Mathematics for Computing Coursework

This document presents solutions to a mathematics coursework assignment. The assignment covers several topics including: geometric properties of quadrilaterals (parallelograms), solving systems of linear equations using Cramer's rule, probability calculations involving Poisson and hypergeometric distributions, regression analysis (finding the equation of a regression line and calculating the correlation coefficient), and descriptive statistics (calculating mean and standard deviation from a frequency table and histogram). Each question is addressed with detailed steps and explanations, providing a comprehensive guide for students. The solutions demonstrate the application of various mathematical concepts and techniques to solve real-world problems.

MATHEMATICS FOR COMPUTING COURSEWORK
Question 1
(i) As per the information given in the question, one point A of the quadrilateral has the coordinates (1,
1). Now we have to take the other three points whose x and y coordinates are greater than 1. So let us
plot the points in such a manner that the quadrilateral looks like a parallelogram.
The points are:
B (4, 2)
C (5, 5)
D (2, 4)
The figure is shown below:

Calculating the slope of the line AB:
m1 = yb – ya / xb – xa
= 2 – 1 / 4 – 1
= 1/3.
Calculating the slope of the line BC:
m2 = yc – yb / xc – xb
= 5 – 2 / 5 – 4
= 3/1
= 3.
Calculating the slope of the line CD:
m3 = yd – yc / xd – xc
= 4 – 5 / 2 – 5
= 1/3.
Calculating the slope of the line DA:
m4 = yd – ya / xd – xa
= 4 – 1 / 2 – 1
= 3.
We see that the opposite lines have the same slope thus the opposite sides are parallel.
We can also calculate the mid points of the diagonals Ac and BD which comes out to be M (3, 3) hence,
we see that AM = MC = 2.828 and BM = MD = 2.828.
In this case, the distances i.e. AM = BM but not always as it depends on selection of points.
Thus, the diagonals bisect each other.
Therefore, we can say that the Quadrilateral ABCD is a Parallelogram.

(ii) Scaling down the matrix M by 1/3 means that all the coordinates of the points will be scaled down by
1/3.
Therefore, the new points are:
A’ (1/3, 1/3)
B’ (4/3, 2/3)
C’ (5/3, 5/3)
D’ (2/3, 4/3)
Now rotate the points about origin by an angle of 55o clockwise one by one using the transformation:
x ' '
y ' ' = [ cos 55 sin55
−sin 55 cos 55 ] x '
y '
For A’’:
x ' '
y ' ' = [ cos 55 sin55
−sin 55 cos 55 ]1/3
1/3
= 0.46
−0.08
For B’’:
x ' '
y ' ' = [ cos 55 sin55
−sin 55 cos 55 ] 4 /3
2/ 3
= 1.3
−0.7
For C’’:
x ' '
y ' ' = [ cos 55 sin55
−sin 55 cos 55 ] 5/ 3
5/ 3
= 2.3
−0.4
For D’’:
x ' '
y ' ' = [ cos 55 sin55
−sin 55 cos 55 ] 2/ 3
4 /3

= 1.46
0.18
Thus, the new points after rotation by 55o are:
A” (0.46, -0.08)
B” (1.3, -0.7)
C” (2.3, -0.4)
D” (1.46, 0.18)
Reflection of the parallelogram about y axis gives us the new coordinates of the points as:
A”’ (-0.46, -0.08)
B”’ (-1.3, -0.7)

C”’ (-2.3, -0.4)
D”’ (-1.46, 0.18)
Question 2
The determinant of coefficient is:
Δ =
8 −1 9
7 5 1
−2 −5 −3
= {8(5*-3 – 1*-5) + (7*-3 – 1*-2) + 9(7*-5 – 5*-2)}
= {-80-19-225}
= -324.

Applying Cramer’s rule:
x = 1
Δ
55 −1 9
49 5 1
0 −5 −3
= 1
−324 (55(5*-3 – 1*-5) – 49(-1*-3+5*9))
= −550−2352
−324
= −2902
−324
= 8.96.
y = 1
−324
8 55 9
7 49 1
−2 0 −3
= 1
−324 (-55(7*-3-1*-2)+49(8*-3-9*-2))
= 1045−294
−324
= 751
−324
= - 2.32.
z = 1
−324
8 −1 55
7 5 49
−2 −5 0
= 1
−324 (55(7*-5-5*-2) – 49(8*-5-(-2*-1))
= −1375+2058
−324
= 683
−324

= -2.11.
Hence, the values are:
x = 8.96
y = -2.32
z = -2.11
Question 3
(i)
Dice 1 (Six Sided) Dice 2 (Eight Sided) Total
1 1 2
1 3 4
1 3 4
1 4 5
1 5 6
1 5 6
1 6 7

1 7 8
1 1 2
1 3 4
1 3 4
1 4 5
1 5 6
1 5 6
1 6 7
1 7 8
2 1 3
2 3 5
2 3 5
2 4 6
2 5 7
2 5 7
2 6 8
2 7 9
3 1 4
3 3 6
3 3 6
3 4 7
3 5 8
3 5 8
3 6 9
3 7 10
4 1 5
4 3 7
4 3 7
4 4 8
4 5 9
4 5 9
4 6 10
4 7 11
4 1 5
4 3 7
4 3 7
4 4 8
4 5 9
4 5 9
4 6 10
4 7 11
(ii)
The most likely score as that can be seen from the table is 7 which occur 9 times.

