Statistics · The Normal Distribution · Year 2
Finding μ and σ
Turning known probabilities into equations for the unknown parameters of a normal distribution.
Recovering unknown parameters
You might need to find an unknown mean or standard deviation for a normally distributed variable, given one or more probabilities. Since you don't know $\mu$ or $\sigma$ directly, you work with the standard normal distribution instead, and then convert back using the coding formula $Z=\dfrac{X-\mu}{\sigma}$.
Worked example 1
Finding μ
The random variable $X\sim \mathrm{N}(\mu, 5^2)$. Given that $\mathrm{P}(X>50)=0.1587$, find the value of $\mu$.
Problem-solvingStandardise first, since μ is unknown
1
$\mathrm{P}(X>50)=0.1587 \Rightarrow \mathrm{P}\left(Z>\dfrac{50-\mu}{5}\right)=0.1587$.
Use $Z=\dfrac{X-\mu}{\sigma}$.
2
Using the inverse normal function with $\mu=0$, $\sigma=1$ and $p=0.8413$: $z=1.0000$.
You don't know $\mu$, so use the standard normal distribution to find the value of $z$ such that $\mathrm{P}(Z>z)=0.1587$.
3
$1 = \dfrac{50-\mu}{5} \Rightarrow \mu = 45.0$.
You know one value of $X$ and the corresponding value of $Z$, so use the coding formula to find $\mu$.
▪ solved
Worked example 2
Finding σ, then a percentile
A machine fills cans with volume, $V$ ml, modelled as a normal distribution such that $V\sim \mathrm{N}(330, \sigma^2)$. a Given that $\mathrm{P}(V<323)=0.1217$, find the value of $\sigma$. b Find the 85th percentile of the volumes.
a
1
$\mathrm{P}(V<323)=0.1217 \Rightarrow \mathrm{P}\left(Z<\dfrac{323-330}{\sigma}\right)=0.1217$.
Use $Z=\dfrac{X-\mu}{\sigma}$.
2
Using the inverse normal function with $p=0.1217$: $z=-1.1665\ldots$, so $\dfrac{323-330}{\sigma}=-1.1665$, giving $\sigma=6.00$ (3 s.f.).
Note $z$ is negative here, since $323$ is below the mean $330$.
▪ solved
b
Now that you have calculated $\sigma$, you can write out the full distribution: $V\sim \mathrm{N}(330, 6.00^2)$.
1
Let $a$ be the 85th percentile. $\mathrm{P}(V<a)=0.85$.
2
$a = 336.22$ ml (2 d.p.).
Use the inverse normal distribution function with $\mu=330$, $\sigma=6.00$ and $p=0.85$.
▪ solved
Worked example 3
Finding both μ and σ · Challenge
The random variable $X\sim \mathrm{N}(\mu,\sigma^2)$. Given that $\mathrm{P}(X>90)=0.0478$ and $\mathrm{P}(X<60)=0.2023$, find the value of $\mu$ and the value of $\sigma$.
Problem-solvingForm two simultaneous equations in μ and σ
1
$\mathrm{P}(Z>z_1)=0.0478 \Rightarrow z_1=1.6666$, and $\mathrm{P}(Z<z_2)=0.2023 \Rightarrow z_2=-0.8334$.
Find the $z$-values corresponding to a right tail of $0.0478$ and a left tail of $0.2023$, using the inverse normal function on the standard normal distribution.
2
$1.6666 = \dfrac{90-\mu}{\sigma} \Rightarrow 1.6666\sigma+\mu=90$ (1)
Use $\dfrac{X-\mu}{\sigma}$ to link the $X$ and $Z$ values.
3
$-0.8334 = \dfrac{60-\mu}{\sigma} \Rightarrow -0.8334\sigma+\mu=60$ (2)
4
(1) $-$ (2): $2.5000\sigma = 30 \Rightarrow \sigma = 12.0$
Subtracting eliminates $\mu$.
5
Substituting into (2): $\mu = 60+0.8334\times12.0 = 70.00$
6
So $\sigma = 12.0$ and $\mu = 70.0$ (3 s.f.).
▪ solved
Check yourself
The random variable $X\sim \mathrm{N}(\mu, 6^2)$. Given that $\mathrm{P}(X<40)=0.3085$, find the value of $\mu$.
1
$\mathrm{P}(Z<-0.5)=0.3085 \Rightarrow \dfrac{40-\mu}{6}=-0.5$.
2
$\mu = 40+0.5\times6 = 43.0$.