HSC April Holidays Lecture
Standard Mathematics
Presented by:
KRYSTELLE VELLA
Who’s behind the screen today?
o I’m Krystelle!
o Graduated from Bede Polding College in 2017
Ø 99 in General Maths 2
Ø ATAR 98.85
o In my 3rd year of Occupational Therapy at
Western Sydney University (& loving it!)
o Wrote Standard Maths Notes & Topic-Tests for
new 2019 syllabus for ATAR Notes
o Management role at TuteSmart for the Maths
team
o I love being adventurous and spending time with
my nieces and nephews!!
By the end of today’s lecture…
First Topic: Algebra
ü Solving simultaneous equations graphically
ü Determining the break-even point
ü Modelling non-linear relationships
Second Topic: Statistical Analysis
ü
ü
ü
ü
Scatterplots
Correlation coefficient
Line of best fit, Least squares line of best fit
Normal distribution
FINALLY: Year 11 content that you should know
There will be Q&A opportunities right throughout the lecture.
Stay tuned for more information!
www.slido.com
Event code: HMS
Passcode: ATAR
Topic 1:
Algebra
SOLVING SIMULTANEOUS
EQUATIONS BY…
Drawing a
graph!
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
Example Question:
Solve the simultaneous equations 𝟐𝒙 − 𝒚 = 𝟒 and 𝟔𝒙 + 𝟐𝒚 = 𝟏𝟐 graphically.
Solution:
To answer this question, we must graph both of these equations as straight lines
using the formula 𝑦 = 𝑚𝑥 + 𝑐. We must rearrange each equation to match this
formula.
1. 2𝑥 − 𝑦 = 4
2. 6𝑥 + 2𝑦 = 12
Who can tell me how to
rearrange these formulas?
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
2𝑥 − 𝑦 = 4
−𝟐𝒙
−𝟐𝒙
−𝑦 = 4 − 2𝑥
6𝑥 + 2𝑦 = 12
−𝟔𝒙
−𝟔𝒙
2𝑦 = 12 − 6𝑥
−𝑦 = 4 − 2𝑥
2𝑦 = 12 − 6𝑥
𝒚 = 𝟐𝒙 − 𝟒
𝒚 = −𝟑𝒙 + 𝟔
×−𝟏 ×−𝟏 ×−𝟏
÷𝟐
÷𝟐
÷𝟐
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
I’m going to show you
step-by-step working of
this question.
Do it with me!
STEP 1: Set up your page
STEP 2: Plot the y-intercept of (1)
and use gradient to draw points
STEP 3: Rule up line (1) using a
ruler!
STEP 4: Plot y-intercept of (2), use
gradient to draw points and rule line
STEP 5: Determine point of
intersection
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝟐, 𝟎)
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
Solution:
Now, using this information, we can substitute these numbers into our current
formulas to double check they solve each equation.
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝒙, 𝒚)
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝟐, 𝟎)
1. 𝟐𝒙 − 𝒚 = 𝟒
(2 × 2) − 0 = 4
2. 𝟔𝒙 + 𝟐𝒚 = 𝟏𝟐
(6 × 2) + (2 × 0) = 12
We have solved this question! (x is equal to 2, and y is equal to 0)
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
Now its your turn!
Example Question:
Solve the simultaneous equations 𝒙 + 𝒚 = 𝟏𝟎 and 𝒚 = 𝟐 + 𝟑𝒙 graphically.
Solution:
KEY STEPS
1. Rearrange formulas to fit y = mx + c.
2.
Graph them using the gradient and y-intercept.
3.
Find the point of intersection. (your solution!)
4. Check your answer.
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
KEY STEPS
1. Rearrange formulas to fit y = mx + c.
1. 𝑥 + 𝑦 = 10
𝒚 = 𝟏𝟎 − 𝒙
2. 𝑦 = 2 + 3𝑥
𝒚 = 𝟑𝒙 + 𝟐
(𝑛𝑜 𝑛𝑒𝑒𝑑 𝑡𝑜 𝑐ℎ𝑎𝑛𝑔𝑒)
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
KEY STEPS
2. Graph them
using the
gradient and yintercept.
Take about 5
mins to do this.
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
KEY STEPS
3. Find the
point of
intersection.
(your solution!)
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝟐, 𝟖)
SOLVING SIMULTANEOUS
EQUATIONS USING A GRAPH
KEY STEPS
4. Check your answer!
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝒙, 𝒚)
𝑷𝒐𝒊𝒏𝒕 𝒐𝒇 𝒊𝒏𝒕𝒆𝒓𝒔𝒆𝒄𝒕𝒊𝒐𝒏: (𝟐, 𝟖)
1. 𝑥 + 𝑦 = 10
2 + 8 = 10
2. 𝑦 = 2 + 3𝑥
8 = 2 + (3×2)
QUICK POLL!
