ 3β₯ 4β₯ 5β₯ 6β₯ 7β₯ 8β₯ 9β₯ 10β₯ Jβ₯ Qβ₯ Kβ₯ Aβ₯
Clubs (Black):
2β£ 3β£ 4β£ 5β£ 6β£ 7β£ 8β£ 9β£ 10β£ Jβ£ Qβ£ Kβ£ Aβ£
Spades(Black):
2β 3β 4β 5β 6β 7β 8β 9β 10β Jβ Qβ Kβ Aβ
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 βnot a red cardβ
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 βnot a picture cardβ
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.
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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
π(πππ‘ πΆ) = 1 β π(πΆ)
1
π(πππ‘ πΆ) = 1 β
4
3
π(πππ‘ πΆ) =
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?
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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. π(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 βWrite down all the possible outcomesβ. 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)
π(π‘πππ πππ 3) =
ii)
π(πππ πππ βππππ ) =
=
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
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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:
π
ππππ‘ππ£π πππππ’ππππ¦ =
ππ’ππππ ππ π π’ππππ π ππ’π π‘πππππ
πππ‘ππ ππ’ππππ ππ π‘πππππ
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
βRelative Frequencyβ can also be called βExperimental Probabilityβ 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
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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.
π¬πππππππ
πππππππππ = π·ππππππππππ × π΅πππππ ππ ππππππ
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
=> π. π + π + π. π + π. ππ + π. ππ = π
=> π. π + π = π
=> π = π β π. π
=> π = π. π
Expected frequency = expected number of 5βs
= 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
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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,
π·(π¨ ππ π©) = π·(π¨) + π·(π©)
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)
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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:
π·(π¨ ππ π©) = π·(π¨) + π·(π©) β π·(π¨ πππ
π©)
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.
π·(< πππ ππππ) = π·(< π) + π·(ππππ) β π·(π)
π
π
π
π
π
π
= + β
=
π
=
π
π
π
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
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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)
π(πΌπππ β) =
=
iii)
58
70
29
35
=
iv)
π(πΉππππβ ππππ¦) =
=
v)
5
π(ππππ‘βππ πΌπππ β ππ πΉππππβ) =
14
7
70
1
10
π(πππ‘β πΉππππβ πππ πΌπππ β) =
=
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70
1
40
70
4
7
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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
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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:
π·(π¨ πππ
π©) = π·(π¨) × π·(π©)
Example 1
Two unbiased dice are thrown. What is the probability of rolling two 5s?
Solution
1
6
1
π(5 ππ π‘βπ 2ππ ππππ) =
6
π(π‘π€π 5π ) = π(5 ππ π‘βπ 1π π‘ ππππ) × π(5 ππ π‘βπ 2ππ ππππ)
1 1
= ×
6 6
1
=
36
π(5 ππ π‘βπ 1π π‘ ππππ) =
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
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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:
π(π π’ππππ π ) =
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
= × ×
π(πππππ’ππ) =
=
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
π(π π’ππππ π ) =
Failure means a card numbered 1, 3 or 4 is selected so
π(πππππ’ππ) =
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
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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
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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
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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
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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
πΈπ₯ππππ‘ππ ππππ’π = (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 β¬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
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15. Past and probable exam questions
Question 1
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Question 2
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Question 3
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Question 4
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Question 5
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Question 6
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Question 7
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Question 8
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Question 9
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Question 10
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Question 11
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Question 12
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Question 13
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Question 14
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Question 15
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Question 16
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16. Solutions to Probability
Question 1.1
Dublin
Mullingar
Ballinasloe
β Mullingar:
β Ballinasloe:
β 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.
π΄π΅
×
6
×
5
×
4
2! ×
×
6!
