Normal Distribution (AA HL)
The normal distribution is the bell-shaped curve behind heights, test scores, measurement errors, and countless other continuous quantities. This sub-topic focuses on the core skill: turning a raw value into a \(z\)-score and reading off a probability from the GDC. It's part of the broader Probability Distributions topic.
21 questions on this sub-topic.
The key formulas
Covered under IB syllabus reference SL4.9: the normal distribution and its curve, with roughly 68% of data within 1 standard deviation of the mean, 95% within 2, and 99.7% within 3. Every normal probability - and its reverse, inverse normal - is found using the GDC rather than by hand.
Standardizing
\(z=\dfrac{x-\mu}{\sigma}\)
Converts any \(X\sim N(\mu,\sigma^2)\) value into the equivalent \(z\)-score on the standard normal curve \(N(0,1)\).
Inverse normal
Given a probability, find the corresponding \(x\)-value - always found using the GDC's inverse normal function, working from the area to the LEFT of the value.
Need the full syllabus wording and formula-booklet reference table? See Probability Distributions. For calculator steps, see the parent topic's GDC guidance.
Worked examples
\(X\sim N(50, 8^2).\) Find \(P(X<58).\)
Worked solution
\(z = \dfrac{58 - 50}{8}\) M1
\(= 1.\) A1
\(P(Z<1)\approx 0.841.\) A1
\(X\sim N(100, 15^2).\) Find \(P(85
Worked solution
\(z_1 = \dfrac{85-100}{15} = -1,\ z_2\) M1
\(= 1.\) A1
\(P(-1<Z<1)\) M1
\(\approx 0.683.\) A1
Heights of men: \(N(175, 7^2)\); women: \(N(162, 6^2)\) (cm).
Find \(P(\text{a random man} > 180)\) and \(P(\text{a random woman} > 180).\)
Worked solution
\(z = \dfrac{180-175}{7} \approx 0.714\); \(P(X>180)\) M1
\(z\approx0.714\) A1
\(\approx 0.238.\) A1
\(z = \dfrac{180-162}{6} = 3\); \(P(X>180)\) M1
\(z=3\) A1
\(\approx 0.00135.\) A1
The mass of apples is \(X\sim N(150, 20^2)\) g.
(a) Find the probability an apple weighs more than 180 g.
(b) In a crate of 200 apples, estimate how many weigh more than 180 g.
Worked solution
(a) \(P(X>180)\). Standardise: \(z=\dfrac{180-150}{20}=1.5;\) M1
\(P(X>180)=P(Z>1.5)\approx 0.0668.\) A1
(b) In 200 apples. Expected number \(=200\times0.0668\) M1
\(\approx 13\) apples. A1
Common mistakes
- Using the wrong bound for "less than" or "greater than". normalcdf needs a numerical lower AND upper bound - for "less than \(k\)" use a very large negative number (e.g. \(-1\times10^{99}\)) as the lower bound, and for "greater than \(k\)" use a very large positive number as the upper.
- Forgetting the two-tail shortcut. When a probability is symmetric about the mean, such as \(P(85
- Reporting the \(z\)-value instead of the actual \(x\)-value. If a question asks for a probability, give the probability from the normalcdf output, not the intermediate \(z\)-score used to get there.
Ready to practise properly?
21 normal-distribution questions, marked instantly like the real exam.
Quick answers
How do you find a normal probability on the GDC?
Use the normal cdf function with the lower and upper bounds, the mean, and the standard deviation. For "less than" use a very large negative number as the lower bound; for "greater than" use a very large positive number as the upper bound.
What does standardizing a normal variable mean?
It converts a value \(x\) from \(X\sim N(\mu,\sigma^2)\) into a \(z\)-score using \(z=\dfrac{x-\mu}{\sigma}\), measuring how many standard deviations \(x\) is from the mean on the standard normal curve \(N(0,1)\).