I have learned that 17% of the vehicles within a certain population drive less than 7. 500 km per year. On average, a given vehicle in that population will drive 13.300 km per year.
Assuming a normal distribution, how (mathematically, using a graphic calculator, or using Excel) can I calculate the standard deviation for the population at hand1?
On
Following @lulu's Comment, a 'grid search' for $\sigma$ would work, but there is a way to get an approximate solution by solving an explicit equation.
You know that $P(X < 7500) = 0.17$ Thus
$$P\left(\frac{X-\mu}{\sigma} < \frac{7500-13300}{\sigma}\right) = 0.17$$
and using your calculator or printed tables you can find that
$\frac{-5800}{\sigma} = -0.9541653$ and solve for $\sigma.$
I used R statistical software (where the inverse of the standard normal CDF
is denoted qnorm) as follows:
qnorm(.17)
## -0.9541653
Addendum: If you suspect that $\sigma$ is between 1000 and 10,000,
then here's how a grid search in R for $\sigma$, denoted sg, might work
to give 6079.
sg = 1000:10000 # list of integers from 1000 to 10000
pr=pnorm(7500, 13300, sg) # P(X <= 7500) for X ~ NORM(13300, sg)
sg[abs(pr-.17) == min(abs(pr-.17))] # search for sg with pr closest to .17
6079
First of all the equation is
$$0.17=P\left( Z\leq z\right)=P\left( Z\leq \frac{7500-13300}{\sigma}\right)=\Phi\left( \frac{7,500-13,300}{\sigma}\right)$$
The random variable $Z$ is standard normal distributed as $Z\sim \mathcal N(0,1)$. $\Phi(z)$ is the notation for the cdf of the standard normal distribution. Now we take the inverse function.
$$\Phi ^{-1}\left( 0.17\right)= \frac{7,500-13,300}{\sigma}\quad (*)$$
Now you can evaluate the value of $ z$ where $\Phi(z)=0.17$ by using
$a)$ an online calculator
Type in $0.174$ at the input field "Cummulative probability"
$b)$ Excel
Type in a cell: $\texttt{=norminv(0.17,0,1)}$
Finally replace $\Phi ^{-1}\left( 0.17\right)$ by the returned value $z$ and solve the equation (*) for $\sigma$