Properties of the Normal Distribution $X \sim N(\mu, \sigma^2)$
A continuous random variable $X$ follows a normal distribution with mean $\mu$ and variance $\sigma^2$ (standard deviation $\sigma$):
- Symmetry: Symmetrical bell shape centered at the mean, median, and mode: $x = \mu$.
- Total Area: The total area under the probability density curve equals exactly $1$ ($100\%$).
- Points of Inflexion: Located at $x = \mu - \sigma$ and $x = \mu + \sigma$.
- Continuity: For any single point $a$, $P(X = a) = 0$. Hence $P(X \le a) = P(X < a)$.
Standard Normal Distribution & $z$-Scores
The standard normal distribution $Z$ has mean $0$ and variance $1$: $Z \sim N(0, 1)$. Any normal variable $X$ is standardized using:
The $z$-score measures how many standard deviations an observation lies above (positive) or below (negative) the mean.
GDC Workflows: Finding Probabilities & Inverse Normal
In IB Math AI SL, calculations are conducted directly via calculator distribution menus:
1. Probability from bounds: Use normCDF(lower, upper, μ, σ)
- For $P(X < b)$: lower $= -1\text{E}99$, upper $= b$.
- For $P(X > a)$: lower $= a$, upper $= 1\text{E}99$.
2. Boundary value from probability (percentile): Use invNorm(area, μ, σ)
- Input area is always the cumulative area to the left (unless your GDC allows specifying RIGHT or CENTER tail).
Determining Unknown $\mu$ and $\sigma$
When the parameters $\mu$ or $\sigma$ are unknown, standard normal $z$-scores must be used:
- Find the corresponding $z$-value using
invNorm(probability, 0, 1). - Set up the linear equation: $z = \frac{x - \mu}{\sigma} \iff x = \mu + z\sigma$.
- If both $\mu$ and $\sigma$ are unknown, two equations from two given probabilities are formed and solved simultaneously.
Check a normal model and condition on selection
- Inspect symmetry, outliers and a roughly bell-shaped distribution. A normal model has unbounded tails, so consider whether predictions below0or beyond physical limits are negligible. A sample mean alone does not establish normality.
- For b>a, $P(X>b\mid X>a)=P(X>b)/P(X>a)$. The denominator represents the selected group, not the entire population.
- Calculator workflow: enter standard deviation, not variance; sketch the requested tail before selecting CDF or inverse CDF. Recheck the boundary by putting it back into the CDF.
A random variable $X$ is normally distributed with mean $\mu = 50$ and standard deviation $\sigma = 8$. Find $P(42 < X < 62)$.
Solution:
Using GDC command $\text{normCDF}(42, 62, 50, 8)$: [M1]
$P(42 < X < 62) = 0.774537...$ [A1]
To 3 significant figures: $0.775$ (or $77.5\%$). [A1]
The weights of packages are modeled by $W \sim N(3.5, 0.4^2)$ kg. Find the probability that a randomly chosen package weighs more than $4.1\text{ kg}$.
Solution:
$P(W > 4.1) = \text{normCDF}(4.1, 1\text{E}99, 3.5, 0.4)$ [M1]
$= 0.066807...$ [A1]
To 3 sig figs: $0.0668$ (or $6.68\%$). [A1]
Exam scores follow $S \sim N(68, 12^2)$. The top $10\%$ of students are awarded a grade 7 certificate. Find the minimum score required to receive a grade 7.
Solution:
Let $k$ be the threshold score. $P(S > k) = 0.10 \implies P(S < k) = 0.90$. [M1]
Using GDC: $k = \text{invNorm}(0.90, 68, 12)$ [M1]
$k = 83.378...$ [A1]
Minimum score is $83.4$ (or integer score 84). [A1]
Convert an observed score of $x = 118$ from a distribution with $\mu = 100$ and $\sigma = 15$ to a standardized $z$-score, and interpret its meaning.
Solution:
$z = \frac{x - \mu}{\sigma} = \frac{118 - 100}{15} = \frac{18}{15} = 1.20$ [M1][A1]
Interpretation: The observation lies $1.2$ standard deviations above the population mean. [R1]
A variable $X \sim N(\mu, 5^2)$ has $P(X < 32) = 0.12$. Find the value of the population mean $\mu$.
Solution:
Find the standard normal $z$-value: $P(Z < z) = 0.12$ [M1]
$z = \text{invNorm}(0.12, 0, 1) = -1.174986...$ [A1]
Apply the transformation: $z = \frac{32 - \mu}{5}$ [M1]
$-1.174986 \times 5 = 32 - \mu \implies -5.8749 = 32 - \mu$ [M1]
$\mu = 32 + 5.8749 = 37.87... \approx 37.9$ [A1]
A machine fills soda cans. The volume of liquid $V$ follows $V \sim N(\mu, \sigma^2)$. If $P(V < 325) = 0.05$ and $P(V > 335) = 0.10$, calculate $\mu$ and $\sigma$.