The probability of the score 7 = Number of times 7 occurs/Total number of outcomes
= 9/6*8
= 9/48.
(iii)
The number of times both dice show (1, 1) is 2.
The number of times both dice show (3, 3) is 2.
The number of times both dice show (4, 4) is 2.
Therefore, total number of times both dice will show same number is 2+2+2 = 6.
Hence, the probability of both dice showing the same number
= total number of times both dice will show same number/ Total number of outcomes
= 6/6*8
=6/48
=1/8.
Question 4
(i) As we can see from the sample of 70 algorithms, the least number is 0.37 and the greatest number is
20.89.
Therefore, the first class interval should start from 0 and the last class interval should end with 21.
It is given that we have to present the frequency table using five equally sized classes.
The frequency table is as follows:
Class interval Frequency Percent Cumulative Percent

0 – 4.2 10 14.28 14.28
4.2 – 8.4 20 28.56 42.84
8.4 – 12.6 19 27.15 69.99
12.6 – 16.8 12 17.15 87.14
16.8 – 21.0 9 12.86 100.0
Total 70 100.0
(ii)
The histogram:
0-4.2 4.2-8.4 8.4-12.6 12.6-16.8 16.8-21.0 21.0-25.2
0
5
10
15
20
25
Histogram
Run times (in minutes)
Frequency
(iii) The frequency table:
Class interval Frequency(f) Mid-point (Xm) (f)*(Xm)
0 – 4.2 10 2.4 24
4.2 – 8.4 20 6.3 126
8.4 – 12.6 19 10.5 199.5
12.6 – 16.8 12 15.7 188.4
16.8 – 21.0 9 18.9 170.1
Total 70 708

Estimated sample of the mean
X̅ = ∑(f)(Xm) / ∑f
= 708 / 70
= 10.11
(iv) From the previous calculation, X̅ = 10.11
Class
interval
Frequency(f) Mid-point
(Xm)
(f)*(Xm) Xm - X̅ f*(Xm-X̅ )2
0 – 4.2 10 2.4 24 -7.71 594.44
4.2 – 8.4 20 6.3 126 -3.81 290.32
8.4 – 12.6 19 10.5 199.5 0.39 2.9
12.6 – 16.8 12 15.7 188.4 5.59 374.98
16.8 – 21.0 9 18.9 170.1 8.79 695.38
Total 70 708 6258.42
Estimate of the sample standard deviation
sd = √ f∗(X m−X̅ ) 2
n−1
where n = ∑f
sd = √ 6258.42 / 69
= 9.52
Question 5
(i)
Mean of the time spent practicing (x̅ ) = (∑x) / n = 88/9 = 10.67
Mean of the time spent practicing (y̅ ) = (∑y) / n = 267/9 = 29.67
Time spent
practicing
Number of
errors
(xi-x̅ ) (xi-x̅ )2 (yi-y̅ ) (yi-y̅ )2 (xi-x̅ ) (yi-y̅ )

(x) (y)
3 39 -7.67 58.83 9.33 87.05 -71.56
6 42 -4.67 21.81 12.33 152.03 -57.58
9 36 -1.67 2.79 6.33 40.07 -10.57
8 35 -2.67 7.13 5.33 28.41 -14.23
10 36 -0.67 0.45 6.33 40.07 -4.24
12 25 1.33 1.77 -4.67 21.81 -6.21
14 21 3.33 11.11 -8.67 75.17 -28.87
16 18 5.33 28.41 -11.67 136.19 -62.2
18 15 7.33 53.73 -14.67 215.21 -107.53
∑x = 96 ∑y = 267 ∑ = 186 ∑ = 796 ∑ = 363
This gives us the coefficient of correlation as
r = ∑ (xi-x̅ ) (yi-y̅ ) /[ √∑(xi-x̅ )2 . √∑(yi-y̅ )2]
= 363 / [√186. √796]
= 0.94
(ii)
Time spent practicing
(x)
Number of errors
(y)
x*y x2
3 39 117 9
6 42 252 36
9 36 324 81
8 35 280 64
10 36 360 100
12 25 300 144
14 21 294 196