= 1,440
©The Dublin School of Grinds
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3
×
2
×
1
1
Question 4.1
i)
4 Aces in a deck:
4
52
1
=
13
π(ππ π΄ππ) =
ii)
13 diamonds in a deck:
13
52
1
=
4
π(π πππππππ) =
iii)
26 black cards in a deck:
26
52
1
=
2
π(πππππ) =
Question 4.2
There are 60 students in total:
i)
30 girls:
30
60
1
=
2
π(ππππ) =
ii)
15 boys like football:
15
60
1
=
4
π(πππ¦ π€βπ πππππ ππππ‘ππππ) =
iii)
22 students favourite sport is basketball:
22
60
11
=
30
π(πππ£ππππ‘π π ππππ‘ ππ πππ ππππππ) =
Question 5.1
There are 7 sweets in total:
i)
There 2 lemon sweets
π(πππππ) =
ii)
Not lemon
©The Dublin School of Grinds
2
7
π(πππ‘ πππππ) = 1 β π(πππππ)
2
= 1β
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
π(πππ‘β ππ£ππ) =
12
1
=
6
ii)
Both odd: (1, 5), (1, 7), (3, 5) & (3, 7)
4
12
1
=
3
π(πππ‘β πππ) =
iii)
Both numbers add up to seven (1, 6), (2, 5) & (3, 4)
3
12
1
=
4
π(πππ π’π π‘π π ππ£ππ) =
Question 7.1
Experimental probability is another name for relative frequency:
ππ’ππππ ππ π π’ππππ π ππ’π π‘πππππ
π
ππππ‘ππ£π πππππ’ππππ¦ =
πππ‘ππ ππ’ππππ ππ π‘πππππ
i)
a) ππππππ‘ππ£π πππππ’ππππ¦ ππ 1 =
30
300
=
b) πππππ‘ππ£π πππππ’ππππ¦ ππ 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
π(πππβπππ π€πππ ) =
ii)
Expected wins =probability ×number of trials:
ππ₯ππππ‘ππ π€πππ = 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
π(πππ) =
ii)
Numbers divisible by 4 from 1-12:
4, 8 & 12
3
12
1
=
4
π(πππ£ππ ππππ ππ¦ 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:
π(πππ ππ πππ£ππ ππππ ππ¦ 4) = π(πππ) + π(πππ£ππ ππππ ππ¦ 4)
6
3
=
+
12 12
9
=
12
3
=
4
Question 9.2
i)
13
52
1
=
4
π(π ππππ) =
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)
π(π½πππ) =
iii)
iv)
π(π½πππ ππ π π ππππ) = π(π½πππ) + π(π ππππ) β π(π½πππ ππ π πππππ )
4 13 1
=
+
β
52 52 52
16
=
52
4
=
13
26
52
1
=
2
π(πππ) =
v)
4
52
1
=
13
π(π‘ππ) =
vi)
©The Dublin School of Grinds
π(πππ ππ π π‘ππ) = π(πππ) + π(π‘ππ) β π(πππ πππ π‘ππ)
26 4
2
=
+
β
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
π(ππππ¦ π£ππππ πππππ ) =
ii)
This is everyone in the intersection between the sets V and S:
27 + 5
π(π ππππ‘π & π£ππππ πππππ ) =
100
32
=
100
8
=
25
iii)
None of the activities is everything outside the sets R, S and V:
π(ππ ππππ ππ π‘βππ π πππ‘ππ£ππ‘ππ ) =
1
100
iv)
Read only is what is in the set R and not the intersections:
2
π(ππππ ππππ¦) =
100
1
=
50
v)
Reads or plays sports is everything in the sets R and S:
2 + 20 + 5 + 12 + 30 + 27
π(πππππ ππ ππππ¦π π ππππ‘π ) =
100
96
=
100
24
=
25
vi)
At least two of these activities is all the intersections all of the set:
12 + 4 + 20 + 27
π(ππ‘ πππππ π‘ π‘π€π π ππππ‘π ) =
100
64
=
100
16
=
25
©The Dublin School of Grinds
Page 147
Question 11.1
Note: O is the letter not zero:
2
i)
π(π) =
=
6
1
3
π(π π‘π€πππ) = π(π πππ π)
= π(π) × π(π)
1 1
= ×
3 3
1
=
9
ii)
π(π πππ πΏ) = π(π) × π(πΏ)
1 1
= ×
6 6
1
=
36
iii)
π(πΆ π‘π€πππ) = π(πΆ πππ πΆ)
= π(πΆ) × π(π)
1 1
= ×
6 6
1
=
36
Question 12.1
Success (S) = a six rolled π(π) =
5
1
6
Failure (F) = not a six π(πΉ) =
6
i)
If a six is rolled on the third attempt that means the first two rolls are failures.
