S1: Normal distribution practice question

(I haven't included the main question as it isn't relevant, it's just giving some context about a guy who is travelling to his work that opens at 08:30am but he starts at 09:00am). I'll happily post it and the other question parts if anybody would like to see them.

So I am a bit confused by the question and it's wording.

I didn't really know what it meant by 'maximise his chance of arriving between ...', I just assumed it meant the probability being as close to 1 as possible.

Therefore I found the z value for a p of 0.999, and then rearranged the formula z=(x-mean)/standard deviation, to find the corresponding x value.

I got a value of 52.7 (53 minutes), so concluded that he has a 99.9% chance of arriving to work within 53 minutes, and I took that away from 09:00am (I was unsure at this part, I just guessed it would be at 09:00am) to get a final answer of 08:07am.

This appears to be wrong though, so I'd love some advice on where I went wrong/how to approach it?

I'm not sure if this is sufficiently rigorous, but here's how I'd think about it:

Imagine the normal distribution bell-curve. You want as much of it to be between 8:30 and 9:00 as possible. The curve is symmetrical, so where do you think the middle/mean will be?

Ohhhhh right, I didn't think about it literally enough.

28 minutes is the average time it would take, with the s.d. of 8 minutes. Therefore the most likely travel time is literally 28 minutes (I believe?), as with the normal distribution curve the highest probability density is around the mean.

Since you want it so that x+28 is most comfortably between 08:30 and 09:00, it makes sense for it to be 08:45. Therefore x=08:17am which is correct.

Ohhhhh right, I didn't think about it literally enough.

28 minutes is the average time it would take, with the s.d. of 8 minutes. Therefore the most likely travel time is literally 28 minutes (I believe?), as with the normal distribution curve the highest probability density is around the mean.

Since you want it so that x+28 is most comfortably between 08:30 and 09:00, it makes sense for it to be 08:45. Therefore x=08:17am which is correct.

Yep. The mean is the most likely travel time because that's the peak of the normal curve.

So I just had a thought.

Does this not contradict the statement that the probability of X being any particular value on the normal-distribution curve is 0?

Or is it just the idea that the particular value that X is most likely to be is where the probability density is the greatest (the peak)?