Shoreham Academy
A-Level Statistics
Statistics 1B
Dean Hubbard
AQA
Dean Hubbard ©
Contents
Formulae Booklet
Formulae to Learn
Numerical Measures
Probability
Binomial Distribution
Normal Distribution
Estimation
Correlation and Regression
Each topic will include
Exam Tips
How to use a Calculator/Graphical Calculator
Worked Examples
Formulae Booklet
On this page you will find Formulae
so you can refer to them
This also includes
Cumulative Binomial Distribution Tables
Normal Distribution Tables
Percentage Points of the Normal Distribution
From Page 10 and onwards the booklet is used for Statistics
Whilst up to page 8 there are useful formulae for Core Maths
These are the formulae which
you don’t need to learn but
can refer to and are needed in
the Stats 1B Exam.
The Binomial Distribution
Tables are found on Page 15 –
Page 20.
This includes up to “n=50”
The Normal Distribution
Tables and Percentage Point
of the Normal Distribution are
located on pages 24 and 25
respectively.
Formulae to Learn
These are useful formulae that you may want to learn as they could end up being the exam.
Some of them will need to be learned, but if not you can always use your calculator
to work the answer out. The calculator method will be used in the topic later on.
Samples
∑𝑥
𝑥̅ =
Mean
̅𝑥 =
Variance
s² =
s² =
raw data
𝑛
∑ 𝑓𝑥
∑𝑓
∑(𝑥−𝑥̅ )2
grouped or tabled data (where x is the midpoint)
raw data
𝑛−1
1
{∑ 𝑓𝑥 ²
𝑛−1
−
(∑ 𝑓𝑥)²
}
∑𝑓
grouped or tabled data (where x is the midpoint)
s = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
Standard Deviation
Populations
Mean
μ =
μ=
Variance
σ² =
∑𝑥
∑ 𝑓𝑥
∑𝑓
grouped data (where x is the midpoint)
∑(𝑥−𝑥̅ )2
raw data
𝑛
1
σ² = 𝑛 {∑ 𝑓𝑥 ² −
Standard Deviation
raw data
𝑛
(∑ 𝑓𝑥)²
}
∑𝑓
grouped data (where x is the midpoint)
σ = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
(Refer to page 12 of this booklet for the meaning for these probabilities)
P (not A) = P (𝐴′ ) = 1 − 𝑃(𝐴) (complementary events)
P (A or B) = P (A ⋃ 𝐵) = P (A) + P (B) – P (A ⋂ 𝐵) if not mutually exclusive
P (A or B) = P (A ⋃ 𝐵) = P(A) + P(B)
If P (A or B) = 1
-
if mutually exclusive
the events A and B are exhaustive
P (A and B) = P (A ⋂ 𝐵) = P(A) x P(B)
if independent events
If events are independent:
P(A|B) = P(A|𝐵′ ) = P(A)
P(B|A) = P(B|𝐴′ ) = P(B)
Scientific Calculator
Statistics Mode
[MODE] [2] (STAT) [1](1-VAR)
Graphical Calculator
To get to Statistics option
[MENU] [2] (STAT)
1 Variable Data
1 Variable Data
Enter your Data
E.g. 25, 34, 13, 25, 34, 13
Enter your Data
E.g. 25, 34, 13, 25, 34, 13
Work out the mean
̅) [=]
[AC] [SHIFT] [1] (STAT) [5] (Var) [2] (𝒙
= 24
To work out all the information needed
[F2] (CALC) [F6] (SET) [DOWN] [F1] (1) [EXIT]
Option should say 1Var Freq : 1
(this makes sure the calculator knows it is
dealing with one variable)
Population Standard Deviation
[AC] [SHIFT] [1] (STAT) [5] (Var) [3] (𝒙𝝈𝒏) [=]
= 8.602325267
Sample Standard Deviation
[AC] [SHIFT] [1] (STAT) [5] (Var) [4] (x𝝈𝒏 − 𝟏)
[=]
=9.42447519
Number in Sample Group
[AC] [SHIFT] [1] (STAT) [5] (Var) [1] (𝒏) [=]
=6
[F1] (1VAR)
(If you were to enter the e.g data you will get
this as a list)
𝑥̅ = 24
∑ 𝑥 = 144
∑ 𝑥 2 = 3900
𝜎𝑥 = 8.60232526
𝑠𝑥 = 9.42337519
𝑛=6
To work out ∑ 𝒙
[AC] [SHIFT] [1] (STAT) [4] (Sum) [2] (∑ 𝒙) [=]
= 144
To work out ∑ 𝒙𝟐
[AC] [SHIFT] [1] (STAT) [4] (Sum) [1] (∑ 𝒙𝟐 )[=]
=3900
2 Variable Data
[MODE] [2] (STAT) [2] (A+BX)
2 Variable Data
[MENU] [2] (STAT)
Enter your data in X and Y
E.g.
X = 25, 34, 13, 25, 34, 13
Y = 17, 3, 4, 25, 30, 32
Enter your data in List 1 and List 2
E.g.
