Higher Nationals
Internal verification of assessment decisions – BTEC (RQF)
INTERNAL VERIFICATION – ASSESSMENT DECISIONS
Programme title
BTEC Higher National Diploma in Computing
Assessor
Unit(s)
Assignment title
Internal Verifier
Unit 11 : Maths for Computing
Importance of Maths in the Field of Computing
Student’s name
List which assessment
criteria the Assessor has
awarded.
Pass
Merit
Distinction
INTERNAL VERIFIER CHECKLIST
Do the assessment criteria awarded match
those shown in the assignment brief?
Is the Pass/Merit/Distinction grade awarded
justified by the assessor’s comments on the
student work?
Has the work been assessed
accurately?
Y/N
Y/N
Y/N
Is the feedback to the student:
Give details:
• Constructive?
• Linked to relevant assessment
criteria?
Y/N
Y/N
• Identifying opportunities for
improved performance?
Y/N
• Agreeing actions?
Y/N
Does the assessment decision need
amending?
Y/N
Assessor signature
Date
Internal Verifier signature
Date
Programme Leader signature (if
required)
Date
Confirm actioncompleted
Remedial action taken
Give details:
Assessor signature
Date
Internal Verifier
signature
Date
Programme Leader
signature (if required)
Date
2
Higher Nationals - Summative Assignment Feedback Form
Student Name/ID
Unit Title
Unit 11 : Maths for Computing
Assignment Number
1
Assessor
Submission Date
Date Received 1st
submission
Re-submission Date
Date Received 2nd
submission
Assessor Feedback:
LO1 Use applied number theory in practical computing scenarios.
Pass, Merit & Distinction P1
P2
M1
D1
Descripts
LO2 Analyse events using probability theory and probability distributions
Pass, Merit & Distinction
Descripts
P3
P4
M2
D2
LO3 Determine solutions of graphical examples using geometry and vector methods
Pass, Merit & Distinction
P5
P6
M3
D3
Descripts
LO4 Evaluate problems concerning differential and integral calculus
Pass, Merit & Distinction
Descripts
Grade:
P7
P8
M4
D4
Assessor Signature:
Date:
Assessor Signature:
Date:
Resubmission Feedback:
Grade:
Internal Verifier’s Comments:
Signature & Date:
* Please note that grade decisions are provisional. They are only confirmed once internal and external moderation has taken place and grades decisions have been
agreed at the assessment board.
Pearson
Higher Nationals in
Computing
Unit 11: Maths for Computing
General Guidelines
1. A Cover page or title page – You should always attach a title page to your assignment. Use previous page as
your cover sheet and be sure to fill the details correctly.
2. This entire brief should be attached in first before you start answering.
3. All the assignments should prepare using word processing software.
4. All the assignments should print in A4 sized paper, and make sure to only use one side printing.
5. Allow 1” margin on each side of the paper. But on the left side you will need to leave room for binging.
Word Processing Rules
1. Use a font type that will make easy for your examiner to read. The font size should be 12 point, and should
be in the style of Time New Roman.
2. Use 1.5 line word-processing. Left justify all paragraphs.
3. Ensure that all headings are consistent in terms of size and font style.
4. Use footer function on the word processor to insert Your Name, Subject, Assignment No, and Page
Number on each page. This is useful if individual sheets become detached for any reason.
5. Use word processing application spell check and grammar check function to help edit your assignment.
Important Points:
1. Check carefully the hand in date and the instructions given with the assignment. Late submissions will not be
accepted.
2. Ensure that you give yourself enough time to complete the assignment by the due date.
3. Don’t leave things such as printing to the last minute – excuses of this nature will not be accepted for failure
to hand in the work on time.
4. You must take responsibility for managing your own time effectively.
5. If you are unable to hand in your assignment on time and have valid reasons such as illness, you may apply
(in writing) for an extension.
6. Failure to achieve at least a PASS grade will result in a REFERRAL grade being given.
7. Non-submission of work without valid reasons will lead to an automatic REFERRAL. You will then be asked to
complete an alternative assignment.
8. Take great care that if you use other people’s work or ideas in your assignment, you properly reference
them, using the HARVARD referencing system, in you text and any bibliography, otherwise you may be guilty
of plagiarism.
9. If you are caught plagiarising you could have your grade reduced to A REFERRAL or at worst you could be
excluded from the course.
Student Declaration
I hereby, declare that I know what plagiarism entails, namely to use another’s work and to present it as my own
without attributing the sources in the correct way. I further understand what it means to copy another’s work.
1. I know that plagiarism is a punishable offence because it constitutes theft.
2. I understand the plagiarism and copying policy of the Edexcel UK.
3. I know what the consequences will be if I plagiaries or copy another’s work in any of the assignments for this
program.
4. I declare therefore that all work presented by me for every aspects of my program, will be my own, and where
I have made use of another’s work, I will attribute the source in the correct way.
5. I acknowledge that the attachment of this document signed or not, constitutes a binding agreement
between myself and Edexcel UK.
6. I understand that my assignment will not be considered as submitted if this document is not attached to the
attached.
Student’s Signature:
(Provide E-mail ID)
Date:
(Provide Submission Date)
Assignment Brief
Student Name /ID Number
Unit Number and Title
Unit 11: Maths for Computing
Academic Year
2017/2018
Unit Tutor
Assignment Title
Importance of Maths in the Field of Computing
Issue Date
Submission Date
IV Name & Date
Submission Format:
This assignment should be submitted at the end of your lesson, on the week stated at the front of this
brief. The assignment can either be word-processed or completed in legible handwriting.
If the tasks are completed over multiple pages, ensure that your name and student number are present
on each sheet of paper.
Unit Learning Outcomes:
LO1
Use applied number theory in practical computing scenarios
LO2
Analyse events using probability theory and probability distributions
LO3
Determine solutions of graphical examples using geometry and vector Methods
LO4
Evaluate problems concerning differential and integral calculus.
Assignment Brief and Guidance:
Activity 01
Part 1
1. Mr. Steve has 120 pastel sticks and 30 pieces of paper to give to his students.
a) Find the largest number of students he can have in his class so that each student gets
equal number of pastel sticks and equal number of paper.
b) Briefly explain the technique you used to solve (a).
2. Maya is making a game board that is 16 inches by 24 inches. She wants to use square tiles. What
is the largest tile she can use?
Part 2
3. An auditorium has 40 rows of seats. There are 20 seats in the first row, 21 seats in the second
row, 22 seats in the third row, and so on. Using relevant theories, find how many seats are there
in all 40 rows?
4. Suppose you are training to run an 8km race. You plan to start your training by running 2km a
week, and then you plan to add a ½km more every week. At what week will you be running 8km?
5. Suppose you borrow 100,000 rupees from a bank that charges 15% interest. Using relevant
theories, determine how much you will owe the bank over a period of 5 years.
Part 3
6. Find the multiplicative inverse of 8 mod 11 while explaining the algorithm used.
Part 4
7. Produce a detailed written explanation of the importance of prime numbers within the field of
computing.
