SPM High-Yield Core Topic (Paper 1 & Paper 2)

Binomial & Normal Distributions Mastery

Probability distributions bridge discrete trial experiments and continuous physical measurements. Every SPM examination tests Binomial probabilities $P(X=r)$ and Normal standardizations $Z = \frac{X-\mu}{\sigma}$. Master reverse z-lookups, simultaneous equations for $\mu$ and $\sigma$, and industrial quality assurance modeling.

1. Core Mathematical Framework

Binomial Distribution $X \sim B(n, p)$

Discrete

Applies when trials $n$ are fixed, outcomes are strictly dichotomous (success $p$ or failure $q = 1-p$), and trials are mutually independent.

$P(X = r) = {^nC_r} p^r q^{n-r}, \quad r = 0, 1, 2, \dots, n$
Mean $\mu = np \qquad \text{Variance } \sigma^2 = npq \qquad \sigma = \sqrt{npq}$
SPM Phrasing Traps:
  • "At least 1 success": $P(X \ge 1) = 1 - P(X = 0)$
  • "At most 2 successes": $P(X \le 2) = P(X=0) + P(X=1) + P(X=2)$
  • "More than $n-1$ successes": $P(X > n-1) = P(X = n)$

Normal Distribution $X \sim N(\mu, \sigma^2)$

Continuous

Symmetric bell curve centered at mean $\mu$. Standardized to standard normal $Z \sim N(0, 1)$ with total area equal to 1.0.

$Z = \frac{X - \mu}{\sigma}$
Symmetry: $P(Z < -z) = P(Z > z) = Q(z)$
Standard Normal Table Strategy:
  • The SPM formula booklet table gives the upper tail area $Q(z) = P(Z \ge z)$.
  • If $P(Z > c) > 0.5$, then $c$ must be negative: $c = -z$ where $P(Z > -z) = 1 - Q(z)$.
  • Between two bounds: $P(a \le X \le b) = P(Z \ge z_a) - P(Z \ge z_b)$ where $z_a < z_b$.
Critical Exam Pitfalls 1. When using $Z = \frac{X - \mu}{\sigma}$, do not square $\sigma$ in the denominator; SPM questions specify variance $\sigma^2 = 25 \implies \sigma = 5$.
2. For Binomial distributions, verify that $p + q = 1$. If $p = 0.35$, ensure $q = 0.65$.
3. For continuous normal distribution, $P(X \le k) = P(X < k)$ because $P(X = k) = 0$.

2. Progressive SPM Worked Examples with Marking Codes

Example 1 • SPM Paper 1 Standard [4 Marks]

A machine produces light bulbs where $8\%$ of the bulbs are defective. A random sample of $10$ bulbs is selected for testing.

(a) Find the probability that exactly 2 bulbs are defective.
(b) Find the probability that at least 1 bulb is defective.
Step-by-Step Marking Solution
Identify parameters: $n = 10, p = 0.08, q = 1 - 0.08 = 0.92$.
(a) Exactly 2 defective:
$P(X = 2) = {^{10}C_2} (0.08)^2 (0.92)^{10 - 2}$
$= 45 \times 0.0064 \times 0.5132188 = 0.1478$
[K1]
[N1] 0.1478
(b) At least 1 defective:
$P(X \ge 1) = 1 - P(X = 0) = 1 - {^{10}C_0} (0.08)^0 (0.92)^{10}$
$= 1 - (1)(1)(0.434388) = 1 - 0.434388 = 0.5656$
[K1]
[N1] 0.5656
Example 2 • SPM Paper 2 Section A/B [6 Marks]

The masses of honeydew melons harvested from an orchard follow a normal distribution with a mean of $1.8\text{ kg}$ and a standard deviation of $0.25\text{ kg}$.

(a) A melon is chosen at random. Find the probability that its mass is between $1.5\text{ kg}$ and $2.1\text{ kg}$.
(b) Melons with a mass greater than $k\text{ kg}$ are classified as "Grade AAA Premium". If $15\%$ of the harvest is graded as "Grade AAA Premium", find the value of $k$.
Step-by-Step Marking Solution
Parameters: $\mu = 1.8, \sigma = 0.25$.
(a) Mass between $1.5$ and $2.1$:
$z_1 = \frac{1.5 - 1.8}{0.25} = \frac{-0.3}{0.25} = -1.20$
$z_2 = \frac{2.1 - 1.8}{0.25} = \frac{0.3}{0.25} = +1.20$
$P(1.5 \le X \le 2.1) = P(-1.20 \le Z \le 1.20) = 1 - 2 \times P(Z > 1.20)$
From standard normal table: $P(Z > 1.20) = 0.1151$
$\implies P(1.5 \le X \le 2.1) = 1 - 2(0.1151) = 1 - 0.2302 = 0.7698$
[K1] standardize
[K1] subtract tails
[N1] 0.7698
(b) Grade AAA threshold $k$:
$P(X > k) = 0.15 \implies P\left(Z > \frac{k - 1.8}{0.25}\right) = 0.15$
From normal table (inverse area $0.15$): $z = 1.036$ (or $1.0364$)
$\frac{k - 1.8}{0.25} = 1.036$
$k - 1.8 = 0.25 \times 1.036 = 0.259$
$k = 1.8 + 0.259 = 2.059\text{ kg}$
[P1] $z = 1.036$
[K1] equate $Z$
[N1] $2.059$
Example 3 • KBAT SPM Simultaneous Distribution [5 Marks]

The monthly electric power consumption of households in a residential district follows a normal distribution with mean $\mu\text{ kWh}$ and standard deviation $\sigma\text{ kWh}$. A consumer census discovers that $5\%$ of households consume less than $180\text{ kWh}$, while $10\%$ consume more than $420\text{ kWh}$.

Calculate the mean $\mu$ and the standard deviation $\sigma$.

Step 1: Express conditions in standard normal $Z$ notation:

$P(X < 180) = 0.05 \implies P\left(Z < \frac{180 - \mu}{\sigma}\right) = 0.05$
Since area is in the lower tail ($< 0.5$), $z_1 = -1.645$:
$\frac{180 - \mu}{\sigma} = -1.645 \implies 180 - \mu = -1.645\sigma \quad \text{--- (1)}$ [K1]
$P(X > 420) = 0.10 \implies P\left(Z > \frac{420 - \mu}{\sigma}\right) = 0.10$
Upper tail lookup gives $z_2 = +1.282$:
$\frac{420 - \mu}{\sigma} = 1.282 \implies 420 - \mu = 1.282\sigma \quad \text{--- (2)}$ [K1]

Step 2: Solve simultaneous linear equations:

Subtract (1) from (2):
$(420 - \mu) - (180 - \mu) = 1.282\sigma - (-1.645\sigma)$
$240 = 2.927\sigma \implies \sigma = \frac{240}{2.927} \approx 82.00\text{ kWh}$ [K1, N1]
Substitute $\sigma = 82.00$ into (2):
$420 - \mu = 1.282(82.00) = 105.12 \implies \mu = 420 - 105.12 = 314.88\text{ kWh}$ [N1]

SPM Examiner Advice & Target Strategy

Standard Normal Table Reading Always sketch the bell curve! Shade the required area. Remember that the SPM table provides $P(Z \ge z)$. If you are looking for area between $0$ and $z$, calculate $0.5 - P(Z \ge z)$.
Binomial Shortcut Check For large $n$ questions, look out for phrases like "at least once" ($1 - P(X=0)$) or "all of them" ($P(X=n)$) which only require one single calculation rather than calculating a sum of 8 or 9 terms.