X = 25, 34, 13, 25, 34, 13
Y = 17, 3, 4, 25, 30, 32
Regression
[AC] [SHIFT] [1] (STAT) [7] (Reg)
(This is for y= a + bx, if you were looking
for y = ax + b you’re a + b values would flip)
[AC] [SHIFT] [1] (STAT) [7] (Reg) [1] (A) [=]
= 19.7972973
[AC] [SHIFT] [1] (STAT) [7] (Reg) [2] (B) [=]
= -0.05405405405
[AC] [SHIFT] [1] (STAT) [7] (Reg) [3] (r) [=]
= -0.04003237259
Regression
[F2] (CALC) [F6] (SET) – (Change Settings to
these)
1Var XList: List1
1Var Freq: 1
2Var XList : List1
2Var YList: List2
2Var Freq : 1
[F2] (CALC) [F3] (REG) [F1] (X) –
For regression Line ax + b [F1] (aX+b)
For regression Line a+bx [F2] (a+bx)
Using the example data and if y = ax + b
Your calculator should display this;
a = -0.54054
b = 19.7972972
r = -0.0400323
𝑟 2 = 1.6025𝐸 − 03
Mse = 202.050675
Using example data and if y= a +bx
Your calculator should display this;
a = 19.7972972
b = -0.54054
r = -0.0400323
𝑟 2 = 1.6025𝐸 − 03
Mse = 202.050675
2 Variable Variance
The Sample Number
[AC] [SHIFT] [1] (STAT) [5] (Var) [1] (n) [=]
=6
The Mean of X Variable
̅) [=]
[AC] [SHIFT] [1] (STAT) [5] (Var) [2] (𝒙
= 24
Population Standard Deviation X Variable
[AC] [SHIFT] [1] (STAT) [5] (Var) [3] (𝒙𝝈𝒏) [=]
= 8.602325267
Sample Standard Deviation X Variable
[AC] [SHIFT] [1] (STAT) [5] (Var) [4] (x𝝈𝒏 − 𝟏)
[=]
= 9.42447519
The Mean of Y Variable
2 Variable Variance
[F2] (CALC) [F2]
Using the example data, the calculator shall
display this
2 - Variable
𝑥̅ = 24
∑ 𝑥 = 144
∑ 𝑥 2 = 3900
𝜎𝑥 = 8.60232526
𝑠𝑥 = 9.42337519
𝑛=6
𝑦̅ = 18.5
∑ 𝑦 = 111
∑ 𝑦 2 = 2863
𝜎𝑦 = 11.6153633
𝑠𝑦 = 12.723993
∑ 𝑥𝑦 = 2640
̅) [=]
[AC] [SHIFT] [1] (STAT) [5] (Var) [5] (𝒚
= 18.5
Population Standard Deviation Y Variable
[AC] [SHIFT] [1] (STAT) [5] (Var) [6] (𝒚𝝈𝒏) [=]
= 11.61536339
Sample Standard Deviation Y Variable
[AC] [SHIFT] [1] (STAT) [5] (Var) [7] (y𝝈𝒏 − 𝟏)
[=]
= 12.72399308
To Work out ∑ 𝒙𝟐
[AC] [SHIFT] [1] (STAT) [4] (Sum) [1] (∑ 𝒙𝟐 ) [=]
= 3900
To Work out ∑ 𝒙
[AC] [SHIFT] [1] (STAT) [4] (Sum) [2] (∑ 𝒙) [=]
= 144
To Work out ∑ 𝒚𝟐
[AC] [SHIFT] [1] (STAT) [4] (Sum) [3] (∑ 𝒚𝟐 ) [=]
= 2863
To Work out ∑ 𝒚
[AC] [SHIFT] [1] (STAT) [4] (Sum) [4] (∑ 𝒚) [=]
= 111
To Work out ∑ 𝒙𝒚
[AC] [SHIFT] [1] (STAT) [4] (Sum) [5] (∑ 𝒙𝒚) [=]
Binomial Distribution
Probabilities of type P(X = x):
F5 (DIST) F5 (BINM) F1 (Bpd)
eg B( 15, 0.2) P( X = 3)
Data
F2 (Variable)
x
3
EXE
Numtrial 15
EXE
p
0.2 EXE
Execute F1 (calc)
p(x) = 0.25013
Probabilities of type P(X ≤ x):
F5 (DIST) F5 (BINM) F2 (Bcd)
eg B( 15, 0.2) P( X ≤ 3)
Data
F2 (Variable)
x
3
EXE
Numtrial 15
EXE
p
0.2 EXE
Execute F1 (calc)
p(x) = 0.64816
This function can also be used for probabilities
of the type P(X > x). For example, P(X > 3) = 1 P( X ≤ 3) = 1 – 0.64816 = 0.35184
Normal Distribution: Calculation of
probabilities
Probabilities of types P(X ≤ x), P(X < x), P(X > x)
and
P(X ≥ x) can be found directly from :
F5 (DIST) F1 (NORM) F2 (Ncd)
eg X~ Normal mean 135 st deviation 15 P(X
≤ 127) or P(X < 127)
Lower
-EXP 99 EXE (lower end needs to
be an extremely small value)
Upper
127
EXE
σ
15
EXE
μ
135
EXE
Execute F1 (CALC)
prob = 0.2969
eg X~ Normal mean 135 st deviation 15
P(X ≥ 118) or P(X >118)
Lower
118
EXE
Upper
EXP 99 EXE (upper end needs to
be an extremely large value)
σ
15
EXE
μ
135
EXE
Execute F1 (CALC)
prob = 0.87146
Problems involving inverse normal probabilities
: F5 DIST F1 NORM F3 InvN
eg X~ Normal mean 135 st deviation 15 Find
value of x such that P(X< x) = 0.15
( or P(X ≤ x) = 0.15 )
Area ( to the left) 0.15
EXE
σ
15
EXE
μ
135
EXE
Execute F1 (CALC)
x = 119.45
eg X~ Normal mean 135 st deviation 15 Find
value of x such that P(X > x) = 0.30
( or P(X ≥ x) = 0.30 )
Area ( to the left) 0.70
EXE ( area to left
is 1 – 0.30)
σ
15
EXE
μ
135
EXE
Execute F1 (CALC)
x = 142.86
Estimation – Confidence intervals for mean
(known variance or large sample)
In both situations, F4 (INTR)
F1 (Z ){ since Normal distribution})
F1
(1-S)
eg A random sample of 12 packets of crisps
with nominal weight 35g is obtained. The
weights of such packets can be modelled by a
normal distribution with standard deviation
3.5g. The weights of each of the packets in the
sample are:
34.6 35.9 37.4 36.2 37.5 34.9 35.8
38.2 37.8 37.2 35.1 35.9
Calculate a 95% confidence interval for the
mean weight.
Enter data into MENU STAT List 1
F4 (INTR) F1 (Z) F1 (1-S)
F1 (List) cursor down to
C- Level 0.95 (as 95%)
EXE
σ
3.5
EXE
List
List 1 (F1)
cursor down
Freq
1 (F1)
EXE
Execute
F1
(CALC)
Left = 34.394
Right = 38.355 so the interval is { 34.394 ,
38.355 }
Also given 𝑋̅ = 36.375 and n = 12
information on s ( σn-1 ) should be ignored.