Event code: HMS
Passcode: ATAR
LET’S APPLY SIMULTANEOUS
EQUATIONS TO…
Practical
problems
WHAT DOES ‘BREAK-EVEN’ MEAN?
We refer to the ‘break-even’ point as the point when costs equal
income (typically in a financial situation).
We can use simultaneous equations to find the break-even point
(i.e. the point when the lines intersect!). Above this point, profit
will be earned. Below this point, a loss is incurred.
LET’S DO AN EXAMPLE TOGETHER
Example Question: This example is from the Atar Notes Standard Maths book.
Archie organises a fundraiser for Beyond Blue. The fixed cost is
$2000, as well as an additional $20 per person who attends. Each
person attending pays $60 per ticket. Draw a graph to represent
these equations.
a) What is the break-even point? What does this mean?
b) How much profit is made when 80 people attend the fundraiser?
EXAMPLE QUESTION
Before attempting to draw our graph, we need to find the straightline equation for both sets of information.
Equation 1: (This will be the cost of the fundraiser)
• Set-up cost of $2000 (y-intercept)
• Additional cost of $20 per person (gradient)
𝑪 = 𝟐𝟎𝒙 + 𝟐𝟎𝟎𝟎
Equation 2: (This will be the income earned by the fundraiser)
• Income received per person is $60 (gradient)
𝐈 = 𝟔𝟎𝒙
LET’S DRAW!
I’m going to give you a couple of
minutes to draw up your graph.
Then, I’ll show you my worked
solutions, step by step.
STEP 1: Prepare your page
STEP 2: Find y-intercept & gradient, and
rule up Cost line
STEP 3: Find y-intercept & gradient, and
rule up Income line
Step 4: Find point of intersection
NOW, LET’S ANSWER OUR
QUESTIONS!
a) What is the break-even point? What does this mean?
Who can answer this before I put the
solution up???
The break-even point occurs when there are 50 people attending. When 50
people attend, the cost is $3000 and the income is $3000. Therefore, if more
than 50 people attend, Archie will make a profit. If less than 50 people attend,
Archie will make a loss.
Let’s check this answer using both
equations.
CHECKING OUR ANSWER
TO PART A)
We’ve established on our graph that the break-even point occurs
when there are 50 people attending. By substituting 50 as x for each
equation, we can determine whether this answer is correct or not.
𝐶 = 20 × 50 + 2000
𝑪 = $𝟑𝟎𝟎𝟎
𝐼 = 60 × 50
𝑰 = $𝟑𝟎𝟎𝟎
MOVING ON…
b) How much profit is made when 80 people attend the
fundraiser?
We therefore need to calculate the income
when 80 people attend, and minus the costs
of 80 people attending.
ANSWERING PART B)
Income when 80 people attend:
𝐼 = 60𝑥
𝐼 = 60 × 80
𝑰 = $𝟒𝟖𝟎𝟎
Cost of 80 people attending:
𝐶 = 20𝑥 + 2000
𝐶 = 20 × 80 + 2000
𝑪 = $𝟑𝟔𝟎𝟎
ANSWERING PART B)
Calculate the profit:
𝑷𝒓𝒐𝒇𝒊𝒕 = 𝑰𝒏𝒄𝒐𝒎𝒆 − 𝑪𝒐𝒔𝒕𝒔
𝑃𝑟𝑜𝑓𝑖𝑡 = $4800 − $3600
= $1200
Archie raises $1200 when 80 people attend the fundraiser.
SOLVING BREAK-EVEN QUESTIONS
Should we
do another
one?
EXAMPLE QUESTION
Example Question
A footwear company produces items whose costs are $300 plus $18 for every pair of
shoes produced. The company sells the shoes for $110 each pair.
a) Write an equation to describe the relationship between the:
i) costs (C) and number of items (x)
ii) income (I) and number of items (x)
b) Draw a graph to represent the costs and income for producing the shoes.
c) How many pairs of shoes need to be sold for the company to break-even?
LET’S SOLVE THIS!!
Solutions
A footwear company produces items whose costs are $300 plus $18 for every pair of
shoes produced. The company sells the shoes for $110 each pair.
a) Write an equation to describe the relationship between the:
i) costs (C) and number of items (x)
𝑪 = 𝟏𝟖𝒙 + 𝟑𝟎𝟎
ii) income (I) and number of items (x)
𝐈 = 𝟏𝟏𝟎𝒙
b) Draw a graph to represent the costs and income for producing the shoes.