Solution:
For $P(V < 325) = 0.05$: $z_1 = \text{invNorm}(0.05, 0, 1) = -1.64485$ [A1]
Equation 1: $\frac{325 - \mu}{\sigma} = -1.64485 \implies 325 = \mu - 1.64485\sigma$ [M1]
For $P(V > 335) = 0.10 \implies P(V < 335) = 0.90$: $z_2 = \text{invNorm}(0.90, 0, 1) = 1.28155$ [A1]
Equation 2: $\frac{335 - \mu}{\sigma} = 1.28155 \implies 335 = \mu + 1.28155\sigma$ [M1]
Subtract Eq 1 from Eq 2: $10 = (1.28155 + 1.64485)\sigma = 2.9264\sigma \implies \sigma = \frac{10}{2.9264} \approx 3.42$ [A1]
Substitute $\sigma$: $\mu = 335 - 1.28155(3.417) \approx 330.6 \approx 331\text{ ml}$ [A1]
Explain why $P(X \le 70) = P(X < 70)$ for a normal random variable $X$.
Solution:
The normal distribution is a continuous distribution. [R1]
The probability of any exact single point is zero: $P(X = 70) = 0$. [A1]
Therefore, $P(X \le 70) = P(X < 70) + P(X = 70) = P(X < 70)$. [R1]
In a production run of 500 light bulbs whose lifespans are modeled by $T \sim N(1200, 150^2)$ hours, estimate the expected number of bulbs that will last fewer than 1000 hours.
Solution:
$P(T < 1000) = \text{normCDF}(-1\text{E}99, 1000, 1200, 150)$ [M1]
$= 0.091211...$ [A1]
Expected count $= 500 \times 0.091211 = 45.605...$ [M1]
Approximately $46$ light bulbs. [A1]
Sketch or describe the effect on the normal curve when the standard deviation $\sigma$ is doubled while the mean $\mu$ remains constant.
Solution:
The curve remains centered at the same mean value $x = \mu$. [A1]
The curve becomes wider and more spread out (greater dispersion). [A1]
The peak height at $\mu$ decreases so that the total area remains exactly 1. [R1]
A symmetrical interval around the mean contains $80\%$ of the population for $X \sim N(200, 25^2)$. Find the boundaries of this interval $[200 - d, 200 + d]$.
Solution:
If $80\%$ lies in the center, each outer tail has $\frac{1 - 0.80}{2} = 0.10$ probability. [M1]
Lower boundary $= \text{invNorm}(0.10, 200, 25) = 167.96... \approx 168$ [A1]
Upper boundary $= \text{invNorm}(0.90, 200, 25) = 232.03... \approx 232$ [A1]
Interval: $[168, 232]$ (or $d \approx 32.0$). [A1]
For X∼N(100,10²), find P(X>120|X>110).
Tails are0.0227501and0.1586553. Ratio≈0.143393or14.3%. Do not subtract the two tail probabilities for a conditional question.
Waiting times have mean5minutes andSD4minutes with a strong right skew. Explain a normal model risk.
Normal N(5,16) predicts P(X<0)=Φ(−1.25)≈0.10565, impossible for a waiting time. The skew and substantial negative tail make this model unsuitable.
Core Formulas (Topic 4.10)
| $z$-Score Standardization | $$z = \frac{x - \mu}{\sigma} \iff x = \mu + z\sigma$$ |
| GDC Probability Function | $$P(a < X < b) = \text{normCDF}(a, b, \mu, \sigma)$$ |
| GDC Inverse Normal | $$k = \text{invNorm}(p, \mu, \sigma) \quad \text{where } P(X < k) = p$$ |
| Total Area & Symmetry | $$P(X < \mu) = P(X > \mu) = 0.5$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing Variance with Standard Deviation: When written as $N(50, 16)$, $\sigma^2 = 16$, which means $\sigma = 4$! Do not enter 16 into the GDC standard deviation prompt.
- Trap 2: Feeding upper tail percentage into invNorm: For "top $15\%$", the cumulative area to the left is $1 - 0.15 = 0.85$. Entering $0.15$ finds the bottom $15\%$.
- Trap 3: Forgetting the sign of $z$: Values below the mean have negative $z$-scores. Omitting the minus sign when setting up simultaneous equations causes major errors.
- Trap 4: Rounding intermediate $z$-scores too early: Store full calculator accuracy (e.g. $z = -1.28155...$) to avoid 3-sig-fig rounding errors in $\mu$ and $\sigma$.
- Trap 5: Using pdf instead of cdf: Never use
normPDFfor probabilities! In continuous distributions, probability is area, which requiresnormCDF.
5-Question Concept Check
Score: - / 51. For $X \sim N(100, 25)$, what is the numerical value of the standard deviation $\sigma$?
2. What is $P(X = 45)$ for any continuous normal distribution $X \sim N(\mu, \sigma^2)$?
3. To find the score $k$ exceeded by the top $20\%$ of candidates, what area should be entered into standard left-tail `invNorm`?
4. According to the empirical rule, approximately what percentage of observations lie within 2 standard deviations of the mean?
5. If an observation has $z = -1.5$, what does this indicate?