By using these values you can sub them into
the formula for Product Moment Correlation
Coefficient (As seen below)
Numerical Measures
Mean : - Sum of products ÷ Number of products
It is important as it uses all the data values
If the data is grouped – use the mid-point of each group as your x
Disadvantage – the mean is affected by extreme values
Median: - It is the middle value when the data is in order
Not affected by extreme value
Mode: - This is the most common value in the sample
It is not suitable if the data values are very varied
Range: - The biggest value – smallest value
It is affected greatly by extreme values
Interquartile Range: - Upper Quartile – Lower Quartile
Measures the spread of middle 50% of the data is not affected by extreme values
Lower Quartile: -
1𝑛
4
(n = the sample size) – rounded to the next nearest number
Upper Quartile: -
3𝑛
4
(n = the sample size) – rounded to the next nearest number
Standard Deviation: - Is the spread of data either side of the mean
Scaling Data


Addition if you add a to each number in the list of data:
New Mean = Old Mean + a
New Median = Old Median + a
New Mode = Old Mode + a
Standard Deviation is UNCHANGED
Multiplication if you multiply each number in the list of data by b
New Values = Old values x b (including old standard deviation x b)
EXAMPLE
Find the median and quartiles of the following data:
2, 5, 3, 11, 6, 8, 3, 8, 1, 6, 2, 23, 9, 11, 18, 19, 22, 7
First put the list in order: 1, 2, 2, 3, 5, 6, 6, 7, 8, 8, 9, 11, 11, 18, 19, 22, 23
𝑛
You need to find 𝑄1 , 𝑄2 and 𝑄3 , so work 4 =
1)
𝑛
4
18 𝑛
,
4 2
=
18
3𝑛
, 𝑎𝑛𝑑 4
2
=
54
4
is not a whole number (= 4.5), so round up and take the 5th term: 𝑄1 = 3
2)
3)
𝑛
7+8
is a whole number (= 9), so find the average of the 9th and 10th terms: 𝑄2 =
2
2
3𝑛
th
is not a whole number (= 13.5), so round up and take the 14 term: 𝑄3 = 11
4
= 7.5
a) i) = Modal Value is 23, because as you can see there are 34 days that 23 letters are
collected
ii) = Median is 22, because you divide your sample size by 2. 175 ÷2 = 84.5. Since this is
not a whole number you round up to 85. Therefore you take the 85th value which lies in
22 daily number of letters.
= LQ is 20, due to the fact you divide your sample size by 4. 175 ÷ 4 = 43.75, you then
round it up giving you a value of 44. Therefore you take the 44th value which lies in the
20 daily number of letters.
= UQ is 23, the sample size divided by 4 and multiplied by 3. 175 ÷ 4 x 3 = 131.25, (you
can also work this out by
3(175)
4
if you were to use the formula). Since you can’t get a
value of 131.25 days, you must round up to the next number which is 132. Then you
take the 132nd value which is the 23 daily number of letters.
= IQR = 3 , as you use the formula of UQ – LQ (23 – 20)
b) = Mean = 22.3
= Standard Deviation = 6.37 or 3.39
To get the full 4 marks on the question, you have to get these specific values. To do this
the group data you must take the mid-point and also remember to change your max
value to 54.
Your mid-points should be as listed;
Daily Numbers
of letters
0-9
10-19
20
21
22
23
24
25-29
30-34
35-39
40-49
50 or more
Total( ∑)
Mid-points
(x)
4.5
14.5
20
21
22
23
24
27
32
37
44.5
54
Use this formula ̅𝑥 =
∑ 𝑓𝑥
∑𝑓
Number of days
(f)
5
16
23
27
31
34
16
10
5
3
4
1
175
𝑓(𝑥 2 )
101.25
3364
9200
11907
15004
17986
9216
7290
5120
4107
7921
2916
94132.25
(fx)
22.5
232
460
567
682
782
384
270
160
111
178
54
3902.5
to work out the mean of grouped data (You can also work it
out via the calculator, by inputting the data which is highlighted. Then following the
steps on pages 5 and 6, but unfortunately a scientific calculator cannot deal with a 2
Variable Frequency so you have to use the Sum Option)
So
3902.5
175
= 22.3
1
For Standard deviation use this formula s² = 𝑛−1 {∑ 𝑓𝑥 ² −
1
into the formula so, s = √𝑛−1 {∑ 𝑓𝑥 ² −
(∑ 𝑓𝑥)²
}
∑𝑓
1
(∑ 𝑓𝑥)²
}
∑𝑓
, inputting the data
, √174 {94132.25 −
3902.52
}
175
= 6.39
It does not matter which formula you use, you can use either population or sample data
– the mark scheme allows both answers.
You can get the answers straight from the graphical calculator if you change a few settings to allow
for 2 Variable Frequency, to do this you must
[F2] (CALC) [F6] (SET) – (Change Settings to these)
1Var XList: List 1
1Var Freq: 1
2Var XList : List 1
2Var YList: List 2
2Var Freq: List 2
Then press [F2] (2Var) and it shall display this list as shown
𝑥̅ = 22.3
∑ 𝑥 = 3902.5
∑ 𝑥 2 = 94132.25
𝜎𝑥 = 6.37248549
𝑠𝑥 = 6.390771
𝑛 = 175
𝑦̅ = 23.02171428
∑ 𝑦 = 4063
∑ 𝑦 2 = 110479
𝜎𝑦 = 9.60587575
𝑠𝑦 = 9.63343929
∑ 𝑥𝑦 = 88186.5
As you can see you can see all information is correct, if you were to have the settings as;
1Var XList: List 1
1Var Freq: 1
2Var XList : List 1
2Var YList: List 2
2Var Freq : 1
This would be incorrect since you are taking the 2Variable Frequency from List 1
EXAM TIP: If you were working it out using a graphical calculator and you were unsure whether it
was in the correct settings. Just use the scientific calculator to work it out since there are no settings
needed to be changed.
c) Mean = 23.9
Since the sum of x is now 280. Use the formula, 280 ÷ 175 = 1.6
Now you just add the new mean to the old mean, 22.3 + 1.6 = 23.9
Probability
Mutually Exclusive Events
two or more events that cannot happen at the same time e.g. passing and failing and exam
P (A 𝑼 B) = P(A) + P(B)
Independent Events
One event has no effect on another event occurring e.g. throwing a 1 or a 2 on a fair dice
P (A ∩ B) = P(A) x P(B)
Exhaustive Events
Any events where all possible outcomes are included e.g. throwing a head or a tail on a fair coin
Conditional Probability
When the outcome of the first even affects the outcome of a second event depends on what has
happened
Notation
EXAMPLE
A and B are two events, with P(A) = 0.4, P(B | A) = 0.25, and P(A’ ∩ B) = 0.2
a) Find i) P(A ∩ B), ii) P(A’), iii) P(B’ | A), iv) P(B | A’), v) P(B), vi) P(A | B)
b) Say whether or not A and B are independent
a) i) 𝑃(𝐵 | 𝐴) =
𝑃 (𝐴 ∩𝐵)
𝑃(𝐴)
= 0.25, so by rearranging the formula you get P(A ∩ B) = P(B | A) x P(A) =
0.25 x 0.4 = 0.1
ii) P (A’) = (Use formula above in notation) = 1 – 0.4 = 0.6
iii) P (B’ | A) = 1 – P (B | A) = 1 – 0.25 = 0.75 - (Since P(B’ | A) + P(B | A) = 1, so we rearrange it)
iv) 𝑃(𝐵 | 𝐴′) =
𝑃(𝐵 ∩ 𝐴′ )
𝑃(𝐴′ )
0.2
𝟏
= 0.6 = 𝟑 (Since P (B ∩ A’) = P (A ∩ B’) , need to learn the formula)
1
3
v) P (B) = P (B | A) P (A) + P(B | A’)P(A’) = (0.25 x 0.4) + ( x 0.6) = 0.3
vi) P (A | B) =
𝑃( 𝐴 ∩ 𝐵) 0..1
= 0.3
𝑃(𝐵)
1
=3
b) If P (B | A) = P (B), then A and B are independent
But P(B | A) = 0.25, while P(B) = 0.3, so A and B are not independent
a) i) P(B = 3) =
194
640
= 0.303
Since it says exactly 3 bedrooms you must put ‘=’. You follow the 3 bedrooms on the table to the
total which gives you 194 then divide it by the total of everything which is 640.