SOLUTIONS FINALISED
c) How many pairs of shoes need to be sold for the company to break-even?
According to our graph, the break-even point is approximately 3.25 pairs of shoes.
In this example, we need to round this answer UP to 4 pairs of shoes, as a company
obviously cannot sell ¼ of a pair.
= 𝟒 𝒑𝒂𝒊𝒓𝒔 𝒐𝒇 𝒔𝒉𝒐𝒆𝒔
THE LAST PART OF OUR ALGEBRA
LEARNING…
Exponential functions
Modelling
non-linear
relationships
Quadratic functions
Cubic functions
EXPONENTIAL FUNCTIONS
An exponential function is a non-linear curve whose equation has an
x as the power i.e. 2x.
The general rule for this function is:
𝒚=𝒂
𝒙
Exponential functions
are often used to
solve growth or decay
problems.
EXPONENTIAL FUNCTIONS
Example Question: The growth of a certain bacteria is estimated according to the
formula 𝑏 = 10(2.1)! , where b is the number of bacteria after t hours.
a) Construct a table of values to represent the growth of bacteria up to 12 hours
b) Draw a graph to represent this information.
First, we’ll go through questions a) and b). Then we’ll learn to interpret our graph
and infer its meaning.
Solution:
To construct a table of values, we must recognise the x (independent) and y
(dependent) values in the equation. In this question, we are trying to find the
growth of bacteria, and this is reliant upon the number of hours. Therefore,
bacteria (b) is the dependent variable and time in hours (t) is the independent
variable.
EXPONENTIAL FUNCTIONS
Solution a).
Remember that the formula representing this information is 𝑏 = 10(2.1)! . In my
table of values, I’m going to use intervals of 2 hours. You can do the same, or find
the growth every hour.
For example, to find the growth of bacteria after 2 hours, simply substitute the t for
2 in the above formula:
𝑏 = 10(2.1)" = 44.1
t
b
0
2
4
6
8
10
12
10
44
194
858
3 782
16 680
73 558
Note: I have rounded to the nearest whole number.
Now we can plot these values on a graph to answer part b).
MODELLING EXPONENTIAL
FUNCTIONS
So, I’ve already completed this example on an online tool chart. It would be quite
difficult to accurately plot these points onto a graph, but feel free to insert the data
into an online graph maker to help understand the shape of an exponential graph.
This graph was created using https://www.onlinecharttool.com/graph
UNDERSTANDING OUR GRAPH
Example Question: The growth of a certain bacteria is estimated according to the
formula 𝑏 = 10(2.1)! , where b is the number of bacteria after t hours.
Further questions
c) What was the initial number of bacteria?
d) What is the number of bacteria after 9 hours, correct to the nearest whole
number?
e) Estimate the time taken for the bacteria growth to reach 35 000.
Solutions
c) The initial number of bacteria is the value of b when t equals 0 (hours).
Therefore, the initial number of bacteria is 10.
d) To calculate the number of bacteria after 9 hours, insert 9 as the t value in the
equation.
𝑏 = 10(2.1)#
𝑏 = 10(2.1)# = 7943
UNDERSTANDING OUR GRAPH
e) Estimate the time taken for the number of bacteria to reach 35 000.
Almost exactly 11 hours!
MOVING ON TO…
QUADRATIC FUNCTIONS
A quadratic function is a non-linear curve whose equation has
an x as an x squared (x2).
The general rule for this function is:
𝒚 = 𝒂𝒙𝟐 + 𝒄
where a and c are numbers.
QUADRATIC FUNCTIONS
To distinguish a quadratic function from the other non-linear relationships, we
recognise the shape of the graph as a parabola. A parabola always has a turning
point and an axis of symmetry. These key aspects allow you to recognise a
quadratic equation – therefore helping you remember it more clearly!
Key point to remember:
§ A parabola that is a smiley face J has a positive gradient
§ A parabola with a sad face L has a negative gradient
QUADRATIC FUNCTIONS
Example Question:
Complete the following table of values and graph the quadratic function.
𝒚 = 𝒙𝟐 − 𝟒𝒙
x
-1
0
1
2
3
4
5
y
5
0
-3
-4
-3
0
5
What is the turning point of this graph?
LET’S DRAW:
QUADRATIC FUNCTIONS
LUCKY LAST…
CUBIC FUNCTIONS
A cubic function is a non-linear curve whose equation has an x as an
x cubed i.e. x3.
