How to interpret $P(\bar X \geq 36.2) = 0.0023$ when population distribution is skewed right with mean of 30 and standard deviation of 12.9? Text: Sullivan, Michael I. Fundamentals of Statistics, 5th Edition.
Scenario: The mean weight gain during pregnancy is 30 pounds, with a standard deviation of 12.9 pounds. Weight gain during pregnancy is skewed right. An obstetrician obtains a random sample of 35 low-income patients and determines their mean weight gain during pregnancy was 36.2 pounds. Does this result suggest anything unusual?
I understand that the sampling distribution of $\bar X$ for a sample of size 35 is approximately normal with mean $\mu_{\bar X} = 30$ and standard error $\sigma_{\bar X} = \frac{12.9}{\sqrt{35}}$. 
So to determine whether having a sample of size 35 with a sample mean of 36.2 is unusual I want to find $P(\bar X \geq 36.2)$. I can do this by standardizing and using a z-score table, so I understand that $P(\bar X \geq 36.2) \approx 0.0023$
Interpretation: The text says the following -- We can conclude one of two things based on this result:


*

*The mean weight gain for low-income patients is 30 pounds, and we happened to select women who, on average, gained more weight.

*The mean weight gain for low-income patients is more than 30 pounds.


We are inclined to accept the second explanation over the first since our sample was obtained randomly.

I am having trouble seeing why that is the interpretation we should make in this scenario. For a sample of size 35, it is unusual to see a sample mean of 36.2 or more. How does this translate to the mean weight gain being more than 30 pounds?

 A: If the population we sampled from (low income) would indeed have an expected weight gain of 30 pounds, seeing 36.2 as an average would be very unlikely. Thus, it is more plausible that the population we sampled from has a higher average weight gain. 
The idea is that maybe the unconditional population mean is 30, but that the weight gain is negatively correlated with income. Thus, the conditional expectation of weight gain conditional on being poor might be higher. This could explain the high average in our sample.
A: The distribution being skewed right made me wonder if the idea of the question was to use something different from the normal distribution.
If we apply Cantelli's inequality to this case we get:
$$
Pr(\bar{X} - \mu_X \geq \lambda) \leq \frac{\sigma_{\bar{X}}^2}{\sigma_{\bar{X}}^2 + \lambda^2} \ .
$$
For $\lambda = 36.2-30 = 6.2$ and $\sigma_{\bar{X}} = \frac{12.9}{\sqrt{35}} = 2.18$, we get $Pr \leq 0.1101$.
Hence, in this case we could conclude that nothing significantly unusual happened.
