Normal Distribution and Z-Values | IB Math AA SL

SL 4.9 · Continuous Probability

1. The Normal Distribution Model $X \sim N(\mu, \sigma^2)$

The normal distribution is a continuous probability distribution defined by two parameters: the mean $\mu$ and the variance $\sigma^2$ (standard deviation $\sigma$).

Key Properties of the Normal Curve:
- Symmetrical bell-shaped curve centered at $x = \mu$, where $\text{Mean} = \text{Median} = \text{Mode} = \mu$.
- Total area under the probability density curve equals $1$: $\int_{-\infty}^{\infty} f(x)\,dx = 1$.
- Points of inflection occur at exactly one standard deviation from the mean: $x = \mu \pm \sigma$.
- The horizontal axis is an asymptote in both directions ($y \to 0$ as $x \to \pm\infty$).
- For any exact single value $c$, $P(X = c) = 0$; therefore, $P(X \le c) = P(X < c)$.
The Empirical Rule (68–95–99.7% Rule):
- $P(\mu - \sigma < X < \mu + \sigma) \approx 0.683$ ($68.3\%$)
- $P(\mu - 2\sigma < X < \mu + 2\sigma) \approx 0.954$ ($95.4\%$)
- $P(\mu - 3\sigma < X < \mu + 3\sigma) \approx 0.997$ ($99.7\%$)

Worked Example 1.1 · Symmetry and Area Properties

The scores $X$ on an exam are normally distributed with mean $\mu = 72$ and $\sigma = 8$. Given that $P(X < 64) \approx 0.1587$, determine: (a) $P(X > 80)$, (b) $P(64 < X < 80)$, and (c) $P(X > 72)$.

(a) Notice that $64 = \mu - \sigma$ and $80 = \mu + \sigma$. By symmetry about the mean $\mu = 72$: $$P(X > 80) = P(X < 64) = 0.1587$$
(b) Total area equals 1 minus the two symmetric tails: $$P(64 < X < 80) = 1 - [P(X < 64) + P(X > 80)] = 1 - 2(0.1587) = 1 - 0.3174 = 0.6826$$
(c) By symmetry, exactly half the distribution lies above the mean: $P(X > 72) = 0.50$.

Answer: (a) $0.1587$; (b) $0.6826$; (c) $0.50$.

SL 4.12 · Standardization

2. Standardization and the Standard Normal Distribution $Z$

Standardization converts any normal random variable $X \sim N(\mu, \sigma^2)$ into the standard normal variable $Z \sim N(0, 1)$ with mean $0$ and standard deviation $1$.

The Z-Score Formula: $$Z = \frac{X - \mu}{\sigma}$$ A $z$-score represents the signed number of standard deviations an observation lies from the mean ($z > 0$ above mean, $z < 0$ below mean).
Equivalence of Probabilities: $$P(X < x) = P\left(Z < \frac{x - \mu}{\sigma}\right)$$
Standardization is mandatory on Paper 1 (non-calculator) and when solving for unknown parameters $\mu$ or $\sigma$ on Paper 2.

Worked Example 1.2 · Z-Score Calculation

A marathon runner finishes in $X = 210\text{ minutes}$. The finishing times are normally distributed with $\mu = 240\text{ minutes}$ and $\sigma = 20\text{ minutes}$. Calculate the runner's $z$-score and interpret its meaning.

Apply the formula $Z = \frac{X - \mu}{\sigma}$: $$z = \frac{210 - 240}{20} = \frac{-30}{20} = -1.5$$
Interpretation: The runner finished $1.5$ standard deviations faster (below) the cohort mean.

Answer: $z = -1.5$ (1.5 standard deviations below the mean).

SL 4.12 · Parameter Recovery

3. Inverse Normal and Solving for Unknown $\mu$ and $\sigma$

When given a probability (or percentage) and asked to find a threshold score or the underlying parameters $\mu$ and $\sigma$, we use the inverse normal function $\text{invNorm}(\text{area})$.

