title: "UAS 24-25 FAD1015 Mathematics III — Answering Guide" exam: "UAS 24-25 FAD1015 Mathematics III" tags:

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UAS 24-25 FAD1015 Mathematics III — Answering Guide

Total marks: 80 | Part A: 24 marks (Q1–2) | Part B: 56 marks (Q3–6)

PART A

Question 1 — Distributions, R t-test, Matrix Types

Marks: 12

Q1(a) — True or False on Distributions

(i) The trial in binomial distribution is dependent.

FALSE. Binomial trials are independent — outcome of one trial does not affect another. See FAD1015 L13 — Binomial Distribution.

(ii) In a Poisson distribution, events must occur independently of each other.

TRUE. Poisson requires events to be independent — one event does not influence the next. See FAD1015 L14 — Poisson Distribution.

(iii) The Poisson distribution is a continuous probability distribution.

FALSE. Poisson is a discrete distribution defined on non-negative integers. See FAD1015 L14 — Poisson Distribution.

(iv) The area under the graph of a uniform distribution is always greater than 1.

FALSE. Area under any PDF = 1 exactly, by definition of a probability density function. See FAD1015 L17-L18 — Uniform & Exponential Distributions + R Intro.

(v) The mean and standard deviation of the exponential distribution are equal.

TRUE. For exponential, mean $\mu = \frac{1}{\lambda}$ and std dev $\sigma = \frac{1}{\lambda}$ — so yes, equal. See FAD1015 L17-L18 — Uniform & Exponential Distributions + R Intro.

(vi) The exponential distribution is symmetric.

FALSE. Exponential is right-skewed (positively skewed), not symmetric. See Exponential Distribution.

Answer summary:

(i) (ii) (iii) (iv) (v) (vi)
FALSE TRUE FALSE FALSE TRUE FALSE

Q1(b) — R t-Test: Three Errors

Code:

library(BSDA)

bp_data <- c(122, 121, 119, 118, 125, 124, 126, 130,
             122, 121, 119, 125)

t.test(Bp_Data, mu=120, sigma=15, alternative =
"greater than")

See FAD1015 L25-L26 — Hypothesis Testing in R.

Error 1 — Case mismatch: Bp_Data vs declared bp_data. R is case-sensitive.

Error 2 — sigma=15 invalid: t.test() has no sigma param. Use z.test(..., sigma.x=15) from BSDA for z-test, or drop sigma=15 for t-test.

Error 3 — "greater than" invalid: Valid alternatives: "two.sided", "greater", "less". Should be "greater".

Corrected code (t-test): t.test(bp_data, mu=120, alternative = "greater")

Q1(c) — 3×3 Matrix Examples

See FAD1015 L27-L28 — Matrices (Types & Operations) and FAD1015 L29-L30 — Matrices (Inverse & Systems of Equations).

(i) Lower triangular:

$$L = \begin{pmatrix} 2 & 0 & 0 \ 4 & 1 & 0 \ -3 & 5 & 6 \end{pmatrix}$$

(ii) Identity:

$$I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix}$$

(iii) Diagonal:

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

Question 2 — Sampling Distribution, Hypothesis Test, Sampling Methods

Marks: 12

Q2(a) — Fill in the Blanks

(i) The sample mean ($\bar{x}$) is used to estimate the population mean. See FAD1015 L20 — Sampling Distribution of the Mean.

(ii) The sampling distribution of the mean is the distribution of all possible sample means if you select all samples of a given size.

Q2(b) — Greenbull Energy Drink Hypothesis Test

Setup: $\mu_0 = 42$, $n = 25$, $\bar{x} = 44.5$, $s = 5$, $\alpha = 0.05$

(i) Hypotheses: $H_0: \mu \leq 42$, $H_1: \mu > 42$ (right-tailed)

(ii) Test: t-test ($\sigma$ unknown, $n < 30$). $df = 24$.

$$t = \frac{44.5 - 42}{5 / \sqrt{25}} = \frac{2.5}{1} = 2.5$$

$t_{0.05,,24} = 1.711$. Since $2.5 > 1.711$, reject $H_0$. Sufficient evidence the drink increases attention span.

See FAD1015 Hypothesis Testing Cookbook.

Q2(c) — Sampling Method Examples

Type Examples
Probability Simple random, stratified, cluster, systematic
Non-probability Convenience, quota, purposive, snowball

PART B

Question 3 — Continuous PDF: Fuel Price Model

Marks: 14

$$f(x) = \begin{cases} k(4x - x^2) & \text{if } 2 \leq x \leq 3 \ 0 & \text{otherwise} \end{cases}$$

Q3(a) — Show $k = 3/10$

$$\int_{2}^{3} k(4x - x^2),dx = k\left[2x^2 - \frac{x^3}{3}\right]_{2}^{3} = k\left(\frac{11}{3}\right) = 1 \quad\Rightarrow\quad k = \frac{3}{10}$$

Q3(b) — $P(2.50 < X < 2.80)$

$$= \frac{3}{10}\left[2x^2 - \frac{x^3}{3}\right]_{2.50}^{2.80} = \frac{3}{10}(8.3627 - 7.2917) = \frac{3}{10}(1.0710) = \boxed{0.3213}$$

Q3(c) — $P(X > 2.70)$ (Subsidy probability)

$$= \frac{3}{10}\left[2x^2 - \frac{x^3}{3}\right]_{2.70}^{3} = \frac{3}{10}(9 - 8.019) = \frac{3}{10}(0.981) = \boxed{0.2943}$$

Q3(d) — Three Consecutive High-Price Months

$$(0.2943)^3 = \boxed{0.0255}$$

Only 2.55% chance assuming independence. In reality, fuel prices may be correlated across months, so the true risk could be higher. Government should monitor trends.

