4.12 Standardisation and inverse normal with unknown parametersIB Maths: Analysis and Approaches SL: Revision notes

If $X\sim N(\mu,\sigma^{2})$, the standardised value or $z$-value of $x$ is $z=\frac{x-\mu}{\sigma}.$ It gives the number of standard deviations that $x$ lies from the mean. A positive $z$ means above the mean and a negative $z$ means below it. Rearranged, $x=\mu+z\sigma$.

Standardising turns any normal variable into the standard normal variable $Z\sim N(0,1)$, and $P(X<x)=P\left(Z<\frac{x-\mu}{\sigma}\right).$ For $X\sim N(50,4^{2})$: $P(X>56)=P(Z>1.5)=0.0668$.

$z$-values let you compare performances measured on different scales. Amira scored 72 where $\mu=60$ and $\sigma=8$, so $z=1.5$. Ben scored 81 where $\mu=70$ and $\sigma=5$, so $z=2.2$. Ben's mark is further above his group's mean, relative to the spread, so Ben did relatively better, even though both marks are above average.

To find an equivalent value in another distribution, keep the $z$-value and use $x=\mu+z\sigma$ with the new parameters: $70+1.5\times5=77.5$.

If $\mu$ or $\sigma$ is unknown, you cannot use inverse normal on $X$ directly. Use the standard normal instead:

1. Write the probability statement, e.g. $P(T<40)=0.2$.
2. Use inverse normal on $Z\sim N(0,1)$ (mean 0, sd 1) to find $z$: area 0.2 to the left gives $z=-0.8416$.
3. Set $\frac{x-\mu}{\sigma}=z$ and solve: $\frac{40-\mu}{6}=-0.8416$ gives $\mu=45.0$.

For an upper tail, convert first: $P(X>a)=0.1$ means $P(X<a)=0.9$, so $z=1.2816$.

Two probability conditions give two equations. For bolts with 5% shorter than 49.2 mm and 10% longer than 50.9 mm: $\frac{49.2-\mu}{\sigma}=-1.6449, \qquad \frac{50.9-\mu}{\sigma}=1.2816.$ Rearrange to $\mu-1.6449\sigma=49.2$ and $\mu+1.2816\sigma=50.9$, then solve simultaneously (by subtraction or with the GDC's equation solver): $\sigma=0.581$ and $\mu=50.2$.

For symmetric conditions about the mean, such as '3% differ from $\mu$ by more than $d$', put half the probability in each tail: $P(Z>\frac{d}{\sigma})=0.015$, so $d=2.170\sigma$.