Activity 02
Part 1
1. Define ‘conditional probability’ with suitable examples.
2. A school which has 100 students in its sixth form, 50 students study mathematics, 29 study biology
and 13 study both subjects. Find the probability of the student studying mathematics given that the
student studies biology.
3. A certain medical disease occurs in 1% of the population. A simple screening procedure is
available and in 8 out of 10 cases where the patient has the disease, it produces a positive result. If
the patient does not have the disease there is still a 0.05 chance that the test will give a positive
result. Find the probability that a randomly selected individual:
(a) Does not have the disease but gives a positive result in the screening test
(b) Gives a positive result on the test
(c) Nilu has taken the test and her result is positive. Find the probability that she has the disease.
Let C represent the event “the patient has the disease” and S represent the event “the screening
test gives a positive result”.
4. In a certain group of 15 students, 5 have graphics calculators and 3 have a computer at home (one
student has both). Two of the students drive themselves to college each day and neither of them
has a graphics calculator nor a computer at home. A student is selected at random from the group.
(a) Find the probability that the student either drives to college or has a graphics calculator.
(b) Show that the events “the student has a graphics calculator” and “the student has a computer at
home” are independent.
Let G represent the event “the student has a graphics calculator” H
represent the event “the student has a computer at home”
D represent the event “the student drives to college each day”
Represent the information in this question by a Venn diagram. Use the above Venn diagram to
answer the questions.
5. A bag contains 6 blue balls, 5 green balls and 4 red balls. Three are selected at random without
replacement. Find the probability that
(a) they are all blue
(b)two are blue and one is green
(c) there is one of each colour
Part 2
6. Differentiate between ‘Discrete’ and ‘Continuous’ random variables.
7. Two fair cubical dice are thrown: one is red and one is blue. The random variable M represents the
score on the red die minus the score on the blue die.
(a) Find the distribution of M.
(b) Write down E(M).
(c) Find Var(M).
8. Two 10p coins are tossed. The random variable X represents the total value of each coin lands
heads up.
(a)Find E(X) and Var(X).
The random variables S and T are defined as follows:
S = X-10 and T = (1/2)X-5
(b)Show that E(S) = E(T).
(c)Find Var(S) and Var (T).
(d)
Susan and Thomas play a game using two 10p coins. The coins are tossed and Susan records her
score using the random variable S and Thomas uses the random variable T. After a large number of
tosses they compare their scores.
Comment on any likely differences or similarities.
9. A discrete random variable X has the following probability distribution:
x
1
2
3
P(X=x)
1/3
1/3
k
where k is a constant.
(a) Find the value of k.
(b) Find P(X ≤3).
4
1/4
Part 3
10. In a quality control analysis, the random variable X represents the number of defective
products per each batch of 100 products produced.
Defects (x) 0
1
2
3
4
5
Batches
95
113
87
64
13
(a) Use the frequency distribution above to construct a probability distribution for X.
(b) Find the mean of this probability distribution.
(c) Find the variance and standard deviation of this probability distribution.
8
11. A surgery has a success rate of 75%. Suppose that the surgery is performed on three
patients.
(a) What is the probability that the surgery is successful on exactly 2 patients?
(b) Let X be the number of successes. What are the possible values of X?
(c) Create a probability distribution for X.
(d) Graph the probability distribution for X using a histogram.
(e) Find the mean of X.
(f) Find the variance and standard deviation of X.
12. Colombo City typically has rain on about 16% of days in November.
(a) What is the probability that it will rain on exactly 5 days in November? 15 days?
(b) What is the mean number of days with rain in November?
(c) What is the variance and standard deviation of the number of days with rain in November?
13. From past records, a supermarket finds that 26% of people who enter the supermarket will make
a purchase. 18 people enter the supermarket during a one-hour period.
(a) What is the probability that exactly 10 customers, 18 customers and 3 customers make a
purchase?
(b) Find the expected number of customers who make a purchase.
(c) Find the variance and standard deviation of the number of customers who make a purchase.
14.On a recent math test, the mean score was 75 and the standard deviation was 5. Shan got 93.
Would his mark be considered an outlier if the marks were normally distributed? Explain.
15.For each question, construct a normal distribution curve and label the horizontal axis and answer
each question.
The shelf life of a dairy product is normally distributed with a mean of 12 days and a standard
deviation of 3 days.
(a) About what percent of the products last between 9 and 15 days?
(b) About what percent of the products last between 12 and 15 days?
(c) About what percent of the products last 6 days or less?
(d) About what percent of the products last 15 or more days?
16.Statistics held by the Road Safety Division of the Police shows that 78% of drivers being tested
for their licence pass at the first attempt.
If a group of 120 drivers are tested in one centre in a year, find the probability
that more than 99 pass at the first attempt, justifying the most appropriate distribution to be
used for this scenario.
Part 4
17.Evaluate probability theory to an example involving hashing and load balancing.
Activity 03
Part 1
1. If the Center of a circle is at (2, -7) and a point on the circle (5,6) find the formula of the circle.
2. What surfaces in R3 are represented by the following equations?
z=3
y=5
3. Find an equation of a sphere with radius r and center C(h, k, l).
4. Show that x2 + y2 + z2 + 4x – 6y + 2z + 6 = 0 is the equation of a sphere. Also, find its center
and radius.
Part 2
5. 3y= 2x-5 , 2y=2x+7 evaluate the x, y values using graphical method.
6.
a=(2i+3j+k) , b=(4i-2j+3k) and c=(1i+4j+2k) evaluate the volume of the shape.
Activity 04
Part 1
1. Find the function whose tangent has slope 4x + 1 for each value of x and whose graph passes
through the point (1, 2).
2. Find the function whose tangent has slope 3x2 + 6x − 2 for each value of x and whose
graph passes through the point (0, 6).
Part 2
3. It is estimated that t years from now the population of a certain lakeside community will be
changing at the rate of 0.6t 2 + 0.2t + 0.5 thousand people per year. Environmentalists have
found that the level of pollution in the lake increases at the rate of approximately 5 units per
1000 people. By how much will the pollution in the lake increase during the next 2 years?
4. An object is moving so that its speed after t minutes is v(t) = 1+4t+3t 2 meters per minute.
How far does the object travel during 3rd minute?
Part 3
5. Sketch the graph of f(x) = x − 3x 2/3 , indicating where the graph is increasing/decreasing,
concave up/down, and any asymptotic behavior.
6. Draw the graph of f(x)= 3x4-6X3+3x2 by using the extreme points from differentiation.
Part 4
7. For the function f(x) = cos 2x, 0.1 ≤ x ≤ 6, find the positions of any local minima or maxima
and distinguish between them.
8. Determine the local maxima and/or minima of the function y = x4 −1/3x3
9. By further differentiation, identify lines with minimum y = 12 x 2 − 2x, y = x 2 + 4x +
1, y = 12x − 2x 2 , y = −3x 2 + 3x + 1.
Grading Rubric
Grading Criteria
LO1 : Use applied number theory in practical computing scenarios
P1 Calculate the greatest common divisor and least common multiple
of a given pair of numbers.
P2 Use relevant theory to sum arithmetic and geometric
progressions.
M1 Identify multiplicative inverses in modular arithmetic.
D1 Produce a detailed written explanation of the importance of prime
numbers within the field of computing.