π(πΉ, πΉ, π) = π(πΉ πππ πΉ πππ π)
= π(πΉ) × π(πΉ) × π(π)
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.
π(πππ π ππ₯ ππ π‘βπππ ππ‘π‘ππππ‘π ) = π(π, πΉ, πΉ)ππ π(πΉ, π, πΉ)ππ π(πΉ, πΉ, π)
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) π(πππππ) =
=
6
24
1
4
π(πππ‘ πππππ) =
=
18
24
3
4
i)
π(πππ‘, πππ‘) =
ii)
9
16
π(πππ‘, πππππ) =
iii)
3
16
1
16
The total number of balls for the second event changes by 1 because the ball is not replaced.
π(πππππ, πππππ) =
©The Dublin School of Grinds
Page 149
i)
π(πππππ, πππππ) =
=
ii)
π(πππππ, πππ‘) πππ(πππ‘, πππππ) =
30
552
5
92
108 108
+
552 552
216
552
9
=
23
=
Question 14.1
i)
1
1
1
1
πΈπ₯ππππ‘ππ π£πππ’π = (1) ( ) + (4) ( ) + (5) ( ) + (7) ( )
4
4
4
4
1 4 5 7
= + + +
4 4 4 4
17
=
4
= β¬4.25
The game coast β¬5 to play so, we must take this away from expected value we calculated.
π΄ππ‘π’ππ ππ₯ππππ‘ππ π£πππ’π = β¬4.25 β β¬5.00
= ββ¬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.
πππ‘ππ = 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)
π(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
π(π΄πππ₯ ππ π΅ππππ¦ π€ππ πππ ππππ) =
=
©The Dublin School of Grinds
1/16
1
1
+
16 16
1
8
Page 152
Question 4
a) Option 1:
10,000×1
= β¬10,000
Option 2:
50,000×0.5
= β¬25,000
Option 3:
75,000×0.3
= β¬22,500
Option 4:
100,000×0.2
= β¬20,000
b) Take the β¬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
πΈπ₯ππππ‘ππ π£πππ’π = (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
πΈπ₯ππππ‘ππ π£πππ’π = (1) ( ) + (2) ( ) + (3) ( ) + (4) ( ) + (5) ( ) + (6) ( )
6
6
6
6
6
6
= 3.5
Expected value of the game with the fair die costing β¬3
β¬3.50 β β¬3.00 = β¬0.50
Expected value of the game with the biased die costing β¬3
β¬2.90 β β¬3.00 = ββ¬0.10
50 cent
10 cent
Question 6
a) i)
7
×
6
×
5
×
5
= 210
ii)
1
×
6
= 30
b) i)
ii)
π(βππππ πππ βππππ ) = π(βππππ ) × π(βππππ )
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)
π(ππππ π€βπ πππππππ) =
977
9634
= 0.101
b)
π(ππππππ π€βπ πππππππ) =
581
6743
= 0.086
c) (Using the probability from a)
πΈπ₯ππππ‘ππ π£πππ’π = β¬6108 × 0.101
= β¬616.91
d) (Using the probability from b)
πΈπ₯ππππ‘ππ π£πππ’π = β¬6051 × 0.0.86
= β¬520.39
e)
Male:
(Using the answer from c)
β¬1688 β β¬616.91 = β¬1071.09
Female:
(Using the answer from c)
β¬1024 β β¬520.39 = β¬503.61
Comment: Insurance companies are making more money from male drivers.
