I'm going to assume the normal distribution applies to this problem.
$$P=\int_{lb}^{rb}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x$$
With P being the chance that a certain amount of jawbreaker has a weight between $$lb$$ and $$rb$$, respectively being the left bound and the right bound. In this equation $$\mu$$ is the mean and $$\sigma$$ the standarddeviation.
At this time we have no knowledge of either $$\mu$$ or $$\sigma$$, all we know for sure is that:
$$P(weight>0.4)=0.60$$
Hence the left bound is equal to 0.4 and the right bound to infinity. We get the equation:
$$P(weight>0.4)=\int_{0.4}^{\infty}\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrb{d}x=0.60$$
This still leaves us with two unknown variables: $$\mu$$ and $$\sigma$$.
When analyzing the function we gain the knowledge that when $$\sigma$$ approaches zero, $$\mu$$ approaches the value 0.4. This means:
$$\lim_{\sigma\to0}\Rightarrow\lim_{\mu\to0.4$$
Since the example shows a different outcome to P(weight>0.4):
$$P(weight>0.4)=\frac{470}{800}\approx0.59$$
We can let $$\sigma$$ approach zero and render the outcome unusual. In a similar manner we can pick $$\sigma$$ in such a way that the sample is in line with the distribution.
In short, the question doesn't provide enough information to say for certain if the sample is unusual or not.