LO2 Analyse events using probability theory and
probability distributions
P3 Deduce the conditional probability of different events occurring
within independent trials.
P4 Identify the expectation of an event occurring from a discrete,
random variable.
M2 Calculate probabilities within both binomially distributed and
normally distributed random variables.
D2 Evaluate probability theory to an example involving hashing and
load balancing.
Achieved
Feedback
LO3 Determine solutions of graphical examples using
geometry and vector methods
P5 Identify simple shapes using co-ordinate geometry.
P6 Determine shape parameters using appropriate vector methods.
M3 Evaluate the coordinate system used in programming a simple
output device.
D3 Construct the scaling of simple shapes that are described by
vector coordinates.
LO4 Evaluate problems concerning differential and
integral calculus
P7 Determine the rate of change within an algebraic function.
P8 Use integral calculus to solve practical problems involving area.
M4 Analyse maxima and minima of increasing and decreasing
functions using higher order derivatives.
D4 Justify, by further differentiation, that a value is a minimum.
Maths for Computing
(D/615/1635)
Assignment 1
Importance of Maths in the field of computing
SAMPLE ANSWERS
Esoft Metro Campus
Sample Answers
Maths for Computing (Unit 11)
Assignment
11
Sample Answers
Activity 1
Part 1
Question 1
GCF or Greatest Common Divisor is the largest common factor of two or more numbers.
(a) 120
120 = 2 x 60
120 = 2 x 6 x 10
120 = 2 x 2 x 3 x 2 x 5
30
30 = 2 x 15
30 = 2 x 3 x 5
2
3
5
120
23
31
51
30
21
31
51
GCF
2
3
5
Prime
#
GCF (120, 30) = 2 x 3 x 5 = 30 students
Therefore, Mr. Steve could have 30 students in his class. Each student will receive
1 piece of paper and 4 pastel sticks.
(b) The technique used to solve the problem in (a) above is as
follows: Step 1 - The numbers 120 and 30 are prime factorized.
120
120 = 2 x 60
120 = 2 x 6 x 10
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
2
120 = 2 x 2 x 3 x 2 x 5
30
30 = 2 x 15
30 = 2 x 3 x 5
Step 2 – The common factors of both 120 and 30 are obtained and arranged in a table
as shown.
2
3
5
120
23
31
51
30
21
31
51
GCF
2
3
5
Prime
#
Step 3 - The smallest product from each column is obtained. They are then multiplied
together to obtain the Greatest Common Factor.
GCF (120, 30) = 2 x 3 x 5 = 30
Question 2
Applying GCF principle:
Prime factorization of 16 and 24
16
16 = 8 x 2
16 = 2 x 4 x 2
16 = 2 x 2 x 2 x 2
24
24 = 8 x 3
24 = 2 x 4 x 3
24 = 2 x 2 x 2 x 3
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
3
2
3
16
24
30
24
23
31
GCF
8
1
Prime
#
GCF (16, 24) = 8 x 1 = 8
Therefore, the largest tile that Maya can use is 8 inches by 8 inches.
Part 2
Question 3
Total number of rows of seats = 40
There are 20 seats in the first row and the seats are increasing progressively in the next two
rows.
Therefore:
a1 (first term of the sequence) = 20
n (number of terms in the sequence) = 40
d (difference) = 21 – 20 = 1
Using arithmetic series formula:
Sn = n/2 [ 2 a1 + (n-1) d]
S40 = 40/2 [ 2 x 20 + (40-1) x 1]
S40 =20 [ 40 + 39]
S40 = 1,580
Hence, there are 1,580 seats in all 40 rows.
Sample Answers
Maths for Computing (Unit 11)
Assignment
14
Question 4
Using the arithmetic rule:
an = a1 + (n-1) d
where: a1 = 2 km, d = ½ km, an = 8 km, n = unknown
a8 = 2 + (n-1) 0.5
8 = 2 + (n-1) 0.5
8 = 2 + 0.5n – 0.5
6.5 = 0.5n
n = 13
Therefore, at week 13, 8 km will be run.
Question 5
Using Geometric rule:
an = a1r(n-1)
Where: a1 (first term) = 100,000 rupees
n (number of terms to be determined) = 5 years
r (common ratio between terms) = 115/100 = 1.15
an = a1r(n-1)
a5 = 100,000 x 1.15(5-1)
a5 = 100,000 x 1.15(4)
a5 = 100,000 x 1.74900625
a5 = 174,900.625
Therefore, 174,900.625 rupees will be owed to the bank over a period of 5 years.
Sample Answers
Maths for Computing (Unit 11)
Assignment
15
Part 3
Question 6
Multiplicative inverse of 8 mod 11:
8x mod 11 ≡ 1
x=7
Therefore, the multiplicative inverse of 8 mod 11 = 7
The Euclidean algorithm comprises of a set of instructions for finding the greatest common
divisor of any two positive integers. It makes use of repeated use of integer division.
The Euclidean formula is a = bq + r where a and b are two positive integers, q is the quotient
and r is the remainder. It is assumed that 0 < r < b.
Repeated division is used until the remainder is zero, as follows:
a = bq1 + r1 where 0 < r1 < b
b = r1q2 + r2 where 0< r2 < r1
r 1 = r2q3 + r3 where 0 < r3 < r2
The following process continues until remainder is zero.
Therefore, gcd (a,b) = last non-zero remainder in the division process.
Hence, applying the Euclidean algorithm:
11=8(1)+3
8=3(2)+2
3=2(1)+1
2=1(2)
Re-arranging the equations above:
3 = 11-8(1)
Sample Answers
Maths for Computing (Unit 11)
Assignment 1 6
2 = 8-3(2)
1 = 3-2(1)
Gcd (8,11) = 1
If the above equations are reversed (using extended Euclidean algorithm):
1 = 3-2(1)
1 = 3-(8-3(2)) (1)
= 3-(8-(3(2))
= 3(3)-8
1 = (11-8(1)) (3)-8
= 11(3)-8(4)
= 11(3) +8(-4)
Therefore, 1 = 11(3) + 8(-4) which is 1 = 8(-4) mod 11
Multiplicative inverse would be 1=8(7) mod 11
Part 4
Question 7
Importance Of Prime numbers within the field of Computing
used in the following areas in computing.
Cryptography and Encryption
Prime numbers are used in the RSA system in cryptography to calculate public and private
keys as a means of encrypting data within messages in a computer system.
Hash codes
Prime numbers are also used in calculating hash codes in computing.
Sample Answers
Maths for Computing (Unit 11)
Assignment
17
Activity 2
Part 1
Question 1
Conditional probability is the probability of an event given that another event has occurred.
