title: "UAS 22-23 FAD1015 Mathematics III — Answering Guide" exam: "UAS 22-23 FAD1015 Mathematics III" tags:

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UAS 22-23 FAD1015 Mathematics III — Answering Guide

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

PART A

Question 1 — Random Variables, Uniform Distribution, T/F

Marks: 2 + 4 + 6 = 12

(a) State the difference (2 marks)

Discrete Continuous
Takes countable values (finite/infinite list) Takes any value in an interval
e.g., number of students, coin flips e.g., height, time, temperature

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

(b) Uniform distribution: oil change time, $X \sim U(10, 25)$

(i) $X$ = time (in minutes) to change oil on a car.

(ii) $X \sim U(a=10,; b=25)$ or $X \sim \text{Uniform}(10, 25)$.

(iii) Parameters: $a = 10$ (minimum), $b = 25$ (maximum).

PDF: $f(x) = \frac{1}{b-a} = \frac{1}{15}$ for $10 \le x \le 25$.

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

(c) True or False (6 × 1 mark)

Statement Answer Reason
(i) Set of all possible outcomes = sample space TRUE Definition of sample space
(ii) $P(B'|A) = 1 - P(B|A)$ TRUE $P(B'|A) = \frac{P(B' \cap A)}{P(A)} = \frac{P(A)-P(A\cap B)}{P(A)} = 1 - P(B|A)$
(iii) $A$ independent of $B$ if $P(A|B)=P(B)$ FALSE Independence: $P(A|B) = P(A)$, not $P(B)$
(iv) $(AB)^T = A^T B^T$ FALSE Correct: $(AB)^T = B^T A^T$ (reverse order)
(v) Identical rows/columns → det = 0 TRUE Property of determinants
(vi) $n \times n$ matrix has a determinant TRUE Determinant exists for any square matrix

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

Question 2 — R Code, CI Interpretation, Hypothesis Testing

Marks: 3 + 3 + 2 + 2 + 2 = 12

(a) Descriptive statistics R code: 3 errors

> X=Nile
> Y=Nile[1:20]
> mean(X)
> median(X)
> stdev(X)       # Error 1: should be sd(X)
> mean(y)        # Error 2: y not defined (Y was assigned, case-sensitive)
> range(y)       # Error 3: same — y vs Y

Errors:

  1. stdev() → should be sd() in base R
  2. mean(y) → y undefined; should be mean(Y)
  3. range(y) → same case error

See FAD1015 L19 — Input Data & Descriptive Statistics in R.

(b) Scatter plot: errors in code

plot(sepal.width, sepal.length, col = 'steelblue',
     main = 'Scatterplot',
     xlab= 'sepal width',
     ylab = Sepal Length,    # Error: unquoted — should be 'Sepal Length'
     pch=19)

Errors:

  1. ylab = Sepal Length → unquoted text; should be ylab = "Sepal Length"
  2. Variables sepal.width, sepal.length not attached — need $ or attach() if from CSV

Corrected code:

data <- read.csv("dahlia.csv")
plot(data$sepal.width, data$sepal.length, col = 'steelblue',
     main = 'Scatterplot', xlab = 'Sepal Width',
     ylab = 'Sepal Length', pch = 19)

(c) Interpret 94% CI (7.2, 7.6) oz

We are 94% confident that the true mean amount of coffee dispensed lies between 7.2 and 7.6 oz.

See FAD1015 L21-L22 — Estimation of Population Mean and Confidence Interval.

(d) 95% CI (28, 35), test $H_0: \mu = 36$ at $\alpha = 0.05$

Yes, reject $H_0$. $\mu_0 = 36$ lies outside the 95% CI (28, 35). Since $\mu_0$ is not in the interval, we reject $H_0$ at $\alpha = 0.05$.

(e) $p = 0.041$, $H_1: \mu \neq 25$, $\alpha = 0.01$

Do not reject $H_0$. $p = 0.041 > 0.01$. The p-value is larger than $\alpha$, so there is not enough evidence to reject $H_0$.

See FAD1015 Hypothesis Testing Cookbook.

PART B

Question 3 — Poisson & Exponential Distributions

Marks: 7 + 7 = 14

(a) Meowzy & Co. — Lorries (Poisson)

Demand per day: $X \sim \text{Pois}(\lambda = 4)$. Capacity = 8 lorries.