f)
πΈπ₯ππππ‘ππ πππππ π£πππ’π = β¬3900 × 0.07
= β¬273
π΄πππ’ππ‘ π‘π πβππππ = β¬273 + β¬175
= β¬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)
ππ’ππππ ππ π‘πππππ × ππππππππππ‘π¦ = ππ’ππππ ππ π π’ππππ π ππ
500 × 0.812 = 406 π‘ππππ
iii)
First, find the number of βsuccessesβ for Group A:
100 × 0.7 = 76 π‘ππππ
Then use the total number of success and trial:
π‘ππ‘ππ β²π π’ππππ π ππ β² = 76 + 406
= 482
π‘ππ‘ππ π‘πππππ = 100 + 500
= 600
π΅ππ π‘ πΈπ π‘ππππ‘π =
482
600
= 0.8
Question 9
a)
π(ππππ) =
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
π(2 πππππ , 2 πππππππ ) =
π(πππ 2 πππππ , 2 πππππππ ) = 1 β π(2 πππππ , 2 πππππππ )
3
=1β
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
π(π»πππ) =
c)
P(H) = 0.58
P(T) = 0.42
(P(T)=1-058=0.42)
π(π» π‘βπππ π‘ππππ ) = π(π» πππ π» πππ π»)
= 0.58 × 0.58 × 0.58
= 0.195112
π(π π‘π€πππ) = π(π πππ π)
= 0.42 × 0.42
= 0.1764
Joeβs claim is NOT true.
Question 11
a) i)
π(πππππ) =
2
5
ii)
RRR
RGG
RRG
GRG
RGR
GGR
GRR
GGG
b)
Player winds
β¬0
β¬1
β¬2
β¬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: βsuccessβ
or βfailureβ.
©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
π(π‘βπππ πππ¦π ππ π‘π€π πππ¦π πππ π ππππ) =
Niamh:
7
8
Niamh is more likely to be correct because of the greater probability.
π(ππ‘ ππππ π‘ πππ ππππ) =
(ππππ ππππ£π)
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)
π(ππππ ππ πππ’ππ ) =
ii)
π(πππ’π ππ ππππ‘π’ππ ππππ) =
1
52
10 + 3 + 9
52
=
22
52
=
11
26
iii)
π(πππ‘ π πππ’π ππ π ππππ‘π’ππ ππππ) =
=
30 29
×
52 51
870
2652
= 0.33
©The Dublin School of Grinds
Page 157
Question 14
a) (i) π(πππππ) =
5
24
(ii) π(πππππ ππ πππ) =
=
=
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)
π(1π π‘ πππ ππ π πππ£ππ πππ 2ππ πππ ππ πππππ) = ×
=
(ii)
24
50
23
552
25
=
276
π(1π π‘ πππ ππ πππ πππ 2ππ πππ ππ πππππ) or π(1π π‘ πππ ππ πππππ πππ 2ππ πππ ππ πππ)
9
5
5
9
= ×
+
×
=
=
=
c) π(πππ ππ ππππ ππ) =
=
=
=
=
9
24
9
24
18
24
16
+
+
β
2
3+2+4
9
24
24
β
β
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) π(ππππ πππ π ππππ) = 1 β π(π πππππ )
3
=1β
4
1
=
4
b) π(π πππππ πππ‘β) = π(π πππππ ) πππ π(π πππππ )
3 3
= ×
4 4
9
=
16
c) π(π πππππ ππ₯πππ‘ππ¦ π‘π€πππ)
= π(π ππππ πππ π ππππ πππ πππ π ) ππ π(π ππππ πππ πππ π πππ π ππππ)ππ π(πππ π πππ π ππππ πππ π ππππ)
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) π(π πππππ 5π‘β ππππππ‘π¦) = π(πππ π πππ πππ π πππ πππ π πππ πππ π πππ π ππππ)
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 β 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
6πΆ3 =
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.
As a member of the Irish Mathematics Teachersβ
Association, he is a popular teacher due to his
specialist Project Maths exam focus, perfect notes
and student friendly language which cuts out the
jargon and explains the maths in a clear, concise,
easy manner that students of all abilities can
follow.
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