563
ii) P(T ≥ 2) = 640 = 0.879,
At least 2 bathrooms, this includes 2 and everything above to hence the ‘≥’ sign. You can get this
answer by various ways of adding or subtracting, the way I did it was to subtract 77 from 640 (640 –
77 = 563) then divide by the total. The number 77 was picked due to the fact the answer is more
than 2 and the only number below 2 which is in the table is 1. If you look down the first column the
total number of houses with 1 toilet is 77.
iii) P(B = 3 & T ≥ 2) =
194−7
=
640
0.292,
Again look at what the question asks you and use the correct notation. On this question you can do
it various ways too, the way I did it was to just look at the 3 number of bedrooms row. The total
which is 194,( since we worked it out in part i) but then it asked AND at least 2 toilets, we look at
every value above 2 toilets in the 3 bedroom row. We do not need the value of 1 toilet in a 3
bedroom house, therefore we do 194 – 7 divided by the total.
iv) P(B ≤ 3 | T = 2) =
172−19
172
= 0.889,
The denominator is different in this question, due to what it asks you. “Given that it has exactly 2
toilets”, therefore we must look down the 2 toilets column and the total is 172 (which is your
denominator) Now we must look at the ‘at most 3 bedrooms’, the property must not have any more
than 3 bedrooms, and there are 19 properties with 4 bedrooms which we must not include.
Therefore we must do 172 – 19 = 153 then divide it by the total number of properties that have
exactly 2 toilets.
The answers from the previous 4 questions can be left as decimals, percentages or fractions. But you
will be penalised if you have the answers mixed, this means if you answer the first part as a decimal,
then the rest of the answers should also be decimals.
b) 0.303 x 0.879 ≠0.292 , therefore they’re not independent
P(A and B) = P(A ⋂ 𝐵) = P(A) x P(B)
This is the equation which is needed to get the marks for this question, the equation that is shown
states to be independent then the probability of events A and B happening = the probability of A x
the probability of b. The question asks you to use relevant answers from part a), the P(A) is P(B = 3),
P(B) is P(T ≥ 2) and P(A ⋂ 𝐵) is P(B = 3 & T ≥ 2). Now we have established each part of the equation
we can sub it in, 0.303 x 0.879 (= 0.266) ≠ 0.292 therefore they’re not independent.
c) P(2𝑇 ∩ 3𝑇 ∩ ≥ 4𝑇 | 𝐵 = 3) =
72
194
x
99
193
x
16
192
=0.01586453715 x 6 = 0.0951872229
= 0.0952
The questions states, exactly 3 bedrooms, with one property having 2 toilets, another 3 toilets and
the other at least 4 toilets. The total will be 194 as that is how many properties have 3 bedrooms,
but the total will lower by one each time. This is because you’re choosing 3 properties in a row;
you’d have to take into consideration the ones you’ve picked. As you can see in the method you do
72
99
16
this, 194 x 193 x 192 .
Now you have to multiply your answer by 6, this may seem confusing at first but it is best described
via a probability tree. Without one it’s hard to show but you multiply by 6, as there is 6 different
ways of choosing the 3 properties randomly.
Binomial Distribution
The Binomial Distribution, B(n,p)
A random variable x follows a Binomial Distribution as long as 5 conditions are satisfied.
1) There is a fixed number (n) of trials
2) Each trial involves either “success” or “failure”
3) All trials are independent
4) The probability of “success” (p) is the same in each trial
5) The variable is the total number of success in the n trials
𝑛
Formula: P(X = x) = ( ) x 𝑝 𝑥 x (1 − 𝑝)𝑛−𝑥
𝑥
Mean of Binomial Distribution
If X ~ B (n,p) then :
Mean (or expected value) = 𝜇 = 𝐸(𝑥) = 𝑛𝑝
(It’s a theoretical mean – the mean of experimental results is unlikely to match it exactly)
Variance of Binomial Distribution
If X ~ B(n,p) then:
Variance= Var(x) = 𝜎 2 = 𝑛𝑝(1 − 𝑝) = 𝑛𝑝𝑞
Standard Deviation = 𝜎 = √𝑛𝑝(1 − 𝑝) = √𝑛𝑝𝑞
EXAMPLE
A Student has to take a 50-question multiple-choice exam, where each question has five possible
answers, of which only one is correct. He believes he can pass the exam by guessing answers at
random.
a) How many questions could the student, be expected to guess correctly?
b) If the pass mark is 15, what is the probability that the student will pass the exam?
c) If the examiner decides to set the pass mark, so that it is at least 3 standard deviations above
the expected number of correct guesses. What should the minimum pass mark be?
Let X be the number of correct guesses over the 50 questions. Then X ~ B (50, 0.2)
a) E(X) = np = 50 x 0.2 = 10
b) P(X ≥15) = 1 – P(X < 15) = 1 – P(X ≤ 14) = 1 – 0.9393 = 0.0607
c) Var(X) = np(1 - p) = 50 x 0.2 x 0.8 = 8 – so the standard deviation = √8 = 2.828
So the pass mark needs to be at least 10 + (3x2.828) ≈ 18.5 – i.e. the minimum pass mark should
be 19
6) a) i) = 0.149
U ~ B (30, 0.13)
P (P=2) = (30
)(0.13)2 (0.87)28
2
Unfortunately for the first question you can’t use the binomial tables in the formula booklet.
Due to your p value being 0.13, which is not included on the tables. You have to use the
binomial distribution formula, (𝑛𝑥)𝑝 𝑥 (1 − 𝑝)𝑛−𝑥 , sub your values into the equation,
(30
)(0.13)2 (0.87)28. Use your calculator to work the answer out, which is 0.14890697 then
2
round to 3.d.p
Or if you have a graphical calculator you can work it out via that,
[MENU] [2] (STAT) [F5] (DIST) [F5] (BINM) [F1] (Bpd)
Change the screen to this;
Data : Variable
𝑥∶2
Numtrial: 30
P : 0.13
Save Res: None
Execute
[Calc] Binomial P.D
P = 0.14890697
P = 0.149
ii) p = 0.35
P(R ∪ P > 10) = 1 – 0.5078
= 0.4922
To get your p value of 0.35, you have to add the Value of red which is 0.22 to purple which is
0.13. 0.22 + 0.13 = 0.35. You can look on page 18 of the formula booklet, for the n = 30 table.