See FAD1015 Week 6 — Continuous Random Variables.

Question 4 — Poisson & Normal Distributions

Marks: 14

Q4(a) — Poisson: Car Ownership

$$p = 6/50 = 0.12$$

(i) $n = 100$, $\lambda = 12$

$$P(X = 12) = \frac{e^{-12} \cdot 12^{12}}{12!} \approx \boxed{0.1144}$$

(ii) $n = 200$, $\lambda = 24$

$$P(X \leq 26) \approx \boxed{0.720} \quad \text{(from Poisson table)}$$

See FAD1015 L14 — Poisson Distribution and FAD1015 Statistical Tables — Murdoch & Barnes.

Q4(b) — Normal: Exam Marks

$\mu = 60$, $\sigma = 6$. See FAD1015 L15-L16 — Normal Distribution & Approximation.

(i) Pass mark = 45: $z = \frac{45-60}{6} = -2.5$

$$P(X \geq 45) = P(Z \geq -2.5) = 0.9938$$

(ii) A grade $\geq 76$: $z = \frac{76-60}{6} = 2.667$

$$P(X \geq 76) = 1 - 0.9962 = 0.0038$$

Students: $0.0038 \times 1500 \approx \boxed{6 \text{ students}}$

(iii) 90% pass: need $P(X \geq x) = 0.90$

$$z_{0.10} = -1.282,\quad \frac{x-60}{6} = -1.282,\quad x = 60 - 7.692 = \boxed{52.3}$$

Question 5 — Hypothesis Testing: Food Pack Weight

Marks: 14

Sample 1: $n=6$, data: $77, 69, 78, 69, 88, 74$

$$\bar{x} = 75.833,\quad s = 7.083$$

Sample 2: $n=49$, $\bar{x}=78.03,; s=19.27$

Q5(a) — Traditional Method, $\alpha = 0.10$

$H_0: \mu = 75$, $H_1: \mu \neq 75$ (two-tailed). $\sigma$ unknown, $n < 30$ → t-test, $df = 5$.

$$t = \frac{75.833 - 75}{7.083 / \sqrt{6}} = \frac{0.833}{2.892} = 0.288$$

$t_{0.05,,5} = 2.015$. $|0.288| < 2.015$ → Do not reject $H_0$.

See FAD1015 Hypothesis Testing Cookbook.

Q5(b)(i) — P-Value Method, $\alpha = 0.05$

$n = 49 \geq 30$ → z-test (CLT).

$$z = \frac{78.03 - 75}{19.27 / \sqrt{49}} = \frac{3.03}{2.753} = 1.101$$

$$p = 2 \times P(Z > 1.101) = 2 \times 0.1357 = 0.2714$$

$0.2714 > 0.05$ → Do not reject $H_0$. Same conclusion as (a).

Q5(b)(ii) — 95% Confidence Interval

$$CI = 78.03 \pm 1.96 \times \frac{19.27}{\sqrt{49}} = 78.03 \pm 5.395 = \boxed{(72.635,; 83.425)}$$

$\mu_0 = 75$ falls inside the CI → consistent with failing to reject $H_0$.

See FAD1015 Confidence Interval Cookbook.

Question 6 — Matrices, Cramer's Rule & R Code

Marks: 14

Q6(a) — System of Linear Equations

$$4p + q = 8,\quad p + 2r = -3,\quad 2p + 2q - 2r = 14$$

(i) Matrix form:

$$A = \begin{pmatrix} 4 & 1 & 0 \ 1 & 0 & 2 \ 2 & 2 & -2 \end{pmatrix},\quad X = \begin{pmatrix} p \ q \ r \end{pmatrix},\quad B = \begin{pmatrix} 8 \ -3 \ 14 \end{pmatrix}$$

(ii) Determinant:

$$\det(A) = 4(0-4) - 1(-2-4) + 0 = -16 + 6 = \boxed{-10}$$

(iii) Cramer's rule:

Variable $A_{\text{replacement}}$ $\det$ Value
$p$ $A_p = \begin{pmatrix}8&1&0\-3&0&2\14&2&-2\end{pmatrix}$ $-10$ $p = -10/-10 = 1$
$q$ $A_q = \begin{pmatrix}4&8&0\1&-3&2\2&14&-2\end{pmatrix}$ $-40$ $q = -40/-10 = 4$
$r$ $A_r = \begin{pmatrix}4&1&8\1&0&-3\2&2&14\end{pmatrix}$ $20$ $r = 20/-10 = -2$

$$\boxed{p = 1,; q = 4,; r = -2}$$

Check: $4(1)+4=8$, $1+2(-2)=-3$, $2(1)+2(4)-2(-2)=14$ ✓

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

Q6(b) — R Code

# (i) Create matrix A with column labels
A <- matrix(c(4, 1, 3, -3, 2, 4, 5, 4, 2), nrow = 3, byrow = TRUE)
colnames(A) <- c("col1", "col2", "col3")

# (ii) Inverse of A
solve(A)

# (iii) A^T B^T
t(A) %*% t(B)

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

Key References

Lecture Notes

Synthesis Materials

Concept Pages