4.9 The normal distributionIB Maths: Analysis and Approaches HL: Revision notes

A continuous random variable $X$ with a normal distribution is written $X\sim N(\mu,\sigma^{2})$. Here $\mu$ is the mean and $\sigma^{2}$ is the variance, so $\sigma$ is the standard deviation. Its graph is the normal curve:

• bell-shaped and symmetrical about $x=\mu$, so the mean, median and mode are equal;
• the total area under the curve is 1, and the area between two values is the probability of lying between them;
• a larger $\sigma$ gives a wider, flatter curve.

The normal distribution occurs naturally when a quantity is affected by many small, independent influences. Examples include heights, masses of fruit, measurement errors and machine-filled quantities. Because $X$ is continuous, $P(X=a)=0$, so $P(X<a)$ and $P(X\le a)$ are equal.

For any normal distribution, approximately:

• 68% of values lie within $\mu\pm\sigma$;
• 95% lie within $\mu\pm2\sigma$;
• 99.7% lie within $\mu\pm3\sigma$.

Combine this with symmetry to estimate tail areas without a calculator. For example, 68% lie within one standard deviation, so 32% lie outside it, and 16% lie above $\mu+\sigma$. Similarly, 2.5% lie above $\mu+2\sigma$. For $N(150,12^{2})$, about 95% of apples weigh between 126 g and 174 g.

Use the GDC's normal cdf with a lower bound, an upper bound, $\mu$ and $\sigma$:

• $P(X<a)$: lower bound a very large negative number (e.g. $-10^{99}$), upper bound $a$;
• $P(X>a)$: lower bound $a$, upper bound a very large positive number;
• $P(a<X<b)$: bounds $a$ and $b$.

Write the probability statement first, e.g. $P(140<X<165)=0.692$. Expected numbers work as for any probability: out of 400 batteries, each with probability $0.0599$ of failing before 550 hours, about $400\times0.0599=24.0$ are expected to fail.

When you know a probability and want the value, use inverse normal. Most calculators need the area to the left:

• '10% of journeys take longer than $k$': $P(T>k)=0.1$, so $P(T<k)=0.9$ and $k=48.4$;
• 'exceeded by 99% of bags': $P(M>m)=0.99$, so $P(M<m)=0.01$ and $m=993.4$ g;
• quartiles: $Q_1$ has area 0.25 to the left and $Q_3$ has area 0.75, so $\text{IQR}=Q_3-Q_1$.

At this level $\mu$ and $\sigma$ are always given. In context, round in the direction the situation demands. A guarantee that must affect at most 2% of batteries uses $g=527$, not 528, because $P(X<528)$ is just over 0.02.

Exam questions often use a normal probability as an input to another calculation.

• Binomial: if each bag is underweight with probability $p=0.0668$, the number of underweight bags in a box of 12 is $B(12,p)$, so $P(\text{at least one})=1-(1-p)^{12}=0.564$.
• Conditional probability: $P(M>1020\mid M\ge1000)=\frac{P(M>1020)}{P(M\ge1000)}=0.170$. The event 'more than 1020 g' lies inside 'at least 1000 g', so the intersection is just $P(M>1020)$.