For example,
Bayes’s Theorem
Another formula that can be used to calculate conditional probabilities is the Bayes’s
theorem. For example,
Question 2
Using a Venn diagram,
Biology
Mathematics
37
13 16
34
n (s) = 100
P (B) = Probability that a student studies Biology
P (M) = Probability that a student studies Mathematics
P (B ∩ M) = Probability that a student studies both Biology and
Mathematics Therefore,
P (B) = 29/100
P (M) = 50/100
P (B ∩ M) = 13/100
Sample Answers
Maths for Computing (Unit 11)
Assignment
18
Hence using Conditional probability,
P(M|B) = P (M ∩ B)
P(B)
= 13/100
29/100
= 13/29
Question 3
If:
C represents ‘Patient has disease’
S represents ‘Screening test gives a positive result’
C’ represents ‘Patient does not have a disease’
S’ represents ‘Screening test does not give a positive result’
Then using a tree diagram:
0.8
S
CS
S’
CS’
0.05
S
C’S
0.95
S’
C’S’
C
0.01
0.99
0.2
C’
(a) Probability that a randomly selected individual does not have the disease but gives a
positive result in the screening test is:
P (C’∩ S) = 0.99 x 0.05 = 0.0495
(b) Probability that a randomly selected individual gives a positive test result is:
P(S)=P(C∩S)+P(C’∩S)
= (0.01 x 0.8) + 0.0495
Sample Answers
Maths for Computing (Unit 11)
Assignment
19
= 0.0575
(c) Probability that Nilu has the disease is:
P(C|S) = P (C ∩ S)
P(S)
= 0.008
0.0575
= 0.139
Question 4
If G = Event that the student has a graphics calculator
H = Event that the student has a computer at home
D = Event that the student drives to college everyday
This information can be displayed using a Venn diagram as follows:
G
D
4
2
1
2
H
6
G and D are mutually exclusive events hence P (G ∩ D) = 0
(a)
The probability that a student either drives or has a graphics calculator will be P(G∪D).
P(G∪D)=P(G)+P(D)
= 5/15 + 2/15
= 7/15
= 0.4667
(b) From the Venn diagram above, it can be deduced that:
P(G∩H)=1/15
P (G) = 5/15 = 1/3
P (H) = 3/15 = 1/5
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 10
Therefore, the events G and H are independent as P(G) x P(H) = P (G ∩ H)
Question 5
n (s) = 15
Sample space = {BBB, RRR, GGG, BBG, GGB, BGB, GBB, RRB, BBR, BRB, RBR, GRR,
RRG, GRG, GGR}
(a) Probability that they are all blue:
P (B1B2B3) = 6/15 * 5/14 * 4/13 = 4/91 = 0.0440
(b) Probability that two are blue and one is green:
P (B1B2G3) = 6/15 * 5/14 * 5/13 = 5/91
However, there are three arrangements of two blues and one green as
follows: {B1B2G3, B1G2B3, G1B2B3}
Therefore, probability = 3 * 5/91 = 15/91 = 0.165
(c) Probability that there is one of each color:
P (B1G2R3) = 6/15 * 5/14 * 4/13 = 4/91
However, there are six arrangements of one blue, one green and one red as
follows: {B1G2R3, B1R2G3, G1B2R3, G1R2B3, R1B2G3, R1G2B3}
Therefore, probability = 6 * 4/91 = 24/91 = 0.264
Part 2
Question 6
A random variable is a
Example
There are two types of random variables which are:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 11
Discrete random variable
Continuous random variable
Question 7
Let M = the score on the red die minus the score on the blue die.
The total possible outcomes for the two fair cubical dice would be = 36
Outcomes are tabulated as follows:
Blue die
Red
die
1
2
3
4
5
6
1
0
-1
-2
-3
-4
-5
2
1
0
-1
-2
-3
-4
3
2
1
0
-1
-2
-3
4
3
2
1
0
-1
-2
5
4
3
2
1
0
-1
6
5
4
3
2
1
0
Table 1: Activity 2 (Part 2) - Question 7 - Probability outcomes
(a) The probability distribution of M:
M
P (M = m)
-5
1
36
-4
-3
-2
2
3
4
36
36
36
-1 0
5
36
1
6
36
2
5
36
3
4
36
4
3
36
5
2
36
1
36
Table 3: Activity 2 (Part 2) - Question 7 - Probability distribution
(b) E(M) = -5 * (1/36) + -4 * (2/36) + -3 * (3/36) + -2 * (4/36) + -1 * (5/36) + 0 *
(6/36) + 1 * (5/36) + 2 * (4/36) + 3 * (3/36) + 4 * (2/36) + 5 * (1/36)
E(M) = -0.139 + -0.222 + -0.250 + -0.222 + -0.139 + 0 + 0.139 + 0.222 + 0.250 +
0.222 + 0.139
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 12
E(M) = 0
(c) Variance of M, Var (M) = ∑ M2P
M
-5
P(M = m)
M2 P
-4
1
2
36
36
0.694
-3
-2
3
4
36
0.889
-1
5
36
0.75
0
6
36
0.44
0.139
1
2
5
36
0
3
4
36
0.139
4
3
36
0.44
5
2
36
0.75
1
36
0.889
36
0.694
Table 4: Activity 2 (Part 2) - Question 7 – Variance calculation
Var (M) = 0.694 + 0.889 + 0.75 + 0.44 + 0.139 + 0 + 0.139 + 0.44 + 0.75 + 0.889
+ 0.694
= 5.824
Question 8
(a) Probability distribution of X:
x
0
10
20
P (X = x) 1/4 1/2 1/4
Table 5: Activity 2 (Part 2) - Question 8 - Probability distribution
E(X)=20*¼+10*½+0*¼=10
Var (X) = 202 * ¼ + 102* ½ + 02 * ¼ - 102 = 50
(b) E(S)=E(X)–10=10–10=0 E(T)
= ½ E(X) – 5
= ½*10–5=0
(c) Var (S) = Var (X) = 50
Var (T) = (½)2 Var (X)
= 50/4 = 12.5
(d) Susan and Thomas play a game using two 10p coins. The coins are tossed and Susan
records her score using the random variable S and Thomas uses the random variable
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 13
T. After a large number of tosses they compare their scores. The similarity will be
that each of their total scores should approximately be zero and the difference
is that Susan’s score should be more diverse than Thomas’s score.