Look along the top for the p value of 0.35 then down to where x = 10. The value should be
0.5078, but since the tables show when P( 𝑋 ≤ 𝑥 ) and you want more than. You need to do
1 – 0.5078 to get the correct answer of 0.4922.
Or by using a graphical calculator,
[MENU] [2] (STAT) [F5] (DIST) [F5] (BINM) [F2] (Bcd)
Change screen to this;
Data : Variable
𝑥 ∶ 10
Numtrial: 30
P : 0.35
Save Res: None
Execute
[Calc] Binomial C.D
P = 0.5075821
P = 0.5076
The graphical calculator is more accurate than the binomial distribution tables, but when
you do 1 – 0.5076, the answer falls in the range on the mark scheme.
iii) P(5 ≤ G ≤ 10) =
0.9744 – 0.2552
= 0.719
For this question, we want the values of P(G = 5) + P(G = 6) + P(G = 7) + P(G = 8) + P(G = 9) + P(G = 10)
to get the correct answer but there is a trick for this. We need the values up to 10, so out notation
will be P(G ≤ 10) , in the question it asks at least 5. This eliminates the values of 4 or less, since we
now know this we can do P(G ≤ 10) – P(G ≤ 4). P(G ≤ 10) = 0.9744 , P(G ≤ 4) = 0.2552.
0.9744 – 02552 = 0.719 will give you full marks.
If you did the sum of them as mentioned in the beginning, you would only get 1 mark for the correct
answer.
b)i) Mean = 100 x 0.22 = 22
Variance = 100 x 0.22 x 0.78 = 17.16
In this question, the number of ‘n’ has changed from 30 to 100. This means to get the mean we do
‘𝑛𝑝’, which is 100 x 0.22 to get 22. Then to get the new variance we sub our new numbers into the
formula of’ 𝑛𝑝𝑞’ which is 100 x 0.22 x 0.78 to get 17.16
ii) 22.1 = 22, so reject claim (that p > 0.22)
√17.16 = 4.14 ≈ 4.17
Reject claim that not random samples
The claim suggests that the proportion of red paper clips in the bin is (p > 0.22), but the analysis is
22.1. They are similar to each other, to 1.d.p. But we reject the claim as p is not greater than 0.22
To find out if the Standard deviation we need to Square Root our variance from part b, √17.16 =
4.14. The two standard deviations do not match, therefore we reject the claim that they’re not
random samples, since the claim does not match with the analysis.
EXAM TIP: Numerical Justification is easy marks; just take your time to answer it correctly. You’re
just comparing to items of data, make sure you use the data provided in the question to your
answer.
Normal Distribution
Normal Distribution N (𝜇 , 𝜎 2 ),
𝑋−𝜇
𝜎
=𝑧
If X is normally distributed with mean (𝜇) and variance (𝜎 2 ), it is written as X ~ N (𝜇 , 𝜎 2 )
The standard normal distribution ‘Z’ has mean 0 and variance 1 .e.g.. Z ~ N (0,1)
EXAMPLE
X ~ N ( 4, 0.52 )
a) Find P (X ≤ 4.5)
=P(Z≤
4.5−4
)
0.5
= P ( Z ≤ 1.0 )
= 0.84134
The normal distribution is ‘Bell-Shaped’, most things in real life are most likely to fall ‘somewhere in
the middle’. They are less likely to take the ‘extremely high’ or ‘extremely low’ values.
This kind of situation is called a normal distribution.
There is a peak in the middle at the mean, the graph is symmetrical. This means that the values are
the same distance above and below the mean are equally likely.
The most important normal distribution is the standard normal distribution, or Z – this has a mean of
zero and a variance of 1.
Use tables to work out probabilities of Z
You look up a value of z and these tables labelled ɸ(𝑧)) tells you the probability Z ≤ z
> or ≥ (1 – ‘N’)
< or ≤ ‘N’
If Z score is negative, you have to flip the inequality
EXAMPLE
(Use the tables on page 24 of the formula booklet)
P (Z < 0.1) = 0.53983
P (Z > 0.23) = 1 – P (Z ≤ 0.23) = 1 – 0.59095 = 0.40905
P (Z ≥ -0.42) = P (Z ≤ 0.42) = 0.66276
P (0.12 < Z ≤ 0.82) = P(Z ≤ 0.82) – P(Z ≤ 0.12)
= 0.79389 – 0.54776 = 0.24613
P (Z = 1) = 0
All probabilities in the table of ɸ(z) are greater than 0.5, but can still use the tales with values less
than this. Just use this formula (1 – probability)

Finding Z values – using the table of percentage points
The Table of percentage points is located on page 25 of the formula book
You use the percentage table in a similar way as the normal distribution table, by starting
with a probability and looking up a value for Z.
EXAMPLE
X ~ N (10, 0.5 2 )
Find z such that P(X < z ) = 0.96 – Look up 0.96
ɸ (z) = 0.96 , z = 1.7507
Finding the mean and variance To find one of these, you will need to form an equation as
follows:
Y ~ N(𝜇, 4²)
P (Z ≤
23 − 𝜇
4
23 − 𝜇
4
P (Y ≤ 23) = 0.75
find 𝜇.
) = 0.75
= 0.6745
𝜇 = 20.302

To find both of them, you will need to form a pair of simultaneous equations:
X ~ N( 𝜇, 𝜎²)
P ( X ≤ 1.83) = 0.3
P(Z≤
1.83 − 𝜇
)
𝜎
P ( X ≤ 2.31) = 0.7
P(Z≤
= 0.3
2.31 − 𝜇
)
𝜎
= 0.7
This will be a negative ‘z’ value
This will be a positive value
1.83 − 𝜇
𝜎
2.31 − 𝜇
𝜎
= -0.5244
1.83 - 𝜇 = -0.5244𝜎 (1)
Eliminate either 𝜇 or 𝜎
= 0.5244
2.31 - 𝜇 = 0.5244𝜎 (2)
𝜇 = 2.07m
𝜎 = 0.458
5) a) i) Weight, W ~ N(2.75, 0.152)
P(W < 2.8) = P( 𝑍 <
2.8−2.75
)
0.15
P(Z < 0.33) = 0.629
The formula what we need to use is,
𝑋−𝜇
𝜎
= 𝑧. Now collect the terms that we need, which are the
‘mean’(𝜇)= 2.75, ‘Standard deviation’ (𝜎) = 0.15 and our ‘X’ is 2.8. Since our weight is less than
2.8kg, we write the notation P(W < 2.8). This is then converted to 𝑃 ( 𝑍 <
2.8−2.75
) as
0.15
we want to
know the normal distribution of it, as you can see the terms we have gathered have been subbed
into the correct places.
