title: "UAS 23-24 FAD1015 Mathematics III — Answering Guide" exam: "UAS 23-24 FAD1015 Mathematics III" tags:

  • answering-guide
  • course/FAD1015
  • uas-23-24 status/seedling

UAS 23-24 FAD1015 Mathematics III — Answering Guide

Total marks: 80 | Part A: Q1 (12 marks, Q2 missing) | Part B: Q3–6 (56 marks)

Note: Question 2 was on pages 3–4 of original PDF, which were lost during transcription.

PART A

Question 1 — T/F, Exponential Distribution, Matrix Types

Marks: 12

(a) True or False (3 × 1 mark)

Statement Answer Reason
(i) $X,Y$ independent $\Rightarrow P(X|Y) = P(Y)$ FALSE Independence: $P(X|Y) = P(X)$, not $P(Y)$
(ii) Continuous $X$: $P(a \le X \le b) = P(a < X < b)$ TRUE For continuous RVs, $P(X=c)=0$, so $\le$ and $<$ are equivalent
(iii) Distance from home to school is discrete FALSE Distance is continuous (any non-negative real)

See FAD1015 Week 4 — Discrete Random Variables (PDF & CDF) and FAD1015 Week 6 — Continuous Random Variables.

(b) Gift shopping time: $X \sim \text{Exp}(\lambda)$

Mean $= 10$ minutes.

(i) $X \sim \text{Exponential}(\lambda = \frac{1}{10})$, with PDF $f(x) = \frac{1}{10}e^{-x/10}$ for $x \ge 0$.

(ii) Parameter $\lambda = \frac{1}{10}$ (rate). It represents the rate at which shopping sessions end — 0.1 sessions per minute. The mean is $1/\lambda = 10$ minutes.

See FAD1015 L17-L18 — Uniform & Exponential Distributions + R Intro and Exponential Distribution.

(c) Matrix examples (3 × 2 marks)

(i) Upper triangular: All entries below main diagonal are zero.

$$U = \begin{pmatrix} 2 & 1 & 5 \ 0 & 3 & -1 \ 0 & 0 & 4 \end{pmatrix}$$

(ii) Symmetric: $A = A^T$

$$S = \begin{pmatrix} 1 & 2 & 0 \ 2 & 3 & -1 \ 0 & -1 & 4 \end{pmatrix}$$

(iii) Diagonal: Non-zero only on main diagonal.

$$D = \begin{pmatrix} 2 & 0 & 0 \ 0 & -3 & 0 \ 0 & 0 & 5 \end{pmatrix}$$

See FAD1015 L27-L28 — Matrices (Types & Operations).

Note: Question 2 is missing (pages 3–4 of PDF lost).

PART B

Question 3 — Binomial (Vaccine) & Exponential (Clerk)

Marks: 14

(a) Vaccine success: $p = 0.75$

(i) $n = 5$, $P(X = 4)$ using binomial table:

$$P(X = 4) = C(5,4) \times (0.75)^4 \times (0.25)^1 = 5 \times 0.3164 \times 0.25 = \boxed{0.3955}$$

(ii) $n = 700$, $p = 0.75$. Find $k$ such that $P(X \ge k) = 0.70$.

$\mu = np = 525$, $\sigma = \sqrt{npq} = \sqrt{700 \times 0.75 \times 0.25} = \sqrt{131.25} = 11.46$

Since $n$ is large, use normal approximation with continuity correction:

$$P(X \ge k) = 0.70 \Rightarrow P\left(Z \ge \frac{k - 0.5 - 525}{11.46}\right) = 0.70$$

$$P\left(Z \le \frac{k - 525.5}{11.46}\right) = 0.30 \Rightarrow \frac{k - 525.5}{11.46} = -0.5244$$

$$k - 525.5 = -6.01 \Rightarrow k = 519.49$$

$$\boxed{k \approx 520} \quad (\text{exact value from binomial table})$$

See FAD1015 L13 — Binomial Distribution and FAD1015 Statistical Tables — Murdoch & Barnes.

(b) Clerk time: $X \sim \text{Exp}(\lambda = 1/5)$, mean = 5 min.