Question 9
The probability distribution is given as:
x
1
2
3
4
P (X=x)
1/3
1/3
k
1/4
(a) Since ∑ P (X) = 1
Therefore, P (1) + P (2) + P (3) + P (4) =
1 1/3 + 1/3 + k + ¼ = 1
k = 1 - 1/4 - 1/3 - 1/3
= 1 – 0.25 – 0.33 – 0.33
= 0.09
(b) P(X≤3)=P(1)+P(2)+P(3)
= 1/3 + 1/3 + 0.09
= 0.33 + 0.33 + 0.09
= 0.75
Part 3
Question 10
Defects (x) 0
1
2
3
4
5
Batches
113
87
64
13
8
95
(a) Using the frequency distribution above, the probability distribution for X can be
constructed as follows:
Total number of batches = 95 + 113 + 87 + 64 + 13 + 8 =
380 P(x) can be calculated as follows:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 14
x
0
1
2
3
4
5
P(x)
95/380
113/380
87/380
64/380
13/380
8/380
Table 2:Activity 2 (Part 3) - Question 10 - Probability distribution
Therefore, the probability distribution for X is:
x
0
1
2
3
4
5
P(x)
0.25
0.30
0.23
0.17
0.03
0.02
(b) The mean of this probability distribution can be calculated as follows:
x
P(x)
x*P(x)
0
0.25
0*0.25 = 0
1
0.30
1*0.30 = 0.30
2
0.23
2*0.23 = 0.46
3
0.17
3*0.17 = 0.51
4
0.03
4*0.03 = 0.12
5
0.02
5*0.02 = 0.10
∑P(x) = 1
∑x*P(x) = 1.49
Therefore, mean µ = 1.49
(c) The variance and standard deviation for this probability distribution can be
calculated as follows:
x
0
P(x)
0.25
x2
0
x2*P(x)
0*0.25 = 0
1
0.30
1
1*0.30 = 0.30
2
0.23
4
4*0.23 = 0.92
3
0.17
9
9*0.17 = 1.53
4
0.03
16
16*0.03 = 0.48
5
0.02
25
25*0.02 = 0.50
∑ x2*P(x) = 3.73
∑P(x) = 1
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
15
Therefore, variance
2
= ∑ x2P(x) - µ2
= 3.73 – (1.49)2
= 3.73 – 2.12
= 1.51
Standard deviation = √ 1.51 = 1.23
Question 11
(a) Probability that the surgery is successful on exactly 2 patients. P (2) can be
calculated if the number of successful surgeries, X is represented by a binomial
distribution where:
Number of trials, n = 3
Probability of a successful surgery, p = 0.75
Probability of the surgery failing, q = 1 - p
= 1 - 0.75 = 0.25
Therefore, P (2) = nC2p2qn-2
= 3C2(0.75)2*(0.25)1
= 0.422
(b) If X is the number of successes, then the possible values of X are 0,1,2,3
(c) In order to calculate the probability distribution for X, P (0), P (1), P (2) and P (3)
need to be calculated as follows:
P (0) = nC0p0qn-0
= 3C0(0.75)0*(0.25)3
= 0.016
P (1) = nC1p1qn-1
= 3C1(0.75)1*(0.25)2
= 0.141
P (2) = nC2p2qn-2
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
16
= 3C2(0.75)2*(0.25)1
= 0.422
P (3) = nC3p3qn-3
= 3C3(0.75)3*(0.25)0
= 0.422
Therefore, the probability distribution for X is:
x
0
1
2
3
P (x)
0.016
0.141
0.422
0.422
(d) Probability distribution for X using a histogram is as shown below:
Probability- P(x)
0.5
0.4
0.3
0.2
0.1
0
x - Number of successes
0
1
2
3
Figure 1:Activity 2 (Part 3) - Question 11- Histogram
(e) Mean of X, µ = n * p
= 3 * 0.75
= 2.25
(f) The variance and standard deviation of X are as follows:
Variance, 2 = n*p*q
= 3*0.75*0.25
= 0.5625
Standard deviation,
= √n*p*q
= √0.5625 = 0.75
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
17
Question 12
Colombo City typically has rain on about 16% days in November.
(a) Assuming that the weather is independent from one day to the next, binomial
distribution can be used to calculate the probability that it will rain on exactly 5
days in November and in 15 days in November.
If the number of rainy days is X, then:
Number of days in November, n = 30
Probability of success (rainy day), p = 0.16
Probability of failure (not a rainy day), q = 1 – p = 1-0.16 = 0.84
Hence,
P (5) = nC5p5qn-5
= 30C5(0.16)5(0.84)25
= 0.191
P (15) = nC5p15qn-15
= 30C15(0.16)15(0.84)15
= 0.00001308
(b) The mean number of days with rain in November is:
µ=n*p
= 30*0.16 = 4.80
(c) The variance and standard deviation of the number of days with rain in November
are:
Variance, 2 = n*p*q
= 30*0.16*0.84
= 4.032
= √4.032 = 2.008
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 18
Question 13
From past records, a supermarket finds that 26% of people who enter the supermarket will
make a purchase. 18 people enter the supermarket during a one-hour period.
(a) Assuming that the customers’ decision to purchase are independent, binomial
distribution can be used to calculate the probability that exactly 10 customers, 18
customers and 3 customers will make a purchase.
If the number of customers who make a purchase is X, then:
Total number of customers, n = 18
Probability of successful purchase (customer makes purchase), p = 0.26
Probability of failure to purchase (customer does not purchase), q = 1 – p = 1-0.26 =
0.74
Hence,
P (10) = nC10p10qn-10
= 18C10(0.26)10(0.74)8
= 0.00555
P (18) = nC18p18qn-18
= 18C18(0.26)0(0.74)18
= 0.0000000000295
P (3) = nC3p3qn-3
= 18C3(0.26)3(0.74)15
= 0.157
(b) The expected number of customers who make a purchase are:
µ=n*p
= 18*0.26 = 4.68
(c) The variance and standard deviation of the number of customers who make a
purchase are:
Variance, 2 = n*p*q
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 19
= 18*0.26*0.74
= 3.46
Standard deviation,
= √n*p*q
= √3.46 = 1.86
Question 14
The standardized score or Z score can be calculated as:
Standardized score = Raw score – mean
Standard deviation
= 93–75
5
= 3.6
Therefore, the score is 3.6 standard deviations above the mean indicating that Shan will
be in the 3 percent outside the 3 standard deviations. Hence, it will be an outlier.
Question 15
The shelf life of a dairy product is normally distributed with a mean µ of 12 days and a
standard deviation of 3 days.
Let x = life of a dairy product.
(a) In order to calculate the percent of products that last between 9 and 15 days, the zscore of the two data points should be calculated to standardize the data. The zscore is the number of standard deviations above or below the mean value µ.
z
=
−
= 9 – 12 = -1 (value one standard deviation below the mean)
3
= 15 – 12 = 1 (value one standard deviation above the
mean) 3

Approximately 68% of the distribution will be within one standard deviation
of the mean. Here, the z-scores will be +1 and -1.
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 20

Approximately 95% of the distribution will be within two standard
deviations of the mean. Here, the z-scores will be +2 and -2.


Approximately 99.7% of the distribution will be within three standard
deviations of the mean. Here the z-scores will be +3 and -3.
The graph below shows application of the bell curve according to the empirical rule:
Figure 2: Activity 2 (Part 3) - Question 15 - Graph 1
Therefore, the percent of the products that last between 9 and 15 days will be
68% according to the empirical rule. This is shown graphically as follows:
Figure 3: Activity 2 (Part 3) - Question 15(a) - Graph 2
(b) In order to calculate the percent of products that last between 12 and 15 days, the zscore of the two data points should be calculated:
z
=
−
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 21
= 12–12=0
3
= 15 – 12 = 1 (value one standard deviation above the
mean) 3
Hence the z-score will be in the range 0 ≤ z ≤ 1
Therefore, the percent of the products that last between 12 and 15 days will be
34% as shown in the figure below:
Figure 4: Activity 2 (Part 3) - Question 15(b) - Graph
(c) In order to calculate the percent of products that last 6 days or less, the z-score of
the should be calculated:
z = 6 – 12 = -2 (value two standard deviations below the mean)
3
Hence the z-score will be z ≤ -2
Therefore, the percent of the products that last 6 days or less will
be 2.35% + 0.15% = 2.5%
This is shown graphically in the figure below:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 22
Figure 5: Activity 2 (Part 3) - Question 15(c) - Graph
(d) In order to calculate the percent of products that last 15 or more days, the z-score of
the should be calculated:
z = 15 – 12 = 1 (value one standard deviation above the mean)
3
Hence the z-score will be z ≥ 1
Therefore, the percent of the products that last 15 or more days will
be 13.5% + 2.35% + 0.15% = 16.0%
This is shown graphically in the figure below:
Figure 6: Activity 2 (Part 3) - Question 15(d) – Graph
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 23
Question 16
Using binomial distribution, let the number of who pass the drivers test is X, then:
Total number of drivers, n = 120
Probability of passing the test, p = 0.78
Probability of failing the test, q = 1 – p = 1-0.78 = 0.22
Required = P (X > 99)
Hence using the binomial formula, the following will need to be calculated:
P(X=100) + P(X=101) +…. + P(X=120)
However, this is difficult to calculate as n is too large and p is not close to 0 or 1. Therefore,
it would be advisable to use the normal distribution.