2.8−2.75
0.15
= 0.33, so now we do P(Z < 0.33) for this we can look in the tables on
page 24 of the formula booklet. We go down to 0.3 and along to 0.03 in the tables which gives us the
answer of 0.62930, then round to 3dp.
By using a graphical calculator
[MENU] [2] (STAT) [F5] (DIST) [F1] (NORM) [F2] (Ncd)
A screen with a set of options should appear change to these settings
Normal C.D
Data: Variable
Lower: -100
Upper: 2.8
𝜎 ∶ 0.15
𝜇: 2.75
Save Res: None
Execute [F1] (CALC) This gives us the P value of 0.63055837, round down up 0.631
This falls in the range of the answer in the mark scheme
ii) P ( W > 2.5) = P( 𝑍 >
2.5−2.75
)
0.15
P(Z > -1.67) = P(Z < 1..67)
= 0.952
The rules pretty much apply as they did in the last question. The only difference is that this time
we’re dealing with a ‘more than sign’, nothing changes until after you sub your numbers in. You
should get a value of -1.67, now since it’s a negative number we flip the sign to less than and get rid
of the negative. Now we just treat it normally, look in the Normal Distribution tables on page 24 , to
get your answer as 0.952
By using a graphical calculator
[MENU] [2] (STAT) [F5] (DIST) [F1] (NORM) [F2] (Ncd)
A screen with a set of options should appear change to these settings
Normal C.D
Data: Variable
Lower: 2.5
Upper: 1E + 99
𝜎 ∶ 0.15
𝜇: 2.75
Save Res: None
Execute [F1] (CALC) This gives us the P value of 0.95220964, round down to 0.952
This falls in the range of the answer in the mark scheme
b) Weight, X ~ N(5.25, 0.202)
i) P(5.1 < X < 5.3) = P(Z < 0.25) – P(Z < -0.75)
0.59871 – ( 1 – 0.77337)
= 0.372
For this we do the normal distribution of 5.3 – the normal distribution of 5.1, we’re combining our
two methods worked out in the part a of this question. But we’re dealing with new numbers, again
we use the same formula and sub the numbers in . For P(X < 5.3) P( 𝑍 <
P(X > 5.1), ) P( 𝑍 <
5.1 −5.25
)=
0.2
5.3 −5.25
)
0.2
= P(Z < 0.25) . For
P(Z < -0.75) remember as it’s a negative we flip the sign to make it
P(Z > 0.75). We still need to do another step, as Normal distribution tables on work when (X ≤ x), as
it’s a more than sign remember we now do (1 – P(X > x)). When we look in the tables for the values
we get 0.59871 – (1-0.77337) which is 0.372.
By using a graphical calculator
[MENU] [2] (STAT) [F5] (DIST) [F1] (NORM) [F2] (Ncd)
A screen with a set of options should appear change to these settings
Normal C.D
Data: Variable
Lower: 5.1
Upper: 5.3
𝜎 ∶ 0.2
𝜇: 0.525
Save Res: None
Execute [F1] (CALC) This gives us the P value of 0.37207897, round down to 0.372
This falls in the range of the answer in the mark scheme
ii) P(0 in 4) = [1 – 0.372]4
= 0.6284
= 0.155
We don’t know how many bags of 5kg the store sells, so we do [1 – 0.372]4, as we’re taking away
the random sample of 4 from the original population, we have the power of 4 because that’s how
many bags we’re dealing with. We have to use the probability from part b)i) since the question ask
to calculate the probability of 5.1 and 5.3.
c) Weight, Y ~ N(10.75, 0.502 )
Variance of 𝑌̅6 = 0.52 /6 = 0.0416
P(𝑌̅6 < 10.5) = P(𝑍 <
10.5−10.75
)
√0.0416
P (Z < - 1.22) = 1 – P(Z < 1.22)
1 – (0.88877) = 0.111
For this question, you need to know a little bit of estimation. To work out the sample mean, we do
the standard error which is
𝜎
√𝑛
or
𝜎2
𝑛
(This will be our new or variance or standard deviation) Variance
of 𝑌̅6 = 0.5 /6 = 0.0416 this is our new variance. We can now treat the question normally for a
normal distribution. We sub in the correct values into the formula; we get the negative number
again. We have to flip the sign, making it a more than sign. With a more than sign we do 1 – P(X>x) ,
this means we do 1 – 0.88877 which is 0.111
2
Estimation
Estimating the population mean from a sample mean and finding a confidence interval.
In a simple random sample
- Every person or item in the population has an equal chance of being in the sample
- Each selection is independent of every other selection
Statistics
A statistic is a quantity calculated only from the known observation in a sample.
A statistic is a random variable it takes different values for different samples.
The probability distribution of a statistic is called a sampling distribution.
Statistics are used to estimate population parameters, ‘population parameters’ are the populations
characteristics e.g. mean (𝜇) and variance (𝜎)
For example, you can use the sample mean 𝑥̅ =
∑𝑥
𝑛
to estimate the population mean (𝜇).
The sample mean is called an unbiased estimator of the population mean. This means the expected
value of the sample mean is the same as the population mean.
Formulae for Unbiased Estimators of population variance and standard deviation
Sample Variance = 𝑠 2 =
𝑛 ∑ 𝑥2
[
𝑛−1 𝑛
∑𝑥 2
−(
𝑛
) ]
Sample Standard Deviation = s = √𝑠𝑎𝑚𝑝𝑙𝑒 𝑣𝑎𝑟𝑖𝑎𝑛𝑐𝑒
EXAMPLE
A random sample was taken from a population whose mean (𝜇) and variance (𝜎 2 ) are unknown. If
the observed values were 5.6, 5.8, 5.5, 5.8 and 5.7, find the unbiased estimates of 𝜇 and 𝜎 2 .