(i) $P(2 < X < 4) = (1 - e^{-4/5}) - (1 - e^{-2/5}) = e^{-0.4} - e^{-0.8}$

$$= 0.6703 - 0.4493 = \boxed{0.2210}$$

(ii) Median: $P(X \le m) = 0.50$

$$1 - e^{-m/5} = 0.50 \Rightarrow e^{-m/5} = 0.50 \Rightarrow -\frac{m}{5} = \ln(0.50) = -0.6931$$

$$m = 5 \times 0.6931 = \boxed{3.47 \text{ min}}$$

(iii) For exponential: $\sigma = \mu = \boxed{5 \text{ min}}$

See FAD1015 L17-L18 — Uniform & Exponential Distributions + R Intro.

Question 4 — Sampling Distribution

Marks: 14

(a) Number of children: ${2, 4, 7, 8}$, $\mu = \frac{21}{4} = 5.25$

(i) $n = 2$ without replacement: $C(4,2) = 6$ samples

Sample $\bar{x}$ Sample $\bar{x}$
(2,4) 3.0 (4,7) 5.5
(2,7) 4.5 (4,8) 6.0
(2,8) 5.0 (7,8) 7.5

Mean of sample means: $\frac{31.5}{6} = 5.25 = \mu$. Type: exact sampling distribution (small finite population).

$n = 3$: $C(4,3) = 4$ samples

Sample $\bar{x}$
(2,4,7) 4.333
(2,4,8) 4.667
(2,7,8) 5.667
(4,7,8) 6.333

Mean of sample means: $\frac{21}{4} = 5.25 = \mu$.

(ii) Std dev of sample mean:

$n=2$: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \sqrt{\frac{N-n}{N-1}}$ (finite population correction)

$n=3$: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{3}} \sqrt{\frac{4-3}{3}}$

Since finite population correction shrinks faster for $n=3$, $n=3$ has smaller variability.

See FAD1015 L20 — Sampling Distribution of the Mean.

(b) Hospital stay: $\mu = 5.5$, $\sigma^2 = 2.6$, $\sigma = 1.612$, skewed left, $n = 100$

(i) By CLT: $\bar{X} \sim N(\mu_{\bar{x}} = 5.5,; \sigma_{\bar{x}} = 1.612/\sqrt{100} = 0.1612)$

(ii) $P(\bar{X} \le 4)$:

$$z = \frac{4 - 5.5}{0.1612} = \frac{-1.5}{0.1612} = -9.31$$

$$P(Z \le -9.31) \approx \boxed{0}$$

(iii) $P(\bar{X} \ge 6)$:

$$z = \frac{6 - 5.5}{0.1612} = \frac{0.5}{0.1612} = 3.10$$

$$P(Z \ge 3.10) = 1 - 0.9990 = \boxed{0.0010}$$

See FAD1015 L20 — Sampling Distribution of the Mean.

Question 5 — PASUM GPA Hypothesis Test

Marks: 14

Setup: $\mu_0 = 3.37$, $n = 147$, $\bar{x} = 3.56$, $s = 0.31$, $\alpha = 0.01$

(a) Traditional method (critical value)

$H_0: \mu = 3.37$, $H_1: \mu \neq 3.37$ (two-tailed) $n \ge 30$, $\sigma$ unknown → z-test (CLT).

$$z = \frac{3.56 - 3.37}{0.31 / \sqrt{147}} = \frac{0.19}{0.02557} = 7.43$$

Critical value (two-tailed, $\alpha = 0.01$): $\pm z_{0.005} = \pm 2.576$

$7.43 > 2.576$ → Reject $H_0$.

Conclusion: At $\alpha = 0.01$, there is sufficient evidence that the average GPA has changed.

(b) P-value method

$$p = 2 \times P(Z > 7.43) \approx 2 \times 0 = \boxed{0}$$

$p < 0.01$ → Reject $H_0$. Same conclusion.

(c) 95% Confidence Interval

$$CI = 3.56 \pm 1.96 \times \frac{0.31}{\sqrt{147}} = 3.56 \pm 1.96 \times 0.02557$$

$$= 3.56 \pm 0.0501 = \boxed{(3.510,; 3.610)}$$

Interpretation: We are 95% confident the true mean GPA lies between 3.510 and 3.610.

