Form 5 Chapter 5: Probability Distribution | SPM Exam Worksheet & Marking Scheme

SPM KSSM Diagnostic Drill

Form 5 Chapter 5: Probability Distribution (Taburan Kebarangkalian)

Total Marks: 40 Marks • Time Allowed: 50 Minutes

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A Bahagian A: SPM Paper 1 Format [16 Marks]

Answer all questions

Soalan 1 [4 Marks]

In a university entrance test, the probability that an applicant passes is $0.75$. A group of $8$ applicants is selected at random.

(a) State the mean number of applicants who pass the test. [1 mark]

(b) Calculate the probability that exactly 6 applicants pass the test. [3 marks]

Soalan 2 [4 Marks]

The probability that an archer hits the bullseye on any given shot is $0.4$. Find the minimum number of arrows the archer must shoot so that the probability of hitting the bullseye at least once exceeds $0.98$.

Soalan 3 [4 Marks]

$Z$ is a standard normal random variable such that $Z \sim N(0, 1)$.

(a) Find $P(Z > 1.45)$. [1 mark]

(b) Given that $P(-k \le Z \le k) = 0.9108$, find the value of $k$. [3 marks]

Soalan 4 [4 Marks]

The lifespan of an LED street lamp is normally distributed with a mean of $48\text{ months}$ and a variance of $64\text{ months}^2$.

(a) Find the standard deviation of the lifespan. [1 mark]

(b) In a municipal district with $2\,500$ identical lamps, calculate the expected number of lamps that will fail before $36\text{ months}$. [3 marks]

B Bahagian B: SPM Paper 2 Format [24 Marks]

Detailed solutions required

Soalan 5 [10 Marks]

(a) In a commercial orchard, the probability of an avocado ripening within 4 days after harvest is $p$. When a sample of 5 avocados is chosen at random, the probability that none of them ripen within 4 days is $\frac{1}{32}$.

(i) Find the value of $p$. [2 marks]

(ii) If 8 avocados are inspected, find the probability that at most 2 avocados ripen within 4 days. [3 marks]

(b) The mass of organic fertilizer packets packed by an automated packaging plant is normally distributed with a mean of $\mu\text{ kg}$ and a standard deviation of $\sigma\text{ kg}$. An auditor discovers that $8\%$ of the packets weigh less than $4.82\text{ kg}$ and $3\%$ weigh more than $5.35\text{ kg}$.

Calculate the value of $\mu$ and the value of $\sigma$. [5 marks]

(a)(i) Find $p$:
$P(X = 0) = {^5C_0} p^0 (1-p)^5 = \frac{1}{32} \implies (1-p)^5 = \left(\frac{1}{2}\right)^5$
$1 - p = \frac{1}{2} \implies p = 0.5$ [K1] $(1-p)^5 = 1/32$
[N1] $p = 0.5$

(a)(ii) Sample of 8, $X \sim B(8, 0.5)$:
$P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$
$P(X=0) = {^8C_0}(0.5)^8 = \frac{1}{256} \approx 0.003906$
$P(X=1) = {^8C_1}(0.5)^8 = \frac{8}{256} = 0.031250$
$P(X=2) = {^8C_2}(0.5)^8 = \frac{28}{256} \approx 0.109375$
$P(X \le 2) = \frac{1 + 8 + 28}{256} = \frac{37}{256} \approx 0.1445$ [K1] sum formula
[K1] evaluation
[N1] 0.1445

(b) Finding $\mu$ and $\sigma$:
Condition 1: $P(X < 4.82) = 0.08 \implies P\left(Z < \frac{4.82 - \mu}{\sigma}\right) = 0.08$
Lower tail $z_1 = -1.405$: $\frac{4.82 - \mu}{\sigma} = -1.405 \implies 4.82 - \mu = -1.405\sigma \quad \text{--- (1)}$ [P1, K1]
Condition 2: $P(X > 5.35) = 0.03 \implies P\left(Z > \frac{5.35 - \mu}{\sigma}\right) = 0.03$
Upper tail $z_2 = +1.881$: $\frac{5.35 - \mu}{\sigma} = 1.881 \implies 5.35 - \mu = 1.881\sigma \quad \text{--- (2)}$ [P1, K1]
Subtract (1) from (2):
$(5.35 - 4.82) = 1.881\sigma - (-1.405\sigma) \implies 0.53 = 3.286\sigma \implies \sigma = \frac{0.53}{3.286} \approx 0.1613\text{ kg}$ [K1, N1]
Substitute $\sigma = 0.1613$ into (2):
$\mu = 5.35 - 1.881(0.1613) = 5.35 - 0.3034 = 5.0466 \approx 5.05\text{ kg}$ [N1]