The sample mean (𝑥̅ ) will give an unbiased estimate of the population mean (𝜇)
∑ 𝑥 = 28.4, 𝑠𝑜 𝑥̅ =
∑𝑥
𝑛
=
28.5
5
= 5.68
Unbiased estimate of the population variance (𝜎 2 )
∑ 𝑥 2 = 161.38, 𝑠𝑜 𝑠 2 =
𝑛 ∑ 𝑥2
[
𝑛−1 𝑛
∑𝑥 2
−(
𝑛
) ] = 𝑠2 =
5 161.38
28.4 2
−
(
) ]
[
4
5
5
= 0.017
The standard error and Central Limit Theorem
A standard error is a measure of the reliability of a statistic
𝜎
The standard error of 𝑋̅ is
√𝑛
Suppose you’ve got a random variable 𝑋, with mean 𝜇 and variance 𝜎 2
When you take a sample of n readings from the distribution of 𝑋, you can work out its sample mean
𝑋̅. If you now keep taking samples of size n from that distribution and working out the sample means
then you get a collection of sample means, drawn from the sampling distribution of ̅𝑋
This sampling distribution also has mean 𝜇, but a variance of
standard error of the sample mean
𝜎2
,
𝑛
this standard deviation is called the
If X is normally distributed (X ~ N(𝜇, 𝜎 2 )), then 𝑋̅ ~ 𝑁(𝜇,
𝜎2
)
𝑛
The Central Limit Theorem
The central limit theorem tells you something about 𝑋̅, - even if you know nothing at all about the
distribution of X
Suppose you take a sample of n readings from any distribution with mean and variance
For Large n, the distribution of the sample mean is 𝑋̅, is approximately normal 𝑋̅ ~ 𝑁(𝜇,
𝜎2
)
𝑛
The bigger n is the better the approximation will be (n > 30 it’s pretty good)
A Confidence interval is a type of interval estimation
A confidence interval for the mean of a population is a range of values that you’re fairly sure
contains the true population mean.
E.G – “(11.2, 14.4) is a 95% confidence interval for 𝜇”
This means that there’s a 95% probability that the interval of (11.2, 14.4) includes the true value of
𝜇.
A confidence interval extends “Z x Standard Error” from the sample mean.
Confidence intervals for 𝜇 – Normal distribution and known variance
A confidence interval for the population mean is (𝑋̅ − 𝑧
𝜎
,
√𝑛
𝑋̅ + 𝑧
𝜎
)
√𝑛
The value of Z depends on the level of confidence you need for example
For a 95% confidence interval, choose z so that P(-z < Z < z) = 0.95
For a 98% confidence interval, choose z so that P(-z < Z < z) = 0.98, you get the picture.
7) a) 𝑥̅ =
18.1
36
= 5.05, 98% (0.98) z = 2.32
Cl for 𝜇 𝑖𝑠 𝑥̅ ± 𝑧 ×
𝜎
√𝑛
, 5.05 ± 2.3263 ×
0.075
√36
5.05 ± 0.03 (or 5.05, 5.08)
We want to find the mean first so we divide the total volume of 181.80 by the total number of
measurements which is 36. 𝑥̅ =
18.1
36
= 5.05, For the confidence limit of 98% we look at page 25 in
the formula booklet, you have to take the limit of 99% which is 2.3263 as this will cover the 98%
confidence limits and all decimal places within.
We now need to multiply by the standard error to measure the reliability.
Confidence limit for 𝑖𝑠 𝑥̅ ± 𝑧 ×
0.075
)
√36
𝜎
√𝑛
, sub in the values into the correct position 5.05 ± (2.3263 ×
Which gives you a final answer of 5.05 ± 0.03
Or by using a graphical calculator
[MENU] [2] (STAT) [F4] (INTR) [F1] (Z) [F1] (1-S)
It now should come up with a list of data variables that need to be entered.
Your screen should end up like this
1 – Sample ZInterval
Data : Variable
C – Level : 0.98
𝜎 ∶ 0.075
𝑥̅ : 5.05
𝑛 ∶ 36
Save Res: None
Execute [F1] (CALC)
A screen should appear with this on
1 – Sample ZInterval
Left = 5.02092065
Right = 5.07907935
𝑥̅ : 5.05
𝑛 ∶ 36
This leads on to our answer of (5.02, 5.08) taking the information gathered from the left and right.
You write the answer as coordinate, as this is where the limit stands.
b) Claim is 𝜇 > 5L, Confidence limits exceed 5, so the first claim is correct.
8/36, 8 bottles out of 36 have less than 5 litres, the claim is fewer than 10%, 8/36 ≠10%, therefore
we disagree with the second claim.
c) Yes because volumes are not stated as normally distributed
We only use the Central Limit Theorem if original population is not normally distributed, assuming
that the value of n Is greater than 30, in which case it is. Therefore we did make use of the Central
Limit Theorem as n > 30, therefore the original population is not normally distributed.
Correlation and Regression
Correlation
Correlation is a measure of how closely variables are linked.
Draw a scatter diagram to see pattern in data.
Positive correlation has a positive gradient. As one goes up, so does the other variable.
Negative correlation has negative gradient. As one group goes up, the other variable decreases.
No correlation, scattered points, no line of best fit. As the two variables aren’t linked at all, you’d
expect a random scattering.
To measure correlation, we use the ‘The Product-Moment Correlation’
The Product-Moment Correlation Coefficient (In Short PMCC, or , ‘r’) Measures how close to a
straight line the points on a scatter graph lie.
PMCC is always between +1 and -1
r = + 1, if all your points lie exactly on a straight line with a positive correlation.
If r is between ± 0.2 and 0 means there is a weak correlation
If r is between the values of ±0.2 and ±0.7 there is a moderate correlation
If r is between the values of ±0.7 and ±0.9 there is a strong correlation
r = -1, if all your points lie exactly on a straight line with a negative correlation.
You will never expect to get a PMCC of +1 or -1, your scatter points may lie pretty close but unlikely
to be all on it.
If you cannot visualise the correlation by the use of values obtained, create a scatter graph to help
you describe the correlation. Always remember when describing the correlation refer to your
variables e.g. there is some positive correlation between the height and the weight of a person.
To work out correlation you would use
these formulae, to know how to get the
correct parts from your data set. Refer to
pages 6 and 7.
EXAMPLE
Illustrate the following data with a scatter diagram, and find the product-moment correlation
coefficient (r) between the variables x and y.
If p= 4x – 3 and q = 9y + 17, what is the PMCC between p and q?
X
Y
1.6
11.4
2.0 2.1 2.1 2.5 2.8 2.9
11.8 11.5 12.2 12.5 12.0 12.9
3.3
13.4
3.4 3.8 4.1 4.4
12.8 13.4 14.2 14.3
16
14
Just plot the points for the scatter graph
12
Now for the correlation coefficient. From the
scatter diagram, the points lie pretty close to a
straight line with a positive gradient – so if the
correlation doesn’t come out pretty close to
+1, we’d not need to worry.