Using $\text{invNorm}$ on the Standard Normal: $$\text{If } P(Z < z) = p \implies z = \text{invNorm}(p, 0, 1)$$ Critical Rule: $\text{invNorm}$ always requires the cumulative area to the left. If given an upper tail $P(Z > z) = \alpha$, input $p = 1 - \alpha$.
Solving for Unknown $\mu$ or $\sigma$:
1. Translate the given probability into a standard normal $z$-score: $z = \text{invNorm}(\text{left area})$.
2. Substitute into the standardization formula: $$z = \frac{x - \mu}{\sigma} \implies x = \mu + z\sigma$$
3. If both $\mu$ and $\sigma$ are unknown, set up a system of two simultaneous linear equations from two given percentiles!

Figure 4.6: Standardization projects raw scores $X$ onto standard normal scores $Z$, preserving exact percentile areas.

Worked Example 1.3 · Inverse Normal Cutoff

The weights of apples in an orchard follow a normal distribution $W \sim N(150, 18^2)$ in grams.
Apples are sorted into three grades: Small (lightest $15\%$), Medium (middle $70\%$), and Large (heaviest $15\%$).
(a) Find the maximum weight of an apple classified as Small. [3 marks]
(b) Find the minimum weight of an apple classified as Large. [2 marks]

Part (a): Small cutoff $w_1$ where $P(W < w_1) = 0.15$:
Standard normal $Z$-value: $Z_1 = \text{invNorm}(0.15, 0, 1) \approx -1.0364$ [M1].
$$w_1 = \mu + Z_1 \sigma = 150 + (-1.0364)(18) = 150 - 18.656 = 131.34\text{ g} \implies 131\text{ g} \quad [A1]$$
Part (b): Large cutoff $w_2$ where $P(W > w_2) = 0.15 \implies P(W < w_2) = 0.85$:
By symmetry of the normal distribution across $\mu = 150$ [R1]:
$$w_2 = 150 + 18.656 = 168.656\text{ g} \implies 169\text{ g} \quad [A1]$$

Answers: (a) $131\text{ g}$; (b) $169\text{ g}$.

Worked Example 1.4 · Finding $\mu$ and $\sigma$ Simultaneously

The heights of adult trees in a forest follow a normal distribution $X \sim N(\mu, \sigma^2)$. It is known that $10\%$ of trees are shorter than $14\text{ m}$ and $5\%$ of trees are taller than $26\text{ m}$.
Find the mean height $\mu$ and standard deviation $\sigma$ of the trees, giving your answers to 3 significant figures. [7 marks]

Step 1: Standard normal critical values from probabilities: $$P(X < 14) = 0.10 \implies Z_1 = \text{invNorm}(0.10) \approx -1.28155 \quad [M1][A1]$$ $$P(X > 26) = 0.05 \implies P(X < 26) = 0.95 \implies Z_2 = \text{invNorm}(0.95) \approx +1.64485 \quad [M1][A1]$$
Step 2: Set up simultaneous equations using $X = \mu + Z\sigma$: $$\mu - 1.28155\sigma = 14 \quad \text{--- (1)}$$ $$\mu + 1.64485\sigma = 26 \quad \text{--- (2)} \quad [M1]$$
Step 3: Subtract equation (1) from (2): $$(1.64485 - (-1.28155))\sigma = 26 - 14$$ $$2.9264\sigma = 12 \implies \sigma = \frac{12}{2.9264} \approx 4.1006 \implies \sigma \approx 4.10\text{ m} \quad [A1]$$
Step 4: Back-substitute into equation (1) to find $\mu$: $$\mu = 14 + 1.28155(4.1006) = 14 + 5.255 = 19.255 \implies \mu \approx 19.3\text{ m} \quad [A1]$$

Answers: $\mu = 19.3\text{ m}, \sigma = 4.10\text{ m}$.

Worked Example 1.5 · Two-Tier Normal to Binomial Model

The lifespan $T$ of a mechanical component is normally distributed with mean $\mu = 800$ hours and standard deviation $\sigma = 60$ hours.
A component is defined as "defective" if its lifespan is less than $700$ hours.
(a) Find the probability that a randomly chosen component is defective. [2 marks]
(b) A random sample of $25$ components is tested. Find the probability that exactly $2$ components in the sample are defective. [3 marks]

Part (a): Normal probability calculation: $$P(T < 700) = \text{normalcdf}(-1\text{E}99, 700, 800, 60) \quad [M1]$$ $$Z = \frac{700 - 800}{60} = -\frac{100}{60} \approx -1.667$$ $$P(T < 700) \approx 0.04779 \implies 0.0478 \quad [A1]$$
Part (b): Binomial probability $Y \sim B(25, 0.04779)$: $$P(Y = 2) = \binom{25}{2} (0.04779)^2 (1 - 0.04779)^{23} \quad [M1]$$ $$P(Y = 2) = 300 \times (0.0022839) \times (0.32353) \approx 0.2216 \implies 0.222 \quad [M1][A1]$$

Answers: (a) $0.0478$; (b) $0.222$.