$\mu_0 = 3.37$ is outside this interval → confirms GPA has changed over the decade.

See FAD1015 Hypothesis Testing Cookbook and FAD1015 Confidence Interval Cookbook.

Question 6 — R Code & Matrix Algebra

Marks: 14

(a) R code

> a <- c(1,2,3)
> b <- c(10,20,30)
> c <- c(100,200,300)
> d <- c(1000,2000,3000)
> A <- cbind(a, b, c, d)    # 3×4 matrix
> B <- rbind(a, b, c, d)    # 4×3 matrix
> SUM <- sum(A)             # sum of all elements
> MEAN <- mean(A[1:2,])     # mean of first 2 rows

(i) Output:

> SUM
[1] 6666

$$\text{SUM} = \boxed{6666}$$

> MEAN
[1] 416.625

Row 1: (1,10,100,1000), Row 2: (2,20,200,2000)

$$\text{MEAN} = \frac{1+10+100+1000+2+20+200+2000}{8} = \frac{3333}{8} = \boxed{416.625}$$

(ii) $A$ is $3 \times 4$, $B$ is $4 \times 3$. Inner dimensions match (4 = 4) → yes, multiply.

A %*% B    # results in a 3×3 matrix

(b) Matrix $\boldsymbol{S}$

$$\boldsymbol{S} = \begin{pmatrix} 1 & 2 & 3 \ 1 & 3 & 2 \ 2 & 1 & 3 \end{pmatrix}$$

(i) Determinant:

$$\det(\boldsymbol{S}) = 1 \begin{vmatrix} 3 & 2 \ 1 & 3 \end{vmatrix} - 2 \begin{vmatrix} 1 & 2 \ 2 & 3 \end{vmatrix} + 3 \begin{vmatrix} 1 & 3 \ 2 & 1 \end{vmatrix}$$

$$= 1(9-2) - 2(3-4) + 3(1-6) = 7 - 2(-1) + 3(-5) = 7 + 2 - 15 = \boxed{-6}$$

(ii) Find $x$ and $y$ from cofactor matrix:

Cofactor $C_{11}$:

$$C_{11} = (-1)^{1+1} \times M_{11} = 1 \times \det\begin{pmatrix}3&2\1&3\end{pmatrix} = 9 - 2 = 7$$

$$\boxed{x = 7}$$

Cofactor $C_{23}$:

$$C_{23} = (-1)^{2+3} \times M_{23} = -1 \times \det\begin{pmatrix}1&2\2&1\end{pmatrix} = -(1-4) = 3$$

$$\boxed{y = 3}$$

(iii) Adjoint and inverse:

Cofactor matrix:

$$\boldsymbol{C} = \begin{pmatrix} 7 & 1 & -5 \ -3 & -3 & 3 \ -5 & 1 & 1 \end{pmatrix}$$

Adjoint = transpose of cofactor matrix:

$$\text{adj}(\boldsymbol{S}) = \boldsymbol{C}^T = \begin{pmatrix} 7 & -3 & -5 \ 1 & -3 & 1 \ -5 & 3 & 1 \end{pmatrix}$$

Inverse:

$$\boldsymbol{S}^{-1} = \frac{\text{adj}(\boldsymbol{S})}{\det(\boldsymbol{S})} = -\frac{1}{6} \begin{pmatrix} 7 & -3 & -5 \ 1 & -3 & 1 \ -5 & 3 & 1 \end{pmatrix}$$

$$\boxed{\boldsymbol{S}^{-1} = \begin{pmatrix} -\frac{7}{6} & \frac{1}{2} & \frac{5}{6} \[2pt] -\frac{1}{6} & \frac{1}{2} & -\frac{1}{6} \[2pt] \frac{5}{6} & -\frac{1}{2} & -\frac{1}{6} \end{pmatrix}}$$

See FAD1015 L29-L30 — Matrices (Inverse & Systems of Equations) and Cramer's Rule.

Key References