Normal Distribution (AA SL)
The normal distribution is the familiar bell-shaped curve used to model continuous data that clusters around a mean, such as heights, exam marks, or measurement error. This page covers the shape of the curve, the empirical rule for standard deviations, and reading off probabilities on the GDC, with worked examples and the mistakes that lose marks. It's part of the broader Random Variables & Distributions topic.
14 questions on this sub-topic.
Notation and the empirical rule
Covered under IB syllabus reference SL4.9: the normal distribution and curve, and the properties of the normal distribution. Normal probability and inverse normal calculations are found using technology.
Notation
\(X\sim N(\mu,\sigma^2)\)
\(\mu\) is the mean and \(\sigma^2\) is the variance. This is standard notation rather than a formula, so it isn't listed in the formula booklet.
68-95-99.7 rule
\(\approx 68\%\) within \(\mu\pm\sigma\), \(95\%\) within \(\mu\pm2\sigma\), \(99.7\%\) within \(\mu\pm3\sigma\).
Handy for a quick sanity check on a normalcdf answer, and for questions phrased around whole numbers of standard deviations.
Need the full syllabus wording and formula-booklet reference table? See Random Variables & Distributions.
Worked examples
\(X \sim N(100, 15^2)\).
Find \(P(85 < X < 115)\).
Worked solution
\(85 = 100-15,\ 115 = 100+15,\) so \(z=\pm1.\) R1
\(P(85<X<115) = 0.6827\ldots\) A1
\(\approx 0.683.\) A1
On the GDC: \(\texttt{normalcdf(85, 115, 100, 15)}\approx 0.683.\)
\(X\sim N(50, 8^2)\).
Find \(P(X>58)\), to 3 significant figures.
Worked solution
using \(z=\dfrac{x-\mu}{\sigma}\) with \(\mu=50,\ \sigma=8:\) \(z=\frac{58-50}{8}=1,\) so \(P(X>58)=P(Z>1).\) M1
Since \(P(Z<1)=0.8413,\) \(P(Z>1)=1-0.8413=0.159\) (to 3 significant figures). A1
Faster on the GDC: \(\texttt{normalcdf(58, 1E99, 50, 8)}\approx 0.159\) - no standardising needed, using a large upper bound for "greater than".
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
- Entering variance instead of standard deviation. The GDC's normalcdf and invNorm functions want \(\sigma\), not \(\sigma^2\) - if a question states \(X\sim N(50,8^2)\), the value to type in is 8, not 64.
- Forgetting the curve is symmetric. \(P(X<\mu)=0.5\) and areas either side of a symmetric interval like \(\mu\pm\sigma\) are equal - spotting this shortcut saves a full calculator entry.
- Using the wrong sign of infinity for one-sided regions. "Less than" needs a very negative lower bound (e.g. \(-1\text{E}99\)); "greater than" needs a very large upper bound (e.g. \(1\text{E}99\)) - mixing these up silently gives the complement of the intended answer.
Ready to practise properly?
14 normal distribution questions, marked instantly like the real exam.
Quick answers
What percentage of data lies within one standard deviation of the mean in a normal distribution?
Approximately 68% of the data lies between \(\mu-\sigma\) and \(\mu+\sigma\), 95% within two standard deviations, and 99.7% within three.
How do you find a normal probability on the GDC?
Use \(\texttt{normalcdf(lower, upper, mean, standard deviation)}\). Use a very large or very small bound like \(1\text{E}99\) or \(-1\text{E}99\) for one-sided regions such as "greater than" or "less than".