The general rule for this function is:
𝟑
𝒚 = 𝒂𝒙 + 𝒄
EXPONENTIAL
CUBIC FUNCTIONS
FUNCTIONS
Example Question: An ant population is predicted using the formula
𝑁 = 100𝑡 $ , where N is the number of ants and t is the time in days.
a) Draw the graph of 𝑁 = 100𝑡 $ using a table of values.
b) How many ants were present after one week? Use the graph drawn to
estimate.
Solution:
a) We construct a table of values like we did previously. The independent value (x)
is the time in days (t), and the dependent value (y) is the number of ants.
E.g. To find the number of ants after 2 days:
100 × (2)$ = 𝟖𝟎𝟎 𝒂𝒏𝒕𝒔
t
N
0
0
2
800
4
6 400
6
21 600
8
51 200
10
100 000
CUBIC FUNCTIONS
Solution:
a) We now need to use this information to construct a graph.
t
N
0
0
2
800
4
6 400
6
21 600
8
51 200
10
100 000
CUBIC FUNCTIONS
Solution:
b) Using our graph, we can answer question b). To estimate the number of ants
after one week (7 days), we need to determine where the non-linear curve passes
7 days, and match this to the number on the y axis.
Answer should be approximately 35 000 ants.
We can test this value using the equation.
100 × (7)! = 𝟑𝟒 𝟑𝟎𝟎 𝒂𝒏𝒕𝒔
NOTE!
The graph we just drew looks quite similar to our exponential graph. A key
characteristic to highlight is that cubic functions can pass through the point (0,0),
whereas exponential functions do not.
QUICK TEST!!!!
Which graph best represents the equation −2𝑥 ! + 3?
(A)
(B)
It’s Q&A time!!!
Ask me any specific questions you have about the topic we just
completed. I’ll allow approximately 3 questions per topic, so we
don’t run out of time J
www.slido.com
Event code: HMS
Passcode: ATAR
BREAK HERE
Topic 2:
Statistical Analysis
OVERVIEW
Bivariate Data Analysis
o
o
o
o
Scatterplots
Correlation coefficient
Line of best fit
Least-squares line of best fit
Normal Distribution
o Properties of a normal distribution
o Calculating z-scores
o Interpreting z-scores
LOTS OF CONTENT TO GET THROUGH!!!
64
SCATTERPLOTS
Scatterplots are used to determine
the relationship between two numerical variables.
• When completing bivariate data analysis, data is collected for each
variable and displayed in a table of ordered pairs.
• For example, we could test students’ height and weight to
determine if a relationship was present.
• From this information displayed in a scatterplot, we are able to
determine if there are dependent or independent variables, the
strength of the relationship and other key features.
SCATTERPLOTS
When constructing scatterplots,
each ordered pair is a dot on the graph.
SCATTERPLOTS
To describe patterns we see, we use the terminology:
•
•
•
•
Positive linear
Negative linear
Non-linear pattern
No correlation
To determine a linear relationship,
the dots should approximate a
straight line.
To determine a non-linear
relationship, the dots should
approximate a curve.
PATTERNS OF SCATTERPLOTS
positive linear pattern
negative linear pattern
PATTERNS OF SCATTERPLOTS
no correlation
non-linear pattern
STRENGTH OF CORRELATION
We can also describe the strength of the correlation as:
• Strong
• Moderate
• Weak
strong correlation
moderate correlation
weak correlation
LET’S DO SOME QUESTIONS
Example Question:
What is the strength of the linear relationship displayed below?
A.
B.
C.
D.
no linear relationship
strong positive correlation
strong negative correlation
weak positive correlation
Answer: B
LET’S DO SOME QUESTIONS
Example Question:
The table below compares a father’s height and his daughter’s height in
centimetres, measured over 10 years.
Father’s height
(cm)
Daughter’s
height (cm)
157
160
162
165
169
170
171
174
174
174
95
103
108
116
122
130
135
141
146
152
i)
Draw a scatterplot to represent this data.
ii)
Describe the correlation between these two variables.
LET’S DO SOME QUESTIONS
Solutions:
a)
b) I would say this relationship is a strong positive correlation, meaning that as the father’s height increases, so
does the daughter’s height. This however becomes stagnant as the father reaches his maximum height.
THERE’S A BETTER WAY TO DESCRIBE THE
STRENGTH OF CORRELATION…
WHAT IS THE CORRELATION
COEFFICIENT?
