Transcript Document

“Teach A Level Maths”
Vol. 3: S1
Demo Disc
© Christine Crisp
Volume 3 of “Teach A level Maths” covers the work
on Probability and Statistics for the A/AS level
Option Module S1.
All topics for the 4 specifications offered by the
English examining bodies are covered. Where a
topic relates to some specifications only, this is
indicated in a contents file and also at the start of
the presentation.
Explanation of Clip-art images
An important result, example or
summary that students might want to
note.
It would be a good idea for students
to check they can use their
calculators correctly to get the
result shown.
An exercise for students to do
without help.
The slides that follow are samples from 9 of the 40
presentations.
4:
Box and Whisker
Diagrams
23: Binomial Problems
6:
Histograms
26: Hypothesis Testing
10: Introduction to
Probability
28: Standardizing to Z
14: Discrete Random
Variables
36: Calculating Residuals
16: Linear Functions of a
Discrete Random Variable
4: Box and Whisker Diagrams
Demo version note: The S1 specifications require
students to be familiar with topics covered in Data
Handling at GCSE.
The first few presentations revise and extend the
GCSE work.
By the time the students reach this 4th presentation
they have been reminded about cumulative frequency
diagrams and have met the age data referred to on
the slide.
Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative
frequency diagram.
I’ll use the age data
that we met earlier.
The box can
be any depth.
The projected population of the
U.K. for 2005
( by age )
minimum
One
The
box
median
upper
age lower
whisker
quartile
quartile
The other
maximum
whisker
age
Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative
frequency diagram.
I’ll use the age data
that we met earlier.
minimum
age
The projected population of the
U.K. for 2005
( by age )
lower median upper
quartile
quartile
maximum
age
Box and Whisker Diagrams
The diagram can easily be drawn using a cumulative
frequency diagram.
I’ll use the age data
that we met earlier.
The projected population of the
U.K. for 2005
( by age )
We need a
scale.
0
50
Age (years)
100
Histograms
6: Histograms
Demo version note: As well as explaining theory,
the presentations show worked examples and set
introductory exercises.
The 6th presentation reminds students about the
rules for drawing Histograms. The exercise shown
here reinforces these rules without the students
needing to spend time drawing a diagram.
Histograms
Exercise
95 components are tested until they fail. The table gives
the times taken ( hours ) until failure.
Time to
failure (hours)
Number of
components
0-19
5
Find 3 things wrong
with the histogram
which represents the
data in the table.
20-29 30-39 40-44 45-49 50-59 60-89
8
16
22
18
16
10
Histograms
Answer:
Time to
failure (hours)
Number of
components
0-19
20-29 30-39 40-44 45-49 50-59 60-89
5
• Frequency has been
plotted instead of
frequency density.
• There is no title.
• There are no units on
the x-axis.
8
16
22
18
16
10
Histograms
Incorrect
diagram
Time taken for 95 components to fail
Correct
diagram
Introduction to Probability
10: Introduction to Probability
Demo version note: This presentation covers the
introductory ideas of probability and leads to a
later one on conditional probability.
Summaries are given from time to time which
teachers may want students to note down. This
slide shows an example of a summary.
Introduction to Probability
SUMMARY
 Outcomes are the results of trials or experiments.
 An event is a particular result or set of results.
 A possibility space is the set of all possible outcomes.
 For equally likely outcomes, the probability of an
event, E, is given by
P (E) = number of ways E can occur
number of possible outcomes
Discrete Random Variables
14: Discrete Random Variables
Demo version note: The presentations all contain
worked examples of the straightforward type of
questions found in exams.
This is the first of the examples in the
presentation on Discrete Random Variables.
Discrete Random Variables
e.g. 1. A random variable X has the probability
distribution
x
5
10
1
1
1
p
P (X = x) 4
2
Find (a) the value of p and (b) the mean of X.
Solution: (a) Since X is a discrete r.v.,


