4.9 Normal distributionIB Maths: Applications and Interpretation SL: Revision notes

A normal distribution models a continuous variable whose values cluster symmetrically around a mean. We write $X\sim N(\mu,\sigma^2)$, where $\mu$ is the mean and $\sigma$ is the standard deviation. The graph of the normal curve is a bell shape. Its properties are:

• it is symmetrical about the vertical line $x=\mu$, so the mean, median and mode are all equal to $\mu$;
• the total area under the curve is $1$, because the area represents probability;
• the curve approaches the horizontal axis on either side of the mean but never touches it;
• a larger $\sigma$ gives a wider, flatter curve and a smaller $\sigma$ a narrower, taller one. The normal distribution occurs naturally in many measurements, such as heights, masses of packets, measurement errors and exam scores, where many small independent factors combine. When you sketch it, draw a symmetric bell, mark $\mu$ on the axis and shade the area that matches the probability asked for. Because the variable is continuous, the probability of any single exact value is zero, so $P(X=a)=0$ and $P(X<a)=P(X\leq a)$.

For any normal distribution, the proportion of the data within a given number of standard deviations of the mean is fixed:

• about $68\%$ lies between $\mu-\sigma$ and $\mu+\sigma$;
• about $95\%$ lies between $\mu-2\sigma$ and $\mu+2\sigma$;
• about $99.7\%$ lies between $\mu-3\sigma$ and $\mu+3\sigma$. By symmetry, the area splits equally between the two tails. Example: masses are $N(500,8^2)$. Between $484$ and $516$ is $\mu\pm2\sigma$, so about $95\%$ of masses lie there. Above $508=\mu+\sigma$ lies half of the remaining $32\%$, which is $16\%$. The rule gives quick estimates and checks. For exact values you use technology.

In the exam you find probabilities for $X\sim N(\mu,\sigma^2)$ with your GDC (normal cdf). Enter the lower bound, the upper bound, $\mu$ and $\sigma$.

• $P(a<X<b)$: lower bound $a$, upper bound $b$.
• $P(X<b)$: lower bound a very small number such as $-10^{99}$, upper bound $b$.
• $P(X>a)$: lower bound $a$, upper bound a very large number such as $10^{99}$. Equivalently $1-P(X<a)$. Example: $T\sim N(42,6^2)$. Then $P(T>50)=0.0912$ and $P(40<T<45)=0.322$. If a question asks for a number of items, multiply the probability by the total, so with $240$ eggs and $P=0.0478$ the expected number is $240\times0.0478=11.5$. Always give answers to $3$ significant figures unless told otherwise, and sketch a curve with the area shaded to check that your answer is sensible, for example less than $0.5$ for a tail beyond the mean.

An inverse normal calculation goes the other way: you know a probability and want the value of the variable. The mean and standard deviation are given, and you use the GDC inverse normal function. You do not need to convert to a standard normal variable $z$. The GDC works with the area to the left of the value. So:

• if $P(X<k)=p$, enter area $p$;
• if $P(X>k)=p$, enter area $1-p$. Example: $T\sim N(42,6^2)$ and the longest $10\%$ of journeys take more than $k$ minutes. Then $P(T<k)=0.9$ and $k=49.7$. Example: $M\sim N(62,3^2)$ and the lightest $5\%$ of eggs are called small. The area to the left is $0.05$, so the greatest mass of a small egg is $57.1$ g. Check the value is on the correct side of the mean: a top percentage gives a value above $\mu$, a bottom percentage gives one below $\mu$.

Modelling questions give a scenario and ask you to interpret results in context. Use the same plan each time: define the variable, write $X\sim N(\mu,\sigma^2)$, sketch and shade, then use your GDC.

• To compare two distributions with the same mean, the one with the larger $\sigma$ is more spread out, so it has more values in both tails.
• To describe where about $95\%$ of values lie, use $\mu\pm2\sigma$.
• To find the proportion above or below a limit, use normal cdf; to find a limit for a given proportion, use inverse normal. Example: Brand A battery lifetimes are $N(120,15^2)$ and Brand B are $N(120,25^2)$. The probability of a lifetime above $150$ hours is $0.0228$ for A and $0.115$ for B, so B has more very long-lasting batteries but also more very short-lived ones. A normal model is only an approximation, so state conclusions as estimates: 'about $11$ eggs per box', not exactly $11$.