Soalan 6 • KBAT SPM Paper 2 [14 Marks]

High-Tech Semiconductor Manufacturing QA

Silicon Wafer Thickness Calibration & Batch Acceptance Sampling

A semiconductor fabrication facility in Penang manufactures silicon wafers. The process engineers monitor two critical quality metrics:

Part I (Continuous Normal Dimension): The wafer thickness, $T$, is normally distributed with a mean $\mu = 725\text{ }\mu\text{m}$ and a standard deviation $\sigma = 12\text{ }\mu\text{m}$.
• A wafer is classified as "In-Spec Target" if its thickness is between $705\text{ }\mu\text{m}$ and $745\text{ }\mu\text{m}$.
• Wafers with thickness below $705\text{ }\mu\text{m}$ are too brittle and must be scrapped at an immediate loss of RM 45 per wafer.
• Wafers with thickness above $745\text{ }\mu\text{m}$ can be mechanically repolished at a rework cost of RM 15 per wafer.

(a) Find the probability that a randomly chosen wafer is "In-Spec Target". [4 marks]

(b) In a daily production lot of $20\,000$ wafers, calculate the total expected financial loss arising from scrap and rework costs. [4 marks]

Part II (Discrete Acceptance Sampling Rule): Before shipping a consignment of $500$ packaged boxes of finished microchips, a client tests a random sample of $12$ boxes. Historically, $6\%$ of boxes produced by the automated packaging line have minor seal defects.
• The consignment is accepted immediately if no more than $1$ defective box is found.
• If more than $2$ defective boxes are found, the consignment is unconditionally rejected.
• If exactly $2$ defective boxes are found, a second independent sample of $10$ boxes is drawn.

(d) Calculate the probability that a second sample will be required. [3 marks]

(a) Probability In-Spec $(705 \le T \le 745)$:
Standardize bounds with $\mu = 725, \sigma = 12$:
$z_1 = \frac{705 - 725}{12} = \frac{-20}{12} = -1.67$
$z_2 = \frac{745 - 725}{12} = \frac{20}{12} = +1.67$
From standard normal table: $P(Z > 1.67) = 0.0475$
$P(705 \le T \le 745) = 1 - 2 \times P(Z > 1.67) = 1 - 2(0.0475) = 1 - 0.0950 = 0.9050$ [K1] $z = \pm 1.67$
[K1] $1 - 2(0.0475)$
[N1] 0.9050

(b) Expected financial loss for 20,000 wafers:
Scrap probability (thickness $< 705$): $P(T < 705) = P(Z < -1.67) = 0.0475$
Expected scrapped wafers $= 20\,000 \times 0.0475 = 950$ wafers.
Scrap loss $= 950 \times \text{RM } 45 = \text{RM } 42\,750$ [K1]
Rework probability (thickness $> 745$): $P(T > 745) = P(Z > 1.67) = 0.0475$
Expected rework wafers $= 20\,000 \times 0.0475 = 950$ wafers.
Rework cost $= 950 \times \text{RM } 15 = \text{RM } 14\,250$ [K1]
Total Expected Financial Loss $= 42\,750 + 14\,250 = \text{RM } 57\,000$ [K1] sum costs
[N1] RM 57,000

(c) First sample acceptance:
$n = 12, p = 0.06, q = 0.94$. Condition: $P(X \le 1) = P(X = 0) + P(X = 1)$.
$P(X = 0) = {^{12}C_0} (0.06)^0 (0.94)^{12} = (0.94)^{12} \approx 0.47592$
$P(X = 1) = {^{12}C_1} (0.06)^1 (0.94)^{11} = 12 \times 0.06 \times 0.50630 \approx 0.36454$
$P(X \le 1) = 0.47592 + 0.36454 = 0.8405$ [K1] $P(X=0)+P(X=1)$
[N1] 0.8405

(d) Probability second sample required:
Second sample required if exactly 2 defective boxes are found in the first sample: $P(X = 2)$.
$P(X = 2) = {^{12}C_2} (0.06)^2 (0.94)^{10}$
$= 66 \times 0.0036 \times 0.538615 = 0.1280$ [K1] formula
[N1] 0.1280