Using the normal distribution, the mean and standard deviation can be calculated as follows:
Mean, µ = n * p
= 120*0.78 = 93.6
Standard deviation, ,
= √n*p*q
= √20.592 = 4.538
Part 4
Question 17
Use of probability theory in hashing and load balancing
Activity 3
Part 1
Question 1
Using the distance formula, the distance between two points (x, y) and (h, k) on a circle given
as:
D = √ (x-h)2+(y-k)2
(x, y)
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
24
r
(h, k)
Where D is the radius of a circle, hence r = √ (x-h)2+(y-k)2
Therefore, if the center of the circle is (2, -7) and a point on the circle is (5, 6) as shown in the
figure below, the radius of the circle can be obtained as follows:
(x, y)
r
(2, -7)
r2 = (5 – 2 )2 + (6 – (-7))2
r2 = 32 + 132
r = √ 9 + 169 = √ 178 = 13.34
If the original equation of the circle is expanded, it would be as follows:
r = √ (x-h)2+(y-k)2
r2 = x2 – 2xh + h2 + y2 – 2yk + k2
If r and (h, k) are substituted into the equation, the formula of the circle can be obtained as
follows:
r2 = x2 – 2xh + h2 + y2 – 2yk + k2
13.342 = x2 – 2x*(2) + 22 + y2 – 2y*(-7) + (-7)2
178 = x2 – 4x + 4 + y2 + 14y + 49
178 = x2 – 4x + y2 + 14y + 53
178 - 53 = x2 – 4x + y2 + 14y
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 25
Therefore, formula of the circle with center (2, -7) and a point on the circle (5 ,6) will be:
x2 – 4x + y2 + 14y = 125
Question 2
{(x, y, z}, z = 3} represents that set of all points in R3 whose z co-ordinate is 3. This is the
horizontal plane that is parallel to the xy plane and it is 3 units above as shown in the
figure below.
z
3
0
x
y
Figure 7: Activity 3 (Part 1) - Question 2 – z = 3, a plane in R3
y = 5 represents the set of all points in R3 where the y co-ordinate is 5. This is the vertical
plane that is parallel to the xz plane and it is 5 units to the right of this plane as shown in
the figure below.
z
0
x
5
y
Sample Answers
Maths for Computing (Unit 11)
Assignment 1 26
Figure 8: Activity 3 (Part 1) - Question 2 – y = 5, a plane in R3
Question 3
The equation of a sphere is given as:
r2 = (x – a)2 + (y – b)2 + (z – c)2
Where the center is (a, b, c) and r is the radius,
Hence, when the center is (h, k, l) and r is the radius, the equation of the sphere will be:
r2 = (x – h)2 + (y – k)2 + (z – l)2
Question 4
Using the equation of a sphere r2 = (x – a)2 + (y – b)2 + (z – c)2
x2+y2+z2+4x-6y+2z+6 = 0
x2+y2+z2+4x-6y+2z = -6
x2+4x+4+ y2 - 6y+ 9 + z2 +2z + 1 = -6 + 4 + 9
+1 (x2+4x+4) + (y2 - 6y+ 9) + (z2 +2z + 1) = 8
(x + 2) 2 + (y – 3) 2 + (z + 1) 2 = 8
Therefore, center of the sphere (x, y, z) = (-2, 3, -1)
Radius of the sphere, r = √8 = 2√2
Part 2
Question 5
The x and y values in the equations 3y = 2x - 5 and 2y = 2x + 7 can be graphically evaluated
as follows:
Letting x be zero in the first equation 3y = 2x – 5
gives 3y = 2 *0 – 5
y = -5/3
y = -1.67
Hence, one co-ordinate would be (0, -1.67)
If y = 0 in the first equation:
3y= 2x – 5
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
27
3 * 0 = 2x – 5
-2x = -5
x = -5/-2
x = 2.25
Hence, another co-ordinate would be (2.25,0)
x
y
0
-1.67
2.250
For the second equation 2y = 2x + 7:
If x = 0,
2y = 2 * 0 + 7
2y = 0+7
y = 7/2
y = 3.5
Hence, one co-ordinate would be (0, 3.5)
If y = 0,
2*0 = 2x + 7
0 = 2x + 7
x = -7/2
x = -3.5
Hence, another co-ordinate would be (-3.5,0)
x
y
0
3.5
-3.5 0
When the co-ordinates of both equations 3y = 2x - 5 and 2y = 2x + 7 are plotted on the xy
Cartesian co-ordinate system, they are graphically represented as shown in the figure below:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 28
Figure 9: Activity 3 (Part 2) - Question 5 – Graph 1
The intersection point between the two lines 2y = 2x + 7 and 3y = 2x – 5 can be determined
as follows:
Therefore, the two lines intersect at the point ( -15.5, -12)
This can be illustrated graphically as shown in the figure below:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 29
Figure 10:Activity 3 (Part 2) - Question 5 – Graph 2
Question 6
A tetrahedron’s volume is 1/6 of the volume of a parallelepiped which is formed by vectors a,
b, c.