(i) Cannot meet demand on any day: $P(X > 8)$

From Poisson table ($\lambda = 4$): $$P(X \le 8) \approx 0.9787$$

$$P(X > 8) = 1 - 0.9787 = \boxed{0.0213}$$

(ii) Less than 6 lorries rented in 3 days: $Y \sim \text{Pois}(\lambda = 12)$

$$P(Y < 6) = P(Y \le 5)$$

From Poisson table ($\lambda = 12$): $P(Y \le 5) \approx \boxed{0.020}$

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

(b) Time between cars (Exponential)

Mean $= 9$ min → $\lambda = \frac{1}{9}$ per min.

(i) $P(X < 7) = 1 - e^{-7/9} = 1 - e^{-0.7778} = 1 - 0.4595 = \boxed{0.5405}$

(ii) $P(X > t) = 0.80$:

$$e^{-t/9} = 0.80 \quad\Rightarrow\quad -\frac{t}{9} = \ln(0.80) = -0.2231$$

$$t = 9 \times 0.2231 = \boxed{2.01 \text{ min}}$$

(iii) For exponential: $\sigma = \mu = \boxed{9 \text{ min}}$

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

Question 4 — Sampling Distribution of the Mean

Marks: 8 + 6 = 14

(a) Marriage duration data

Population: ${2, 4, 7, 8, 10, 11}$, $\mu = \frac{42}{6} = 7$

(i) $n = 2$, without replacement:

All $C(6,2) = 15$ samples:

Sample Mean Sample Mean
2,4 3.0 4,10 7.0
2,7 4.5 4,11 7.5
2,8 5.0 7,8 7.5
2,10 6.0 7,10 8.5
2,11 6.5 7,11 9.0
4,7 5.5 8,10 9.0
4,8 6.0 8,11 9.5
— — 10,11 10.5

Mean of sample means $= \frac{\sum \bar{x}}{15} = \frac{105}{15} = 7 = \mu$ ✓

(ii) $n = 5$: $C(6,5) = 6$ samples, each leaves out one value.

Sample (omit) Mean
Omit 2: {4,7,8,10,11} 8.0
Omit 4: {2,7,8,10,11} 7.6
Omit 7: {2,4,8,10,11} 7.0
Omit 8: {2,4,7,10,11} 6.8
Omit 10: {2,4,7,8,11} 6.4
Omit 11: {2,4,7,8,10} 6.2

Mean of sample means $= \frac{42}{6} = 7 = \mu$ ✓

(iii) Shape: For $n=2$, distribution is roughly symmetric but not truly normal. For $n=5$, distribution is more concentrated. Neither is exactly normal due to small finite population — but both show the sampling distribution centered at $\mu$.

(iv) $n=5$ has smaller variance. Larger $n$ → standard error $\sigma/\sqrt{n}$ is smaller → less variability in the sampling distribution.

See FAD1015 L20 — Sampling Distribution of the Mean.

(b) Tire lifespan

Population: $\mu = 65,!000$ mi, $\sigma = 9,!000$ mi.

(i) By CLT: $\bar{X} \sim N\left(\mu = 65,!000,; \frac{\sigma}{\sqrt{n}} = \frac{9,!000}{\sqrt{n}}\right)$

(ii) Assumptions: random sample, $n$ large enough for CLT ($n \ge 30$).

(iii) $n = 81$:

$$\bar{X} \sim N\left(65,!000,; \frac{9,!000}{9} = 1,!000\right)$$

$$z = \frac{90,!000 - 65,!000}{1,!000} = 25$$

$$P(\bar{X} \ge 90,!000) = P(Z \ge 25) \approx \boxed{0}$$

A sample mean of 90,000 miles is 25 SEs above the claimed mean — essentially impossible. This would be strong evidence against the manufacturer's claim.

See FAD1015 L20 — Sampling Distribution of the Mean.

Question 5 — Normal Distribution & Hypothesis Testing

Marks: 5 + 9 = 14

(a) FAD1015 scores: $N(\mu=84,; \sigma=3)$

Empirical rule:

  • 68% within $\mu \pm \sigma$: $84 \pm 3$ = (81, 87)
  • 95% within $\mu \pm 2\sigma$: $84 \pm 6$ = (78, 90)
  • 99.7% within $\mu \pm 3\sigma$: $84 \pm 9$ = (75, 93)

(i) 68% of scores: 81 to 87

(ii) Score 90: $z = \frac{90-84}{3} = \boxed{2}$ standard deviations above mean.