(b) mean,
1
4

1
2
 p=1
p = 14
 P( X = x) = 1
 =  xP ( X = x ) = 1  14  5  12  10  14 =
21
4
Tip: Always check that your value of the mean lies
within the range of the given values of x. Here, 21
4 or
5·25, does lie between 1 and 10.
Linear Functions of a Discrete Random Variable
16: Linear Functions of a Discrete
Random Variable
Demo version note: Some topics are not required by
all the specifications. The contents file shows which
topics are needed by each of the specifications and
contains hyperlinks to the files.
The topic Linear Functions of a Discrete Random
Variable is only required in S1 by Edexcel.
Linear Functions of a Discrete Random Variable
The results we have found can be generalised to
give
E(aX + b) = aE(X) + b
e.g. The probability distribution for the r.v. X is
given by
x
2
4
6
8
10
P( X = x)
1
12
1
4
5
12
1
6
1
12
Find (a) E(X), (b) Hence find E(2X - 3)
Solution:
( X ) of
= the
xP(question
X = x ) means that we
“Hence” in (a)
partE(b)
1part (a)1 rather than1using
35
must use the answer
to
= 2   4   . . .  10 
=
= 5 56
12 of 42X - 3.
12 6
the values and probabilities
26
 35 
(b) E ( 2 X - 3) = 2 E ( X ) - 3 = 2  - 3 =
= 8 23
 6 
3
Binomial Problems
23: Binomial Problems
Demo version note: The Binomial Distribution is
covered by AQA, MEI/OCR and OCR.
Having learnt to carry out Binomial Calculations,
students practise recognising the conditions for
using the model and also learn the importance of
defining a random variable and writing down its
distribution.
Binomial Problems
e.g. 1. A factory produces a particular type of computer
chip. Over a long period the number that are defective has
been found to be 15%. What is the probability that in a
sample of 20 taken at random, 19 are perfect?
Are the conditions met for using the Binomial model?
• A trial has 2 possible outcomes, success and failure.
Yes: Each chip is either defective or not.
•
The trial is repeated n times.
Yes: 20 chips are selected so n = 20.
•
The probability of success in one trial is p and p is
constant for all the trials.
•
Yes: We are given 15% (so p = 0·15 ) and we can
assume it is constant.
The trials are independent.
Yes: The probability of selecting a defective chip does
not depend on whether one has already been selected.
Binomial Problems
e.g. 1. A factory produces a particular type of computer
chip. Over a long period the number that are defective has
been found to be 15%. What is the probability that in a
sample of 20 taken at random, 19 are perfect?
Solution: Let X be the r.v. “number of defective chips”
We must never miss out this stage since it reminds us that
(i) X represents a number ( that can be 0, 1, 2, . . . n ), and
(ii) we have to make the decision as to whether to count
the number of defective chips or perfect ones.
So,
X ~ B( 20, 0  15)
Writing the distribution of X in this way makes us check
that we have the p that fits our definition of the r.v.,
defective rather than perfect.
Binomial Problems
e.g. 1. A factory produces a particular type of computer
chip. Over a long period the number that are defective has
been found to be 15%. What is the probability that in a
sample of 20 taken at random, 19 are perfect?
Solution: Let X be the r.v. “number of defective chips”
So, X ~ B( 20, 0  15)
The solution is now straightforward. We want P ( X = 1 ).
We need to be very careful here and not use P ( X = 19) by
mistake.
I had set up the Binomial for the number of defective chips,
because I had the proportion for defective. However, the
question asked for the probability of 19 perfect ones.
P ( X = 1)= 20C 1 (0  15)(0  85)19 = 0  137 ( 3 d . p. )
If I had written
Let X be the r.v. “ number of perfect chips”
Then, X ~ B( 20, 0  85) and I would want P ( X = 19)
Hypothesis Testing
26: Hypothesis Testing
Demo version note: In the presentations extensive
use is made of snapshots from the software
package “Autograph”.
Here Autograph is used to illustrate an example on
Hypothesis Testing in the presentation for the
MEI/OCR specification.
Hypothesis Testing
e.g. 2. In a trial, 16 seeds are sown and only 11 germinate.
Use a 10% significance level to test the supplier’s claim
that 85% germinate. Find the critical region for the test.
Solution:
Let X be the random variable ”the number of seeds
that germinate”
X ~ B(16, p)
Test at 10% level
H 0 : p = 0  85
of significance.
H
 0 supplier’s
 85
To1 :testpthe
claim, the alternative
hypothesis
is that
fewerathan
85% test
This
is again
1-tailed
but 
this
P( X
 11
) = germinate.
0  0791
0  1time we need to
test the bottom end of the distribution.
There is a probability of 0·0791 ( less than 10% ) that 11
or fewer seeds will germinate.
We reject the null hypothesis at the 10% level of
significance and conclude that the germination rate
is below 85% .
Hypothesis Testing
The Autograph illustration is as follows:
X ~ B(16, 0  85)
P ( X  11) = 0  0791= 7  91%
The probability of 12 or fewer germinating is 0·2101
( 21·01% ), so the critical region for the test is
0, 1, 2, . . . 10, 11.
Standardizing to Z
28: Standardizing to Z
Demo version note: Students are encouraged to
use their Formulae and Statistical Tables even
when worked examples are being developed.
This presentation is part of a series to be used by
AQA and Edexcel students on the Normal
Distribution.
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (a)
z=
So,
x-