Volume of a parallelepiped (scalar triple product) = | (a * b) * c |
Hence, the volume of a tetrahedron with a, b. c as co-terminal edges is obtained as:
Volume = 1/3 * (Area of base) * (height)
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 30
Where height = ||c|| * |cos θ|
Thus,
Volume = 1/3 * (½* ||a*b||) * ||c||* |cos θ|
= 1/6 * ||a*b|| *|| c ||* |cos θ| Since | (a
* b) * c | = ||a*b|| *|| c ||* |cos θ|, Volume of the
tetrahedron = 1/6 * | (a * b) * c |
= 1/6 * [(2i + 3j+k) * (4i-2j-3k) * (1i+4j-2k)]
a = (2i + 3j + k)
b = (4i – 2j -3k)
c = (1i + 4j - 2k)
2
3
1
Volume = 1/6 |4
−2
−3 |
1
4
−2
2
3
Determinant of |4
−2
1
4
1
−3 | = 2 [(-2)(-2)-4(-3)]-3[4(-2)-1(-3)]+1[(4)(4)-1(2)]
−2
= 65
Therefore, volume of tetrahedron = 1/6 * 65 = 65/6
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 31
Activity 4
Part 1
Question 1
The slope of the tangent is the derivative of the function f. Hence, f’(x) = 4x + 1 and f(x) is
the indefinite integral f(x) = ∫ f’(x)
Therefore, to find the function whose tangent has slope 4x + 1 for each value of x and whose
graph passes through the point (1,2) the integral of the function f(x) will be:
f(x) = ∫ f ’(x)
f(x) = ∫ 4 + 1
= 2x2 + x + c
Using the fact that the graph of f passes through the point (1,2), to find c, x = 1 and f (1) = 2
is substituted into the equation, as the graph passes through the point (1,2)
f(x) = 2x2 + x + c
2 = 2*(1)2 + 1 + c
2=3+c
c = -1
Therefore, the desired function is f(x) = 2x2 + x – 1
Question 2
The slope of the tangent is the derivative of the function f. Hence, f’(x) = 3x2 + 6x – 2 and
f(x) is the indefinite integral f(x) = ∫ f’(x)
Therefore, to find the function whose tangent has slope 3x2 + 6x - 2 for each value of x and
whose graph passes through the point (0,6) the integral of the function f(x) will be:
f(x) = ∫ f ’(x)
f(x) = ∫ 3x2 + 6x – 2
= x3+ 3x2 – 2x + c
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 32
Using the fact that the graph of f passes through the point (0,6), to find c, x = 0 and f (0) = 6
is substituted into the equation as follows:
f(x) = x3+ 3x2 – 2x + c
f (0) = 03+3*02 – 2 *0 + c
6=0+c
c=6
Therefore, the desired function is f(x) = x3 + 3x2 – 2x + 6
Part 2
Question 3
It is estimated that t years from now the population of a certain lakeside community will be
changing at the rate of 0.6t2 + 0.2t + 0.5 thousand people per year. Environmentalists have
found that the level of pollution in the lakeside increases at the rate of approximately 5 units
per 1000 people.
Let P(t) denote the population of the community t years from now.
Then the rate of change of the population with respect to time is the derivative
= P’ (t) = 0. 6t2 + 0.2t + 0.5
However, the population function P(t) is an anti-derivative of 0. 6t2 + 0.2t + 0.5 which is
given by:
P(t) = ∫ P’(t) dt
= ∫ (0. 6t2 + 0.2t + 0.5) dt
= 0.2t3 + 0.1t2 + 0.5t + C
Where C is a constant
During the subsequent 2 years, the population will grow by:
P (2) – P (0) = 0.2 * (2)3+ 0.1 * (2)2 + 0.5*2 + C –
C =1.6+0.4+1
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 33
= 3 thousand people
Therefore, the pollution in the lake will increase by 5 units * 3 = 15 units
Question 4
If speed of the moving object after t minutes is v(t) = 1 + 4t + 3t2 meters per
minute, let s(t) be the displacement of the car after t minutes, Then
v(t) = ds/dt = s’(t)
s(t) = ∫ ( )
= ∫ ( 1 + 4t + 3t2 )
= t + 2t2 +t3 + c
During the third minute, the object travels,
s (3) – s (2) = 3 + 2 * (3) 2 + (3)3 + c – 2 + 2*(2) 2 – (2)3
= 30 meters
Therefore, the object travels 30 meters during the third minute.
Part 3
Question 5
The graph of f(x) = x – 3x2/3 can be sketched by first identifying all the intercepts where f(x) > 0 and f(x) < 0 . This is where f(x) = 0.
f(x) = x – 3x2/3
0 = x – 3x2/3
x2/3 (x1/3 – 3) = 0
Hence when f(x) = 0 (y co-ordinate will be 0),
x2/3 = 0
x=0
x1/3 = 3
x = ∛3 = 27
Therefore, the x-intercepts are (0,0) and (27,0)
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 34
Using the power rule, the first derivative of f(x) is calculated as:
f ’(x) =
(x - 3x2/3)
= 1 – 2x-1/3
= x1/3 – 2
x1/3
The intercepts are identified where f ‘(x) = 0
f ’(x) = x1/3 – 2
x1/3
0 = x1/3 – 2
x1/3
x1/3 = 2
x = 23 = 8
Also, when f ‘(x) is undefined, x = 0
As f(x) is continuous without any vertical asymptotes, f(x) = x – 3x2/3 should
have A local maximum at (0, f (0)) which is (0,0) and a local minimum at (8, f
(8)) which is
(8, (8-3*(82/3) = (8, -4)
To check the concavity of the graph, the second derivative of f(x) which is f “(x) is calculated
as follows:
f “(x) =
(1 – 2x-1/3)
= 23 x-4/3
=
2
3 4/3
When f ‘ (0) does not exist, x = 0 but x ≠ 0 when f ’’(x) > 0, so the curve is a concave up
(slope increases) on any interval that does not contain x = 0.
Hence the graph can be sketched as shown in the figure below:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 35
Figure 11:Activity 4 - Question 5 - Graph
Also, it can be noted that this function does not have any vertical or horizontal asymptotes, however
through concavity it can be seen that as x → ∞, f(x) → ∞ and while x → −∞, f(x) → −∞.