To quantify the strength of a linear association (i.e. determine a number which
represents the strength of the relationship between two variables), we use
Pearson’s correlation coefficient, denoted by the pronumeral r.
The correlation coefficient can have a value between -1 and 1
(these values being perfect negative or perfect positive).
To correctly quantify the strength, use the following table as a guide:
Strong
Moderate
Weak
No correlation
Coefficient (r)
Positive
Negative
0.8 %& 1
−0.8 %& − 1
0.5 %& 0.8
−0.5 %& − 0.8
0.3 %& 0.5
−0.3 %& − 0.5
0 %& 0.3
0 %& − 0.3
CORRELATION COEFFICIENT
KEY POINT: You are only expected to determine the correlation coefficient
using your calculator! Again, this is a very advanced formula students are not
expected to know to calculate by hand.
We must also note the meaning of the relationships we determine:
§ positive correlation (0 to +1) – both variables increase or decrease at the
same time
§ zero (no correlation) (0) – no relationship between the variables
§ negative correlation (-1 to 0) – one variable increases while the other variable
decreases.
CALCULATING THE CORRELATION
COEFFICIENT
To calculate the correlation coefficient given a data set:
CALCULATOR STEPS:
2: STAT
ð
MODE
ð
2: A + Bx
Screenshot this to help
you with the next question
𝑒𝑛𝑡𝑒𝑟 𝑑𝑎𝑡𝑎
𝑖𝑛 𝒙 𝑎𝑛𝑑 𝒚
𝑐𝑜𝑙𝑢𝑚𝑛𝑠
AC
ð
SHIFT
3: r
ð
=
ð
1
ð
5: reg
LET’S DO MORE QUESTIONS
Example Question:
A teacher measured her students’ heart rate in an examination and their height
in cm, and presented the information in a table.
Heart
rate (r)
Height
(h)
61
63
63
67
69
71
154 172 151 163 180 165
72
75
78
168 170 158
Calculate the value of the correlation coefficient, correct to four decimal
places.
Solution: 0.2556 (weak positive correlation)
NOW LET’S LEARN ABOUT…
WHAT IS THE LINE OF BEST FIT?
After drawing a scatterplot using the value of ordered pairs, we may
be able to draw a line of best fit.
If the points have a linear correlation, we can approximate a straightline graph that relates the two variables.
This process can be termed linear regression.
NOTE!
As with a normal straight-line graph, the aim of linear regression is to
model the gradient-intercept formula 𝒚 = 𝒎𝒙 + 𝒄.
HOW DO WE DRAW THE LINE OF
BEST FIT?
Before we can draw a line of best fit, we need a set of data to plot onto a
scatterplot. We use these points to ‘guestimate’ a line that passes through the
most points – hence being the ‘line of best fit’.
Example Question: Using the points below, draw a scatterplot and line of best fit.
(0, 6) (2, 24) (3, 39) (4, 44) (5, 59) (6, 64) (7, 79) (8, 84)
Solutions: First, let’s plot our points!
SOLUTIONS – EXAMPLE QUESTION
SOLUTIONS CONT.
As we can see, the points do not exactly line up. Therefore, we will draw a
line of best fit.
NOTE!
Students often find it difficult to determine exactly where the line should be
placed.
It is important to remember:
ü the line should pass through as many points as possible.
ü if the line does not pass through a point, it must have an even amount of
points on each side. This ensures it evenly crosses the data, and assumes its
role as line of best fit.
In our example, the line of best fit does not pass through many points. However,
there are approximately 4 points on either side of the line. Therefore, this is the
line of best fit.
SOLUTIONS CONT.
REMEMBER to always use a ruler to accurately draw your line of best fit.
NEXT WE NEED TO LEARN ABOUT:
WHAT IS THE LEAST-SQUARES LINE
OF BEST FIT?
Unlike the line of best fit, the least-squares line of best fit is a formula which
represents the straight-line equation formed when trying to find the ‘line of best fit’.
To find the least-squares line of best fit, we use the formula:
𝒚 = 𝒎𝒙 + 𝒄
where:
§
𝒎 − 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡
§
𝒄 − 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡
HOW DO WE FIND THE LEASTSQUARES LINE OF BEST FIT?
This syllabus only requires us to find the least-squares line of best fit using our
calculators (yay for us!!).
In order to find the least-squares line, we must use the given data set (in the
question), and a particular Statistics mode to calculate the gradient and y-intercept.
LEAST-SQUARES EXAMPLE
Example Question: Find the equation of the least-squares line of best fit for the
data set provided, correct to two decimal places where necessary.