10
8
6
4
2
0
0
1
2
3
4
5
There are 12 pairs of readings, n = 12. Now you have to work out a load of sums, if you
working it out manually add extra rows to your table. But you can work parts using a
calculator.
x
y
𝑥2
𝑦2
xy
1.6
2.0
2.1
2.1
2.5
2.8
2.9
3.3
3.4
3.8
11.4
11.8
11.5
12.2
12.5
12.0
12.9
13.4
12.8
13.4
2.56
4
4.41
4.41
6.25
7.84
8.41
10.89
11.56
14.44
129.96 139.24 132.25 148.84 156.25 144 166.41 179.56 163.84 179.56
18.24
23.6
24.15
25.62
31.25 33.6 37.41
44.22
43.52
50.92
Note: Answers in blue are the sums of each row, which are needed for the PMCC
4.1
14.2
16.81
201.64
58.22
4.4
14.3
19.36
204.49
62.92
Stick all these in the formula to get
𝑟=
[453.67 −
√[110.94 −
=
9.17
√8.857 ×10.56
35 × 152.4
]
12
352
152.42
×
−
]
[1946.04
12
12 ]
= 0.948
Correlation coefficients aren’t greatly affected by linear scaling – you can multiply either variables by
positive numbers (or multiply them both by negative numbers) and add numbers to them, and you
won’t change the PMCC between them.
So if p and q are given by p = 4x – 3 and q = 9y + 17 , then the PMCC between p and q is also 0.948
35
152.4
110.94
1946.04
453.67
1) a) r =
𝑆𝑥𝑦
√𝑆𝑥𝑥 ×𝑠𝑦𝑦
=
−0.410
=
√2.030 ×1.498
-0.235
The question asks you to find the value of the product moment correlation coefficient, all you need
to do is look up the equation in the formula booklet. Then sub in your values into the correct places
in the formula. Use a calculator to get your answer of -0.235.
b) There is a weak correlation between width and thickness of lengths of steel
Since the r value is very close to ± 0.2, then the correlation is a weak one. To get the full marks you
must refer to the two variables.
Linear Regression
Linear Regression is in the simplest of words, finding the line of best fit.
Decide which the independent variable is and which the dependent is. The independent variable is
the variable along the x-axis, it is the variable you can control or the one you think is affecting the
other. Whilst the dependent variable, goes up the y-axis, it is the response variable the one you
think is being affected.
The equation for the regression line is y = a + bx ,
Where ‘a’ is the intercept, and ‘b’ is the gradient.
Before plotting a line of best fit, you must work out at least two coordinates that lie on the line by
substituting in value of x
Linear Scaling is a linear transformation or scaling of one or both of the variables. It doesn’t affect
the ‘r’ value.
A Residual is the difference between an observed y-value and the y-value predicted by the
regression line.
Residual = Observed y-value – Estimated y-value
1) Residuals show the experimental error between the y-values that’s observed and the y-value your
regression line says it should be
2) Residuals are shown by a vertical from the actual point to the regression line.
3) You would like your residuals to be small, as this would show your regression line fits the data
well. If they’re large then your regression line on your model won’t be a very reliable one
Outliers have a big effect on the regression line; they are able to drag the line of best fit away from
the rest of the data values. When you indicate an outlier, you should try to explain it, e.g. it could be
an error that you are told to correct.
EXAMPLE
Find the equation of the regression line of y on x for the data below, the data shows the load on a
lorry, 𝑥 (in tonnes), and the fuel efficiency, 𝑦 (in km per litre)
𝑥
5.1
5.6
5.9
6.3
6.8
7.4
7.8
8.5
9.1
9.8
72.3 = ∑ 𝑥
𝑦
𝑥2
9.6
26.01
9.5
31.36
8.6
34.81
8
39.69
7.8
46.24
6.8
54.76
6.7
60.84
6
72.25
5.4
82.81
5.4
96.04
𝑥𝑦
48.96
53.2
50.74
50.4
52.04
50.32
52.26
51
49.14
52.92
73.8 = ∑ 𝑦
544.81 =
∑ 𝑥2
511.98 =
∑ 𝑥𝑦
Then work out 𝑆𝑥𝑦 , is given by 𝑆𝑥𝑦 = ∑(𝑥 − 𝑥̅ )(𝑦 − 𝑦̅)= ∑ 𝑥 −
2
2
and 𝑆𝑥𝑥, is given by: 𝑆𝑥𝑥 =∑(𝑥 − 𝑥̅ ) = ∑ 𝑥 −
(∑ 𝑥)(∑ 𝑦)
𝑛
(∑ 𝑥)2
𝑛
𝑆𝑥𝑦
The gradient (b) of your regression line is given by b = 𝑆 and the intercept (a) is given by: a = 𝑦̅ - b𝑥̅
𝑥𝑥
1) Work out the sums: ∑ 𝑥 = 72.3 , ∑ 𝑦 = 73.8 , ∑ 𝑥 2 = 544.81, ∑ 𝑥𝑦 = 511.98
2) Then work out 𝑆𝑥𝑦 and 𝑆𝑥𝑥 ∶ 𝑆𝑥𝑦 = 511.98 −
𝑆𝑥𝑥 = 544.81 −
72.32
10
72.3 ×73.8
10
= −21.594
= 22.081
3) So the gradient of the regression line is: b =
4) And the intercept is: 𝑎 =
∑𝑦
𝑛
−𝑏
∑𝑥
𝑛
=
−21.594
=
22.081
73.8
−
10
-0.978
(−0.978) ×
72.3
10
= 14.451 = 14.5
5) This all means that your regression line is: 𝑦 = 14.5 − 0.978𝑥
This tells you: (i) for every extra tonne carried, you’d expect the lorry’s fuel efficiency to fall by
0.978km per litre, and (ii) with no load (x = 0), you’d expect the lorry to do 14.5kkm per litre of fuel.
3) a) b (gradient) = 2.27
a (intercept) = 4.17
To get these answers, use a calculator. Input the data values on your calculator, and then
follow the step by step on page 6. Double check that your values are correct, then print your
values on the page. ‘A’ = 4.16981131 and ‘B’ = ‘2.27’
b) Correct Straight line
(You will get a graph for this question, so you don’t need to draw one out)
200
180
160
140
120
100
80
60
40
20
0
0
20
40
60
80
100
c) i) Correct Straight line
It will look different to the line above, as it’s a different regression line. Work out the
first and last coordinate. Sub x = 10 into equation, 𝑦𝑏 = 60.1 + 0.255(10) = 62.65
First Coordinate = (10, 62.65) then sub x = 80 into equation, 𝑦𝑏 = 60.1 + 0.255(80) =
80.5. Last coordinate = (80, 80.5) The mark scheme is lenient of your coordinates.
ii) 27 to 29
The mark scheme is lenient on this as well, not everyone’s line of best fit on part i) will
be the same. But all you do is look at y = 100, then go across to the regression line and
read the X coordinate of it.
iii) At low temperatures more B (than A) Dissolves or At high temperatures more A
dissolves than B.
Amount increases more rapidly for A than B or Amount increases more slowly for B than
A