Disagree with Greg solution

GRE Quant Problem Solving
1

Hello,
The question below is from Greg’s video on “Normal Distribution Calculations” in PrepSwift

Let W be a random variable that is normally distributed. If the probability of W being less than 40 is 1/6 and 70 is two standard deviations above the mean, which of the following is the best estimate for the mean of W?

In the solution, Greg says that 1/6 is ~16% (which is correct, ofcourse), but actually calculating 1/6 gives us 16.66%. Wouldn’t this mean that the probability of 40 being less than 1/6 is slightly more than right at -1SD from the mean? That is, -1SD from the mean would be something like 38 or 39? Because of that, wouldn’t the mean also shift slightly less?

I understand I’m splitting hairs here with the 16% vs 16.66%, but I assumed this is one way ETS will try to trick us (given that 16.66% is slightly more than the 16% of data that falls below -1SD).

Is my thinking correct?

2

I don’t have PrepSwift nor watched the video so missing the context and his solution, but note that the 3-\sigma rule is not exactly 68-95-99. More precisely for the first deviation it is
P(\mu-\sigma \leq W \leq \mu+\sigma) \approx 0.6827
This implies the left tail’s probability is P(W \leq \mu-\sigma) \approx 0.15865 which is less than 1/6. This then implies that \mu-\sigma < 40 since P(W \leq 40) = 1/6 > 0.15865.

Together with \mu + 2\sigma = 70, we know that 2\sigma = 70-\mu and \sigma > \mu -40. Solving this system then gives 70-\mu > 2\mu - 80, which implies that \mu < 50.

But \mu also can’t be smaller than 40, since P(W \leq \mu) = 0.5 > 1/6 as it’s a symmetric distribution. Therefore we know that \mu \in (40, 50) and likely leaning towards the right of this interval.