Height (h) cm
Foot length (l) cm
150 152 155 160 165 167 169
25 23 28 26 30 29 27
First, we need to insert the data as x and y values.
(Note that height forms the x values, and foot length form the y values)
CALCULATOR STEPS:
MODE
ð
2: STAT ð
2: A + BX
𝑒𝑛𝑡𝑒𝑟 𝑑𝑎𝑡𝑎
𝑖𝑛 𝒙 𝑎𝑛𝑑 𝒚
𝑐𝑜𝑙𝑢𝑚𝑛𝑠
Be sure to accurately insert values into table!
EXAMPLE CONT.
Now that our data is in the table, we need to continue our calculator steps:
CALCULATOR STEPS continued:
AC
ð
SHIFT
ð
1
ð
5: reg
Once we’ve reached this step, we need to take careful note of the values we select.
There are different letters which represent the gradient and y-intercept.
The gradient is represented by the letter B.
2: B
The y-intercept is represented by the letter A.
1: A
Solutions:
B = gradient = 0.22
A = y-intercept = -8.35
ð
ð
=
=
FINISHED EXAMPLE
Therefore, our least-squares line of best fit for this data is equal to:
𝒚 = 𝒎𝒙 + 𝐜
𝒇𝒐𝒐𝒕 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟎. 𝟐𝟐𝒉 − 𝟖. 𝟑𝟓
NOTE!
You must also change the x and y pronumerals into their actual measurements. (In this
case, y refers to foot length and x refers to height).
We can now use the equation found to answer subsequent questions.
For example: What is the expected foot length of a student given their height is 180cm? Answer
correct to two decimal places.
𝑓𝑜𝑜𝑡 𝑙𝑒𝑛𝑔𝑡ℎ = 0.22(180) − 8.35
𝒇𝒐𝒐𝒕 𝒍𝒆𝒏𝒈𝒕𝒉 = 𝟑𝟏. 𝟐𝟓𝐜𝐦
NEXT TOPIC!
WHAT IS THE NORMAL
DISTRIBUTION?
• In a normal distribution, the distribution of data is symmetrical
about the mean.
• The mean and median are approximately equal.
• A bell-shaped curve represents a normal distribution.
PROPERTIES OF A NORMAL DISTRIBUTION
The graph below is a representation of a normal distribution.
When observing this curve, we must note:
§ approximately 68% of data will have z-scores between -1 and 1
§ approximately 95% of data will have z-scores between -2 and 2
§ and approximately 99.7% of data will have z-scores between -3 and 3
This is known as the empirical rule.
34%
34%
13.5%
13.5%
2.35%
2.35%
0.15%
0.15%
z-scores
THE EMPIRICAL RULE
The empirical rule aims to help students remember the percentages!
That is why this rule is also known as the:
𝟔𝟖 − 𝟗𝟓 − 𝟗𝟗. 𝟕
𝑟𝑢𝑙𝑒
NOTE: if you do not remember the empirical rule, you can always remember the percentages
of the bell-curve on one side. As the curve is symmetrical, remembering the values of 34%,
13.5%, 2.35% and 0.15% can also help you apply this concept to z-score questions. Either
way, it is vital that you have memorised at least one way of doing these questions, as
examiners will not provide the percentages for you.
BUT WHAT IS A Z-SCORE??
The z-score (standardised score) is used as a tool of comparison
in a normal distribution.
It is the number of standard deviations the score
is from the mean.
To calculate the z-score:
]
𝒙− 𝒙
𝒛=
𝒔
where:
§ z is the z-score or standardised score
§ x is the score
§ 𝑥̅ is the mean of a set of scores
§ 𝑠 is the standard deviation
LET’S TEST YOUR UNDERSTANDING!
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KEY POINTS TO REMEMBER!
A z-score of 0 indicates the score is equal to the mean
A z-score of 1 is one standard deviation above the mean
A z-score of -1 is one standard deviation below the mean
KEY POINTS TO REMEMBER!
The LARGER the z-score, the FURTHER it
is from the mean of the data (the
centre). This applies to both negative
and positive numbers – i.e. -4 is further
away from the centre than -3.
LET’S DO A Z-SCORE QUESTION
Example Question 1.
After sitting an examination, a university student is told they have a z-score of 0. The marks
in the test were normally distributed with a mean of 65.5 and a standard deviation of 4.8.
What mark did the student achieve?
Answer: 65.5
Example Question 2.
Willow is in Year 11 and has completed 3 out of 4 English assessment tasks. A summary of
her results are shown below:
Willow’s mark
Mean
Standard deviation
Task 1 (out of 20) Task 2 (out of 30) Task 3 (out of 25)
15
24
19
14
X
23
3
4
2
What is Willow’s z-score in Task 3?