Topic 4.9 & 4.12 Practice Exercises

10 tiered problems covering normal probabilities, z-scores, inverse normal percentiles, conditional probabilities, and solving for unknown $\mu$ and $\sigma$.

Question 1

Let $X \sim N(50, 16)$ (so variance $\sigma^2 = 16$). Calculate the $z$-score corresponding to:
(a) $x = 58$
(b) $x = 44$

Show Step-by-Step Solution

Step 1: Identify standard deviation: $\sigma = \sqrt{16} = 4$. Mean $\mu = 50$.

Part (a): $z = \frac{x - \mu}{\sigma} = \frac{58 - 50}{4} = \frac{8}{4} = 2.0$.

Part (b): $z = \frac{44 - 50}{4} = \frac{-6}{4} = -1.5$.

Final Answer: (a) $z = 2.0$; (b) $z = -1.5$.

Question 2

The mass $M$ of bags of sugar is distributed as $M \sim N(1000, 25)$ in grams. Write down the percentage of bags that have mass between $995\text{ g}$ and $1005\text{ g}$ using the empirical rule.

Show Step-by-Step Solution

Step 1: Mean $\mu = 1000\text{ g}$, $\sigma = \sqrt{25} = 5\text{ g}$.

Step 2: Notice that $995 = \mu - \sigma$ and $1005 = \mu + \sigma$.

Step 3: By the empirical rule, the area within $1$ standard deviation of the mean is approximately $68.3\%$.

Final Answer: Approximately $68.3\%$.

Question 3

Let $Z \sim N(0, 1)$. Find $k$ such that $P(Z < k) = 0.825$ to 3 significant figures.

Show Step-by-Step Solution

Step 1: Apply the inverse normal function:

$$k = \text{invNorm}(0.825, 0, 1) \approx 0.934589 \approx 0.935$$

Final Answer: $k \approx 0.935$

Question 4

Explain why $P(X \le 25) = P(X < 25)$ for any normally distributed random variable $X$.

Show Step-by-Step Solution

Explanation: The normal distribution is a continuous random variable. In continuous probability models, the probability of obtaining any single exact value $c$ is zero ($P(X = c) = 0$), because area over a point has zero width. Therefore, $P(X \le 25) = P(X < 25) + P(X = 25) = P(X < 25) + 0 = P(X < 25)$.

Final Answer: Because $X$ is continuous and the probability of any exact value is zero ($P(X = 25) = 0$).

Question 5

The height $H$ of adult males in a city follows $H \sim N(\mu, 36)$ with $\sigma = 6\text{ cm}$. Given that $10\%$ of adult males are taller than $185\text{ cm}$, find the mean height $\mu$.

Show Step-by-Step Solution

Step 1: Translate upper tail into left area:

$$P(H > 185) = 0.10 \implies P(H \le 185) = 0.90$$

Step 2: Find standard normal $z$-value:

$$z = \text{invNorm}(0.90, 0, 1) \approx 1.28155$$

Step 3: Solve for $\mu$:

$$z = \frac{185 - \mu}{\sigma} \implies 1.28155 = \frac{185 - \mu}{6}$$ $$185 - \mu = 6(1.28155) = 7.6893 \implies \mu = 185 - 7.6893 = 177.31 \approx 177\text{ cm}$$

Final Answer: $\mu \approx 177\text{ cm}$ (or $177.3\text{ cm}$).

Question 6

The lifetime $T$ of batteries is normally distributed with mean $\mu = 800\text{ hours}$. It is known that $5\%$ of batteries last fewer than $718\text{ hours}$. Find the standard deviation $\sigma$.