Question 6
The graph of f(x) = 3x4 – 6x3 + 3x2 can be drawn by first calculating the first derivative of
the function to locate the stationary points on the graph as follows: f(x) = 3x4 – 6x3 + 3x2
Factorization of f(x) = 3x2(x2-1) (x2-1)
f ‘(x) = 12x3 – 18x2 + 6x
According to Fermat’s theorem, if f(x) has a local extremum at x = a and f is differentiable at
a, then f ‘(a) = 0
So when f ‘(x) = 0,
0 = 12x3 – 18x2 + 6x
By factoring: 0 = 6x (2x2 – 3x + 1)
x=0
Using the quadratic formula to solve (2x2 – 3x + 1) results in:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 36
− ±√
−4
2
= −(−3) ± √(−3)2 − (4 ∗ 2 ∗ 1)
2∗2
=
2
3±√9−8
=
4
3±√1
=
4
3+
√1
=
= 1 and
4
3−
√1
=
=½
4
Therefore, the stationary points on the graph will be at x = 0, x = 1 and x = ½
1 3
The maximum or minimum points lie at (1,0) , ( , ) , (0,0)
1
2 6
Using the first derivative test:
f ‘(x)
x (points neighboring the
Result
stationary point)
x = -1
f ‘(-1) = 12 * (-1)3 – 18*(-1)2
+6*(-1)
= -36
x = 0.25
f ‘(0.25) = 12 * (0.25)3 –
18*(0.25) + 6 * (0.25)
= 0.5625
18*(0.75) + 6 * (0.75)
= - 0.5625
Sample Answers
0.5625 > 0
Function is increasing
(curving upwards)
f ‘(0.75) = 12 * (0.75)3 –
2
x=2
Function is decreasing
(curving downwards)
2
x = 0.75
-36<0
f ‘(2) = 12 * (2)3 – 18*(2)2 +
6*(2)
-0.5625 < 0
Function is decreasing
(curving downwards)
36>0
Function is increasing
Maths for Computing (Unit 11)
Assignment 1
37
= 36
(curving upwards)
However, if the second derivative test is applied:
d2x
dx2 = 36x2 − 36x + 6
When x = 1, y = 6
6 > 0 , It is a positive. Thus, the local minimum is at (1,0)
If x = ½ y = -3
−3 < 0 , It is a negative. Thus, the local maximum is at (12 , 163)
If x = 0 y = 6
6 > 0 , It is a positive. Thus, the local minimum is at (0, 0)
In order to identify the points of inflection, the second derivative is solved for f ’’(x) = 0 as
follows:
d2x
2
dx2 = 36x − 36x + 6
−b ± √b2 − 4ac
2a
−(−36) ± √(−36)2 − 4 ∗ 36 ∗ 6
2∗36
−(−36) ± √1296 − 864
72
36 ± √432
72
3 ± 20.78
72
x = 0.788 or x = 0.211
Therefore, the points of inflection are (0.211, 0.577) and (0.788, -0.577)
Hence, the graph is drawn as shown in the figure below:
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 38
3 4−6 3+3 2
0.2
0.18
0.16
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
1.2
1.4
Figure 12:Activity 4 (Part 3) - Question 6 – Graph
Part 4
Question 7
In order to find the positions of any local minima or maxima for the function f(x) = cos 2 ,
0.1≤ x ≤ 6, the first derivative of the function is calculated in order to locate stationary
points, as follows:
f(x) = cos 2
ⅆxⅆf = - 2 sin 2x
Therefore, the stationary points are at values of x in the range 0.1≤ x ≤ 6 for which sin 2x =
0 which will be at:
2x = π or 2x = 2π or 2x = 3π
The stationary points are at x = 2 , x = π and x = 3 2
The second derivative of the function is then calculated as:
ⅆ2
f
ⅆ = - 4 cos 2
x2
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
39
The second derivative is evaluated at each stationary point to locate the local minima and
maxima as follows:
ⅆ2
f
|
= - 4 cos = 4 > 0
ⅆx π
x2 =
2
Therefore , x = 2 indicates a local minimum.
ⅆ2f
= - 4 cos 2 = - 4 < 0
ⅆx2|x=π
Therefore, x = π indicates a local maximum.
ⅆ2f
= - 4 cos 3 = 4 > 0
ⅆx2|x=3π
Therefore, x =
indicates a local minimum.
Question 8
To determine the local maxima and/ or minima of the function y = 4 − 1/3 3 , the positions of the stationary points should be calculated by
finding the first derivative of the function as follows:
f(x) =
ⅆx
ⅆf
3
4 − 1/3 3
2
= 4x – x
= x2 (4x – 1)
Therefore the stationary points are when ⅆxⅆf = 0, which are at x = 0 or when x = ¼
The second derivative of the function is then calculated as:
ⅆ2 f
= 12x2 – 2x
ⅆx2
The second derivative is evaluated at each stationary point to locate the local minima and
maxima as follows:
ⅆ2f
ⅆx2|x=0=
ⅆ2f
ⅆx2
0 This is inconclusive
|
=
1
x=
12
16
–
1
=
2
1
>0
4
4
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
40
Therefore , x = indicates a local minimum.
Question 9
By further differentiation,
y =12x2 – 2x
By taking the derivative of the slope:
f ‘(y) = 24x – 2
By factoring 2(12x − 1) = 0
By dividing all by 4(
12x −
12
1
)=0
12
Therefore, the stationary points when y = 0 are at x = 0 and x = 121
Using the first derivative test:
f ‘(x)
x (points neighboring the
Result
stationary point)
f ‘(1) = 24 * (1) – 2
x = 1/12
= 22
22>0
Function is
increasing(curving upwards)
f ‘(-1) = 24 * (-1) – 2
= -26
-26<0
Function is
decreasing(curving
downwards)
x=
1
indicates a local minimum.
12
Using the second derivative test:
ⅆ2 x
= 24
ⅆx2
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 41
Using the second derivative test, the second derivative at x = 121 when f(x) =0 is 24 which is a positive value. According to this test, if the second derivative is positive, then
this is a local minimum. Hence, x = 121 is a local minimum.
Local minimum at (
,−
)
y = x2 + 4x + 1
By taking the derivative of the slope:
f ‘(y) = 2x + 4
By factoring
2(x + 2) = 0
Therefore, the stationary points when y = 0 are at x = 0 and x = -2.
Using the first derivative test:
f ‘(x)
x (points neighboring the
Result
stationary point)
x=-2
f ‘(1) = 2 * (1) + 4
=6
6>0
Function is
increasing(curving upwards)
f ‘(-3) = 2 * (-3) + 4
= -2
-2<0
Function is
decreasing(curving
downwards)
x = - 2 indicates a local minimum.
Using the second derivative test:
ⅆ2x
ⅆx2
=2
Using the second derivative test, the second derivative at x = -2 when f(x) =0 is 2 which is a
positive value. According to this test, if the second derivative is positive, then this is a local
minimum. Therefore, x = -2 is a local minimum.
Local minimum (− , − )
Sample Answers
Maths for Computing (Unit 11)
Assignment 1
42
y = 12x – 2x2
By taking the derivative of the slope:
f ‘(y) = 12x – 4x
By factoring 4(3 − x) = 0
Therefore, the stationary points when y = 0 are at x = 0 and x = 3.
Using the first derivative test:
f ‘(x)
x (points neighboring the
Result
stationary point)
f ‘(-1) = 12 – 2 * (-1)2
= 10
x=3
10>0
Function is
increasing(curving upwards)
f ‘(3) = 12 – 2 * (3)2
=-6
-6<0
Function is
decreasing(curving
downwards)
x = 3 indicates a local maximum.
Using the second derivative test:
ⅆ2 x
= −4
ⅆx2
Using the second derivative test, the second derivative at x = 3 when f(x) =0 is - 4 which is a
negative value. According to this test, if the second derivative is negative, then this is a local
maximum. Therefore, x = 3 is a local maximum.
Local maximum ( , − )
y = -3x2 + 3x + 1
By taking the derivative of the slope:
f ‘(y) = -6x + 3
3(−2x + 1)
By factoring
Sample Answers
=0
Maths for Computing (Unit 11)
Assignment
1 43
By dividing all by 4 (
−2x
+
−2
1)
=0
−2
Therefore, the stationary points when y = 0 are at x = 0 and x = ½
Using the first derivative test:
f ‘(x)
x (points neighboring the
Result
stationary point)
x=½
f ‘(-1) = -6 * (-1) + 3
=9
9>0
Function is
increasing(curving upwards)
f ‘(1) = -6 * (1) + 3
= -3
-3<0
Function is
decreasing(curving
downwards)
x = ½ indicates a local maximum.
Using the second derivative test:
ⅆ2x
ⅆx2
= −6
Using the second derivative test, the second derivative at x = ½ when f(x) =0 is - 6 which is a
negative value. According to this test, if the second derivative is negative, then this is a local
maximum. Therefore, x = ½ is a local maximum.
Local maximum at ( , )
Sample Answers
Maths for Computing (Unit 11)
Assignment
1 44