𝑧=
𝑥 − 𝑥̅
𝑠
𝑧=
19 − 23
2
𝒛 = −𝟐
FINALLY, LET’S LEARN TO…
We can use z-scores to compare scores in a normally distributed data set. We do this by
obtaining the z-scores of two values, and compare their meaning against the set of data.
Example Question 1.
The HSC marks for Standard Mathematics are normally distributed with a mean of 67 and a
standard deviation of 7.
Zac’s mark in this course was 94. Barbara’s z-score was 3. Barbara claims that she
achieved a higher mark than Zac.
Is Barbara’s claim correct? Use mathematical calculations to justify your answer. (2 marks)
ANSWERING OUR EXAMPLE QUESTION
Solution:
We already know that Barbara achieved a z-score of 3. Now, we must calculate Zac’s zscore, using his score of 94:
𝑧=
94 − 67
7
𝒛 = 𝟑. 𝟖𝟓𝟕 …
Therefore, Zac’s z-score is higher than Barbara’s, meaning that he achieved a score
further above the mean than Barbara. Therefore, Barbara’s claim is incorrect.
Alternatively, you could complete this question by actually finding Barbara’s score instead
of Zac’s z-score.
Remembering that she obtained a z-score of 3:
3=
3=
𝑥 − 67
× 7
𝑥 − 67
7
21 = 𝑥 − 67
+67
+67
88 = 𝑥
𝑩𝒂𝒓𝒃𝒂𝒓𝒂! 𝒔 𝒔𝒄𝒐𝒓𝒆 = 𝟖𝟖
Barbara achieved a score of 88, which is lower than Zac’s score of 94. Therefore, her claim
is incorrect.
It’s Q&A time!!!
Ask me any specific questions you have about the topic we just
completed. I’ll allow approximately 3 questions per topic, so we
don’t run out of time J
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BREAK HERE
KRYSTELLEHSC
Topic 3:
Year 11 Content to
revise!
YEAR 11 KNOWLEDGE IS
ASSESSABLE!
Maybe you didn't realise Year 11 content
could be examined in the HSC – but it
definitely can be, and it WILL BE!
Let’s recap Measurement Year 11 content.
LET’S BRAINSTORM!
Event code: HMS
Passcode: ATAR
In Year 11, you would have learned
how to:
ü convert between metric units (length, capacity, area etc)
ü calculate errors in measurement
ü use significant figures and scientific notation
ü calculate perimeter, area & volume
ü use Pythagoras’ theorem & similar figures
ü use Trapezoidal rule
ü work with time (latitude, longitude)
CONVERSIONS
How to convert between metric units:
CONVERSIONS
CONVERSIONS
CONVERTING MEASUREMENTS
Question 1:
Convert 1096cm to m:
10.96m
Question 2:
Convert 55m2 to cm2:
550 000cm2
Question 3:
Convert 250KL to L:
250 000L
ERRORS IN MEASUREMENT
Remember that:
LIMIT OF READING = smallest unit on measuring instrument
ABSOLUTE ERROR = measured value – actual value
(or ½ x limit of reading)
RELATIVE ERROR = ±
MNOPQRST TUUPU
VTMORUTVTWS
PERCENTAGE ERROR = ±
MNOPQRST TUUPU
VTMORUTVTWS
× 100%
QUESTION
Question 8 – 2019 HSC Exam:
SIGNIFICANT FIGURES
Significant figures are the digits that contribute to the meaning
and accuracy of a number.
For example:
– 37.30 has three significant figures: 3, 7, and 3.
– 0.00032 has two significant figures: 3 and 2. (The zeros must remain
there as they are placeholders – indicating the size of the number)
SIGNIFICANT FIGURES
Example Question 1: Round 234 789 to two significant figures.
Solution:
𝟐𝟑4 789
When rounding to two significant figures in this number, we only need the first two digits to be
significant. We can therefore round the following digits to zeros, which act as placeholders to indicate
the size of the number.
𝟐𝟑0 000
Example Question 2: Round 0.006989 to one significant figure.
Solution:
Again, we must remember the value of the zeros in this number. To round to one significant figure, we
can only look at the ‘6’ and beyond, as these contribute to the accuracy of the number.
0.00𝟔989
In this example, the 6 is the first significant figure, with 9 being the deciding figure. As 9 is larger than 5
(the halfway point), we must round the 6 up to 7, therefore correctly rounding to 1 significant figure.