(iii) Scores 78 to 87:

  • 78 = $\mu - 2\sigma$ (2.5th percentile)
  • 87 = $\mu + \sigma$ (84th percentile)

$$P(78 < X < 87) = 84% - 2.5% = \boxed{81.5%}$$

See FAD1015 L15-L16 — Normal Distribution & Approximation.

(b) Yummylicious cereal: $N(16,; 0.12)$

(i) $P(X \ge 15.95)$:

$$z = \frac{15.95 - 16}{0.12} = \frac{-0.05}{0.12} = -0.417$$

$$P(Z \ge -0.417) = P(Z \le 0.417) = \boxed{0.662}$$

(ii) Hypothesis test: manager concerned about lower weight

$H_0: \mu = 16$, $H_1: \mu < 16$ (left-tailed) $n = 50$, $\bar{x} = 15.95$, $\sigma = 0.12$, $\alpha = 0.05$ $\sigma$ known, $n \ge 30$ → z-test

$$z = \frac{15.95 - 16}{0.12 / \sqrt{50}} = \frac{-0.05}{0.01697} = -2.946$$

Critical value (left-tailed, $\alpha = 0.05$): $-z_{0.05} = -1.645$

Since $-2.946 < -1.645$, reject $H_0$.

Conclusion: At $\alpha = 0.05$, sufficient evidence that the mean weight is less than 16 oz. The production manager's concern is justified.

See FAD1015 Hypothesis Testing Cookbook.

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

Marks: 7 + 7 = 14

(a) R code: matrix A

> A <- cbind(c(1,2,0), c(2,3,-1), c(1,-1,3))
> rownames(A) <- c("row1","row2","row3")
> colnames(A) <- c("Col1","Col2","Col3")
> A

(i) Output:

     Col1 Col2 Col3
row1    1    2    1
row2    2    3   -1
row3    0   -1    3

(ii) Multiply A and B? Yes — both are $3 \times 3$, so inner dimensions match.

A %*% B

(b) Cake prices (Cramer's Rule)

Let $x$ = strawberry, $y$ = cheese, $z$ = chocolate.

$$3x + 2y + 5z = 267$$ $$2x + 3y + z = 145$$ $$x + 5y + 4z = 230$$

$$A = \begin{pmatrix} 3 & 2 & 5 \ 2 & 3 & 1 \ 1 & 5 & 4 \end{pmatrix},\quad B = \begin{pmatrix} 267 \ 145 \ 230 \end{pmatrix}$$

Determinant of $A$:

$$\det(A) = 3 \begin{vmatrix} 3 & 1 \ 5 & 4 \end{vmatrix} - 2 \begin{vmatrix} 2 & 1 \ 1 & 4 \end{vmatrix} + 5 \begin{vmatrix} 2 & 3 \ 1 & 5 \end{vmatrix}$$

$$= 3(12-5) - 2(8-1) + 5(10-3) = 3(7) - 2(7) + 5(7) = 21 - 14 + 35 = 42$$

Cramer's rule:

Variable Determinant Value
$x$ $\det(A_x) = \begin{vmatrix}267&2&5\145&3&1\230&5&4\end{vmatrix} = 267(7) - 2(350) + 5(35) = 1869-700+175 = 1344$ $x = \frac{1344}{42} = 32$
$y$ $\det(A_y) = \begin{vmatrix}3&267&5\2&145&1\1&230&4\end{vmatrix} = 3(350) - 267(7) + 5(315) = 1050-1869+1575 = 756$ $y = \frac{756}{42} = 18$
$z$ $\det(A_z) = \begin{vmatrix}3&2&267\2&3&145\1&5&230\end{vmatrix} = 3(-35) - 2(315) + 267(7) = -105-630+1869 = 1134$ $z = \frac{1134}{42} = 27$

$$\boxed{x = \text{RM}32,\quad y = \text{RM}18,\quad z = \text{RM}27}$$

Check: $3(32)+2(18)+5(27) = 96+36+135 = 267$ ✓

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

Key References