statistics z score

Question

I'm curious what the phrase "on average" means. Here is an example:

On average, 30% were further than ___ kilometers away when they had their accident.

Is $30\%$ a $z$-score or is it a mean? Transport Canada was investigating accident records to find out how far from their residence people were to when they got into a traffic accident. They took the population of accident records from Ontario and measured the distance the drivers were from home when they had their accident in kilometers (km). The distribution of distances was normally shaped, with $µ = 30$ kilometers and $σ = 8.0$ kilometers.

Answer 1

Calculation by using the z-score

The equation is $P(X > x)=0.3$

We know, that $P(X > x)=1-P(X \leq x)$

Thus the equation with the standard normal distribution is:

$1-\Phi \left( \frac{x-\mu}{\sigma}\right)=0.3$, with $Z=\frac{X-\mu}{\sigma}$

Z is standard normal distributed: Z $\sim \mathcal N(0,1)$

$1-\Phi \left( \frac{x-30}{8}\right)=0.3 \quad | -1$

$-\Phi \left( \frac{x-30}{8}\right)=-0.7 \quad | \cdot (-1)$

$\Phi \left( \frac{x-30}{8}\right)=0.7 \quad$

Taking the inverse function.

$ \frac{x-30}{8}=\Phi^{-1}(0.7) \quad$

Now you have to find the corresponding z-score for $p=0.7$. Then solve the equation for x.

Answer 2

I don't know what "on average" means in this context, and I have a Ph.D. in statistics. The sentence is one that looks perfectly sensible until you think about what it says. Probably what was intended was "$30\%$ of the drivers were more than how many kilometers from home?". How an "average" gets involved I don't know.

PS: One could say that $0.3$ is an average of $0$s and $1$s. Assign to each accident a $1$ or a $0$ according as it was or was not in excess of the stated distance from home. But the statement that $30\%$ is the proportion of the population involved is something that involves only one number: $30\%$, not a average of various percentages.