P ( X  400)
X
 = 350
x = 400, so

400 - 350
z=
110

z = 0  45
400
Tables only give 2
d.p. for z so this is
all we need.
P ( X  400)  P ( Z  0  45)
= (0  45 )
= 0  6736
Z
0  45
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (b) P ( 250  X  400)
There are 2 values to
convert so we use
subscripts for z.
X
250
350 400
N.B. This is left of
the mean so the z
value will be negative.
So,
250 - 350
z1 =
= - 0  91
110
400 - 350
z2 =
= 0  45
110
P (250  X  400)  P (-0  91  Z  0  45)
Standardizing to Z
e.g.1 If X is a random variable with distribution
X ~ N ( 350 , 110 2 )
find (a) P ( X  400)
(b) P ( 250  X  400)
Solution: (b)
P (250  X  400)  P (-0  91  Z  0  45)
= (0  45) - ( -0  91)
Z
( -0  91) = 1 - (0  91)
- 0  91
= 1 - 0  8186
= 0  1814
0  45
(0  45) - ( -0  91) = 0  6736 - 0  1814
= 0  4922
Calculating Residuals
36: Calculating Residuals
Demo version note: Throughout the module,
students are encouraged to use their calculators
efficiently and this is particularly important in the
topic for AQA, Edexcel and OCR on Least Squares
Regression.
In the following slides, however, the emphasis is on
the effect of outliers on the equation of a
regression line rather than on calculating the line
itself.
Calculating Residuals
e.g. This is a scatter diagram of the data shown in the
table.
If we were to draw the line “by eye”,
the 1st point . . . would lie well
away from the line we would want to
draw.
x
y
1
5
2
18
3
12
4
14
5
12
6
11
7
7
8
3
However, the calculation of the regression line includes
the 1st point and distorts the position of the line.
Calculating Residuals
e.g. This is a scatter diagram of the data shown in the
table.
y = 14  21 - 0  88 x
The diagram shows the y on x
regression line for all the data. The
residuals are shown by the red lines.
The left-hand end of the line is
further down than it would be
without the 1st point.
x
y
1
5
2
18
3
12
4
14
5
12
6
11
7
7
8
3
Calculating Residuals
e.g. This is a scatter diagram of the data shown in the
table.
y = 14  21 - 0  88 x
Removing the 1st point . . .
x
y
1
5
2
18
3
12
4
14
5
12
6
11
7
7
8
3
Calculating Residuals
e.g. This is a scatter diagram of the data shown in the
table.
y = 14  21 - 0  88 x
Removing the 1st point gives
y = 21  36 - 2  07 x
x
y
1
5
2
18
3
12
4
14
5
12
6
11
7
7
8
3
Calculating Residuals
e.g. This is a scatter diagram of the data shown in the
table.
The sum of the squares
of the residuals,
2
R
 = 139
y = 14  21 - 0  88 x
Removing the 1st point gives
The sum of the squares
of the residuals,
2
R
 = 19  9
y = 21  36 - 2  07 x
Without the 1st point,
we have a regression
line that is a much
better fit.
Full version available from:
Chartwell-Yorke Ltd.
114 High Street,
Belmont Village,
Bolton,
Lancashire,
BL7 8AL
England
tel (+44) (0)1204 811001,
fax (+44) (0)1204 811008
[email protected]
www.chartwellyorke.com