0.00𝟕
SCIENTIFIC NOTATION
We use scientific notation to write very large or very small numbers in a more
convenient and understandable way. It involves us determining how many times a
number has been multiplied by 10, and expressing this as its power.
To express a number in scientific notation:
7 400 000 000
1. Put a decimal point in between the first two numbers – i.e. 7.4. These numbers
contribute to the meaning and accuracy of the large number.
2. Next, count the number of times the value has been multiplied by 10 – this will
become the power of ten. We can do this by counting the placeholders from the
decimal point.
7. 400 000 000
Here, we have 9 places after the decimal point. Therefore, to express this number in
scientific notation:
𝟕. 𝟒 × 𝟏𝟎𝟗
SCIENTIFIC NOTATION
Continuing from previous slide…
7.4 × 10#
meaning and accuracy of
number
to the power of how many times it
is multiplied by 10
multiplied by 10
NOTE!
o Large numbers (i.e. anything above zero) have a positive power of 10.
o Small numbers (i.e. anything below zero) have a negative power of 10.
Example Question: Express 0.0009432 in scientific notation.
0.0009.432
𝟗. 𝟒𝟑𝟐 × 𝟏𝟎&𝟒
WHY ARE WE LEARNING THESE
TWO METHODS OF ROUNDING??
•
Often, HSC examiners will not write a question specifically about significant
figures or scientific notation, but will include these concepts as part of
‘rounding’.
•
Therefore, we NEED TO UNDERSTAND how to apply these concepts correctly,
to ensure that we maximise our marks when answering exam questions.
REVISING PYTHAGORAS’ THEOREM
Pythagoras’ theorem
𝒂Z + 𝒃Z = 𝒄Z
c
a
b
key tip: c is always the longest side (hypotenuse) and is found opposite the right angle – the
decision where to place a and b is entirely up to you!
THIS THEOREM CAN ONLY BE APPLIED TO RIGHT-ANGLED
TRIANGLES.
REVISING RIGHT-ANGLED TRIGONOMETRY
Trigonometric ratios – AGAIN only apply to right-angled triangles
Hypotenuse
Opposite
𝜽
Adjacent
𝑠𝑖𝑛 𝜃 =
())(*+!,
-.)(!,/0*,
(SOH)
𝑐𝑜𝑠 𝜃 =
12314,/!
-.)(!,/0*,
(CAH)
𝑡𝑎𝑛 𝜃 =
())(*+!,
12314,/!
(TOA)
RIGHT-ANGLED TRIGONOMETRY
Question 12 – 2019 HSC Exam:
RIGHT-ANGLED TRIGONOMETRY
Example Question:
Find the length of the unknown side x in the triangle below, correct to two
decimal places.
24˚5’
x
40cm
x = 17.88cm
RIGHT-ANGLED TRIGONOMETRY
Example Question:
Find the unknown angle 𝜃 in the triangle below, correct to the nearest
degree.
𝜃
19.9mm
17.5mm
𝜃 = 28˚
TRAPEZOIDAL RULE
The trapezoidal rule is used to estimate an area bounded by a curved edge. The area is
approximated by replacing the curved edge with a straight line, thus creating a
trapezium.
To use the Trapezoidal rule for a single application, we use the formula:
𝑨 ≈
𝒉
(𝒅 + 𝒅𝒍 )
𝟐 𝒇
𝒅𝒇
𝒅𝒍
𝒉
where 𝑑! (first length) and 𝑑"
(last length) are the lengths of
the parallel sides of the
trapezium and h is the
perpendicular distance
between them
TRAPEZOIDAL RULE
Example Question:
Use two applications of the Trapezoidal rule to estimate the area of the following field.
ℎ
𝐴 ≈ (𝑑$ + 𝑑% )
2
ℎ
𝐴 ≈ (𝑑$ + 𝑑% )
2
15
𝐴 ≈
(63 + 42)
2
𝐴 ≈
𝐴 ≈ 787.5𝑚&
63𝑚
42𝑚
15𝑚
37𝑚
15𝑚
𝑻𝒐𝒕𝒂𝒍 𝒂𝒓𝒆𝒂 = 𝟏𝟑𝟖𝟎𝒎𝟐
15
(42 + 37)
2
𝐴 ≈ 592.5𝑚&
It’s Q&A time!!!
Ask me any specific questions you have about the topic we just
completed. I’ll allow approximately 3 questions per topic, so we
don’t run out of time J
www.slido.com
Event code: HMS
Passcode: ATAR
THANKYOU YEAR 12!