Mathematical Modelling & Analysis II (ENGF0004) Coursework 4 - UCL

This document presents the complete solution to Coursework 4 for the Mathematical Modelling & Analysis II (ENGF0004) module at University College London (UCL). The assignment covers topics in decision statistics, data analysis, and discrete statistics. The solution begins with an analysis of data from electronic and electrical engineering and mechanical engineering graduates, estimating proportions and standard errors, and performing hypothesis testing to compare the competence of the two groups. It then proceeds to solve probability problems involving conditional probability, the calculation of probabilities related to dice rolls, and the application of Bayes' theorem. Finally, the solution explores Poisson distribution and determines the probabilities for different events, and the assignment concludes with a discussion on how to calculate the mean from the variance given the mean is not provided. The solution provides detailed workings and explanations for each step, demonstrating a strong understanding of statistical concepts and their application to real-world scenarios.

1)
a)
Number of students from Electronic and Electrical Engineering scored more than 70 % is = 37
Total students from Electronic and Electrical Engineering = n1 = 114
Proportion = p1= 37/114 = 0.325
σ 1= √ p1 ( 1− p1 )
n1
= √ 0.325 ( 1−0.325 )
114 =0.0438
b)
Number of students from Mechanical Engineering scored more than 70 % is = 48
Total students from Mechanical Engineering = n2 = 131
Proportion = p2 = 48/131 = 0.366
σ 2= √ p2 ( 1− p2 )
n2
= √ 0.3 66 (1−0.3 66 )
131 =0.04 20
c)
Pooled sample proportion
p= p1 × n1 + p2 × n2
n1 +n2
p=0.347
σ =
√ p ( 1− p )
1
1
n1
+ 1
n2
=0.0609
Hypothesis is
H0 : p1= p2
H1 : p1 ≠ p2
z= p1− p2
σ =−0.673
Cumulative probability: P (Z <-0.673)
= 0.025
Since the P value is less than 5 % then we reject the null hypothesis and we can infer that

Electronic and Electrical Engineering applicants does not have the same level of competence in
mathematics as the Mechanical Engineering applicants
2)
a)
i) Therefore, the probability that the first die shows 4 given that the sum is 7 is
P (A|B) = P (A ∩ B)/ P (B) = (1/36)/ (6/36) = 1/6.
ii)
Two dice are less than 4 and one is 4 or above 4 = (3/6)*(3/6)*(3/6) = 1/8
All the three dice are less than 4 = (3/6)*(3/6)*(3/6) = 1/8
Therefore total probability = (1/8) + (1/8) = (1/4)
b)
P ( A |H ) = P( A ∩ H )
P ( H) = P ( H | A ) P ( A )
P ( H | A ) P ( A ) +P ( H |B ) P( B)
P ( H |A ) P ( A )=0.6 ×0.5=0.3
P ( H |B ) P ( B )=0. 5 ×0.5=0.25
P ( A |H )= 0.3
0.3+0.25 =0.545
c)
P ( Defective| A ) =0.01
P ( Defective|B ) =0.02
P ( A )=0.6
P ( B )=0.4
P ¿
¿ 0.01× 0.6
0.01× 0.6+0.02 ×0.4 =0.4285
d)
Since the mean is not given we cannot calculate the mean of a normal distribution from Variance. So
we are going to proceed with the Poisson distribution. We cannot apply Binomial also as the number
of trials are not known initially so we cannot calculate the mean also.
i)

f ( x )= λx e−λ
x !
Now,
λ=4
Therefore,
Putting this in to the f(x)
f ( x )= 4x e−4
x !
Maximum occurs at,
x=3∧4
Ii)
P ( X ≥1 ) =1−P ( X=0 )=0.981