Show Step-by-Step Solution

Step 1: Lower tail condition:

$$P(T < 718) = 0.05$$

Step 2: Find $z$-value:

$$z = \text{invNorm}(0.05, 0, 1) \approx -1.64485$$

Step 3: Solve for $\sigma$:

$$-1.64485 = \frac{718 - 800}{\sigma} = \frac{-82}{\sigma}$$ $$\sigma = \frac{-82}{-1.64485} \approx 49.852 \approx 49.9\text{ hours}$$

Final Answer: $\sigma \approx 49.9\text{ hours}$.

Question 7

Exam scores $X$ follow $X \sim N(68, 100)$ ($\mu = 68, \sigma = 10$).
(a) Find $P(60 \le X \le 82)$.
(b) Find the symmetric interval about the mean that contains the middle $80\%$ of students' scores.

Show Step-by-Step Solution

Part (a): GDC normal CDF:

$$P(60 \le X \le 82) = \text{normCdf}(60, 82, 68, 10) \approx 0.7071 \approx 0.707$$

Part (b): Central 80% interval:
If middle area is $0.80$, the two remaining tails each have area: $$\frac{1 - 0.80}{2} = 0.10$$ Lower tail cutoff ($10^{\text{th}}$ percentile): $$z_1 = \text{invNorm}(0.10, 0, 1) \approx -1.28155 \implies x_1 = 68 - 1.28155(10) = 68 - 12.82 = 55.18 \approx 55.2$$ Upper tail cutoff ($90^{\text{th}}$ percentile): $$z_2 = +1.28155 \implies x_2 = 68 + 12.82 = 80.82 \approx 80.8$$ Interval: $[55.2, 80.8]$.

Final Answer: (a) $0.707$; (b) $[55.2, 80.8]$.

Question 8

The weights of apples are distributed as $W \sim N(150, 144)$ in grams ($\sigma = 12\text{ g}$).
(a) Calculate $P(W > 165)$.
(b) Given that an apple weighs more than $140\text{ g}$, find the probability that it weighs more than $165\text{ g}$.

Show Step-by-Step Solution

Part (a):

$$P(W > 165) = \text{normCdf}(165, 10^{99}, 150, 12) \approx 0.10565 \approx 0.106$$

Part (b): Conditional probability:

$$P(W > 165 | W > 140) = \frac{P(W > 165 \cap W > 140)}{P(W > 140)} = \frac{P(W > 165)}{P(W > 140)}$$ $$P(W > 140) = \text{normCdf}(140, 10^{99}, 150, 12) \approx 0.79767$$ $$P(W > 165 | W > 140) = \frac{0.10565}{0.79767} \approx 0.13245 \approx 0.132$$

Final Answer: (a) $0.106$; (b) $0.132$.

Question 9

The masses $X$ of melons harvested on an agricultural farm are normally distributed with unknown mean $\mu$ and standard deviation $\sigma$.
• $12\%$ of the melons weigh less than $1.40\text{ kg}$.
• $18\%$ of the melons weigh more than $2.20\text{ kg}$.

(a) Set up a system of two linear equations in $\mu$ and $\sigma$.
(b) Solve the equations to find $\mu$ and $\sigma$ to 3 significant figures.

Show Step-by-Step Solution

Part (a): System of equations:
Condition 1: $P(X < 1.40) = 0.12$. $$z_1 = \text{invNorm}(0.12, 0, 1) \approx -1.17499$$ $$\frac{1.40 - \mu}{\sigma} = -1.17499 \implies 1.40 - \mu = -1.17499\sigma \implies \mu - 1.17499\sigma = 1.40 \quad (1)$$ Condition 2: $P(X > 2.20) = 0.18 \implies P(X \le 2.20) = 0.82$. $$z_2 = \text{invNorm}(0.82, 0, 1) \approx 0.91537$$ $$\frac{2.20 - \mu}{\sigma} = 0.91537 \implies 2.20 - \mu = 0.91537\sigma \implies \mu + 0.91537\sigma = 2.20 \quad (2)$$

Part (b): Solve simultaneously:
Subtract equation (1) from equation (2): $$(\mu + 0.91537\sigma) - (\mu - 1.17499\sigma) = 2.20 - 1.40$$ $$(0.91537 + 1.17499)\sigma = 0.80 \implies 2.09036\sigma = 0.80$$ $$\sigma = \frac{0.80}{2.09036} \approx 0.3827 \approx 0.383\text{ kg}$$ Substitute $\sigma$ into equation (2): $$\mu = 2.20 - 0.91537(0.3827) = 2.20 - 0.3503 = 1.8497 \approx 1.85\text{ kg}$$

Final Answer: (a) System established; (b) Mean $\mu \approx 1.85\text{ kg}$, standard deviation $\sigma \approx 0.383\text{ kg}$.

Question 10

The volume of beverage $V$ dispensed by an automated machine is normally distributed with mean $250\text{ mL}$ and standard deviation $5\text{ mL}$. Cups that contain less than $242\text{ mL}$ are deemed underfilled.
(a) Find the probability that a randomly chosen cup is underfilled.
(b) A tray of 12 independent cups is prepared. Find the probability that at least 2 cups on the tray are underfilled.
(c) The machine is recalibrated by adjusting the mean $\mu$, while $\sigma$ remains $5\text{ mL}$. Determine the new setting for $\mu$ so that no more than $0.5\%$ of cups are underfilled.

Show Step-by-Step Solution

Part (a): Underfilled probability:
$V \sim N(250, 25)$. $$z = \frac{242 - 250}{5} = \frac{-8}{5} = -1.6$$ $$P(V < 242) = \text{normCdf}(-10^{99}, 242, 250, 5) \approx 0.054799 \approx 0.0548$$

Part (b): Binomial connection:
Let $Y$ be the number of underfilled cups in the tray of $12$: $Y \sim B(12, p)$ with $p = 0.054799$. $$P(Y \ge 2) = 1 - P(Y \le 1) = 1 - \text{binomCdf}(12, 0.054799, 1)$$ $$P(Y = 0) = (1 - 0.0548)^{12} \approx (0.9452)^{12} \approx 0.5073$$ $$P(Y = 1) = 12(0.0548)(0.9452)^{11} \approx 12(0.0548)(0.5367) \approx 0.3529$$ $$P(Y \le 1) \approx 0.5073 + 0.3529 = 0.8602$$ $$P(Y \ge 2) = 1 - 0.8602 = 0.1398 \approx 0.140$$

Part (c): Recalibrating mean:
We require $P(V < 242) \le 0.005$.
$z = \text{invNorm}(0.005, 0, 1) \approx -2.57583$. $$\frac{242 - \mu_{\text{new}}}{5} = -2.57583$$ $$242 - \mu_{\text{new}} = 5(-2.57583) = -12.879$$ $$\mu_{\text{new}} = 242 + 12.879 = 254.879 \approx 255\text{ mL}$$

Final Answer: (a) $0.0548$; (b) $0.140$; (c) $\mu_{\text{new}} \ge 255\text{ mL}$.

IB Formula Booklet References (Topic 4)

Standardized Normal Variable:

$$z = \frac{x - \mu}{\sigma}$$

Must-Memorize Symmetry & Inverse Properties

Symmetry Across the Mean:

$$P(Z < -z) = P(Z > z) = 1 - P(Z < z)$$

Standard Deviation from Variance:

$$\sigma = \sqrt{\sigma^2} \quad \text{Always check whether given } N(\mu, \sigma) \text{ or } N(\mu, \sigma^2)$$

Inverse Normal Transformation:

$$x = \mu + z\sigma, \quad \text{where } z = \text{invNorm}(\text{area to the left})$$

Critical IB Examination Pitfalls

Dividing by Variance $\sigma^2$ instead of $\sigma$: In $Z = \frac{X - \mu}{\sigma}$, the denominator is standard deviation $\sigma$. If the distribution is written $N(40, 25)$, then $\sigma = 5$, not $25$!
Feeding Upper Tail into $\text{invNorm}$: If the question states "the top $15\%$ exceed score $k$", the input to $\text{invNorm}$ must be $1 - 0.15 = 0.85$, not $0.15$. Inputting $0.15$ produces the negative lower-tail cutoff!
Forgetting Continuity Equality: Writing $P(X \le k) \neq P(X < k)$ is a serious conceptual error in continuous distributions; both are strictly equal because $P(X = k) = 0$.
Sign Errors in Simultaneous Equations: When setting up simultaneous equations for two tails, the lower tail $z$-score is negative ($z_1 < 0$) and the upper tail $z$-score is positive ($z_2 > 0$). Dropping the negative sign on $z_1$ makes solving for $\sigma$ impossible.

Topic 4.9 & 4.12 Diagnostic Mini Test

Section A: 4 Diagnostic MCQs with immediate feedback. Section B: 2 Exam-style Free Response Questions with full markschemes.

Section A: Diagnostic Multiple Choice

Question 1: For a normal variable $X \sim N(80, 25)$, what is the $z$-score corresponding to $x = 70$?

Question 2: What is the probability $P(X = 50)$ for a normal random variable $X \sim N(50, 9)$?

Question 3: If $P(Z > k) = 0.05$ where $Z \sim N(0, 1)$, which calculator command yields the correct value of $k$?

Question 4: The points of inflection on the normal probability curve $X \sim N(\mu, \sigma^2)$ occur at:

Section B: Examination-Style Free Response Questions

Question 5 (6 marks)

The journey times $T$ of commuter trains are distributed as $T \sim N(45, 16)$ in minutes ($\sigma = 4\text{ min}$).
(a) Find the probability that a randomly chosen train journey takes more than $50\text{ minutes}$. [2 marks]
(b) The railway authority promises a refund if a journey is in the longest $2.5\%$ of journey times. Find the threshold journey time that qualifies for a refund. [4 marks]

Show Full Markscheme

(a) Probability over 50 minutes:
$z = \frac{50 - 45}{4} = 1.25$. [M1]
$P(T > 50) = P(Z > 1.25) \approx 0.1056 \approx 0.106$. [A1]

(b) Threshold time for top 2.5%:
Upper tail area $= 0.025 \implies$ left cumulative area $= 0.975$. [M1]
$z = \text{invNorm}(0.975, 0, 1) \approx 1.95996 \approx 1.96$. [A1]
$T_{\text{threshold}} = \mu + z\sigma = 45 + 1.96(4) = 45 + 7.84 = 52.84 \approx 52.8\text{ minutes}$. [M1 A1]

Question 6 (8 marks)

The volume of liquid $L$ in soft drink cans is normally distributed with mean $\mu$ and standard deviation $\sigma$.
• $10\%$ of cans contain less than $325\text{ mL}$
• $15\%$ of cans contain more than $338\text{ mL}$

(a) State the $z$-scores corresponding to $325\text{ mL}$ and $338\text{ mL}$. [2 marks]
(b) Find the values of $\mu$ and $\sigma$ to 3 significant figures. [4 marks]
(c) A can is selected at random. Given that it contains more than $330\text{ mL}$, find the probability that it contains more than $338\text{ mL}$. [2 marks]

Show Full Markscheme

(a) Z-scores:
For $325\text{ mL}$: $P(L < 325) = 0.10 \implies z_1 = \text{invNorm}(0.10) \approx -1.28155$. [A1]
For $338\text{ mL}$: $P(L > 338) = 0.15 \implies P(L \le 338) = 0.85 \implies z_2 = \text{invNorm}(0.85) \approx +1.03643$. [A1]

(b) Solve for $\mu$ and $\sigma$:
$325 - \mu = -1.28155\sigma \implies \mu - 1.28155\sigma = 325$
$338 - \mu = 1.03643\sigma \implies \mu + 1.03643\sigma = 338$ [M1]
Subtract: $(1.03643 + 1.28155)\sigma = 338 - 325 \implies 2.31798\sigma = 13$.
$\sigma = \frac{13}{2.31798} \approx 5.608 \approx 5.61\text{ mL}$. [A1]
$\mu = 338 - 1.03643(5.608) \approx 338 - 5.812 \approx 332.188 \approx 332\text{ mL}$. [A2]

(c) Conditional probability:
$P(L > 330) = \text{normCdf}(330, 10^{99}, 332.19, 5.608) \approx 0.6520$.
$P(L > 338 | L > 330) = \frac{P(L > 338)}{P(L > 330)} = \frac{0.15}{0.6520} \approx 0.230$. [M1 A1]